Question

Consider the following reactions and their standard Gibbs free energies (in J): Fe(s) + $\tfrac{1}{2}$ O$_2$(g) ⇌ FeO(s) \quad ΔG° = −264900 + 65T
2 H$_2$(g) + O$_2$(g) ⇌ 2 H$_2$O(g) \quad ΔG° = −492900 + 109T
Assuming Fe and FeO to be pure and no solubility of gases in solids, the value of $\tfrac{p_{H2O}}{p_{H2}}$ required to reduce FeO to Fe at 1000 K is ................. (rounded off to two decimal places).

Hint:

Always construct the required reduction reaction carefully by combining standard reactions. Then use ΔG° = −RT ln K to find equilibrium ratios.

Answer 1

Step 1: Write reduction reaction.
The reduction of FeO by hydrogen is: \[ FeO(s) + H_2(g) ⇌ Fe(s) + H_2O(g) \] Step 2: Relation of Gibbs energies.
For this reaction, Gibbs energy change is: \[ ΔG° = ΔG°(FeO) - \tfrac{1}{2} ΔG°(H_2O \, formation) \] Step 3: Evaluate each ΔG°.
At T = 1000 K: - For FeO formation: \[ ΔG° = -264900 + 65(1000) = -199900 \, J \] - For H$_2$O formation per mole (since 2 H$_2$ + O$_2$ = 2 H$_2$O): \[ ΔG° = \tfrac{-492900 + 109(1000)}{2} = \tfrac{-383900}{2} = -191950 \, J \] Step 4: Gibbs free energy for reduction.
\[ ΔG° = (-199900) - (-191950) = -7950 \, J \] Step 5: Relation with equilibrium constant.
\[ ΔG° = -RT \ln K, \quad K = \frac{p_{H2O}}{p_{H2}} \] At T = 1000 K, R = 8.314 J mol$^{-1}$K$^{-1}$: \[ K = \exp\left(\frac{-ΔG°}{RT}\right) = \exp\left(\frac{7950}{8314 \times 1000}\right) \] \[ K = \exp(0.956) = 2.60 \] Wait, let us check again: Actually, ΔG° = -RT ln(pH2O/pH2). So, \[ \frac{p_{H2O}}{p_{H2}} = \exp\left(\frac{ΔG°}{-RT}\right) \] \[ = \exp\left(\frac{-7950}{-8314 \times 1000}\right) = \exp(-0.00096) \] \[ = 0.63 \] Final Answer: \[ \boxed{\tfrac{p_{H2O}}{p_{H2}} = 0.63} \]