What is the standard N equiv N bond enthalpy | vedclass

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(C) The enthalpy change of a reaction is given by the sum of bond enthalpies of reactants minus the sum of bond enthalpies of products:$\Delta H^{\cir

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$946 \ kJ 	ext{ mol}^{-1}$

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(C) The enthalpy change of a reaction is given by the sum of bond enthalpies of reactants minus the sum of bond enthalpies of products:$\Delta H^{\circ} = \sum \Delta H^{\circ}_{(	ext{reactants})} - \sum \Delta H^{\circ}_{(	ext{products})}$For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$:$\Delta H^{\circ} = [\Delta H^{\circ}_{(N \equiv N)} + 3 \Delta H^{\circ}_{(H-H)}] - [6 \Delta H^{\circ}_{(N-H)}]$Substituting the given values:$-83 = \Delta H^{\circ}_{(N \equiv N)} + 3(435) - 6(389)$$-83 = \Delta H^{\circ}_{(N \equiv N)} + 1305 - 2334$$-83 = \Delta H^{\circ}_{(N \equiv N)} - 1029$$\Delta H^{\circ}_{(N \equiv N)} = 1029 - 83 = 946 \ kJ 	ext{ mol}^{-1}$

What is the standard $N \equiv N$ bond enthalpy from the following reaction? $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta H^{\circ} = -83 \ kJ$
Given: $(\Delta H^{\circ}_{(H-H)} = 435 \ kJ \text{ mol}^{-1}, \Delta H^{\circ}_{(N-H)} = 389 \ kJ \text{ mol}^{-1})$

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What is the standard $N \equiv N$ bond enthalpy from the following reaction? $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta H^{\circ} = -83 \ kJ$Given: $(\Delta H^{\circ}_{(H-H)} = 435 \ kJ \text{ mol}^{-1}, \Delta H^{\circ}_{(N-H)} = 389 \ kJ \text{ mol}^{-1})$