Question

The heat of formation of $NH_{3}\left(\right.g\left.\right)$ is $-46 \, kJmol^{- 1}$ . The $\Delta H$ (in kJ) of the reaction is

$2NH_{3}\left(\right.g\left.\right) \rightarrow N_{2}\left(\right.g\left.\right)+3H_{2}\left(\right.g\left.\right)$

• A. 46
• B. $-$ 46
• C. 92
• D. $-$ 92

Answer 1

Answer: (C)

$\frac{1}{2}N_{2}\left(\right.g\left.\right)+\frac{3}{2}H_{2}\left(\right.g\left.\right) \rightarrow NH_{3}\left(\right.g\left.\right)$

$\Delta H_{r}=-46kJ/mol$

$2NH_{3}\left(\right.g\left.\right) \rightarrow N_{2}\left(\right.g\left.\right)+3H_{2}\left(\right.g\left.\right)$

$\Delta H_{r}=-2\Delta H_{r}$

$=-2\left(\right.-46\left.\right)=92kJ$