Question

Given,

The heat formation of $N{Cl}_3\left(g\right)$ 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 these

Answer 1

Answer: (D)

none of these

$\frac{1}{2}{N}_2+\frac{3}{2}C{l}_2⟶NC{l}_3$ $\Delta H_f$

$\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\rightleftharpoons N{H}_3$, $\frac{-\Delta H_2}{2}.....\left(1\right)$

$\frac{3}{2}{H}_2+\frac{3}{2}C{l}_2\rightleftharpoons 3HCl$, $\frac{-3}{2}\Delta H_3.....\left(2\right)$

$N{H}_3+3C{l}_2\rightleftharpoons NC{l}_3+3HCl$, $-\Delta H_1.....\left(3\right)$

$\left(1\right)-\left(2\right)+\left(3\right)$

$\Delta H_f=\frac{-\Delta H_2}{2}+\frac{3}{2}\Delta H_3-\Delta H_1$

$=-\Delta H_1-\frac{\Delta H_2}{2}+\frac{3}{2}\Delta H_3$