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

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.

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

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.