Question

Given,
$ NH_3(g) + 3Cl_2(g) \rightleftharpoons NCl_3(g) + 3HCl(g) - \Delta H_1 $
$ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) - \Delta H_2 $
$ H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g) - \Delta H_3 $
The heat of formation of $ NCl_3(g) $ in terms of $ \Delta H_1, \Delta H_2 $ and $ \Delta H_3 $ is

• A. $ \Delta H_{f} = -\Delta H_{1} + \frac{1}{2} \Delta H_{2} - \frac{3}{2} \Delta H_{3} $
• B. $ \Delta H_{f} = -\Delta H_{1} + \frac{1}{2} \Delta H_{2} + \frac{3}{2} \Delta H_{3} $
• C. $ \Delta H_{f} = -\Delta H_{1} - \frac{1}{2} \Delta H_{2} - \frac{3}{2} \Delta H_{3} $
• D. None of the above

Answer 1

Answer: (D)

$NH_3(g) + 3Cl_2( g ) \rightleftharpoons NCl_3(g) + 3HCl(g)$;
$-\Delta H_1 ...(i)$
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g);$
$ -\Delta H_2...(ii)$
$H_2(g) + Cl_2( g ) \rightleftharpoons 2HCl( g )$ ;
$ + \Delta H_3...(iii)$
Multiply Eq. $(i)$ by $2$ and Eq. $(iii)$ by $3$, add
Eq. $(i)$ to Eq. $(ii)$ and subtract Eq. $(iii)$ to get
the heat of formation of $NCl_3(g)$;
$N_2 + 3Cl_2 \rightleftharpoons 2NCl_3$
$2\Delta H_f = 2 \times ( -\Delta H_1) + ( -\Delta H_2) - 3 \times ( +\Delta H_3)$
or $\Delta H_f = -\Delta H_1 -\frac{1}{2} \Delta H_2 - \frac{3}{2} \Delta H_3$