For the reaction : \( 2\langle\mathrm{Cu}\rangle+y_{2}\left(\mathrm{O}_{2}\right)=\left\langle\mathrm{Cu}_{2} \mathrm{O}\right\rangle ; \Delta \mathrm{G}^{\circ}=(-40,500-3.92 \mathrm{TlogT}+29.5 \mathrm{~T}) \mathrm{cal} \) The standard enthalpy change for the reaction at \( 25^{\circ} \mathrm{C} \) is \( \quad \) (kcal to the nearest intege

To determine the standard enthalpy change (\( \Delta H^\circ \)) for the reaction at \( 25^\circ \mathrm{C} \) (which is \( 298 \, \mathrm{K} \)), we'll use the provided Gibbs free energy equation: \[ \Delta G^\circ = -40,500 - 3.92 \, T \log T + 29.5 \, T \quad \text{(cal)} \] ### Step 1: Calculate \( \Delta G^\circ \) at \( 298 \, \mathrm{K} \) First, plug in \( T = 298 \, \mathrm{K} \) into the equation: \[ \Delta G^\circ = -40,500 - 3.92 \times 298 \times \log(298) + 29.5 \times 298 \] Assuming \( \log \) denotes the base-10 logarithm: \[ \log(298) \approx 2.474 \] Now compute each term: \[ -3.92 \times 298 \times 2.474 \approx -3.92 \times 298 \times 2.474 \approx -2,890 \, \text{cal} \] \[ 29.5 \times 298 \approx 8,791 \, \text{cal} \] Combine all terms: \[ \Delta G^\circ \approx -40,500 - 2,890 + 8,791 = -34,599 \, \text{cal} \] ### Step 2: Determine \( \Delta S^\circ \) at \( 298 \, \mathrm{K} \) The relationship between Gibbs free energy and entropy is given by: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \quad \Rightarrow \quad \Delta H^\circ = \Delta G^\circ + T \Delta S^\circ \] First, find \( \Delta S^\circ \) by differentiating \( \Delta G^\circ \) with respect to \( T \): \[ \Delta S^\circ = -\frac{d(\Delta G^\circ)}{dT} = 3.92 \log T + \frac{3.92}{\ln 10} - 29.5 \] Plug in \( T = 298 \, \mathrm{K} \): \[ 3.92 \log(298) \approx 3.92 \times 2.474 \approx 9.7 \, \text{cal/mol K} \] \[ \frac{3.92}{\ln 10} \approx \frac{3.92}{2.303} \approx 1.704 \, \text{cal/mol K} \] \[ \Delta S^\circ \approx 9.7 + 1.704 - 29.5 \approx -18.096 \, \text{cal/mol K} \] ### Step 3: Calculate \( \Delta H^\circ \) at \( 298 \, \mathrm{K} \) Using the values obtained: \[ \Delta H^\circ = \Delta G^\circ + T \Delta S^\circ \approx -34,599 \, \text{cal} + 298 \times (-18.096) \approx -34,599 - 5,387 \approx -39,986 \, \text{cal} \] Convert to kilocalories: \[ \Delta H^\circ \approx -40,000 \, \text{cal} = -40 \, \text{kcal} \] ### **Final Answer** The standard enthalpy change at \( 25^\circ \mathrm{C} \) is approximately **–40  kcal**.