Q30 / 120 · VITEEE 2017 · Chemistry

Consider the reaction : $\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}$ carried out at constant temperature and pressure. If $\Delta \mathrm{H}$ and $\Delta \mathrm{U}$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

  • A
    $\Delta \mathrm{H}>\Delta \mathrm{U}$
  • Correct answer
    B
    $\Delta \mathrm{H} < \Delta \mathrm{U}$
  • C
    $\Delta \mathrm{H}=\Delta \mathrm{U}$
  • D
    $\Delta \mathrm{H}=0$

Solution

$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{nRT}$ for $\begin{array}{l} \mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3} \\ \Delta \mathrm{n}_{\mathrm{g}}=2-4=-2 \end{array}$ $\therefore \Delta \mathrm{H}=\Delta \mathrm{U}-2 \mathrm{RT}$ or $\Delta \mathrm{U}=\Delta \mathrm{H}+2 \mathrm{R} \mathrm{T} \therefore \Delta \mathrm{U}>\Delta \mathrm{H}$

Hence, the correct answer is option: 'B'.

Relevant Questions