The enthalpy of formation of NH_3 is -46 kJ | vedclass

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Question

(C) The reaction is $2 NH_{3(g)} \longrightarrow N_{2(g)} + 3 H_{2(g)}$.For any reaction,$\Delta H_r = \sum \Delta H_f^{\circ}(	ext{products}) - \sum

Vedclass · Accepted Answer

$+92 \ kJ$

Vedclass · Answer

(C) The reaction is $2 NH_{3(g)} \longrightarrow N_{2(g)} + 3 H_{2(g)}$.For any reaction,$\Delta H_r = \sum \Delta H_f^{\circ}(	ext{products}) - \sum \Delta H_f^{\circ}(	ext{reactants})$.Since the elements $N_2$ and $H_2$ are in their standard states,their enthalpy of formation is $0$.$\Delta H_r = [1 	imes \Delta H_f^{\circ}(N_2) + 3 	imes \Delta H_f^{\circ}(H_2)] - [2 	imes \Delta H_f^{\circ}(NH_3)]$.$\Delta H_r = [0 + 0] - [2 	imes (-46 \ kJ \ mol^{-1})]$.$\Delta H_r = +92 \ kJ$.

The enthalpy of formation of $NH_3$ is $-46 \ kJ \ mol^{-1}$. The enthalpy change for the reaction $2 NH_{3(g)} \longrightarrow N_{2(g)} + 3 H_{2(g)}$ is

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