The enthalpy of formation of ammonia is -46.0 | vedclass

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(B) The enthalpy of formation of ammonia is given by the reaction: $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} 	o NH_{3(g)}$ with $\Delta H_f = -46.0

Vedclass · Accepted Answer

$92$

Vedclass · Answer

(B) The enthalpy of formation of ammonia is given by the reaction: $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} 	o NH_{3(g)}$ with $\Delta H_f = -46.0 \ kJ \ mol^{-1}$.To find the enthalpy change for the reaction $2NH_{3(g)} 	o N_{2(g)} + 3H_{2(g)}$,we first multiply the formation reaction by $2$:$N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$,where $\Delta H = 2 	imes (-46.0) = -92.0 \ kJ \ mol^{-1}$.Next,we reverse the reaction:$2NH_{3(g)} 	o N_{2(g)} + 3H_{2(g)}$,where $\Delta H = -(-92.0) = +92.0 \ kJ \ mol^{-1}$.Thus,the enthalpy change for the reaction is $92.0 \ kJ \ mol^{-1}$.

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

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