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: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$ with $\Delt

Vedclass · Accepted Answer

$92$

Vedclass · Answer

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

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

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