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Chemistry · THERMODYNAMICS
हिन्दीConsider the combustion of liquid benzene: \( 2\text{C}_6\text{H}_6(l) + 15\text{O}_2(g) \rightarrow 12\text{CO}_2(g) + 6\text{H}_2\text{O}(l) \) at \( 298\text{ K} \). The difference between the enthalpy of reaction and internal energy change, \( (\Delta H - \Delta U) \), per mole of benzene combusted is: (Take \( R = 8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} \))
Correct answer · Option 4
\( -3.72\text{ kJ} \)
Explanation
Per mole of \( \text{C}_6\text{H}_6(l) \): \( \Delta n_g = 6 - 7.5 = -1.5 \). \( \Delta H - \Delta U = \Delta n_g RT = (-1.5) \times 8.314 \times 298 = -3716.3\text{ J} \approx -3.72\text{ kJ} \).
