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Chemistry · THERMODYNAMICS

हिन्दी

The thermochemical equation for the combustion of ethanol is: \( \text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \); \( \Delta_r H^\ominus = -1367.0\text{ kJ mol}^{-1} \) at 298 K. The value of \( \Delta_r U^\ominus \) for this reaction at 298 K is: (Take \( R = 8.314\text{ J mol}^{-1}\text{ K}^{-1} \))

  1. A
    -1369.48 kJ mol⁻¹
  2. B
    -1362.04 kJ mol⁻¹
  3. C
    -1364.52 kJ mol⁻¹
  4. D
    -1371.96 kJ mol⁻¹
View Answer & Solution

Correct answer · Option C

-1364.52 kJ mol⁻¹

Explanation

\( \Delta n_g = 2 - 3 = -1 \). \( \Delta_r U^\ominus = \Delta_r H^\ominus - \Delta n_g RT = -1367.0 - (-1 \times 8.314 \times 298 \times 10^{-3}) = -1367.0 + 2.478 = -1364.52\text{ kJ mol}^{-1} \).
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