Question

Consider the following reactions and their standard Gibbs free energies (in J): \[ {Fe(s)} + \frac{1}{2} {O}_2(g) \rightleftharpoons {FeO(s)} \quad \Delta G^\circ = -264900 + 65T \] \[ 2 {H}_2(g) + {O}_2(g) \rightleftharpoons 2 {H}_2{O(g)} \quad \Delta G^\circ = -492900 + 109T \] Assuming Fe and FeO to be pure and no solubility of gases in the solids, the value of \( \frac{p_{H_2O}}{p_{H_2}} \) required to reduce solid FeO to solid Fe at 1000 K is _________ (rounded off to two decimal places). Given: Ideal gas constant \( R = 8.314 \, {J mol}^{-1} {K}^{-1} \).

Hint:

When solving for equilibrium in reactions, remember that the Gibbs free energy can be used to relate the reaction quotient to the equilibrium constant. At equilibrium, the standard Gibbs free energy change is zero, and the ratio of partial pressures is important for determining the required conditions.

Answer 1

To solve this problem, we use the Gibbs free energy equation for the reduction of FeO to Fe. At equilibrium, the standard Gibbs free energy change \( \Delta G^\circ \) is related to the ratio of partial pressures by: \[ \Delta G^\circ = -RT \ln Q \] where \( Q = \frac{p_{{Fe}} p_{{O}_2}}{p_{{FeO}}} \) is the reaction quotient. At standard conditions, this equation becomes: \[ \Delta G^\circ = -RT \ln \left( \frac{p_{{FeO}}}{p_{{Fe}} p_{{O}_2}} \right) \] For this reaction, we need to use the Gibbs free energy change at 1000 K. At this temperature, we calculate: For the reduction reaction: \[ \Delta G^\circ_{{FeO to Fe}} = -264900 + 65 \times 1000 = -199900 \, {J/mol} \] Now, we calculate the value of \( \frac{p_{{H}_2O}}{p_{{H}_2}} \) that would result in a similar equilibrium, using the second reaction's Gibbs free energy expression: \[ \Delta G^\circ_{{H}_2O} = -492900 + 109 \times 1000 = -383900 \, {J/mol} \] Thus, by equating the two reactions, the required ratio \( \frac{p_{{H}_2O}}{p_{{H}_2}} \) at 1000 K is: \[ \frac{p_{{H}_2O}}{p_{{H}_2}} \approx 0.37 \quad {to} \quad 0.39 \]