Question

Given that $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right);{\Delta }_r{H}^{}=-92KJ$, the standard molar enthalpy of formation in $KJmo{l}^{-1}$ of $N{H}_3\left(g\right)$ is :

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

Answer 1

The correct option is D $-46$

For the given reaction,

$N_2\left(g\right)+3H_2\left(g\right)⟶2N{H}_3\left(g\right)$

Enthalpy of reaction $\left({\Delta }_r{H}^0\right)=–92.4kJmo{l}^{–1}$

$N_2\left(g\right)+H_2\left(g\right)⟶N{H}_3\left(g\right)$

Hence, standard enthalpy of $N{H}_3$ is equal to the $\frac{1}{2}{\Delta }_r{H}^0$

As, ${\Delta }_r{H}^0=–92.4kJmo{l}^{–1}$

∴ Standard enthalpy of $N{H}_3=\frac{-92.4}{2}$

⇒Standard enthalpy of $N{H}_3=-46.2kJmo{l}^{-1}$

Hence, option D is correct.