MCQ
The enthalpy change $(\Delta H)$ for the reaction, $N_2 (g) + 3H_2 (g) \to 2NH_3(g)$ is $-92.38\, kJ$ at $298\, K$. The internal energy change $\Delta U$ at $298\, K$ is .............. $\mathrm{kJ}$
- A$-92.38$
- ✓$-87.42$
- C$-97.34$
- D$-89.9$
✓
Answer
Correct option: B.
$-87.42$
b
$\Delta E = \Delta H - \Delta nRT$
$\Delta E = \Delta H - \Delta nRT$
$ = - 92.38 - \left( {\frac{{ - 2 \times 8.31 \times 298}}{{1000}}} \right)$
$ = - 92.38 + 4.95$
$ = - 87.43\,kJ$
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