Question

Given
$N_{2}\left(g\right)+3H_{2}\left(g\right)=2NH_{3}\left(g\right);\Delta H^{\circ}=-22 kcal.$. The standard enthalpy of formation of $NH_3$ gas is

• A. $- 11 \,kcal/mol$
• B. $11 \,kcal/mol$
• C. $- 22\, kcal/mol$
• D. $22\, kcal/mol$

Answer 1

Answer: (A)

$\Delta H^{^{\circ}}=\Sigma n\Delta H^{^{\circ}}_{f}(\text{product})-\Sigma n\Delta H^{^{\circ}}_{f}(\text{reactant})$ $=2\Delta H^{\circ}_{f}\left(NH_{3}\right)-\left(\Delta H^{\circ}_{f}\left(N_{2}\right)+3\Delta H^{\circ}_{f}\left(H_{2}\right),\right)\Delta H^{\circ}=22$ kcal
$=\Delta H^{\circ}_{f}\left(NH_{3}\right)-\left(0+0\right)$
So, $\Delta H^{\circ}_{f}=-11$ kcal