For the following reaction H2 (g) + ½ O2 (g) → H2O (l) The value of enthalpy change and free energy change are −68.32 and −56.69 kcal respectively at 25 °C. Calculate the value of free energy change at 30 °C.

Question

For the following reaction

[latex]H_2 (g) + ½ O_2 (g) → H_2O (l)[/latex]

The value of enthalpy change and free energy change are −68.32 and −56.69 kcal respectively at 25 °C. Calculate the value of free energy change at 30 °C.

Accepted Answer

[latex]\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}\left(\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right)_{\mathrm{p}}[/latex]

Here [latex]\Delta \mathrm{G}=-56.69 \mathrm{~kcal} ; \Delta \mathrm{H}=-68.32 \mathrm{~kcal} \quad \mathrm{T}=273+25=298 \mathrm{~K}\[/latex]

[latex]\therefore\left(\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right)_{\mathrm{P}}=\frac{-56.69+68.32}{298}=0.039[/latex]

Assuming that [latex]\left(\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right)_{\mathrm{p}}[/latex] remains constant over this range of temperature, at 30 °C we can write

[latex]\begin{aligned}\Delta \mathrm{G} & =-68.32+303 \times 0.039 \& =-68.32+11.81 \& =-56.51 \mathrm{~kcal}\end{aligned}[/latex]

Question 15.9.2: For the following reaction H2 (g) + ½ O2 (g) → H2O (l) The......

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