MCQ
The enthalpy change $(\Delta H)$ for the reaction,
$N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)$ is
$-92.38\, KJ$ at $298\, K$. The internal energy change $\Delta U$ at $298\, K$ is ......$KJ$
$N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)$ is
$-92.38\, KJ$ at $298\, K$. The internal energy change $\Delta U$ at $298\, K$ is ......$KJ$
- A$-92.38$
- ✓$-87.42$
- C$-97.34$
- D$-89.9$
