Question

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?

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

Short answer

The increase of temperature and pressure of the equilibrium is

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 left

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

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 right

d. (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

Step-by-step solution

Le Chatelier’s principle

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.

The increase and decrease of equilibrium

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.

The increase and decrease of equilibrium

(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

The increase and decrease of equilibrium

(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.

 Step 5: The increase and decrease of equilibrium

(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.