ThermochemistryHard

Question

The difference between enthalpies of reaction at constant pressure and constant volume for the reaction

2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l) at 298 K in kJ is

Options

A.−7.43
B.+3.72
C.−3.72
D.+7.43

Solution

$\Delta H - \Delta E = \Delta n_{g}RT = (12 - 15) \times \frac{8.314}{1000} \times 298 = - 7432.7\text{ J}$

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