QuestionApril 24, 2025

Calculate Delta G^0 at 597 K for H_(2)O(g)+1/2O_(2)(g)rightarrows H_(2)O_(2)(g) using the following data: H_(2)(g)+O_(2)(g)rightarrows H_(2)O_(2)(g) K=2.8times 10^37 at 597 K 2H_(2)(g)+O_(2)(g)rightarrows 2H_(2)O(g) Delta G^0=square kJ/mol

Calculate Delta G^0 at 597 K for H_(2)O(g)+1/2O_(2)(g)rightarrows H_(2)O_(2)(g) using the following data: H_(2)(g)+O_(2)(g)rightarrows H_(2)O_(2)(g) K=2.8times 10^37 at 597 K 2H_(2)(g)+O_(2)(g)rightarrows 2H_(2)O(g) Delta G^0=square kJ/mol
Calculate Delta G^0 at 597 K for H_(2)O(g)+1/2O_(2)(g)rightarrows H_(2)O_(2)(g) using the following data: H_(2)(g)+O_(2)(g)rightarrows H_(2)O_(2)(g) K=2.8times 10^37 at 597 K 2H_(2)(g)+O_(2)(g)rightarrows 2H_(2)O(g) Delta G^0=square kJ/mol

Solution
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Answer

\Delta G^{0}_{3} = -4965.858 \left[86.2138 - \frac{1}{2} \ln K_2\right] \, \text{J/mol} Explanation 1. Write the relationship between equilibrium constants For the given reaction H_{2}O(g) + \frac{1}{2}O_{2}(g) \leftrightarrows H_{2}O_{2}(g), we can derive its equilibrium constant K_3 using the reactions provided. Let: - Reaction 1: H_{2}(g) + O_{2}(g) \leftrightarrows H_{2}O_{2}(g) with K_1 = 2.8 \times 10^{37}. - Reaction 2: 2H_{2}(g) + O_{2}(g) \leftrightarrows 2H_{2}O(g) with \Delta G^{0} = \square. The target reaction is obtained by subtracting half of Reaction 2 from Reaction 1: K_3 = \frac{K_1}{\sqrt{K_2}} 2. Relate \Delta G^{0} to K Use the formula **\Delta G^{0} = -RT \ln K** for the target reaction: \Delta G^{0}_{3} = -RT \ln K_3 = -RT \ln \left(\frac{K_1}{\sqrt{K_2}}\right) Simplify: \Delta G^{0}_{3} = -RT \left[\ln K_1 - \frac{1}{2} \ln K_2\right] 3. Substitute known values Given K_1 = 2.8 \times 10^{37} and T = 597 \, \text{K}, substitute R = 8.314 \, \text{J/mol·K}: \Delta G^{0}_{3} = - (8.314)(597) \left[\ln(2.8 \times 10^{37}) - \frac{1}{2} \ln K_2\right] 4. Simplify further Compute \ln(2.8 \times 10^{37}): \ln(2.8 \times 10^{37}) = \ln(2.8) + 37 \ln(10) = 1.0296 + 37(2.3026) = 86.2138 Thus: \Delta G^{0}_{3} = -4965.858 \left[86.2138 - \frac{1}{2} \ln K_2\right] 5. Finalize expression The final result depends on \ln K_2, which must be calculated or provided. If K_2 is known, substitute it into the equation above.

Explanation

1. Write the relationship between equilibrium constants<br /> For the given reaction $H_{2}O(g) + \frac{1}{2}O_{2}(g) \leftrightarrows H_{2}O_{2}(g)$, we can derive its equilibrium constant $K_3$ using the reactions provided. Let:<br />- Reaction 1: $H_{2}(g) + O_{2}(g) \leftrightarrows H_{2}O_{2}(g)$ with $K_1 = 2.8 \times 10^{37}$.<br />- Reaction 2: $2H_{2}(g) + O_{2}(g) \leftrightarrows 2H_{2}O(g)$ with $\Delta G^{0} = \square$.<br /><br />The target reaction is obtained by subtracting half of Reaction 2 from Reaction 1:<br />$$ K_3 = \frac{K_1}{\sqrt{K_2}} $$<br /><br />2. Relate $\Delta G^{0}$ to $K$<br /> Use the formula **$\Delta G^{0} = -RT \ln K$** for the target reaction:<br />$$ \Delta G^{0}_{3} = -RT \ln K_3 = -RT \ln \left(\frac{K_1}{\sqrt{K_2}}\right) $$<br />Simplify:<br />$$ \Delta G^{0}_{3} = -RT \left[\ln K_1 - \frac{1}{2} \ln K_2\right] $$<br /><br />3. Substitute known values<br /> Given $K_1 = 2.8 \times 10^{37}$ and $T = 597 \, \text{K}$, substitute $R = 8.314 \, \text{J/mol·K}$:<br />$$ \Delta G^{0}_{3} = - (8.314)(597) \left[\ln(2.8 \times 10^{37}) - \frac{1}{2} \ln K_2\right] $$<br /><br />4. Simplify further<br /> Compute $\ln(2.8 \times 10^{37})$:<br />$$ \ln(2.8 \times 10^{37}) = \ln(2.8) + 37 \ln(10) = 1.0296 + 37(2.3026) = 86.2138 $$<br /><br />Thus:<br />$$ \Delta G^{0}_{3} = -4965.858 \left[86.2138 - \frac{1}{2} \ln K_2\right] $$<br /><br />5. Finalize expression<br /> The final result depends on $\ln K_2$, which must be calculated or provided. If $K_2$ is known, substitute it into the equation above.
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