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Chapter 13: Q39E (page 753)

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?

a. \(2{\rm{N}}{{\rm{H}}_3}(g)\rightleftharpoons {{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g)\) \({\rm{\Delta }}H = 92{\rm{kJ}}\)

b. \({{\rm{N}}_2}(g) + {{\rm{O}}_2}(g)\rightleftharpoons 2{\rm{NO}}(g)\) \({\rm{\Delta }}H = 181{\rm{kJ}}\)

c. \(2{{\rm{O}}_3}(g)\rightleftharpoons 3{{\rm{O}}_2}(g)\) \({\rm{\Delta }}H = - 285{\rm{kJ}}\)

d.\({\rm{CaO(s) + C}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}\rightleftharpoons {\rm{CaC}}{{\rm{O}}_{\rm{3}}}{\rm{(s)}}\) \({\rm{\Delta }}H = - 176{\rm{kJ}}\)

Short Answer

Expert verified

a)

(i) Increase in temperature (heat) will move the equilibrium to the right.

(ii) Increase in pressure (decrease in volume) will move the equilibrium to the left.

b)

(i) Increase in temperature (heat) will move the equilibrium to the right.

(ii) Increase in pressure (decrease in volume) will have no effect on the equilibrium.

c)

(i) Increase in temperature (heat) will move the equilibrium to the left.

(ii) Increase in pressure (decrease in volume) will move the equilibrium to the left.

d)

(i) Increase in temperature (heat) will move the equilibrium to the left.

(ii) Increase in pressure (decrease in volume) will move the equilibrium to the right.

Step by step solution

01

Understanding change in equilibria for parts (a) and (b)

Let us find how will an increase in temperature, and decrease in the volume of the reaction vessel (in other words, increase in pressure), affect the equilibria.

a. \(2{\rm{N}}{{\rm{H}}_3}(g)\rightleftharpoons {{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g)\) \({\rm{\Delta }}H = 92{\rm{kJ}}\)

Since \({\rm{\Delta }}H > 0\)the reaction is endothermic, and hence

\({\rm{2N}}{{\rm{H}}_{\rm{3}}}{\rm{(g) + \;heat\;}}\rightleftharpoons {{\rm{N}}_{\rm{2}}}{\rm{(g) + 3}}{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\)

(i) Increase in temperature (heat) will move the equilibrium to the right.

(ii) Since the number of moles of gas on the product side is higher than the number of moles of gas on the reactant side, the increase in pressure (decrease in volume) will move the equilibrium to the left.

\({{\rm{N}}_2}({\rm{g}}) + {{\rm{O}}_2}({\rm{g}}) + {\rm{\;heat\;}}\rightleftharpoons 2{\rm{NO}}({\rm{g}})\) \({\rm{\Delta }}H = 181{\rm{kJ}}\)

Since \({\rm{\Delta }}H > 0\)the reaction is endothermic, and hence

\({{\rm{N}}_2}({\rm{g}}) + {{\rm{O}}_2}({\rm{g}}) + {\rm{\;heat\;}}\rightleftharpoons 2{\rm{NO}}({\rm{g}})\)

(i) Increase in temperature (heat) will move the equilibrium to the right.

(ii) Since the number of moles of gas on the product side is equal to the number of moles of gas on the reactant side, the increase in pressure (decrease in volume) will have no effect on the equilibrium.

02

Understanding change in equilibria for parts (c) and (d)

\(c)2{{\rm{O}}_3}(g)\rightleftharpoons 3{{\rm{O}}_2}(g)\) \({\rm{\Delta }}H = - 285{\rm{kJ}}\)

Since \({\rm{\Delta }}H > 0\), the reaction is exothermic, and hence

\(2{{\rm{O}}_3}({\rm{g}})\rightleftharpoons 3{{\rm{O}}_2}({\rm{g}}) + {\rm{\;heat\;}}\)

(i) Increase in temperature (heat) will move the equilibrium to the left.

(ii) Since the number of moles of gas on the product side is higher than the the number of moles of gas on the reactant side, the increase in pressure (decrease in volume) will move the equilibrium to the left.

\({\rm{d)CaO}}(s) + {\rm{C}}{{\rm{O}}_2}(g)\rightleftharpoons {\rm{CaC}}{{\rm{O}}_3}(s)\) \({\rm{\Delta }}H = - 176{\rm{kJ}}\)

Since \({\rm{\Delta }}H > 0\), the reaction is exothermic, and hence

\({\rm{CaO}}({\rm{s}}) + {\rm{C}}{{\rm{O}}_2}({\rm{g}})\rightleftharpoons {\rm{CaC}}{{\rm{O}}_3}({\rm{s}}) + {\rm{\;heat\;}}\)

(i) Increase in temperature (heat) will move the equilibrium to the left.

(ii) Since the number of moles of gas on the reactant side is higher than the the number of moles of gas on the product side, the increase in pressure (decrease in volume) will move the equilibrium to the right.

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Most popular questions from this chapter

Carbon reacts with water vapor at elevated temperatures

\(C(s) + {H_2}O(g) \rightleftharpoons CO(g) + {H_2}(g)\)

\({K_c} = 0.2at100{0^o}C\)

What is the concentration of CO in an equilibrium mixture with \(\left[ {{H_2}O} \right] = 0.500M\)at \(100{0^0}C\)?

Question : A 0.010Msolution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 掳C. A 0.010Msolution 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[HA(aq) 鈬� A鈭�(aq) + H+(aq)]or HB[HB(aq) 鈬� H+(aq) + B鈭�(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鈭� or B鈭�), 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.)

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 H2S with an initial pressure of 0.824 atm.

\(2{H_2}S(g) \rightleftharpoons 2{H_2}(g) + {S_2}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_p} = 2.2 \times {10^{( - 6)}}\)

(b) Show that the change is small enough to be neglected.

Convert the values of Kc to values of Kp or the values of Kp to values of Kc .

\((a)\,{N_2}\left( g \right) + 3{H_2}\left( g \right)\rightleftharpoons 2N{H_3}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_C} = 0.50\,\,at\,\,400^\circ C\)

\((b){{\rm{H}}_2}(g) + {{\rm{I}}_2}(g)\rightleftharpoons 2HI(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_c} = 50.2\,at\,{448^\circ }{\rm{C}}\)

\((c)N{a_2}{\rm{S}}{{\rm{O}}_4} \cdot 10{{\rm{H}}_2}O(s)\rightleftharpoons N{a_2}{\rm{S}}{{\rm{O}}_4}(s) + 10{{\rm{H}}_2}O(g){K_P} = 4.08 \times {10^{ - 25}}at\,{25^\circ }{\rm{C}}\)

\((d){{\rm{H}}_2}{\rm{O}}(l)\rightleftharpoons {{\rm{H}}_2}{\rm{O}}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_P} = 0.122\,at\,{50^\circ }{\rm{C}}\)

Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl2produced when a sample of NOCl with a pressure of 10.0 atm comes to equilibrium according to this reaction:

\(2NOCl(g) \rightleftharpoons 2NO(g) + C{l_2}(g)\quad {K_P} = 4.0 \times 1{0^{ - 4}}\)
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