for is –113 kJ/mol. Bond energy for the bond is 273.5 kJ/mol. The bond energies of and , if their ma — Thermodynamics and Thermochemistry Chemistry Question
Question
$\Delta_f H^\circ$ for $\text{NF}_3\text{(g)}$ is –113 kJ/mol. Bond energy for the $\text{N}-\text{F}$ bond is 273.5 kJ/mol. The bond energies of $\text{N}_2$ and $\text{F}_2$, if their magnitudes are in the ratio 6:1, are, respectively,
Answer: C
💡 Solution & Explanation
Formation of $\text{NF}_3$: $1/2 \text{N}_2 + 3/2 \text{F}_2 \to \text{NF}_3$. $\Delta H_f = 0.5\text{BE(N}_2\text{)} + 1.5\text{BE(F}_2\text{)} - 3\text{BE(N-F)}$. Let $\text{BE(F}_2\text{)} = x$, $\text{BE(N}_2\text{)} = 6x$. $-113 = 0.5(6x) + 1.5x - 3(273.5) = 4.5x - 820.5 \implies 4.5x = 707.5 \implies x = 157.22\text{ kJ/mol}$. Thus $\text{BE(N}_2\text{)} = 6x = 943.32\text{ kJ/mol}$.
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