How will an increase in temperature affect each of the following equilibrium? How will a decrease in the volume of the reaction vessel affect each?

\(\begin{gathered} (a)2{H_2}O(g) \rightleftharpoons 2{H_2}(g) + {O_2}(g) \hfill \\ \Delta H = 484kJ \hfill \\ (b){N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g) \hfill \\ \Delta H = - 92.2kJ \hfill \\ (c)2Br(g) \rightleftharpoons B{r_2}(g) \hfill \\ \Delta H = - 224kJ \hfill \\ (d){H_2}(g) + {I_2}(s) \rightleftharpoons 2HI(g) \hfill \\ \Delta H = 53kJ \hfill \\ \end{gathered}\)

According to the Le Chatelier’s principle, if a system in equilibrium is disturbed, the system tries to restore its equilibrium. We will use the Le Chatelier’s principle to predict the direction of the reaction.

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

(a) \(2{{\text{H}}_2}{\text{O}}({\text{g}}) \rightleftharpoons 2{{\text{H}}_2}({\text{g}}) + {{\text{O}}_2}({\text{g}})\)

\({\rm{\Delta }}H = 484{\rm{kJ}}\)

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

\(2{{\text{H}}_2}{\text{O}}({\text{g}}) + {\text{heat}} \rightleftharpoons 2{{\text{H}}_2}({\text{g}}) + {{\text{O}}_2}({\text{g}})\)

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

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

(b) \({{\text{N}}_2}({\text{g}}) + 3{{\text{H}}_2}({\text{g}}) \rightleftharpoons 2{\text{N}}{{\text{H}}_3}({\text{g}})\)

\({\rm{\Delta }}H = - 92.2{\rm{kJ}}\)

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

\({{\text{N}}_2}({\text{g}}) + 3{{\text{H}}_2}({\text{g}}) \rightleftharpoons 2{\text{N}}{{\text{H}}_3}({\text{g}}) + {\text{heat}}\)

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

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

(c) \(2{\text{Br}}({\text{g}}) \rightleftharpoons {\text{B}}{{\text{r}}_2}({\text{g}})\)

\({\rm{\Delta }}H = - 224{\rm{kJ}}\)

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

\(2{\text{Br}}({\text{g}}) \rightleftharpoons {\text{B}}{{\text{r}}_2}({\text{g}}) + {\text{heat}}\)

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

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

(d) \({{\text{H}}_2}({\text{g}}) + {{\text{I}}_2}({\text{s}}) \rightleftharpoons 2{\text{HI}}({\text{g}})\)

\({\rm{\Delta }}H = 53{\rm{kJ}}\)

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

\({{\text{H}}_2}({\text{g}}) + {{\text{I}}_2}({\text{s}}) + {\text{heat}} \rightleftharpoons 2{\text{HI}}({\text{g}})\)

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

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