13.E: Fundamental Equilibrium Concepts (Exercises)
13.1: Chemical Equilibria Exercises
Q13.1.1
What does it mean to describe a reaction as “reversible”?
S13.1.1
The reaction can proceed in both the forward and reverse directions.
Q13.1.2
When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?
Q13.1.3
If a reaction is reversible, when can it be said to have reached equilibrium?
S13.1.3
When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the reactions continue to occur, but at equivalent rates.
Q13.1.4
Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?
Q13.1.5
If the concentrations of products and reactants are equal, is the system at equilibrium?
S13.1.5
The concept of equilibrium does not imply equal concentrations, though it is possible.
13.2: Equilibrium Constant Exercises
Q13.2.1
Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.
Q13.2.2
Explain why an equilibrium between Br 2 ( l ) and Br 2 ( g ) would not be established if the container were not a closed vessel shown below:
S13.2.2
Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle because the system is not closed; one of the components of the equilibrium, the Br 2 vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.
Q13.2.3
If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO 2 or with pure N 2 O 4 ?
\[\ce{ 2NO2 }(g) \ rightleftharpoons \ce{ N2O4 }(g)\]
Q13.2.4
Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg 2 Cl 2 , AgCl, PbCl 2 , and CuCl.
Q13.2.5
- (a) Write the expression for the equilibrium constant for the reaction represented by the equation \(\ce{AgCl}(s) \rightleftharpoons \ce{Ag+}(aq)+\ce{Cl-}(aq)\). Is K c > 1, < 1, or ≈ 1? Explain your answer.
- (b) Write the expression for the equilibrium constant for the reaction represented by the equation \(\ce{Pb^2+}(aq)+\ce{2Cl-}(aq) \rightleftharpoons \ce{PbCl2}(s)\). Is K c > 1, < 1, or ≈ 1? Explain your answer.
S13.2.5
(a) K c = [Ag + ][Cl − ] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M ; (b) \(K_c=\ce{\dfrac{1}{[Pb^2+][Cl- ]^2}}\) > 1 because PbCl 2 is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1 M ).
Q13.2.6
Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble.
- Write the expression for the equilibrium constant for the reaction represented by the equation \(\ce{CaCO3}(s) \rightleftharpoons \ce{Ca^2+}(aq)+\ce{CO3-}(aq)\). Is K c > 1, < 1, or ≈ 1? Explain your answer.
- Write the expression for the equilibrium constant for the reaction represented by the equation \(\ce{3Ba^2+}(aq)+\ce{2PO4^3-}(aq) \rightleftharpoons \ce{Ba3(PO4)2}(s)\). Is K c > 1, < 1, or ≈ 1? Explain your answer.
Q13.2.7
Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: \(\ce{3C2H2}(g)⟶\ce{C6H6}(g)\). Which value of K c would make this reaction most useful commercially? K c ≈ 0.01, K c ≈ 1, or K c ≈ 10. Explain your answer.
S13.2.7
Since \(K_c=\ce{\dfrac{[C6H6]}{[C2H2]^3}}\), a value of K c ≈ 10 means that C 6 H 6 predominates over C 2 H 2 . In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.
Q13.2.8
Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation \(\ce{KI}(aq)+\ce{I2}(aq) \rightleftharpoons \ce{KI3}(aq)\) give the same expression for the reaction quotient. KI 3 is composed of the ions K + and I 3 − .
Q13.2.9
For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is K c > 1, < 1, or ≈ 1 for a titration reaction?
S13.2.9
K c > 1
Q13.2.10
For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is K c > 1, < 1, or ≈ 1 for a useful precipitation reaction?
Q13.2.11
Write the mathematical expression for the reaction quotient, Q c , for each of the following reactions:
- \(\ce{CH4}(g)+\ce{Cl2}(g) \rightleftharpoons \ce{CH3Cl}(g)+\ce{HCl}(g)\)
- \(\ce{N2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2NO}(g)\)
- \(\ce{2SO2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2SO3}(g)\)
- \(\ce{BaSO3}(s) \rightleftharpoons \ce{BaO}(s)+\ce{SO2}(g)\)
- \(\ce{P4}(g)+\ce{5O2}(g) \rightleftharpoons \ce{P4O10}(s)\)
- \(\ce{Br2}(g) \rightleftharpoons \ce{2Br}(g)\)
- \(\ce{CH4}(g)+\ce{2O2}(g) \rightleftharpoons \ce{CO2}(g)+\ce{2H2O}(l)\)
- \(\ce{CuSO4⋅5H2O}(s) \rightleftharpoons \ce{CuSO4}(s)+\ce{5H2O}(g)\)
S13.2.11
(a) \(Q_c=\ce{\dfrac{[CH3Cl][HCl]}{[CH4][Cl2]}}\); (b) \(Q_c=\ce{\dfrac{[NO]^2}{[N2][O2]}}\); (c) \(Q_c=\ce{\dfrac{[SO3]^2}{[SO2]^2[O2]}}\); (d) \(Q_c\) = [SO 2 ]; (e) \(Q_c=\ce{\dfrac{1}{[P4][O2]^5}}\); (f) \(Q_c=\ce{\dfrac{[Br]^2}{[Br2]}}\); (g) \(Q_c=\ce{\dfrac{[CO2]}{[CH4][O2]^2}}\); (h) \(Q_c\) = [H 2 O] 5
Q13.2.12
Write the mathematical expression for the reaction quotient, Q c , for each of the following reactions:
- \(\ce{N2}(g)+\ce{3H2}(g) \rightleftharpoons \ce{2NH3}(g)\)
- \(\ce{4NH3}(g)+\ce{5O2}(g) \rightleftharpoons \ce{4NO}(g)+\ce{6H2O}(g)\)
- \(\ce{N2O4}(g) \rightleftharpoons \ce{2NO2}(g)\)
- \(\ce{CO2}(g)+\ce{H2}(g) \rightleftharpoons \ce{CO}(g)+\ce{H2O}(g)\)
- \(\ce{NH4Cl}(s) \rightleftharpoons \ce{NH3}(g)+\ce{HCl}(g)\)
- \(\ce{2Pb(NO3)2}(s) \rightleftharpoons \ce{2PbO}(s)+\ce{4NO2}(g)+\ce{O2}(g)\)
- \(\ce{2H2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2H2O}(l)\)
- \(\ce{S8}(g) \rightleftharpoons \ce{8S}(g)\)
S13.2.12
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
- \(\ce{2NH3}(g) \rightleftharpoons \ce{N2}(g)+\ce{3H2}(g) \hspace{20px} K_c=17\); [NH 3 ] = 0.20 M , [N 2 ] = 1.00 M , [H 2 ] = 1.00 M
- \(\ce{2NH3}(g) \rightleftharpoons \ce{N2}(g)+\ce{3H2}(g) \hspace{20px} K_P=6.8×10^4\); initial pressures: NH 3 = 3.0 atm, N 2 = 2.0 atm, H 2 = 1.0 atm
- \(\ce{2SO3}(g) \rightleftharpoons \ce{2SO2}(g)+\ce{O2}(g) \hspace{20px} K_c=0.230\); [SO 3 ] = 0.00 M , [SO 2 ] = 1.00 M , [O 2 ] = 1.00 M
- \(\ce{2SO3}(g) \rightleftharpoons \ce{2SO2}(g)+\ce{O2}(g) \hspace{20px} K_P=16.5\); initial pressures: SO 3 = 1.00 atm, SO 2 = 1.00 atm, O 2 = 1.00 atm
- \(\ce{2NO}(g)+\ce{Cl2}(g) \rightleftharpoons \ce{2NOCl}(g) \hspace{20px} K_c=4.6×10^4\); [NO] = 1.00 M , [Cl 2 ] = 1.00 M , [NOCl] = 0 M
- \(\ce{N2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2NO}(g) \hspace{20px} K_P=0.050\); initial pressures: NO = 10.0 atm, N 2 = O 2 = 5 atm
S13.2.13
(a) \(Q_c\) 25 proceeds left; (b) Q P 0.22 proceeds right; (c) \(Q_c\) undefined proceeds left; (d) Q P 1.00 proceeds right; (e) Q P 0 proceeds right; (f) \(Q_c\) 4 proceeds left
Q13.2.14
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
- \(\ce{2NH3}(g) \rightleftharpoons \ce{N2}(g)+\ce{3H2}(g) \hspace{20px} K_c=17\); [NH 3 ] = 0.50 M , [N 2 ] = 0.15 M , [H 2 ] = 0.12 M
- \(\ce{2NH3}(g) \rightleftharpoons \ce{N2}(g)+\ce{3H2}(g) \hspace{20px} K_P=6.8×10^4\); initial pressures: NH 3 = 2.00 atm, N 2 = 10.00 atm, H 2 = 10.00 atm
- \(\ce{2SO3}(g) \rightleftharpoons \ce{2SO2}(g)+\ce{O2}(g) \hspace{20px} K_c=0.230\); [SO 3 ] = 2.00 M , [SO 2 ] = 2.00 M , [O 2 ] = 2.00 M
- \(\ce{2SO3}(g) \rightleftharpoons \ce{2SO2}(g)+\ce{O2}(g) \hspace{20px} K_P=\mathrm{6.5\:atm}\); initial pressures: SO 2 = 1.00 atm, O 2 = 1.130 atm, SO 3 = 0 atm
- \(\ce{2NO}(g)+\ce{Cl2}(g) \rightleftharpoons \ce{2NOCl}(g) \hspace{20px} K_P=2.5×10^3\); initial pressures: NO = 1.00 atm, Cl 2 = 1.00 atm, NOCl = 0 atm
- \(\ce{N2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2NO}(g) \hspace{20px} K_c=0.050\); [N 2 ] = 0.100 M , [O 2 ] = 0.200 M , [NO] = 1.00 M
Q13.2.15
The following reaction has K P = 4.50 × 10 −5 at 720 K.
\[\ce{N2}(g)+\ce{3H2}(g) \ rightleftharpoons \ce{2NH3}(g)\]
If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? P (NH 3 ) = 93 atm, P (N 2 ) = 48 atm, and P (H 2 ) = 52
S13.2.15
The system will shift toward the reactants to reach equilibrium.
Q13.2.16
Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?
\(\ce{SO2Cl2}(g) \ rightleftharpoons \ce{SO2}(g)+\ce{Cl2}(g)\)
[SO 2 Cl 2 ] = 0.12 M , [Cl 2 ] = 0.16 M and [SO 2 ] = 0.050 M . K c for the reaction is 0.078.
Q13.2.17
Which of the systems described in Exercise give homogeneous equilibria? Which give heterogeneous equilibria?
S13.2.17
(a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous
Q13.2.18
Which of the systems described in Exercise give homogeneous equilibria? Which give heterogeneous equilibria?
Q13.2.19
For which of the reactions in Exercise does K c (calculated using concentrations) equal K P (calculated using pressures)?
S13.2.19
This situation occurs in (a) and (b).
Q13.2.19
For which of the reactions in Exercise does K c (calculated using concentrations) equal K P (calculated using pressures)?
Q13.2.20
Convert the values of K c to values of K P or the values of K P to values of K c .
- \(\ce{N2}(g)+\ce{3H2}(g) \rightleftharpoons \ce{2NH3}(g) \hspace{20px} K_c=\textrm{0.50 at 400°C}\)
- \(\ce{H2 + I2 \rightleftharpoons 2HI} \hspace{20px} K_c=\textrm{50.2 at 448°C}\)
- \(\ce{Na2SO4⋅10H2O}(s) \rightleftharpoons \ce{Na2SO4}(s)+\ce{10H2O}(g) \hspace{20px} K_P=4.08×10^{−25}\textrm{ at 25°C}\)
- \(\ce{H2O}(l) \rightleftharpoons \ce{H2O}(g) \hspace{20px} K_P=\textrm{0.122 at 50°C}\)
S13.2.20
(a) K P = 1.6 × 10 −4 ; (b) K P = 50.2; (c) K c = 5.31 × 10 −39 ; (d) K c = 4.60 × 10 −3
Q13.2.21
Convert the values of K c to values of K P or the values of K P to values of K c .
- \(\ce{Cl2}(g)+\ce{Br2}(g) \rightleftharpoons \ce{2BrCl}(g) \hspace{20px} K_c=4.7×10^{−2}\textrm{ at 25°C}\)
- \(\ce{2SO2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2SO3}(g) \hspace{20px} K_P=\textrm{48.2 at 500°C}\)
- \(\ce{CaCl2⋅6H2O}(s) \rightleftharpoons \ce{CaCl2}(s)+\ce{6H2O}(g) \hspace{20px} K_P=5.09×10^{−44}\textrm{ at 25°C}\)
- \(\ce{H2O}(l) \rightleftharpoons \ce{H2O}(g) \hspace{20px} K_P=\textrm{0.196 at 60°C}\)
Q13.2.22
What is the value of the equilibrium constant expression for the change \(\ce{H2O}(l) \rightleftharpoons \ce{H2O}(g)\) at 30 °C?
S13.2.22
Q13.2.23
Write the expression of the reaction quotient for the ionization of HOCN in water.
Q13.2.24
Write the reaction quotient expression for the ionization of NH 3 in water.
S13.2.24
\[Q_c=\ce{\dfrac{[NH4+][OH- ]}{[HN3]}}\]
Q13.2.25
What is the approximate value of the equilibrium constant K P for the change \(\ce{C2H5OC2H5}(l) \rightleftharpoons \ce{C2H5OC2H5}(g)\) at 25 °C. (Vapor pressure was described in the previous chapter on liquids and solids; refer back to this chapter to find the relevant information needed to solve this problem.)
13.3: Shifting Equilbria Exercises
Q13.3.1
The following equation represents a reversible decomposition:
\(\ce{CaCO3}(s)\rightleftharpoons\ce{CaO}(s)+\ce{CO2}(g)\)
Under what conditions will decomposition in a closed container proceed to completion so that no CaCO 3 remains?
S13.3.1
The amount of CaCO 3 must be so small that \(P_{\ce{CO2}}\) is less than K P when the CaCO 3 has completely decomposed. In other words, the starting amount of CaCO 3 cannot completely generate the full \(P_{\ce{CO2}}\) required for equilibrium.
Q13.3.2
Explain how to recognize the conditions under which changes in pressure would affect systems at equilibrium.
Q13.3.3
What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?
S13.3.3
The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants' side.
Q13.3.4
What would happen to the color of the solution in part (b) of Figure if a small amount of NaOH were added and Fe(OH)3 precipitated? Explain your answer.
Q13.3.5
The following reaction occurs when a burner on a gas stove is lit:
\(\ce{CH4}(g)+\ce{2O2}(g)\ rightleftharpoons \ce{CO2}(g)+\ce{2H2O}(g)\)
Is an equilibrium among CH 4 , O 2 , CO 2 , and H 2 O established under these conditions? Explain your answer.
S13.3.5
No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.
Q13.3.6
A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO3, from sulfur dioxide, SO2, and oxygen, O2, shown here. At high temperatures, the rate of formation of \(\ce{SO3 }\)is higher, but the equilibrium amount (concentration or partial pressure) of SO3 is lower than it would be at lower temperatures.
\[\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g)\]
- (a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?
- (b) Is the reaction endothermic or exothermic?
Q13.3.7a
Suggest four ways in which the concentration of hydrazine, N 2 H 4 , could be increased in an equilibrium described by the following equation:
\[\ce{N2}(g)+\ce{2H2}(g)\ rightleftharpoons \ce{N2H4}(g) \hspace{20px} ΔH=\ce{95\:kJ}\]
S13.3.7a
Add N 2 ; add H 2 ; decrease the container volume; heat the mixture.
Q13.3.7b
Suggest four ways in which the concentration of PH 3 could be increased in an equilibrium described by the following equation:
\[\ce{P4}(g)+\ce{6H2}(g)\rightleftharpoons\ce{4PH3}(g) \hspace{20px} ΔH=\mathrm{110.5\:kJ}\]
Q13.3.8
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
- \(\ce{2NH3}(g)\rightleftharpoons\ce{N2}(g)+\ce{3H2}(g) \hspace{20px} ΔH=\mathrm{92\:kJ}\)
- \(\ce{N2}(g)+\ce{O2}(g)\rightleftharpoons\ce{2NO}(g) \hspace{20px} ΔH=\mathrm{181\:kJ}\)
- \(\ce{2O3}(g)\rightleftharpoons\ce{3O2}(g) \hspace{20px} ΔH=\mathrm{−285\:kJ}\)
- \(\ce{CaO}(s)+\ce{CO2}(g)\rightleftharpoons\ce{CaCO3}(s) \hspace{20px} ΔH=\mathrm{-176\:kJ}\)
S13.3.8
(a) Δ T increase = shift right, Δ P increase = shift left; (b) Δ T increase = shift right, Δ P increase = no effect; (c) Δ T increase = shift left, Δ P increase = shift left; (d) Δ T increase = shift left, Δ P increase = shift right.
Q13.3.9
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
- \(\ce{2H2O}(g)\rightleftharpoons\ce{2H2}(g)+\ce{O2}(g) \hspace{20px} ΔH=\ce{484\:kJ}\)
- \(\ce{N2}(g)+\ce{3H2}(g)\rightleftharpoons\ce{2NH3}(g) \hspace{20px} ΔH=\mathrm{-92.2\:kJ}\)
- \(\ce{2Br}(g)\rightleftharpoons\ce{Br2}(g) \hspace{20px} ΔH=\mathrm{-224\:kJ}\)
- \(\ce{H2}(g)+\ce{I2}(s)\rightleftharpoons\ce{2HI}(g) \hspace{20px} ΔH=\ce{53\:kJ}\)
Q13.3.10
Water gas is a 1:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction:
\[\ce{H2O}(g)+\ce{C}(s)\rightleftharpoons\ce{H2}(g)+\ce{CO}(g).\]
Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure in the presence of a suitable catalyst.
- Write the expression for the equilibrium constant (\(K_c\)) for the reversible reaction \[\ce{2H2}(g)+\ce{CO}(g)\rightleftharpoons\ce{CH3OH}(g) \hspace{20px} ΔH=\mathrm{-90.2\:kJ}\]
- What will happen to the concentrations of \(\ce{H2}\), \(\ce{CO}\), and \(\ce{CH3OH}\) at equilibrium if more H 2 is added?
- What will happen to the concentrations of H\(\ce{H2}\), \(\ce{CO}\), and \(\ce{CH3OH}\) at equilibrium if CO is removed?
- What will happen to the concentrations of \(\ce{H2}\), \(\ce{CO}\), and \(\ce{CH3OH}\) at equilibrium if CH 3 OH is added?
- What will happen to the concentrations of H\(\ce{H2}\), \(\ce{CO}\), and \(\ce{CH3OH}\) at equilibrium if the temperature of the system is increased?
- What will happen to the concentrations of \(\ce{H2}\), \(\ce{CO}\), and \(\ce{CH3OH}\) at equilibrium if more catalyst is added?
S13.3.10
- \(K_c=\ce{\dfrac{[CH3OH]}{[H2]^2[CO]}}\);
- [H 2 ] increases, [CO] decreases, [CH 3 OH] increases;
- [H 2 ] increases, [CO] decreases, [CH 3 OH] decreases;
- [H 2 ] increases, [CO] increases, [CH 3 OH] increases;
- [H 2 ] increases, [CO] increases, [CH 3 OH] decreases;
- no changes.
Q13.3.11
Nitrogen and oxygen react at high temperatures.
- Write the expression for the equilibrium constant (Kc) for the reversible reaction \[\ce{N2}(g)+\ce{O2}(g)\rightleftharpoons\ce{2NO}(g)\hspace{20px}ΔH=\ce{181\:kJ}\]
- What will happen to the concentrations of N 2 , O 2 , and NO at equilibrium if more O 2 is added?
- What will happen to the concentrations of N 2 , O 2 , and NO at equilibrium if N 2 is removed?
- What will happen to the concentrations of N 2 , O 2 , and NO at equilibrium if NO is added?
- What will happen to the concentrations of N 2 , O 2 , and NO at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?
- What will happen to the concentrations of N 2 , O 2 , and NO at equilibrium if the temperature of the system is increased?
- What will happen to the concentrations of N 2 , O 2 , and NO at equilibrium if a catalyst is added?
Q13.3.12
Water gas, a mixture of H 2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.
- Write the expression for the equilibrium constant for the reversible reaction \[\ce{C}(s)+\ce{H2O}(g)\rightleftharpoons\ce{CO}(g)+\ce{H2}(g)\hspace{20px}ΔH=\mathrm{131.30\:kJ}\]
- What will happen to the concentration of each reactant and product at equilibrium if more C is added?
- What will happen to the concentration of each reactant and product at equilibrium if H 2 O is removed?
- What will happen to the concentration of each reactant and product at equilibrium if CO is added?
- What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
 In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.
Q13.3.13
Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.
- Write the expression for the equilibrium constant ( K c ) for the reversible reaction \[\ce{Fe2O3}(s)+\ce{3H2}(g)\rightleftharpoons\ce{2Fe}(s)+\ce{3H2O}(g) \hspace{20px} ΔH=\mathrm{98.7\:kJ}\]
- What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
- What will happen to the concentration of each reactant and product at equilibrium if H 2 O is removed?
- What will happen to the concentration of each reactant and product at equilibrium if H 2 is added?
- What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?
- What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
Q13.3.14
Ammonia is a weak base that reacts with water according to this equation:
\[\ce{NH3}(aq)+\ce{H2O}(l)\ rightleftharpoons \ce{NH4+}(aq)+\ce{OH-}(aq)\]
Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water and why?
- Addition of NaOH
- Addition of HCl
- Addition of NH 4 Cl
S13.3.14
Only (b)
Q13.3.15
Acetic acid is a weak acid that reacts with water according to this equation:
\[\ce{CH3CO2H}(aq)+\ce{H2O}(aq)\ rightleftharpoons \ce{H3O+}(aq)+\ce{CH3CO2-}(aq)\]
Will any of the following increase the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion?
- Addition of HCl
- Addition of NaOH
- Addition of NaCH 3 CO 2
Q13.3.16
Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of Na + , Cl − , Ag + , and \(\ce{NO3-}\), in contact with solid AgCl.
\(\ce{Na+}(aq)+\ce{Cl-}(aq)+\ce{Ag+}(aq)+\ce{NO3-}(aq)\rightleftharpoons \ce{ AgCl }(s)+\ce{Na+}(aq)+\ce{NO3-}(aq)\)
S13.3.16
Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).
Q13.3.17
How can the pressure of water vapor be increased in the following equilibrium?
Q13.3.18
Additional solid silver sulfate, a slightly soluble solid, is added to a solution of silver ion and sulfate ion at equilibrium with solid silver sulfate.
\[\ce{2Ag+}(aq)+\ce{SO4^2-}(aq)\ rightleftharpoons \ce{Ag2SO4}(s)\]
Which of the following will occur?
- Ag + or \(\ce{SO4^2-}\) concentrations will not change.
- The added silver sulfate will dissolve.
- Additional silver sulfate will form and precipitate from solution as Ag + ions and \(\ce{SO4^2-}\) ions combine.
- The Ag + ion concentration will increase and the \(\ce{SO4^2-}\) ion concentration will decrease.
S13.3.18
(a)
Q13.3.19
The amino acid alanine has two isomers, α-alanine and β-alanine. When equal masses of these two compounds are dissolved in equal amounts of a solvent, the solution of α-alanine freezes at the lowest temperature. Which form, α-alanine or β-alanine, has the larger equilibrium constant for ionization \(\ce{(HX \rightleftharpoons H+ + X- )}\)?
13.4: Equilibrium Calculations Exercises
Q13.4.1
A reaction is represented by this equation: \(\ce{A}(aq)+\ce{2B}(aq)⇌\ce{2C}(aq) \hspace{20px} K_c=1×10^3\)
- Write the mathematical expression for the equilibrium constant.
- Using concentrations ≤1 M , make up two sets of concentrations that describe a mixture of A, B, and C at equilibrium.
S13.4.1
\(K_c=\ce{\dfrac{[C]^2}{[A][B]^2}}\). [A] = 0.1 M , [B] = 0.1 M , [C] = 1 M ; and [A] = 0.01, [B] = 0.250, [C] = 0.791.
Q13.4.2
A reaction is represented by this equation: \(\ce{2W}(aq)⇌\ce{X}(aq)+\ce{2Y}(aq) \hspace{20px} K_c=5×10^{−4}\)
- Write the mathematical expression for the equilibrium constant.
- Using concentrations of ≤1 M , make up two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.
Q13.4.3
What is the value of the equilibrium constant at 500 °C for the formation of NH 3 according to the following equation?
\[\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g)\]
An equilibrium mixture of NH 3 ( g ), H 2 ( g ), and N 2 ( g ) at 500 °C was found to contain 1.35 M H 2 , 1.15 M N 2 , and 4.12 × 10 −1 M NH 3 .
S13.4.3
K c = 6.00 × 10 −2
Q13.4.4
Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.
\[\ce{CH4}(g)+\ce{H2O}(g)⇌\ce{3H2}(g)+\ce{CO}(g)\]
What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH 4 , 0.126 M ; H 2 O, 0.242 M ; CO, 0.126 M ; H 2 1.15 M , at a temperature of 760 °C?
A 0.72-mol sample of PCl 5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl 3 ( g ) and 0.40 mol of Cl 2 ( g ). Calculate the value of the equilibrium constant for the decomposition of PCl 5 to PCl 3 and Cl 2 at this temperature.
S13.4.4
K c = 0.50
Q13.4.5
At 1 atm and 25 °C, NO 2 with an initial concentration of 1.00 M is 3.3 × 10 −3 % decomposed into NO and O 2 . Calculate the value of the equilibrium constant for the reaction.
\[\ce{2NO2}(g)⇌\ce{2NO}(g)+\ce{O2}(g)\]
Q13.4.6
Calculate the value of the equilibrium constant K P for the reaction \(\ce{2NO}(g)+\ce{Cl2}(g)⇌\ce{2NOCl}(g)\) from these equilibrium pressures: NO, 0.050 atm; Cl 2 , 0.30 atm; NOCl, 1.2 atm.
S13.4.6
The equilibrium equation is K P = 1.9 × 10 3
Q13.4.7
When heated, iodine vapor dissociates according to this equation:
\[\ce{I2}(g)⇌\ce{2I}(g)\]
At 1274 K, a sample exhibits a partial pressure of I 2 of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, K P , for the decomposition at 1274 K.
Q13.4.8
A sample of ammonium chloride was heated in a closed container.
\[\ce{NH4Cl}(s)⇌\ce{NH3}(g)+\ce{HCl}(g)\]
At equilibrium, the pressure of NH 3 ( g ) was found to be 1.75 atm. What is the value of the equilibrium constant K P for the decomposition at this temperature?
S13.4.8
K P = 3.06
Q13.4.9
At a temperature of 60 °C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant K P for the transformation at 60 °C?
\[\ce{H2O}(l)⇌\ce{H2O}(g)\]
Q13.4.10
Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
(a)
\(\begin{alignat}{3}
&\ce{2SO3}(g)⇌\:&&\ce{2SO2}(g)+\:&&\ce{O2}(g)\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&+x\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&0.125\:M
\end{alignat}\)
(b)
\(\begin{alignat}{3}
&\ce{4NH3}(g)+\:&&\ce{3O2}(g)⇌\:&&\ce{2N2}(g)+\:&&\ce{6H2O}(g)\\
&\underline{\hspace{40px}} &&3x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\\
&\underline{\hspace{40px}} &&0.24\:M &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\)
(c) Change in pressure:
\(\begin{alignat}{3}
&\ce{2CH4}(g)⇌\:&&\ce{C2H2}(g)+\:&&\ce{3H2}(g)\\
&\underline{\hspace{40px}} &&x &&\underline{\hspace{40px}}\\
&\underline{\hspace{40px}} &&\textrm{25 torr} &&\underline{\hspace{40px}}
\end{alignat}\)
(d) Change in pressure:
\(\begin{alignat}{3}
&\ce{CH4}(g)+\:&&\ce{H2O}(g)⇌\:&&\ce{CO}(g)+\:&&\ce{3H2}(g)\\
&\underline{\hspace{40px}} &&x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\\
&\underline{\hspace{40px}} &&\textrm{5 atm} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\)
(e)
\(\begin{alignat}{3}
&\ce{NH4Cl}(s)⇌\:&&\ce{NH3}(g)+\:&&\ce{HCl}(g)\\
& &&x &&\underline{\hspace{40px}}\\
& &&1.03×10^{−4}\:M &&\underline{\hspace{40px}}
\end{alignat}\)
(f) change in pressure:
\(\begin{alignat}{3}
&\ce{Ni}(s)+\:&&\ce{4CO}(g)⇌\:&&\ce{Ni(CO)4}(g)\\
& &&4x &&\underline{\hspace{40px}}\\
& &&\textrm{0.40 atm} &&\underline{\hspace{40px}}
\end{alignat}\)
S13.4.10
- −2 x , 2 x , −0.250 M , 0.250 M ;
- 4 x , −2 x , −6 x , 0.32 M , −0.16 M , −0.48 M ;
- −2 x , 3 x , −50 torr, 75 torr;
- x , − x , −3 x , 5 atm, −5 atm, −15 atm;
- x , 1.03 × 10 −4 M ; (f) x , 0.1 atm.
Q13.4.11
Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
(a)
\(\begin{alignat}{3}
&\ce{2H2}(g)+\:&&\ce{O2}(g)⇌\:&&\ce{2H2O}(g)\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&+2x\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&1.50\:M
\end{alignat}\)
(b)
\(\begin{alignat}{3}
&\ce{CS2}(g)+\:&&\ce{4H2}(g)⇌\:&&\ce{CH4}(g)+\:&&\ce{2H2S}(g)\\
&x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\\
&0.020\:M &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\)
(c) Change in pressure:
\(\begin{alignat}{3}
&\ce{H2}(g)+\:&&\ce{Cl2}(g)⇌\:&&\ce{2HCl}(g)\\
&x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\\
&\textrm{1.50 atm} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\)
(d) Change in pressure:
\(\begin{alignat}{3}
&\ce{2NH3}(g)+\:&&\ce{2O2}(g)⇌\:&&\ce{N2O}(g)+\:&&\ce{3H2O}(g)\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&x\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\textrm{60.6 torr}
\end{alignat}\)
(e)
\(\begin{alignat}{3}
&\ce{NH4HS}(s)⇌\:&&\ce{NH3}(g)+\:&&\ce{H2S}(g)\\
& &&x &&\underline{\hspace{40px}}\\
& &&9.8×10^{−6}\:M &&\underline{\hspace{40px}}
\end{alignat}\)
(f) Change in pressure:
\(\begin{alignat}{3}
&\ce{Fe}(s)+\:&&\ce{5CO}(g)⇌\:&&\ce{Fe(CO)4}(g)\\
& &&\underline{\hspace{40px}} &&x\\
& &&\underline{\hspace{40px}} &&\textrm{0.012 atm}
\end{alignat}\)
Q13.4.12
Why are there no changes specified for Ni in Exercise , part (f)? What property of Ni does change?
S13.4.12
Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.
Q13.4.13
Why are there no changes specified for NH 4 HS in Exercise , part (e)? What property of NH 4 HS does change?
Q13.4.14
Analysis of the gases in a sealed reaction vessel containing NH 3 , N 2 , and H 2 at equilibrium at 400 °C established the concentration of N 2 to be 1.2 M and the concentration of H 2 to be 0.24 M .
\[\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g) \hspace{20px} K_c=\textrm{0.50 at 400 °C}\]
Calculate the equilibrium molar concentration of NH 3 .
S13.4.14
[NH 3 ] = 9.1 × 10 −2 M
Q13.4.16
Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H 2 and 1.25 mol of I 2 in a 5.00−L flask at 448 °C.
\(\ce{H2 + I2 ⇌ 2HI} \hspace{20px} K_c=\textrm{50.2 at 448 °C}\)
Q13.4.17
What is the pressure of BrCl in an equilibrium mixture of Cl 2 , Br 2 , and BrCl if the pressure of Cl 2 in the mixture is 0.115 atm and the pressure of Br 2 in the mixture is 0.450 atm?
\[\ce{Cl2}(g)+\ce{Br2}(g)⇌\ce{2BrCl}(g) \hspace{20px} K_P=4.7×10^{−2}\]
S13.4.17
P BrCl = 4.9 × 10 −2 atm
Q13.4.18
What is the pressure of CO 2 in a mixture at equilibrium that contains 0.50 atm H 2 , 2.0 atm of H 2 O, and 1.0 atm of CO at 990 °C?
\[\ce{H2}(g)+\ce{CO2}(g)⇌\ce{H2O}(g)+\ce{CO}(g) \hspace{20px} K_P=\textrm{1.6 at 990 °C}\]
Q13.4.12
Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.
\(\ce{CoO}(s)+\ce{CO}(g)⇌\ce{Co}(s)+\ce{CO2}(g) \hspace{20px} K_c=4.90×10^2\textrm{ at 550 °C}\)
What concentration of CO remains in an equilibrium mixture with [CO 2 ] = 0.100 M ?
S13.4.12
[CO] = 2.0 × 10 −4 M
Q13.4.13
Carbon reacts with water vapor at elevated temperatures.
\(\ce{C}(s)+\ce{H2O}(g)⇌\ce{CO}(g)+\ce{H2}(g) \hspace{20px} K_c=\textrm{0.2 at 1000 °C}\)
What is the concentration of CO in an equilibrium mixture with [H 2 O] = 0.500 M at 1000 °C?
Q13.4.14
Sodium sulfate 10−hydrate, \(\ce{Na2SO4 \cdot 10H2O}\), dehydrates according to the equation
\[\ce{Na2SO4⋅10H2O}(s)⇌\ce{Na2SO4}(s)+\ce{10H2O}(g) \hspace{20px} \]
with \(K_p=4.08×10^{−25}\) at 25°C. What is the pressure of water vapor at equilibrium with a mixture of \(\ce{Na2SO4·10H2O}\) and \(\ce{NaSO4}\)?
S13.4.14
\(P_{\ce{H2O}}=3.64×10^{−3}\:\ce{atm}\)
Q13.4.15
Calcium chloride 6−hydrate, CaCl 2 ·6H 2 O, dehydrates according to the equation
\(\ce{CaCl2⋅6H2O}(s)⇌\ce{CaCl2}(s)+\ce{6H2O}(g) \hspace{20px} K_P=5.09×10^{−44}\textrm{ at 25 °C}\)
What is the pressure of water vapor at equilibrium with a mixture of CaCl 2 ·6H 2 O and CaCl 2 ?
Q13.4.16
A student solved the following problem and found the equilibrium concentrations to be [SO 2 ] = 0.590 M , [O 2 ] = 0.0450 M , and [SO 3 ] = 0.260 M . How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 °C:
\(\ce{2SO2}(g)+\ce{O2}(g)⇌\ce{2SO3}(g) \hspace{20px} K_c=4.32\)
What are the equilibrium concentrations of all species in a mixture that was prepared with [SO 3 ] = 0.500 M , [SO 2 ] = 0 M , and [O 2 ] = 0.350 M ?
S13.4.16
Calculate Q based on the calculated concentrations and see if it is equal to K c . Because Q does equal 4.32, the system must be at equilibrium.
Q13.4.16
A student solved the following problem and found [N 2 O 4 ] = 0.16 M at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N 2 O 4 in a mixture formed from a sample of NO 2 with a concentration of 0.10 M ?
\[\ce{2NO2}(g)⇌\ce{N2O4}(g) \hspace{20px} K_c=160\]
Assume that the change in concentration of N 2 O 4 is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N 2 O 4 with chloroform as the solvent.
\(\ce{N2O4}(g)⇌\ce{2NO2}(g) \hspace{20px} K_c=1.07×10^{−5}\) in chloroform
(b) Show that the change is small enough to be neglected.
S13.4.16
(a)
- [NO 2 ] = 1.17 × 10 −3 M
- [N 2 O 4 ] = 0.128 M
(b) Percent error \(=\dfrac{5.87×10^{−4}}{0.129}×100\%=0.455\%\). The change in concentration of N 2 O 4 is far less than the 5% maximum allowed.
Q13.4.17
Assume that the change in concentration of COCl 2 is small enough to be neglected in the following problem.
- Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl 2 with an initial concentration of 0.3166 M . \[\ce{COCl2}(g)⇌\ce{CO}(g)+\ce{Cl2}(g) \hspace{20px} K_c=2.2×10^{−10}\]
- Show that the change is small enough to be neglected.
Q13.4.18
Assume that the change in pressure of H 2 S is small enough to be neglected in the following problem.
(a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H 2 S with an initial pressure of 0.824 atm.
\(\ce{2H2S}(g)⇌\ce{2H2}(g)+\ce{S2}(g) \hspace{20px} K_P=2.2×10^{−6}\)
(b) Show that the change is small enough to be neglected.
S13.4.18
(a)
- [H 2 S] = 0.810 atm
- [H 2 ] = 0.014 atm
- S 2 ] = 0.0072 atm
(b) The 2 x is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2 x is \(\dfrac{0.014}{0.824}×100\%=1.7\%\), which is less than allowed by the “5% test.” The error is, indeed, negligible.
Q13.4.19
What are all concentrations after a mixture that contains [H 2 O] = 1.00 M and [Cl 2 O] = 1.00 M comes to equilibrium at 25 °C?
\[\ce{H2O}(g)+\ce{Cl2O}(g)⇌\ce{2HOCl}(g) \hspace{20px} K_c=0.0900\]
Q13.4.20
What are the concentrations of PCl 5 , PCl 3 , and Cl 2 in an equilibrium mixture produced by the decomposition of a sample of pure PCl 5 with [PCl 5 ] = 2.00 M ?
\[\ce{PCl5}(g)⇌\ce{PCl3}(g)+\ce{Cl2}(g) \hspace{20px} K_c=0.0211\]
S13.4.20
[PCl 3 ] = 1.80 M ; [PC 3 ] = 0.195 M ; [PCl 3 ] = 0.195 M .
Q13.4.21
Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl 2 produced when a sample of NOCl with a pressure of 10.0 atm comes to equilibrium according to this reaction:
Q13.4.22
Calculate the equilibrium concentrations of NO, O 2 , and NO 2 in a mixture at 250 °C that results from the reaction of 0.20 M NO and 0.10 M O 2 . (Hint: K is large; assume the reaction goes to completion then comes back to equilibrium.)
\[\ce{2NO}(g)+\ce{O2}(g)⇌\ce{2NO2}(g) \hspace{20px} K_c=2.3×10^5\textrm{ at 250 °C}\]
S13.4.22
- [NO 2 ] = 0.19 M
- [NO] = 0.0070 M
- [O 2 ] = 0.0035 M
Q13.4.23
Calculate the equilibrium concentrations that result when 0.25 M O 2 and 1.0 M HCl react and come to equilibrium.
\[\ce{4HCl}(g)+\ce{O2}(g)⇌\ce{2Cl2}(g)+\ce{2H2O}(g) \hspace{20px} K_c=3.1×10^{13}\]
Q13.4.24
One of the important reactions in the formation of smog is represented by the equation
\[\ce{O3}(g)+\ce{NO}(g)⇌\ce{NO2}(g)+\ce{O2}(g) \hspace{20px} K_P=6.0×10^{34}\]
What is the pressure of O 3 remaining after a mixture of O 3 with a pressure of 1.2 × 10 −8 atm and NO with a pressure of 1.2 × 10 −8 atm comes to equilibrium? (Hint: K P is large; assume the reaction goes to completion then comes back to equilibrium.)
S13.4.24
\(P_{\ce{O3}}=4.9×10^{−26}\:\ce{atm}\)
Q13.4.24
Calculate the pressures of NO, Cl 2 , and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl 2 . (Hint: K P is small; assume the reverse reaction goes to completion then comes back to equilibrium.)
\(\ce{2NO}(g)+\ce{Cl2}(g)⇌\ce{2NOCl}(g) \hspace{20px} K_P=2.5×10^3\)
Q13.4.25
Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H 2 and 63.5 g of iodine at 448 °C.
\(\ce{H2 + I2 ⇌ 2HI} \hspace{20px} K_c=\textrm{50.2 at 448 °C}\)
S13.4.25
507 g
Q13.4.26
Butane exists as two isomers, n −butane and isobutane.
K P = 2.5 at 25 °C
What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?
Q13.4.27
What is the minimum mass of CaCO 3 required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant ( K c ) is 0.050 for the decomposition reaction of CaCO 3 at that temperature?
\(\ce{CaCO3}(s)⇌\ce{CaO}(s)+\ce{CO2}(g)\)
S13.4.27
330 g
Q13.4.28
The equilibrium constant ( K c ) for this reaction is 1.60 at 990 °C:
\[\ce{H2}(g)+\ce{CO2}(g)⇌\ce{H2O}(g)+\ce{CO}(g)\]
Calculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H 2 , 2.00 mol of CO 2 , 0.750 mol of H 2 O, and 1.00 mol of CO to a 5.00-L container at 990 °C.
Q13.4.29
At 25 °C and at 1 atm, the partial pressures in an equilibrium mixture of N 2 O 4 and NO 2 are \(P_{\ce{N2O4}}=0.70\:\ce{atm}\) and \(P_{\ce{NO2}}=0.30\:\ce{atm}\).
- Predict how the pressures of NO 2 and N 2 O 4 will change if the total pressure increases to 9.0 atm. Will they increase, decrease, or remain the same?
- Calculate the partial pressures of NO 2 and N 2 O 4 when they are at equilibrium at 9.0 atm and 25 °C.
S13.4.29
(a) Both gases must increase in pressure.
(b) \(P_{\ce{N2O4}}=\textrm{8.0 atm and }P_{\ce{NO2}}=1.0\:\ce{atm}\)
Q13.4.30
In a 3.0-L vessel, the following equilibrium partial pressures are measured: N 2 , 190 torr; H 2 , 317 torr; NH 3 , 1.00 × 10 3 torr.
\[\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g)\]
- How will the partial pressures of H 2 , N 2 , and NH 3 change if H 2 is removed from the system? Will they increase, decrease, or remain the same?
- Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.
Q13.4.31
The equilibrium constant ( K c ) for this reaction is 5.0 at a given temperature.
\[\ce{CO}(g)+\ce{H2O}(g) <=> \ce{CO2}(g)+\ce{H2}(g)\)]
- On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H 2 in a liter. How many moles of CO 2 were there in the equilibrium mixture?
- Maintaining the same temperature, additional H 2 was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H 2 in a liter. How many moles of CO 2 were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.
S13.4.31
(a) 0.33 mol.
(b) [CO] 2 = 0.50 M Added H 2 forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H 2 is added.
Q13.4.32a
Antimony pentachloride decomposes according to this equation:
\(\ce{SbCl5}(g)⇌\ce{SbCl3}(g)+\ce{Cl2}(g)\)
An equilibrium mixture in a 5.00-L flask at 448 °C contains 3.85 g of SbCl 5 , 9.14 g of SbCl 3 , and 2.84 g of Cl 2 . How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?
Q13.4.32b
Consider the reaction between H 2 and O 2 at 1000 K
\[\ce{ 2H2 }(g)+\ce{O2}(g)⇌\ce{ 2H2O }(g) \ hspace { 20px } K_P=\ dfrac {(P_{\ce{ H2O }})^2}{(P_{\ce{O2}})(P_{\ce{H2}})^3}=1.33×10^{20}\]
If 0.500 atm of H 2 and 0.500 atm of O 2 are allowed to come to equilibrium at this temperature, what are the partial pressures of the components?
S13.4.32b
\(P_{\ce{H2}}=8.64×10^{−11}\:\ce{atm}\)
\(P_{\ce{O2}}=0.250\:\ce{ atm }\)
\(P_{\ce{ H2O }}=0.500\:\ce{ atm }\)
Q13.4.33
An equilibrium is established according to the following equation
\[\ce{Hg2^2+}(aq)+\ce{NO3−}(aq)+\ce{3H+}(aq)⇌\ce{2Hg^2+}(aq)+\ce{HNO2}(aq)+\ce{H2O}(l) \hspace{20px} K_c=4.6\]
What will happen in a solution that is 0.20 M each in \(\ce{Hg2^2+}\), \(\ce{NO3−}\), H + , Hg 2+ , and HNO 2 ?
- \(\ce{Hg2^2+}\) will be oxidized and \(\ce{NO3−}\) reduced.
- \(\ce{Hg2^2+}\) will be reduced and \(\ce{NO3−}\) oxidized.
- Hg 2+ will be oxidized and HNO 2 reduced.
- Hg 2+ will be reduced and HNO 2 oxidized.
- There will be no change because all reactants and products have an activity of 1.
Q13.4.34
Consider the equilibrium
\[\ce{4NO2}(g)+\ce{6H2O}(g)⇌\ce{4NH3}(g)+\ce{7O2}(g)\]
- What is the expression for the equilibrium constant ( K c ) of the reaction?
- How must the concentration of NH 3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?
- If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the volume of the reaction vessel) affect the pressure of NO 2 ?
- If the change in the pressure of NO 2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O 2 change?
S13.4.34
(a) \(K_c=\ce{\dfrac{[NH3]^4[O2]^7}{[NO2]^4[H2O]^6}}\). (b) [NH 3 ] must increase for Q c to reach K c . (c) That decrease in pressure would decrease [NO 2 ]. (d) \(P_{\ce{O2}}=49\:\ce{torr}\)
Q13.4.35
The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO 2 ), is partially regulated by the concentration of H 3 O + and dissolved CO 2 in the blood. Although the equilibrium is complicated, it can be summarized as
\(\ce{HbO2}(aq)+\ce{H3O+}(aq)+\ce{CO2}(g)⇌\ce{CO2−Hb−H+}+\ce{O2}(g)+\ce{H2O}(l)\)
- (a) Write the equilibrium constant expression for this reaction.
- (b) Explain why the production of lactic acid and CO 2 in a muscle during exertion stimulates release of O 2 from the oxyhemoglobin in the blood passing through the muscle.
Q13.4.36
The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.
\(\ce{C12H22O11}(aq)+\ce{H2O}(l)⟶\ce{C6H12O6}(aq)+\ce{C6H12O6}(aq)\)
Rate = k [C 12 H 22 O 11 ]
In neutral solution, k = 2.1 × 10 −11 /s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C 6 H 12 O 6 , but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 10 5 at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1.
S13.4.36
[fructose] = 0.15 M
Q13.4.37
The density of trifluoroacetic acid vapor was determined at 118.1 °C and 468.5 torr, and found to be 2.784 g/L. Calculate K c for the association of the acid.
Liquid N 2 O 3 is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO 2 . At 25 °C, a value of K P = 1.91 has been established for this decomposition. If 0.236 moles of N 2 O 3 are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N 2 O 3 ( g ), NO 2 ( g ), and NO( g ).
S13.4.37
\(P_{\ce{N2O3}}=\textrm{1.90 atm and }P_{\ce{NO}}=P_{\ce{NO2}}=\textrm{1.90 atm}\)
Q13.4.38
A 1.00-L vessel at 400 °C contains the following equilibrium concentrations: N 2 , 1.00 M ; H 2 , 0.50 M ; and NH 3 , 0.25 M . How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 M ?
Q13.4.39
A 0.010 M solution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010 M solution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions.
(a) Which acid has the larger equilibrium constant for ionization
HA \([\ce{HA}(aq)⇌\ce{A-}(aq)+\ce{H+}(aq)]\) or HB \([\ce{HB}(aq)⇌\ce{H+}(aq)+\ce{B-}(aq)]\)?
(b) What are the equilibrium constants for the ionization of these acids?
(Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A − or B − ), and the hydrogen ion (H + ). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)
S13.4.39
(a) HB ionizes to a greater degree and has the larger K c .
(b) K c (HA) = 5 × 10 −4
K c (HB) = 3 × 10 −3


1.2: Pharmacodynamics
Complex Interactions
So far, we have learned the importance of pharmacokinetics in describing how the body absorbs, moves, processes, and eliminates a medication. Now let’s consider a drug’s impact on the body, a series of complex interactions known as pharmacodynamics .
When considering how the cells of the body respond to medications, it is important to remember that the majority of drugs bind to specific receptors on the surface or interior of cells. However, there are many other cellular components and non-specific sites that can serve as receptor sites where drugs can bind to create a response. For example, did you know that an osmotic laxative like magnesium citrate attracts and binds with water? This medication works to pull water content into the bowel and increases the likelihood of a bowel movement.
Other medications may inhibit specific enzyme binding sites in order to impact the functionality of a cell or tissue. For example, antimicrobial and antineoplastic drugs commonly work by inhibiting enzymes that are critical to the function of the cell. With blockage of the enzyme binding site, the cell microbe or neoplastic cell is no longer viable and cell death occurs.
Agonist and Antagonist Actions
Understanding the mechanism of action , [1] or how a medication functions within the body, is essential to understanding the processes medications go through to produce the desired effect (see Figure 1.6). Drugs have agonistic or antagonistic effects. A drug agonist binds tightly to a receptor to produce a desired effect. A drug antagonist competes with other molecules and blocks a specific action or response at a receptor site. For example, the cardiac medication atenolol (Tenormin) is a beta-1 receptor antagonist used to treat patients with hypertension or heart disease. Beta-1 receptor antagonist medications like atenolol produce several effects by blocking beta-1 receptors: a negative inotropic effect occurs by weakening the contraction of the heart, thus causing less work of the heart muscle; a negative chronotropic effect occurs when the heart rate is decreased; and a negative dromotropic effect occurs when the conduction of the electrical charge in the heart is slowed. Understanding the effects of a beta-1 antagonist medication allows the nurse to anticipate expected actions of the medication and the patient response. Agonistic and antagonistic effects on receptors are further discussed in the “Autonomic Nervous System” chapter. [2]
- "Mechanism of Action" by Dominic Slausen at Chippewa Valley Technical College is licensed under CC BY 4.0 ↵
- This work is a derivative of Principles of Pharmacology by LibreTexts licensed under CC BY-NC-SA 4.0 . ↵


53.12: Altruism
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Learning Objectives
- Explain how altruistic behaviors can benefit populations
Altruistic Behaviors
Behaviors that lower the fitness of the individual engaging in the behavior, but increase the fitness of another individual, are termed altruistic. Examples of such behaviors are seen widely across the animal kingdom. Social insects, such as worker bees, have no ability to reproduce, yet they maintain the queen so she can populate the hive with her offspring. Meerkats keep a sentry standing guard to warn the rest of the colony about intruders, even though the sentry is putting itself at risk. Wolves and wild dogs bring meat to pack members not present during a hunt. Lemurs take care of infants unrelated to them. Although on the surface these behaviors appear to be altruistic, it may not be so simple.
Why Does Altruism Exist?
There has been much discussion over why altruistic behaviors exist. Do these behaviors lead to overall evolutionary advantages for their species ? Do they help the altruistic individual pass on its own genes? And what about such activities between unrelated individuals? One explanation for altruistic-type behaviors is found in the genetics of natural selection. In the 1976 book, The Selfish Gene, scientist Richard Dawkins attempted to explain many seemingly-altruistic behaviors from the viewpoint of the gene itself. Although a gene obviously cannot be selfish in the human sense, it may appear that way if the sacrifice of an individual benefits related individuals that share genes that are identical by descent (present in relatives because of common lineage). Mammal parents make this sacrifice to take care of their offspring. Emperor penguins migrate miles in harsh conditions to bring food back for their young. Selfish gene theory has been controversial over the years and is still discussed among scientists in related fields.
Even less-related individuals (those with less genetic identity than that shared by parent and offspring) benefit from seemingly altruistic behavior. The activities of social insects such as bees, wasps, ants, and termites are good examples. Sterile workers in these societies take care of the queen because they are closely related to it; as the queen has offspring, she is passing on genes from the workers indirectly. Thus, it is of fitness benefit for the worker to maintain the queen without having any direct chance of passing on its genes due to its sterility. The lowering of individual fitness to enhance the reproductive fitness of a relative and, thus, one’s inclusive fitness evolves through kin selection. This phenomenon can explain many superficially-altruistic behaviors seen in animals. However, these behaviors may not be truly defined as altruism in these cases because the actor is actually increasing its own fitness either directly (through its own offspring) or indirectly (through the inclusive fitness it gains through relatives that share genes with it).
Unrelated individuals may also act altruistically to each other; this seems to defy the “selfish gene” explanation. An example of this is observed in many monkey species where a monkey will present its back to an unrelated monkey to have that individual pick the parasites from its fur. After a certain amount of time, the roles are reversed and the first monkey now grooms the second monkey. Thus, there is reciprocity in the behavior. Both benefit from the interaction and their fitness is raised more than if neither cooperated or if one cooperated and the other did not. This behavior is still not necessarily altruism, as the “giving” behavior of the actor is based on the expectation that it will be the “receiver” of the behavior in the future; a concept termed reciprocal altruism. Reciprocal altruism requires that individuals repeatedly encounter each other, often the result of living in the same social group, and that cheaters (those that never “give back”) are punished.
Evolutionary Game Theory and Altruism
According to evolutionary game theory, a modification of classical game theory in mathematics, many of these so-called “altruistic behaviors” are not altruistic at all. The definition of “pure” altruism, based on human behavior, is an action that benefits another without any direct benefit to oneself. Most of the behaviors previously described do not seem to satisfy this definition; game theorists are good at finding “selfish” components in them. Others have argued that the terms “selfish” and “altruistic” should be dropped completely when discussing animal behavior, as they describe human behavior and may not be directly applicable to instinctual animal activity. What is clear, though, is that heritable behaviors that improve the chances of passing on one’s genes or a portion of one’s genes are favored by natural selection and will be retained in future generations as long as those behaviors convey a fitness advantage.
Key Points
- Behaviors that lower the fitness of the individual, but increase the fitness of another individual are termed altruistic; why altruistic behaviors exist has been the topic of some debate.
- One explanation for altruistic-type behaviors is found in the genetics of natural selection and the “selfish gene ” theory: although a gene cannot be selfish in the human sense, it may appear that way if the sacrifice of an individual benefits related individuals that share genes that are identical.
- Even less-related individuals, those with less genetic identity than that shared by parent and offspring, benefit from seemingly-altruistic behavior, such as sterile worker bees protecting the queen.
- Unrelated individuals may also act altruistically to each other, which seems to defy the “selfish gene” explanation; however, this altruism is typically reciprocal, in that both benefit from the interaction.
- Most of the behaviors described when speaking of altruism do not seem to satisfy the definition of “pure” altruism; some evolutionary game theorists suggest that we get rid of the terms “altruistic” and “selfish” altogether since they describe human behavior.
Key Terms
- kin selection : an evolutionary mechanism by which an organism’s behavior benefits the reproductive success of its relatives, including at the expense of its own survival or reproduction
- altruism : devotion to the interests of others; brotherly kindness; opposed to egoism or selfishness
- game theory : a branch of applied mathematics that studies strategic situations in which individuals or organizations choose various actions in an attempt to maximize their returns


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1.1
Psychology and Human Development
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Lifespan development is a subfield of the discipline of psychology; human development adds a multidisciplinary approach to the study of the lifespan.
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The study of lifespan development extends back to the ancient Greeks, through the Child Study Movement in the early 1900s, to today.
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Lifespan development as a scientific discipline is concerned with application—taking findings and putting them to use in people’s lives.
1.2
Themes of Development
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Lifespan development is especially interested in the timing of developmental events and their impact on development.
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Depending on the topic, development can occur continuously (smooth and gradual) or discontinuously (abrupt or sudden).
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Lifespan development addresses the nature and nurture question, which explores the sources of developmental change and the relative contributions of genetics and the environment.
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Critical periods are times when an experience is expected and necessary for development to occur. Sensitive periods are windows of time when certain experiences can lead to an individual’s ideal development.
1.3
Major Theories and Theorists
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Lifespan psychologists use many different theoretical perspectives to study development, including cognitive, biological and evolutionary, and behaviorist perspectives.
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Our understanding benefits from the overlap of these perspectives—that is, the theories do not necessarily compete with one another, but instead complement one another, leading to a more complete and holistic understanding of development.
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The perspectives have a range of emphasis from the biological to the environmental; each perspective has a view on the nature and nurture question.
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The behaviorist perspective has helped scientists identify important learning and behavior concepts including classical conditioning, operant conditioning, and observational learning.
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Theories from the cognitive perspective that have helped us understand various facets of development include psychosocial theory of development, stage theory of cognitive development, sociocultural theory of cognitive development, and information processing theory.
1.4
Contexts and Settings of Development
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The study of the individual across the lifespan has benefitted from an ecological systems model in which each system is made up of contexts (distinct environments) in which development can be influenced.
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Psychological research findings are often broken down and further analyzed by social sub-groupings including gender, sexual orientation, ethnoracial background, socioeconomic status, culture, and religion.
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A major concern for the field of lifespan development is to be more aware of the universality and generalizability of its research, and to make findings as applicable as possible to people from around the world. Criticism of research centered on participants exclusively from Western, Educated, Industrialized, Rich, and Democratic societies is summarized in the acronym “WEIRD.”
1.5
Lifespan Development as a Science: Research Methods
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Psychological science has many scientific tools to explore developmental questions.
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These tools include naturalistic observation, case studies, correlational studies, experiments, natural experiments, and topic-specific apparatuses.
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Developmental researchers can choose a number of developmental research designs—cross-sectional, longitudinal, and cross-sequential.
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Ethical considerations guide all scientific research, especially when studying humans. A set of ethical principles operates in developmental research.
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Informed consent is a key requirement for psychological research.


2.8: Interdisciplinary Influences [3]
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Interdisciplinary refers to more than one branch of knowledge. In the case of Early Childhood Education, the disciplines include medicine, psychology, biology, parent educators and other early childhood professionals who have knowledge that helps to inform our practices with children and families. The collective knowledge we gain from these contributions, gives our field the evidence to support the role of the teacher in providing engaging environments, meaningful curriculum, guidance strategies, etc. This is often referred to as “best practices.” As you continue to engage with this textbook, this will become more apparent to you as it relates to the unique role of an early childhood professional.


3.1: Career Criminals
The United States Sentencing Commission was directed by Congress to set sentencing guidelines for repeat violent offenders or repeat drug trafficking offenders, known as “Career Offenders,” at or near the statutory maximum penalty. Tracking statutory criteria, a defendant qualifies as a Career Offender in the sentencing guidelines if:
the defendant was at least 18 years of age at the time he or she committed the instant offense the instant offense is a felony that is a crime of violence or a controlled substance offense the defendant has at least two prior felony convictions of either a crime of violence or a controlled substance offense.
There were 75,836 federal criminal cases reported to the United States Sentencing Commission during fiscal year 2014. Of the 67,672 cases in which the Commission received complete guideline application information, 2,269 (3.4%) offenders were sentenced as career offenders. Offenders sentenced under the career offender guidelines over the past ten years have consistently accounted for about three percent of federal offenders sentenced each year.
While career offenders account for just 3% of the annual federal caseload, they account for more than 11% of the federal prison population due to their lengthy prison sentences (on average, more than 12 years, or 147 months, in prison).
Career offenders often receive sentences below the guideline range (often at the government’s request), especially when they qualify as career offenders on the basis of drug trafficking offenses alone (“drug trafficking only” pathway).
Although career offenders with a violent instant or prior offense often have more serious criminal histories, the career offender directive has the most significant impact on drug trafficking offenders because they often carry higher statutory maximum penalties than some violent offenses
Despite similar average guideline minimums, “drug trafficking only” career offenders are generally sentenced less severely than other career offenders. In these cases, federal judges impose sentences similar to the sentences recommended in the guidelines for the underlying drug trafficking offense.
While career offenders, as a group, tend to recidiviate at higher rates than non-career offenders, the United States Sentencing Commission found a lower recidivism rate among career offenders qualifying on the basis of drug trafficking offenses alone.
In addition to having a more serious and extensive criminal history, career offenders who have committed a violent offense recidivate at a higher rate and are more likely to commit another violent offense in the future (see the “violent only” and “mixed” groups in the table below).
|
Drug Trafficking |
Only Mixed |
Only Mixed |
|
|
Recidivism Rate: |
54.4% |
69.4% |
69.0% |
|
Median Time to Recidivism |
26 Months |
20 Months |
14 Months |
|
Most Serious Post-Release Event |
Trafficking |
Assault |
Robbery |
|
26.5% |
28.6% |
35.3% |
Table 1. U.S. Sentencing Commission’s Recidivism Study Cohort Followed for Eight Years
Offender Demographics & Offense Characteristics
In fiscal year 2014, Black offenders accounted for more than half (59.7%) of offenders sentenced under the career offender guideline, followed by White (21.6 %), Hispanic (16.0%), and Other races (2.7%). Nearly all offenders sentenced under the career offender guideline were male (97.5%) and U.S. citizens (97.7%). Their average age at sentencing was 38 years.
In 2014, nearly three-quarters (74.1%) of career offenders were convicted of a drug trafficking offense. The remaining offenders would have been sentenced for robbery (11.6%), unlawful receipt, possession, or transportation of firearms (5.4%), aggravated assault (1.6%), and drug offenses occurring near a protected location (1.6%). Other offense types constituted the remaining 5.8 percent.


Blood Vessels and Blood Vocabulary
Acquired immunodeficiency syndrome (AIDS)
: A disease caused by the human immunodeficiency virus (HIV). People with acquired immunodeficiency syndrome are at an increased risk for developing certain cancers and for infections that usually occur only in individuals with a weak immune system (National Cancer Institute, n.d.)
Anaphylaxis
: An acute hypersensitivity reaction due to exposure to a previously encountered antigen.
Anemia
: A condition in which the number of red blood cells or hemoglobin is deficient.
Aneurysm:
Weakening of the wall of a blood vessel, causing it to thin and balloon out, and possibly eventually burst, resulting in internal bleeding.
Angiography
: A procedure to x-ray blood vessels.
Angioplasty
: A procedure in which an occlusion is mechanically widened with a balloon.
Angioscope
: Instrument used for visual examination of blood vessels.
Angioscopy
: Endoscopic examination of blood vessels.
Anti-B antibodies
: Proteins that will mount an immune response against B antigens.
Antibodies
: Proteins made by plasma cells (a type of white blood cell) in response to an antigen (a substance that causes the body to make a specific immune response). Each antibody can bind to only one specific antigen. The purpose of this binding is to help destroy the antigen.
Antigens
: Substances that provokes an immune response. This happens because the immune system sees the antigen as foreign, or ‘non-self” (does not belong in that body).
Aortic stenosis
: A condition in which the aortic valve becomes rigid and may calcify over time.
Artery
: A blood vessel that transports blood away from the heart.
Arteriole
: A very small artery that leads to a capillary.
Arteriogram
: An x-ray of arteries.
Arteriosclerosis
: The generalized loss of compliance; “hardening of the arteries”.
Atherectomy
: Excision of fatty plaque.
Atherosclerosis
: A hardening of the arteries that involves the accumulation of fatty plaque.
Brachial artery
: The large artery in the upper arm near the biceps muscle.
Capillaries
: The smallest type of blood vessel. A capillary connects an arteriole (small artery) to a venule (small vein) to form a network of blood vessels in almost all parts of the body.
Cardiac output
: The measurement of blood flow from the heart through the ventricles and is usually measured in liters per minute. Any factor that causes cardiac output to increase, by elevating heart rate or stroke volume or both, will elevate blood pressure and promote blood flow.
Cardiac tamponade
: A potentially fatal condition in which excess fluid builds within the pericardial space, preventing the heart from beating effectively.
Cardiogenic
: Originating from the heart.
Carotid artery
: Located in the neck, it is one of the three major branches of the aortic arch.
Centrifugation
: Process of using a rotating machine to generate centrifugal force to separate substances of different densities, remove moisture, or simulate gravitational effects.
Chemoreceptors
: Cells that sense changes in chemical levels.
Chemotaxis
: Movement in response to chemicals; a phenomenon in which injured or infected cells and nearby leukocytes emit the equivalent of a chemical “911” call, attracting more leukocytes to the site.
Compliance
: The ability of the blood vessels to dilate and constrict as needed.
Coronary artery bypass graft (CABG)
: Surgery in which a healthy blood vessel taken from another part of the body is used to make a new path for blood around a blocked artery leading to the heart. This restores the flow of oxygen and nutrients to the heart.
Coronary heart disease
: A disease in which there is a narrowing or blockage of the coronary arteries.
Crohn’s disease
: A condition in which the gastrointestinal tract is inflamed over a long period of time.
Diapedesis
: The migration of blood cells through the intact walls of blood vessels into the surrounding tissue.
Diastolic pressure
: The arterial pressure of blood during ventricular relaxation, or diastole.
Edema
: Swelling due to excessive liquid in the tissues.
Embolus
: An obstruction such as a blood clot or plaque that blocks the flow of blood in an artery or vein.
Endarterectomy
: Excision of plaque from within the artery.
Endothelium
: Epithelium that lines vessels in the lymphatic and cardiovascular systems.
Epiphyses
: The wider section at the end of long bones.
Erythrocyte
: A red blood cell.
Erythropoietin (EPO)
: A hormone produced by the kidneys that triggers the production of red blood cells.
Extramedullary h
ematopoiesis
: Hematopoiesis outside the medullary cavity of adult bones.
Heart rate
: The number of times the heart beats within a certain time period, usually a minute.
Hematocrit
: A lab test which measures the percentage red blood cells in a sample of whole blood.
Hematologist
: A doctor who has special training in diagnosing and treating blood disorders.
Hematology
: The study of blood and blood-forming issues.
Hematoma
: A pool of mostly clotted blood that forms in an organ, tissue, or body space.
Hemolysis
: The breakdown of red blood cells.
Hemopoiesis
: The process by which the body produces blood.
Hemopoietic growth factors
: Chemical messengers which promote the proliferation and differentiation of formed elements and include erythropoietin, thrombopoietin, colony-stimulating factors, and interleukins.
Hemorrhage
: Excessive bleeding.
Hemostasis
: The process by which the body seals a ruptured blood vessel to prevent further blood loss.
Homeostasis
: The state of steady internal conditions maintained by living things.
Hypertension
: Abnormally high blood pressure.
Hypothermia
: Abnormally low body temperature.
Hypothyroidism
: The disease state caused by insufficient production of thyroid hormone by the thyroid gland.
Hypovolemic
: An abnormally low volume of blood circulating through the body.
Hypoxemia
: Below-normal level of oxygen saturation of blood (typically <95 percent).
Hypoxia
: Lack of oxygen supply to the tissues.
Immunodeficiency
: The decreased ability of the body to fight infections and other diseases.
Intravenous:
Into or within the vein.
Ischemia
: Lack of blood flow to body tissues.
Leukocyte
: White blood cell(s).
Leukocytopenia
: An abnormal decrease in the number of leukocytes.
Lupus
: A chronic, inflammatory, connective tissue disease that can affect the joints and many organs.
Lymphadenitis
: Inflammation of lymph nodes.
Lymphadenopathy
: Disease or swelling of the lymph nodes.
Lymphoma
: A form of cancer in which masses of malignant T and/or B lymphocytes collect in lymph nodes, the spleen, the liver, and other tissues. These leukocytes do not function properly, and the patient is vulnerable to infection.
Macrophage
: A large cell derived from a monocyte; they participate in innate immune responses.
Medulla oblongata
: A part of the brain stem responsible for control of heart rate and breathing.
Myeloma
: Cancer that arises in plasma cells.
Myelopoiesis
: Formation of bone marrow.
Pancytopenia
: A condition in which there is a lower-than-normal number of red and white blood cells and platelets in the blood.
Perfusion
: Penetration of blood.
Peripheral arterial disease
: Obstruction of vessels in peripheral regions of the body.
pH
: A measure of how acidic or alkaline a substance is, as determined by the number of free hydrogen ions in the substance.
Phagocytized
: The process by which certain cells are able to “eat” other cells or substances by engulfing them.
Phlebitis
: Inflammation of a vein.
Phlebotomist
: A medical professional trained to draw blood, typically by performing a venipuncture of a surface vein of the arm.
Phlebotomy
: A procedure in which a needle is used to take blood from a vein, usually for laboratory testing.
Placenta
: The organ that supplies oxygen and nutrients to the fetus, excretes waste products, and produces and secretes estrogens and progesterone.
Plaque
: A fatty material including cholesterol, connective tissue, white blood cells, and some smooth muscle cells.
Plasma cells
: A type of B lymphocyte that produces antibodies which bind to specific foreign or abnormal antigens, in order to destroy them.
Plasmapheresis
: A procedure in which a machine is used to separate the plasma from the blood cells.
Pneumothorax
: An abnormal collection of air in the space between the thin layer of tissue that covers the lungs and the chest cavity that can cause all or part of the lung to collapse.
Polycythemia vera
: A type of bone marrow disease that causes an excessive production of immature erythrocytes.
Pulmonary embolism
: A blood clot within the lung.
Rheumatoid arthritis
: An autoimmune disorder in which the body mounts an immune response against its own joint tissues, causing inflammation and damage to the joints.
Sepsis
: Organismal-level inflammatory response to a massive infection.
Sickle cell disease
: An inherited disease in which the red blood cells have an abnormal crescent shape, block small blood vessels, and do not last as long as normal red blood cells; also called sickle cell anemia.
Splenectomy
: Excision of the spleen.
Splenomegaly
: Enlarged spleen.
Sphygmomanometer
: A blood pressure cuff attached to a measuring device, or gauge.
Systolic pressure
: The arterial pressure resulting from the ejection of blood during ventricular contraction, or systole.
Thalassemia
: A genetic disorder characterized by abnormal synthesis of globin proteins and excessive destruction of erythrocytes.
Thrombocyte
: Platelets.
Thrombocytopenia
: A condition in which there is an insufficient number of platelets.
Thrombocytosis
: A condition in which there are too many platelets.
Thrombophlebitis
: Inflammation of a vein that occurs when a blood clot forms.
Thrombosis
: The formation of unwanted blood clots.
Thrombus
: Aggregation of fibrin, platelets, and erythrocytes in an intact artery or vein.
Thrombolysis
: The process of breaking up a thrombus that is blocking blood flow.
Thymectomy
: Excision of the thymus gland.
Thymoma
: Tumor of the thymus gland.
Tissue rejection
: The recipient’s immune system recognizes the transplanted tissue as non-self and mounts an immune response against it, ultimately destroying it.
Vasoconstriction
: The physiological narrowing of blood vessels by contraction of the vascular smooth muscle.
Vasodilation
: The physiological widening of blood vessels by relaxing the vascular smooth muscle.
Veins
: Blood vessels that carry blood back to the heart.
Venules
: Small blood vessels that carry blood to a vein.
Viscosity
: A measure of a fluid’s thickness or resistance to flow.
Image Descriptions
Figure 10.1 image description:
The left panel shows the structure of a skeletal muscle vein pump when the muscle is relaxed, and the right panel shows the structure of a skeletal muscle vein pump when the muscle is contracted.
[Return to Figure 10.1].
Figure 10.2 image description:
The top left panel of this figure shows the ultrastructure of an artery (labels read from top: tunica externa, tunica media, tunica intima, smooth muscle, internal elastic membrane, vasa vasorum, external elastic membrane, nervi vasorum, endothelium, elastic fiber), and the top right panel shows the ultrastructure of a vein (labels read from top: tunica externa, tunica media, tunica intima, vasa vasorum, smooth muscle, endothelium). The bottom panel shows a micrograph with the cross-sections of an artery and a vein.
[Return to Figure 10.2].
Figure 10.3 image description:
The major arteries in the human body. Labels read (from the top, clockwise) right common carotid, left common carotid, axillary, pulmonary trunk, descending aorta, diaphragm, renal, superior mesenteric, gonadal, inferior mesenteric, common iliac, internal iliac, deep femoral, femoral, descending genicular, dorsalis pedis, plantar arch, fibular, anterior tibial, posterior tibial, popliteal, palmar arches, external iliac, ulnar, radial, brachial, celiac trunk, ascending aorta, aortic arch, brachiocephalic trunk, right subclavian, vertebral.
[Return to Figure 10.3].
Figure 10.4 image description:
The major veins in the human body. Labels read (from the top, clockwise) internal jugular, brachiocephalic, superior vena cava, intercostal, inferior vena cava, gonadal, lumbar, right and left common iliac, external iliac, internal iliac, deep femoral, femoral, posterior tibial, anterior tibial, dorsal venous arch, plantar venous arch, fibular, small saphenous, popliteal, great saphenous, digital, palmar venous arches, ulnar, median antebrachial, medial cubital, hepatic, basilic, brachial, cephalic, axillary, subclavian, external jugular.
[Return to Figure 10.4].
Figure 10.5 image description:
This diagram shows how oxygenated and deoxygenated blood flows through the major organs in the body. Pulmonary circulation involves the lungs, pulmonary artery and vein, vena cava, and aorta. Systemic circulation involves the upper body, hepatic vein, renal vein, aorta, liver, hepatic artery, hepatic portal vein, stomach, intestines, renal artery, kidneys, and lower body.
[Return to Figure 10.5].
Figure 10.6 image description:
The pulse points as shown on a woman’s body. Labels read (from top) temporal artery, facial artery, common carotid artery, brachial artery, radial artery, femoral artery, popliteal artery, posterior tibial artery, dorsalis pedis artery.
[Return to Figure 10.6].
Figure 10.7 image description:
This figure shows three test tubes with a red and yellow liquid in them. The left panel shows normal blood, the center panel shows anemic blood and the right panel shows polycythemia. Labels indicate plasma (water, proteins, nutrients, hormones et cetera), buffy coat (white blood cells, platelets), and hematocrit (red blood cells).
[Return to Figure 10.7].
Figure 10.8 image description:
This flowchart shows the pathways in which a multipotent hematopoietic stem cell differentiates into the different cell types found in blood. From the top (multipotent hematopoietic stem cells can divide and some cells remain stem cells, while the remaining cell goes down one of two paths depending on the chemical signals received: myeloid stem cell or lymphoid stem cell. A myeloid stem cell then can become either a megakaryoblast (which then turns into a megakaryocyte, then becomes platelets), or it can become a proerythroblast (which then becomes a reticulocyte, then becoming an erythrocyte), or it can become a myeloblast (which then becomes either a basophil, neutrophil, eosinophil), or it can become a monoblast (which then it becomes a monocyte). If the cell becomes a lymphoid stem cell, it then becomes a lymphoblast, which then becomes either a natural killer cell or a small lymphocyte ( either T or B lymphocyte).
[Return to Figure 10.8].
Figure 10.9 image description:
This image shows a microscopic view of erythrocytes (red blood cells). Erythrocytes have the appearance of a disc with a shallow center, which aids their function.
[Return to Figure 10.9].
Figure 10.10 image description:
This image shows a micrographic view of different leukocytes. From left to right: basophil, eosinophil, neutrophil, monocyte, lymphocyte.
[Return to Figure 10.10].
Figure 10.11 image description:
This figure shows how leukocytes respond to chemical signals from injured cells. The top panel shows chemical signals sent out by the injured cells (text labels read: 1) Leukocytes in the blood respond to chemical attractants released by pathogens and chemical signals from nearby injured cells). The middle panel shows leukocytes migrating to the injured cells (text labels read: 2)the leukocytes squeeze between the capillary wall as they follow the chemical signals to where they are most concentrated (positive chemotaxis)). The bottom panel shows macrophages phagocytosing the pathogens (text label reads: 3) Within the damaged tissue, monocytes differentiate into macrophages that phagocytize the pathogens. The eosinophils and neutrophils release chemicals that break apart pathogens. They are also capable of phagocytosis.).
[Return to Figure 10.11].
Figure 10.12 image description:
This figure details the steps in the clotting of blood. Each step is shown along with a detailed text box describing the steps on the left. On the right, a signaling pathway shows the different chemical signals involved in the clotting process. The steps described: 1. Injury: a blood vessel is severed. Blood and blood components (e.g. erythrocytes, white blood cells, et cetera) are leaking out of the breaks. 2. Vascular spasm: the smooth muscle in the vessel wall contracts near the injury point reducing blood loss. 3. Platelet plug formation: platelets are activated by chemicals released from the injury site and by contact with underlying collagen. The platelets become spiked and stick to each other and the wound site. Initial platelets are activated by chemicals released from the injured cells and by contact with broken collagen. Bound platelets release chemicals that activate and attract other platelets. platelets move toward the source of chemical signals and bind. Platelet plug grows in size. 4. Coagulation. In coagulation, fibrinogen is converted to fibrin (see part b), which forms a mesh that traps more platelets and erythrocytes, producing a clot. Part B Fibryn synthesis cascade: Intrinsic pathway (damaged vessel wall), Extrinsic pathway (trauma to extravascular cells), final common pathway (cross-linked fibrin clot).
[Return to Figure 10.12].
Figure 10.13 image description:
This chart shows the ABO blood group types. From left to right, the columns are blood types A, B, AB, and O. In descending order, the rows are: red blood cell type; antibodies in plasma; antigens in red blood cell; and blood types compatible in an emergency.
[Return to Figure 10.13].
Figure 10.15 image description:
This figure shows three different red blood cells with different blood types.
[Return to Figure 10.15].
Figure 10.14 image description:
This figure shows an umbilical artery and vein passing through the placenta on the top left. The top right panel shows the first exposure to Rh+ antibodies in the mother. The bottom right panel shows the response when the second exposure in the form of another fetus takes place. Textboxes detail the steps in each process: First exposure birth of first Rh+ infant: 1. During birth, Rh+ fetal erythrocytes leak into maternal blood after breakage of the embryonic chorion, which normally isolates the fetal and maternal blood. 2) Maternal B cells are activated by the Rh antigen and produce large amounts of anti-Rh antibodies. Second exposure: Rh+ fetus: 3) Rh antibody titer in mother’s blood is elevated after first exposure. 4) Rh antibodies are small enough to cross the embryonic chorion and attach to the fetal erythrocytes.
[Return to Figure 10.14].
Figure 10.16 image description:
The left panel (a) shows the cross-section of a normal and a narrowed artery. A normal artery has no plaque along the artery walls which means there is normal blood flow. In a narrow artery, plague forms on the arterial walls causing abnormal blood flow. The right panel (b) shows a micrograph of an artery with plaque in it.
[Return to Figure 10.16].
Figure 10.17 image description:
This photograph shows varicose veins in the lower leg. Varicose veins are distended, twisted veins that may present in patients with edema.
[Return to Figure 10.17].
Figure 10.18 image description:
This photograph shows the red blood cells of a person suffering from sickle cell anemia.
[Return to Figure 10.18].


1.1: Freely Open and Accessible
Open, Accessible and Free
With the advent of digital materials and learning management systems being integrated in grounded and online classes, course materials and resources are more accessible (in terms of access) and accessible (in terms of WACAG standards) than ever before. The terms 'free' and 'open licensed' as they relate to educational materials are often used interchangeably or in association with OER. That is because the intent of OER materials, by design, is to be open for use and adaption freely.
But just because a material is free, does that make it an OER?
"Open," in OER, relies on the design of the material to be not only free, but to carry a license, placed by the author, that allows them to be freely used in a variety of ways. Other freely available materials, such as those in the public domain, are free and open, but they were not intended to be OER. That does not take away from the value of those resources in classes, as that content could be used within OERs. But there is a difference in the materials because of their initial intent. An OER was designed with the intent to be an educational resource, whereas something that falls within the public domain may have achieved that status by other means, such as just by being government content or having a copyright expire.
Faculty and students have access to other types of zero-cost materials through their institution via the library course reserves, digital collections, and digital journals. While these materials are not necessarily open licensed or editable, they would be free. They can collaborate with library faculty to make select materials available directly in the LMS/Canvas. This not only removes the financial burden of course material costs, it allows students to use the materials immediately.
Ease of access is important, but ensuring that written and visual materials are accessible for students is likely a requirement as well. One of the benefits of finding true OERs is that they can be updated and improved, including making them more accessible for people who use screen readers and providing.
OERs are just one of many methods for the well-rounded educator who is looking to facilitate a more open learning approach with materials.
Enter Creative Commons
Seeing the limitations of copyright and public domain, Creative Commons, an international nonprofit organization, developed a set of licenses and tools to help build and sustain a commons of shared knowledge and culture ( CC:About ). The availability of Creative Commons licensing has made it possible for any creator to create works while retaining attribution. It gives educators the ability to free their content from published textbooks that are difficult to update and customize. As professors improve the content and share it with others, there is the potential for continuous improvement, a kind of "snowball" effect where others who use the text can build on what has been remixed or created.
Pulling up Anchor on Published Textbooks
Using a textbook to frame the content of a course has been a common practice among faculty for decades. This practice is often referred to as "anchoring" a text to a course, and, as long as the professor feels the content aligns with their learning objectives--or is willing to have the textbook influence the course focus--this has been a useful way to lay out content in a course term. Another challenge with this approach is the amount of information may have been designed for a different length of course term. In this traditional course that uses a traditional book, the course is typically structured in the same progression as the chapters of the text being used. But at times, it can also make a course dependent on a textbook and lock the professor into content they would not necessarily choose, in addition to the issues of whether or not it stays relevant, affordable, and available. Editions and changes by publishers can disrupt this organization with chapter shifts or by not updating the content enough to stay relevant. Students are savvy to this, and more and more often, they opt out of purchasing a book if they don't see the value in how it will be used in a course.
A good open educational resource can free the anchor for an educator, giving them the ability to organize, customize, update, and create content that works well for their course goals and outcomes.
Time and Equity
While providing students equitable access to reading and course materials for free is an incredible benefit, the very creation of OERs relies on a significant amount of time and resources by the professor. Similar to writing anything academic, whether it be a chapter, textbook, or article, creating--or even remixing--an OER for a course is a time-consuming process. It doesn't end with pushing "publish" on an OER publishing site. An OER can and should be an organic and iterative process for updates, as feedback on the OER comes back to the author from their students or from others in the community using it. It can leave a new OER author wondering, "How do I find the time to do all this?"
To the author or a team of authors, time is neither free nor necessarily open, as there are only a fixed number of hours in a day and everyone's means are different. There is no universal support structure in higher education for the creation of open educational resources, leading educators to find the time and support on their own or, possibly never have the opportunity. In some cases, or at some institutions, there may be professional development opportunities (like the WWU workshop associated with this text) or grants that provide support or compensation for OER production. This could be one of the many reasons why, despite being widely acknowledged for being beneficial to teaching and learning, the creation and adoption of OER has felt slow to catch on at a wider scale. It has also brought up questions about the subset of authors who are afforded the opportunity to create OERs, how they may present biased content in OER, and how some may remix content in ways they were not intended. These are all real concerns and all things that an OER author today should bring into consideration when they begin to map out the content of their OER.


12.11: Chapter 12 Problems
An underlined problem number or problem-part letter indicates that the numerical answer appears in Appendix I.
12.1
Consider the heterogeneous equilibrium \(\ce{CaCO3}\tx{(s)} \arrows \ce{CaO}\tx{(s)} + \ce{CO2}\tx{(g)}\). Table 12.3 lists pressures measured over a range of temperatures for this system.
12.14
The method described in Prob. 12.13 has been used to obtain high-precision values of the Henry’s law constant, \(\kHB\), for gaseous methane dissolved in water (Timothy R. Rettich, Y. Paul Handa, Rubin Battino, and Emmerich Wilhelm,
J. Phys. Chem.
,
85
, 3230–3237, 1981). Table 12.6 lists values of \(\ln (\kHB/p\st)\) at eleven temperatures in the range \(275\K\)–\(328\K\) and at pressures close to \(1\br\). Use these data to evaluate \(\Delsub{sol,B}H\st\) and \(\Delsub{sol,B}C\st_p\) at \(T=298.15\K\). This can be done by a graphical method. Better precision will be obtained by making a least-squares fit of the data to the three-term polynomial \begin{equation*} \ln (\kHB/p\st) = a + b(1/T) + c(1/T)^2 \end{equation*} and using the values of the coefficients \(a\), \(b\), and \(c\) for the evaluations.
12.15
Liquid water and liquid benzene have very small mutual solubilities. Equilibria in the binary water–benzene system were investigated by Tucker, Lane, and Christian (
J. Solution Chem.
,
10
, 1–20, 1981) as follows. A known amount of distilled water was admitted to an evacuated, thermostatted vessel. Part of the water vaporized to form a vapor phase. Small, precisely measured volumes of liquid benzene were then added incrementally from the sample loop of a liquid-chromatography valve. The benzene distributed itself between the liquid and gaseous phases in the vessel. After each addition, the pressure was read with a precision pressure gauge. From the known amounts of water and benzene and the total pressure, the liquid composition and the partial pressure of the benzene were calculated. The fugacity of the benzene in the vapor phase was calculated from its partial pressure and the second virial coefficient.
At a fixed temperature, for mole fractions \(x\B\) of benzene in the liquid phase up to about \(3\timesten{-4}\) (less than the solubility of benzene in water), the fugacity of the benzene in the equilibrated gas phase was found to have the following dependence on \(x\B\): \[ \frac{\fug\B}{x\B} = \kHB - Ax\B \] Here \(\kHB\) is the Henry’s law constant and \(A\) is a constant related to deviations from Henry’s law. At \(30\units{\(\degC\)}\), the measured values were \(\kHB=385.5\br\) and \(A=2.24\timesten{4}\br\).
(a) Treat benzene (B) as the solute and find its activity coefficient on a mole fraction basis, \(\g\xbB\), at \(30\units{\(\degC\)}\) in the solution of composition \(x\B=3.00\timesten{-4}\).
(b) The fugacity of benzene vapor in equilibrium with pure liquid benzene at \(30\units{\(\degC\)}\) is \(\fug\B^*=0.1576\br\). Estimate the mole fraction solubility of liquid benzene in water at this temperature.
(c) The calculation of \(\g\xbB\) in part (a) treated the benzene as a single solute species with deviations from infinite-dilution behavior. Tucker et al suggested a dimerization model to explain the observed negative deviations from Henry’s law. (Classical thermodynamics, of course, cannot prove such a molecular interpretation of observed macroscopic behavior.) The model assumes that there are two solute species, a monomer (M) and a dimer (D), in reaction equilibrium: \(\ce{2M} \arrows \ce{D}\). Let \(n\B\) be the total amount of C\(_6\)H\(_6\) present in solution, and define the mole fractions \[ x\B \defn \frac{n\B}{n\A+n\B} \approx \frac{n\B}{n\A} \] \[ x\subs{M} \defn \frac{n\subs{M}}{n\A+n\subs{M}+n\subs{D}} \approx \frac{n\subs{M}}{n\A} \qquad x\subs{D} \defn \frac{n\subs{D}}{n\A+n\subs{M}+n\subs{D}} \approx \frac{n\subs{D}}{n\A} \] where the approximations are for dilute solution. In the model, the individual monomer and dimer particles behave as solutes in an ideal-dilute solution, with activity coefficients of unity. The monomer is in transfer equilibrium with the gas phase: \(x\subs{M}=\fug\B/\kHB\). The equilibrium constant expression (using a mole fraction basis for the solute standard states and setting pressure factors equal to 1) is \(K=x\subs{D}/x\subs{M}^2\). From the relation \(n\B=n\subs{M}+2n\subs{D}\), and because the solution is very dilute, the expression becomes \[ K = \frac{x\B-x\subs{M}}{2x\subs{M}^2} \]
Make individual calculations of \(K\) from the values of \(\fug\B\) measured at \(x\B=1.00\timesten{-4}\), \(x\B=2.00\timesten{-4}\), and \(x\B=3.00\timesten{-4}\). Extrapolate the calculated values of \(K\) to \(x\B{=}0\) in order to eliminate nonideal effects such as higher aggregates. Finally, find the fraction of the benzene molecules present in the dimer form at \(x\B=3.00\timesten{-4}\) if this model is correct.
12.16
Use data in Appendix H to evaluate the thermodynamic equilibrium constant at \(298.15\K\) for the limestone reaction \[ \ce{CaCO3}\tx{(cr, calcite)} + \ce{CO2}\tx{(g)} + \ce{H2O}\tx{(l)} \arrow \ce{Ca^2+}\tx{(aq)} + \ce{2HCO3-}\tx{(aq)} \]
12.17
For the dissociation equilibrium of formic acid, \(\ce{HCO2H}\tx{(aq)} \arrows \ce{H+}\tx{(aq)} + \ce{HCO2-}\tx{(aq)}\), the acid dissociation constant at \(298.15\K\) has the value \(K\subs{a}=1.77\timesten{-4}\).
(a) Use Eq. 12.9.7 to find the degree of dissociation and the hydrogen ion molality in a 0.01000 molal formic acid solution. You can safely set \(\G\subs{r}\) and \(\g\subs{\(m\),HA}\) equal to \(1\), and use the Debye–Hückel limiting law (Eq. 10.4.8) to calculate \(\g_{\pm}\). You can do this calculation by iteration: Start with an initial estimate of the ionic strength (in this case 0), calculate \(\g_{\pm}\) and \(\alpha\), and repeat these steps until the value of \(\alpha\) no longer changes.
(b) Estimate the degree of dissociation of formic acid in a solution that is 0.01000 molal in both formic acid and sodium nitrate, again using the Debye–Hückel limiting law for \(\g_{\pm}\). Compare with the value in part (a).
12.18
Use the following experimental information to evaluate the standard molar enthalpy of formation and the standard molar entropy of the aqueous chloride ion at \(298.15\K\), based on the conventions \(\Delsub{f}H\st(\tx{H\(^+\), aq})=0\) and \(S\m\st(\tx{H\(^+\), aq})=0\) (Secs. 11.3.2 and 11.8.4). (Your calculated values will be close to, but not exactly the same as, those listed in Appendix H, which are based on the same data combined with data of other workers.)
-
12.19
The solubility of crystalline AgCl in ultrapure water has been determined from the electrical conductivity of the saturated solution (J. A. Gledhill and G. McP. Malan, Trans. Faraday Soc. , 48 , 258–262, 1952). The average of five measurements at \(298.15\K\) is \(s\B=1.337\timesten{-5}\units{mol dm\(^{-3}\)}\). The density of water at this temperature is \(\rho\A^*=0.9970\units{kg dm\(^{-3}\)}\).(a) From these data and the Debye–Hückel limiting law, calculate the solubility product \(K\subs{s}\) of AgCl at \(298.15\K\).
12.20
The following reaction was carried out in an adiabatic solution calorimeter by Wagman and Kilday ( J. Res. Natl. Bur. Stand. (U.S.) , 77A , 569–579, 1973): \[ \tx{AgNO\(_3\)(s)} + \tx{KCl(aq, } m\B=0.101\units{mol kg\(^{-1}\)}) \arrow \tx{AgCl(s)} + \tx{KNO\(_3\)(aq)} \] The reaction can be assumed to go to completion, and the amount of KCl was in slight excess, so the amount of AgCl formed was equal to the initial amount of AgNO\(_3\). After correction for the enthalpies of diluting the solutes in the initial and final solutions to infinite dilution, the standard molar reaction enthalpy at \(298.15\K\) was found to be \(\Delsub{r}H\st=-43.042\units{kJ mol\(^{-1}\)}\). The same workers used solution calorimetry to obtain the molar enthalpy of solution at infinite dilution of crystalline AgNO\(_3\) at \(298.15\K\): \(\Delsub{sol,B}H^{\infty}=22.727\units{kJ mol\(^{-1}\)}\).(a) Show that the difference of these two values is the standard molar reaction enthalpy for the precipitation reaction \[ \ce{Ag+}\tx{(aq)} + \ce{Cl-}\tx{(aq)} \arrow \ce{AgCl}\tx{(s)} \] and evaluate this quantity.
(b) Evaluate the standard molar enthalpy of formation of aqueous Ag\(^+\) ion at \(298.15\K\), using the results of part (a) and the values \(\Delsub{f}H\st(\tx{Cl\(^-\), aq})=-167.08\units{kJ mol\(^{-1}\)}\) and \(\Delsub{f}H\st(\tx{AgCl, s})=-127.01\units{kJ mol\(^{-1}\)}\) from Appendix H. (These values come from calculations similar to those in Probs. 12.18 and 14.4.) The calculated value will be close to, but not exactly the same as, the value listed in Appendix H, which is based on the same data combined with data of other workers.