The value of $\Delta_{f} H^{o}$ for $NH_3$ is $-91.8 \ kJ \ mol^{-1}$. Calculate the enthalpy change for the following reaction: $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$
The formation reaction for $1 \ mole$ of $NH_3$ is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \to NH_{3(g)}$; $\Delta_f H^o = -91.8 \ kJ \ mol^{-1}$.
For the formation of $2 \ moles$ of $NH_3$,the reaction is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta_r H^o = 2 \times (-91.8 \ kJ \ mol^{-1}) = -183.6 \ kJ \ mol^{-1}$.
The given reaction is the reverse of the formation of $2 \ moles$ of $NH_3$.
Therefore,for the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$,the enthalpy change is: $\Delta_r H^o = -(-183.6 \ kJ \ mol^{-1}) = +183.6 \ kJ \ mol^{-1}$.
For the formation of $2 \ moles$ of $NH_3$,the reaction is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta_r H^o = 2 \times (-91.8 \ kJ \ mol^{-1}) = -183.6 \ kJ \ mol^{-1}$.
The given reaction is the reverse of the formation of $2 \ moles$ of $NH_3$.
Therefore,for the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$,the enthalpy change is: $\Delta_r H^o = -(-183.6 \ kJ \ mol^{-1}) = +183.6 \ kJ \ mol^{-1}$.
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