Given that N_{2(g)} + 3 H_{2(g)} longrightarr | vedclass

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(D) The standard molar enthalpy of formation $(\Delta_f H^{\circ})$ is defined as the enthalpy change when $1 \ mol$ of a substance is formed from its

Vedclass · Accepted Answer

$-46$

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(D) The standard molar enthalpy of formation $(\Delta_f H^{\circ})$ is defined as the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.
For the reaction: $N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$, the enthalpy change is $\Delta_r H^{\circ} = -92 \ kJ$ for the formation of $2 \ mol$ of $NH_{3(g)}$.
To find the enthalpy of formation for $1 \ mol$ of $NH_{3(g)}$, we divide the reaction by $2$:
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
$\Delta_f H^{\circ} = \frac{\Delta_r H^{\circ}}{2} = \frac{-92 \ kJ}{2} = -46 \ kJ \ mol^{-1}$.

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

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