Question

Given that $N _{2}(g)+3 H _{2}(g) \longrightarrow 2 NH _{3}(g)$ $\Delta_{r} H^{\circ}=-92 \,kJ$, the standard molar enthalpy of formation in $kJ mol ^{-1}$ of $NH _{3}(g)$ is

• A. -92
• B. +46
• C. +92
• D. -46

Answer 1

Answer: (D)

Given,

$N _{2}(g)+3 H _{2}(g) \longrightarrow 2 NH _{3}(g) ; \Delta_{r} H^{\circ}=-92\, kJ$

Chemical reaction for molar enthalpy of formation of $NH _{3}$,

$\frac{1}{2} N _{2}(g)+\frac{3}{2} H _{2}(g) \longrightarrow NH _{3}(g)$

$\left(\because H_{f}^{\circ}\right.$ for $N_{2}$ and $\left.H_{2}=0\right)$

Therefore, $\Delta_{f} H^{\circ}=\frac{-\Delta_{r} H^{\circ}}{2}=\frac{-92}{2}$

$ =-46\, kJ \,mol ^{-1}$