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Chemistry · THERMODYNAMICS
हिन्दीGiven that for benzene (\( \text{C}_6\text{H}_6 \)), the standard enthalpy of fusion is \( \Delta_{\text{fus}} H^\ominus = 9.83\text{ kJ mol}^{-1} \) and the standard enthalpy of vaporisation is \( \Delta_{\text{vap}} H^\ominus = 30.80\text{ kJ mol}^{-1} \). The heat required to sublime 0.50 mole of solid benzene under standard conditions is:
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Correct answer · Option D
20.32 kJ
Explanation
\( \Delta_{\text{sub}} H^\ominus = \Delta_{\text{fus}} H^\ominus + \Delta_{\text{vap}} H^\ominus = 9.83 + 30.80 = 40.63\text{ kJ mol}^{-1} \). For 0.50 mol: \( q = 0.50 \times 40.63 = 20.315\text{ kJ} \approx 20.32\text{ kJ} \).
