, and are diatomic molecules. If the bond enthalpies of , and are in the ratio 2:2:1 and enthalpy of — Thermodynamics and Thermochemistry Chemistry Question
Question
$\text{AB}$, $\text{A}_2$ and $\text{B}_2$ are diatomic molecules. If the bond enthalpies of $\text{A}_2$, $\text{AB}$ and $\text{B}_2$ are in the ratio 2:2:1 and enthalpy of formation $\text{AB}$ from $\text{A}_2$ and $\text{B}_2$ is –100 kJ mol$^{-1}$. What is the bond energy of $\text{A}_2$?
Answer: D
💡 Solution & Explanation
Let $\text{BE}(\text{B}_2) = x$, $\text{BE}(\text{AB}) = 2x$, and $\text{BE}(\text{A}_2) = 2x$. Formation reaction: $1/2 \text{A}_2 + 1/2 \text{B}_2 \to \text{AB}$. $\Delta H = 1/2(2x) + 1/2(x) - 2x = -100 \implies x + 0.5x - 2x = -100 \implies -0.5x = -100 \implies x = 200\text{ kJ/mol}$. $\text{A}_2$ bond energy is $2x = 400\text{ kJ/mol}$.
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