Question

Given,
$\ce{NH3_{(g)} <=> 3Cl2_{(g)} +NCl_3{(g)}} + 3HCl_{(g)}$; $-\Delta H_1$
$\ce{N2_{(g)} +3H2_{(g)} <=> 2NH3_{(g)}}$; $-\Delta H_2$
$\ce{H2_{(g)} +Cl2_{(g)} <=> 2HCl_{(g)} }$; $\Delta H_3$
The heat of formation of $NCl_{3(g)}$ in the terms of $\Delta H_1$, $\Delta H_2$ and $\Delta H_3$ is

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

Answer 1

Answer: (C)

Find $\frac{1}{2} N_{2}+\frac{3}{2}Cl_{2} \to NCl_{3} ; \Delta H_{f} $
Multiply Eq. $\left(ii\right)$ by $1/2$ and add to Eq. $\left(i\right)$, Eq.$\left(iii\right)$ by $3/2 $ and subtract from Eq.$\left(i\right)$ ; we get
$\Delta H_{f} =-\Delta H_{1}-\left[-\frac{\Delta H_{2}}{2}+\frac{3}{2}\Delta H_{3}\right]$
$=-\Delta H_{1}+\frac{\Delta H_{2}}{2}-\frac{3}{2}\Delta H_{3}$