MCQ
Consider the reaction : ${N_2} + 3{H_2} \to 2N{H_3}$ carried out at constant temperature and pressure. If $\Delta H$ and $\Delta U$ are the enthalpy and internal energy changes for the reaction, which of the following expression is true
  • A
    $\Delta H = 0$
  • B
    $\Delta H = \Delta U$
  • $\Delta H < \Delta U$
  • D
    $\Delta H > \Delta U$

Answer

Correct option: C.
$\Delta H < \Delta U$
(c)At constant $P$ or $T$
$\Delta H = \Delta U + \Delta nRT$ $==>$ $\Delta n = {n_p} - {n_R} = 2 - 4 = - 2$
$\therefore $ $\Delta H < \Delta U$.

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