The standard enthalpy of formation (Delta_f H | vedclass

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(C) The standard enthalpy of reaction $(\Delta_r H^{\circ})$ is calculated using the formula: $\Delta_r H^{\circ} = \Sigma \Delta_f H^{\circ} (	ext{p

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$-92.4 \ kJ \ mol^{-1}$

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(C) The standard enthalpy of reaction $(\Delta_r H^{\circ})$ is calculated using the formula: $\Delta_r H^{\circ} = \Sigma \Delta_f H^{\circ} (	ext{products}) - \Sigma \Delta_f H^{\circ} (	ext{reactants})$.For the reaction $N_{2(g)} + 3H_{2(g)} ightarrow 2NH_{3(g)}$, the expression is: $\Delta_r H^{\circ} = [2 	imes \Delta_f H^{\circ}(NH_3)] - [\Delta_f H^{\circ}(N_2) + 3 	imes \Delta_f H^{\circ}(H_2)]$.Since $N_2$ and $H_2$ are elements in their standard states, their standard enthalpy of formation is $0 \ kJ \ mol^{-1}$.Substituting the values: $\Delta_r H^{\circ} = [2 	imes (-46.2 \ kJ \ mol^{-1})] - [0 + 3 	imes 0] = -92.4 \ kJ \ mol^{-1}$.

The standard enthalpy of formation $(\Delta_f H^{\circ})$ of ammonia is $-46.2 \ kJ \ mol^{-1}$. What is the $\Delta_r H^{\circ}$ of the following reaction?
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$

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The standard enthalpy of formation $(\Delta_f H^{\circ})$ of ammonia is $-46.2 \ kJ \ mol^{-1}$. What is the $\Delta_r H^{\circ}$ of the following reaction?$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$