The enthalpy of formation of ammonia is -46.0 | vedclass

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Question

(B) The formation reaction of ammonia is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightarrow NH_{3(g)}$; $\Delta H_f = -46 \ kJ \ mol^{-1}$.For

Vedclass · Accepted Answer

$92$

Vedclass · Answer

(B) The formation reaction of ammonia is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} ightarrow NH_{3(g)}$; $\Delta H_f = -46 \ kJ \ mol^{-1}$.For the reaction $N_{2(g)} + 3H_{2(g)} ightarrow 2NH_{3(g)}$,the enthalpy change is $\Delta H = 2 	imes (-46) = -92 \ kJ \ mol^{-1}$.The given reaction $2NH_{3(g)} ightarrow N_{2(g)} + 3H_{2(g)}$ is the reverse of the above reaction.Therefore,the enthalpy change for the reverse reaction is $\Delta H = -(-92) = +92 \ kJ \ mol^{-1}$.

The enthalpy of formation of ammonia is $-46.0 \ kJ \ mol^{-1}$. The enthalpy change for the reaction $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$ is ............... $kJ \ mol^{-1}$.

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