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प्रश्न
The difference between heats of reaction at constant pressure and at constanl volume for the reaction
\[\ce{2C6H6_{(l)} + 15O2_{(g)} -> 12CO2_{(g)} + 6H2O_{(l)}}\] at 25°C in kJ
विकल्प
−7.43
+3.72
−3.72
+7.43
MCQ
उत्तर
−7.43
Explanation:
∆H − ∆U = ∆ngRT
= −3 mol × 8.314 J K−1 mol−1 × 298 K
= −7432 J
= −7.43 kJ
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Enthalpy (H)
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