Consider the following reaction at \( 1000^{\circ} \mathrm{C} \) and 1 atm : \( \mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ; \Delta \mathrm{G}^{0}=12761.2-7.03 \mathrm{~T} / \mathrm{mol} \). The equilibrium partial pressure of hydrogen gas is ___ atm (round off to two decimal plac

To find the equilibrium partial pressure of hydrogen gas (\( P_{\text{H}_2} \)) at \( 1000^{\circ} \mathrm{C} \), we can start by using the given Gibbs free energy equation. The standard Gibbs free energy change at equilibrium \( (\Delta G^{0}) \) is related to the equilibrium constant \( K \) by the equation: \[ \Delta G^{0} = -RT \ln K \] Given the reaction: \[ \mathrm{FeO}(\mathrm{s}) + \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{Fe}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \] At equilibrium, we write the expression for \( K \): \[ K = \frac{P_{H_2O}}{P_{H_2}} \] To find \( \Delta G^{0} \) at \( 1000^{\circ} \mathrm{C} \) (or \( 1273 \, \mathrm{K} \)), substitute \( T = 1273 \, \mathrm{K} \) into the provided equation: \[ \Delta G^{0} = 12761.2 - 7.03 \times 1273 \] Calculating the second term: \[ 7.03 \times 1273 = 8942.39 \] Thus: \[ \Delta G^{0} = 12761.2 - 8942.39 = 3818.81 \, \mathrm{J/mol} \] Now, using \( R = 8.314 \, \mathrm{J/(mol \cdot K)} \): \[ \Delta G^{0} = -RT \ln K \Rightarrow 3818.81 = -8.314 \times 1273 \ln K \] Now calculating \( -RT \): \[ -RT = -8.314 \times 1273 = -10556.562 \] So, \[ 3818.81 = 10556.562 \ln K \Rightarrow \ln K = \frac{3818.81}{10556.562} \approx 0.361 \] Taking the antilogarithm: \[ K = e^{0.361} \approx 1.435 \] Now substituting back into the equilibrium constant expression: \[ K = \frac{P_{H_2O}}{P_{H_2}} \Rightarrow K \cdot P_{H_2} = P_{H_2O} \] Assuming \( P_{H_2O} \) is approximately equal to 1 atm due to water being in the gaseous phase at this high temperature: \[ 1.435 \cdot P_{H_2} = 1 \] Thus, \[ P_{H_2} \approx \frac{1}{1.435} \approx 0.696 \, \text{atm} \] Rounding to two decimal places, the equilibrium partial pressure of hydrogen gas is approximately: **0.70 atm**