Question

The value of $\Delta H-\Delta E$ for the following reaction at $27^{\circ} C$ will be
$2NH_3(g) \to N_2(g) + 3H_2(g)$

• A. $ 8.314\times 273\times (-2) $
• B. $ 8.314\times 300\times (-2) $
• C. $ 8.314\times 27\times (-2) $
• D. $ 8.314\times 300\times (2) $

Answer 1

Answer: (D)

$2 NH _{3} \rightleftharpoons N _{2}+3 H _{2}$
$\because \Delta H=\Delta E+\Delta n_{g} R T$
$\therefore \Delta H-\Delta E=\Delta n_{g} R T$
$\Delta n_{g}=$ gaseous product $-$ gaseous reactant
$=4-2=+2$
At $27^{\circ} C$,
$ \Delta H-\Delta E=2 R T=2 \times 8.314 \times 300$