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......$kJ$
- A$-92.98$
- ✓$-87.42$
- C$-97.34$
- D$-89.9$
✓
Answer
Correct option: B.
$-87.42$
b
$\Delta H=-92.38\, \mathrm{kJ}$ and $\Delta \mathrm{n}=-2$
$\Delta H=-92.38\, \mathrm{kJ}$ and $\Delta \mathrm{n}=-2$
$\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{nRT}$
$-92.38=\Delta E+(-2) \times\left(8.3 \times 10^{-3}\right) \times 298$
$\Delta E=-87.4 \,\mathrm{kJ}$
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