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: $$\begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6} \text { at } 600 . \mathrm{K}\\\ &2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37} \mathrm{at} 600 . \mathrm{K} \end{aligned}$$

We have the following related reactions and their equilibrium constants at \(600\, \mathrm{K}\): 1. \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}_{2}(g)\) with \(K_1 = 2.3 \times 10^{6}\) 2. \(2\, \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2\, \mathrm{H}_{2}\mathrm{O}(g)\) with \(K_2 = 1.8 \times 10^{37}\) We want to find the equilibrium constant for the reaction: \(\mathrm{H}_{2}\mathrm{O}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}_{2}(g)\)

We can manipulate the related reactions to get the desired reaction and its equilibrium constant. From reaction 2, we will make some modifications: $$\frac{1}{2}(2\, \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2\, \mathrm{H}_{2}\mathrm{O}(g))$$ Which gives us the following reaction: $$\mathrm{H}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}(g)$$ For a reaction multiplied by a factor, the new equilibrium constant is the old one raised to the power of that factor, so we will calculate the new equilibrium constant as: $$K'_2 = (K_2)^{\frac{1}{2}} = (1.8 \times 10^{37})^{\frac{1}{2}}$$ Now, we can combine Reaction 1 and the modified Reaction 2 to get the desired reaction: $$ (\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}_{2}(g)) + (\mathrm{H}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}(g)) $$ $$\Rightarrow \mathrm{H}_{2}\mathrm{O}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}_{2}(g)$$ We'll call the equilibrium constant for this reaction \(K'\) and calculate it by multiplying the equilibrium constants of the related reactions: $$K' = K_1 \times K'_2$$

Once we have the equilibrium constant for the desired reaction, we can calculate the standard Gibbs free energy change, \(\Delta G^{\circ}\). The formula connecting the equilibrium constant and Gibbs free energy change is: $$\Delta G^{\circ} = -RT\ln{K'}$$ where \(R\) is the gas constant \((8.314\, \text{J/mol}\cdot\text{K})\) and \(T\) is the temperature \((600\, \mathrm{K})\). Plugging in the values, we have: $$\Delta G^{\circ} = -8.314\, \text{J/mol}\cdot\text{K} \cdot 600\, \mathrm{K} \cdot \ln{(K')}$$ Now, we have all the equations we need to solve for \(\Delta G^{\circ}\).

First, calculate \(K'_2\): $$K'_2 = (1.8 \times 10^{37})^{\frac{1}{2}} = 4.243\times10^{18}$$ Next, calculate \(K'\): $$K' = K_1 \times K'_2 = 2.3\times 10^6 \times 4.243\times10^{18} = 9.758\times10^{24}$$ Finally, calculate \(\Delta G^{\circ}\): $$\Delta G^{\circ} = -8.314\, \text{J/mol}\cdot\text{K} \cdot 600\, \mathrm{K} \cdot \ln{(9.758\times10^{24})}$$ $$\Delta G^{\circ} \approx -14,109\, \text{J/mol}$$ So, the standard Gibbs free energy change for the given reaction at \(600\, \mathrm{K}\) is approximately \(-14,109\, \text{J/mol}\).