Find the bond enthalpy of $N-H$ bond in ammonia by using the change in enthalpy for the reaction given below: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta H = -23 \ kcal$. Given bond energies: $N \equiv N = 226 \ kcal/mol$,$H-H = 103 \ kcal/mol$.
(N/A) The reaction is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$.
$\Delta H = \sum \text{Bond Energy of Reactants} - \sum \text{Bond Energy of Products}$.
$-23 = [1 \times BE(N \equiv N) + 3 \times BE(H-H)] - [2 \times 3 \times BE(N-H)]$.
$-23 = [226 + 3 \times 103] - 6 \times BE(N-H)$.
$-23 = [226 + 309] - 6 \times BE(N-H)$.
$-23 = 535 - 6 \times BE(N-H)$.
$6 \times BE(N-H) = 535 + 23 = 558$.
$BE(N-H) = 558 / 6 = 93 \ kcal/mol$.
$\Delta H = \sum \text{Bond Energy of Reactants} - \sum \text{Bond Energy of Products}$.
$-23 = [1 \times BE(N \equiv N) + 3 \times BE(H-H)] - [2 \times 3 \times BE(N-H)]$.
$-23 = [226 + 3 \times 103] - 6 \times BE(N-H)$.
$-23 = [226 + 309] - 6 \times BE(N-H)$.
$-23 = 535 - 6 \times BE(N-H)$.
$6 \times BE(N-H) = 535 + 23 = 558$.
$BE(N-H) = 558 / 6 = 93 \ kcal/mol$.
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