Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\), using the following data: \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at 600. \(\mathrm{K}\) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at \(600 . \mathrm{K}\)

The given reactions are: 1) \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K_1=2.3 \times 10^{6}\) at 600. \(\mathrm{K}\) 2) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K_2=1.8 \times 10^{37}\) at \(600 . \mathrm{K}\)

To obtain the desired reaction, we can subtract half of reaction 1 from reaction 2: \(\left(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g)\right) - \frac{1}{2}\left(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\right)\) This simplifies to the desired reaction: $\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)$

To find the equilibrium constant for the desired reaction, we'll use the relationship between the equilibrium constants of the manipulated reactions: \(\frac{K}{K_1^{\frac{1}{2}}} = K_2\) We need to find the value of K: \(K = K_2 \times K_1^{\frac{1}{2}} = (1.8 \times 10^{37}) \times (2.3 \times 10^6)^{\frac{1}{2}}\) Calculate K: \(K \approx 2.65 \times 10^{34}\)

Now that we have the equilibrium constant for the desired reaction, we can calculate the standard Gibbs free energy change using the relationship: \(\Delta G^{\circ} = -RT \ln{K}\) Where R is the gas constant, which has the value \(8.314 \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\) and T is the temperature in Kelvin (600 K in this case). Now, plug in the values: \(\Delta G^{\circ} = -(\mathrm8.314 J \cdot K^{-1} \cdot mol^{-1})(600 K) \ln{(2.65 \times 10^{34})}\) Finally, calculate \(\Delta G^{\circ}\): \(\Delta G^{\circ} \approx -4.63 \times 10^{5}\mathrm{J/mol}\) So the standard Gibbs free energy change for the reaction $\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\( at \)600\mathrm{K}\( is approximately \)-4.63 \times 10^{5}\mathrm{J/mol}$.