Question

Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.

(a) Write the expression for the equilibrium constant \(\left( {{K_c}} \right.)\)for the reversible reaction

\(F{e_2}{O_3}(s) + 3{H_2}(g) \rightleftharpoons 2Fe(s) + 3{H_2}O(g)\) \(\Delta H = 98.7kJ\)

(b) What will happen to the concentration of each reactant and product at equilibrium if more \(Fe\)is added?

(c) What will happen to the concentration of each reactant and product at equilibrium if \({H_2}O\) is removed?

(d) What will happen to the concentration of each reactant and product at equilibrium if \({H_2}\) is added?

(e) 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?

(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

Short answer

The solution for the reactant and product at equilibrium is

(a)\({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{{\rm{H}}_2}{\rm{O}}} \right]}^3}}}{{{{\left[ {{{\rm{H}}_2}} \right]}^3}}}\)

(b) The concentrations will remain the same

(c)\(\left. {{\rm{[}}{{\rm{H}}_2}{\rm{O}}} \right],\left[ {{{\rm{H}}_2}} \right]\) will decrease

(d)\(\left. {{\rm{[}}{{\rm{H}}_2}{\rm{O}}} \right],\left[ {{{\rm{H}}_2}} \right]\)will increase

(e)\(\left. {{\rm{[}}{{\rm{H}}_2}{\rm{O}}} \right],\left[ {{{\rm{H}}_2}} \right]\)will remain the same

(f)\(\left. {{\rm{[}}{{\rm{H}}_2}{\rm{O}}} \right]\)will increase

\(\left[ {{{\rm{H}}_2}} \right]\)will decrease

Step-by-step solution

Addition or removal of concentrations of given equation at equilibrium for parts (a), (b) and (c)

\({\text{F}}{{\text{e}}_2}{{\text{O}}_3}({\text{s}}) + 3{{\text{H}}_2}({\text{g}}) \rightleftharpoons 2{\text{Fe}}({\text{s}}) + 3{{\text{H}}_2}{\text{O}}({\text{g}})\)

\({\rm{\Delta H}} = 98.7{\rm{kJ}}\)

(a)Let us write the expression for the equilibrium constant for the given reversible reaction.

Since solids and liquids are not included in the equilibrium constant expression, we get

\({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{{\rm{H}}_2}{\rm{O}}} \right]}^3}}}{{{{\left[ {{{\rm{H}}_2}} \right]}^3}}}\)

(b) Let us see what will happen if more \({\rm{Fe}}\)is added.

Since\({\rm{Fe}}\)is solid, the addition will have no effect on equilibrium, and hence the concentrations will remain the same.

(c) Let us see what will happen if \({{\rm{H}}_2}{\rm{O}}\)is removed.

If some \({{\rm{H}}_2}{\rm{O}}\)is removed, more \({{\rm{H}}_2}\)will react with \({\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3}\)to produce \({{\rm{H}}_2}{\rm{O}}\), in order to achieve a new equilibrium. Therefore, the concentration of \({{\rm{H}}_2}{\rm{O}}\) and \({{\rm{H}}_2}\)will decrease.

Addition or removal of concentrations of given equation at equilibrium for parts (d), (e) and (f)

(d) Let us see what will happen if \({{\rm{H}}_2}\)is added.

If \({{\rm{H}}_2}\)is added, excess of \({{\rm{H}}_2}\)will react with \({\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3}\) to produce \({{\rm{H}}_2}{\rm{O}}\). Therefore, the concentration of \({{\rm{H}}_2}\) and \({{\rm{H}}_2}{\rm{O}}\)will increase.

(e) Let us see what will happen if the pressure on the system is increased. Since the number of moles of gas on the reactant side is equal to the number of moles of gas on the product side, change in pressure (increase) will have no effect on the equilibrium, and therefore the concentration of \({{\rm{H}}_2}\)and \({{\rm{H}}_2}{\rm{O}}\)will remain the same.

(f) Let us see what will happen if the temperature of the system is increased. Since\({\rm{\Delta H}} > 0(98.7{\rm{kJ}} > 0)\), the reaction is endothermic:

\({\text{F}}{{\text{e}}_2}{{\text{O}}_3}({\text{s}}) + 3{{\text{H}}_2}({\text{g}}) + {\text{heat}} \rightleftharpoons 2{\text{Fe}}({\text{s}}) + 3{{\text{H}}_2}{\text{O}}({\text{g}})\)

If the temperature (heat) is increased, the equilibrium will shift to the right, resulting in production of \({{\rm{H}}_2}{\rm{O\;and\;Fe}}{\rm{.\;}}\)Therefore, the concentration of \({{\rm{H}}_2}{\rm{O}}\)will increase, and the concentration of \({{\rm{H}}_2}\)will decrease