For the reaction N_2 + 3 X_2 longrightarrow 2 | vedclass

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(A) The heat of reaction $\Delta H$ is calculated using the formula: $\Delta H = \Sigma BE_{	ext{reactants}} - \Sigma BE_{	ext{products}}$.For the f

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$-226$ and $+467$

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(A) The heat of reaction $\Delta H$ is calculated using the formula: $\Delta H = \Sigma BE_{	ext{reactants}} - \Sigma BE_{	ext{products}}$.For the formation of $2 \ mol$ of $NF_3$: $N_2 + 3 F_2 \longrightarrow 2 NF_3$.$\Delta H = [BE_{N \equiv N} + 3 	imes BE_{F-F}] - [6 	imes BE_{N-F}] = [941 + 3(155)] - [6(272)] = 1406 - 1632 = -226 \ kJ \ mol^{-1}$.Since this is for $2 \ mol$ of $NF_3$,the heat of formation per mole is $\Delta H_f(NF_3) = -226 / 2 = -113 \ kJ \ mol^{-1}$. However,the question asks for the heat of reaction for the given stoichiometry.For the formation of $2 \ mol$ of $NCl_3$: $N_2 + 3 Cl_2 \longrightarrow 2 NCl_3$.$\Delta H = [BE_{N \equiv N} + 3 	imes BE_{Cl-Cl}] - [6 	imes BE_{N-Cl}] = [941 + 3(242)] - [6(200)] = 1667 - 1200 = +467 \ kJ \ mol^{-1}$.The values for the reactions are $-226 \ kJ \ mol^{-1}$ and $+467 \ kJ \ mol^{-1}$ respectively.

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