Given
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta_{r}H^{\theta} = -92.4 \, kJ \, mol^{-1}$
What is the standard enthalpy of formation of $NH_{3}$ gas?
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta_{r}H^{\theta} = -92.4 \, kJ \, mol^{-1}$
What is the standard enthalpy of formation of $NH_{3}$ gas?
(C) The standard enthalpy of formation of a compound is the change in enthalpy that occurs during the formation of $1 \, mol$ of a substance in its standard state from its constituent elements in their standard states.
Rewriting the given equation for the formation of $1 \, mol$ of $NH_{3(g)}$:
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
Therefore,the standard enthalpy of formation of $NH_{3(g)}$ is:
$\Delta_{f}H^{\theta} = \frac{1}{2} \Delta_{r}H^{\theta}$
$\Delta_{f}H^{\theta} = \frac{1}{2} (-92.4 \, kJ \, mol^{-1})$
$\Delta_{f}H^{\theta} = -46.2 \, kJ \, mol^{-1}$
Rewriting the given equation for the formation of $1 \, mol$ of $NH_{3(g)}$:
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
Therefore,the standard enthalpy of formation of $NH_{3(g)}$ is:
$\Delta_{f}H^{\theta} = \frac{1}{2} \Delta_{r}H^{\theta}$
$\Delta_{f}H^{\theta} = \frac{1}{2} (-92.4 \, kJ \, mol^{-1})$
$\Delta_{f}H^{\theta} = -46.2 \, kJ \, mol^{-1}$
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