Question

Consider the reaction : $N_2 + 3H_2 \to 2NH_3$ 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 expressions is true ?

• A. $ \Delta H > \Delta U$
• B. $ \Delta H < \Delta U$
• C. $ \Delta H = \Delta U$
• D. $ \Delta H = 0$

Answer 1

Answer: (B)

$ \Delta H = \Delta U + \Delta n_{g}RT$
$\Delta H = $ enthalpy change (at constant pressure)
$\Delta U =$ internal energy change (at constant volume) (given reaction is exothermic)
Thus $\Delta H < \Delta U.$
$\Delta n_{g} = $ mole of (gaseous products - gaseous reactants)
$= - ve$
Thus $\Delta H < \Delta U.$
Note : Numerical value of $\Delta H < \Delta U$ in exothermic reaction and when $\Delta n_{g} < 0$