Given the bond energies of N equiv N,H - H an | vedclass

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

Vedclass · Accepted Answer

$-93$

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(A) The enthalpy of reaction is calculated using the formula: $\Delta H = \sum BE_{	ext{reactants}} - \sum BE_{	ext{products}}$.For the reaction $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$,the bonds broken are $1$ mole of $N \equiv N$ and $3$ moles of $H - H$,and the bonds formed are $6$ moles of $N - H$.$\Delta H = [BE(N \equiv N) + 3 	imes BE(H - H)] - [6 	imes BE(N - H)]$.Substituting the given values: $\Delta H = [945 + 3(436)] - [6(391)]$.$\Delta H = [945 + 1308] - [2346]$.$\Delta H = 2253 - 2346 = -93 \ kJ$.

Given the bond energies of $N \equiv N$,$H - H$ and $N - H$ bonds are $945$,$436$ and $391 \ kJ \cdot mol^{-1}$ respectively,the enthalpy of the following reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is ...... $kJ$.

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