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# Given three reaction below, $N_2(g) + 3H_2(g) \leftrightharpoons 2NH_3(g); \Delta H_1$, $H_1(g) + Cl_2(g) \leftrightharpoons 2HCl(g) ; \Delta H_2$, $NH_3 (g) + 3Cl_2 \leftrightharpoons NCl_3(g) + 3HCl(g); -\Delta H_3$. In the terms of $\Delta H_1, \Delta H_2$ and $\Delta H_3$, heat of formation of $NCl_3(g)$ is

(a) $\Delta H_f = -\Delta H_3 + \frac{\Delta H_1}{2} - \frac{3}{2} \Delta H_2$
(b) $\Delta H_f = \Delta H_3 + \frac{\Delta H_1}{1} - \frac{3}{2} \Delta H_2$
(c) $\Delta H_f = \Delta H_3 -\frac{\Delta H_1}{2} - \frac{3}{2} \Delta H_2$
(d) None of the above

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A)
Answer: $\Delta H_f = -\Delta H_3 + \frac{\Delta H_1}{2} - \frac{3}{2} \Delta H_2$
The reaction that depicts the formation of $NCl_3$ is
$N_2(g) + 3Cl_2(g) \leftrightharpoons 2NCl_3(g)$
{$NH_3 (g) + 3Cl_2 \leftrightharpoons NCl_3(g) + 3HCl(g)$ } $\times$ 1
{$N_2(g) + 3H_2(g) \leftrightharpoons 2NH_3(g)$ } $\times \frac{1}{2}$
{$H_1(g) + Cl_2(g) \leftrightharpoons 2HCl(g) ; \Delta H_2$ } $\times - \frac{3}{2}$
We have, $\Delta -H_3 + \frac{1}{2}\Delta H_1 - \frac{3}{2} \Delta H_2$
$\frac{1}{2}N_2(g) + \frac{3}{2}Cl_2(g) \leftrightharpoons NCl_3(g)$
$\therefore \Delta H_f =-\Delta H_3 + \frac{1}{2}\Delta H_1 - \frac{3}{2} \Delta H_2$