Q7 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry
For the reaction $\text{A}_2 + \text{B}_2 \rightarrow 2\text{AB}$, $\Delta H_f$ (AB) $= -400 \text{ kJ/mol}$. The bond dissociation enthalpies of $\text{A}_2$, $\text{B}_2$ and $\text{AB}$ are in the ratio 1 : 0.75 : 1. What is the bond dissociation enthalpy of $\text{B}_2$ in kJ/mol?
- A800
- B1600
- Correct answerC2400
- D3200
Solution
$\Delta H_{rxn} = 2 \times \Delta H_f(AB) = 2 \times (-400) = -800$ kJ/mol
Let $BE_{A_2} = x$, $BE_{B_2} = 0.75x$, $BE_{AB} = x$
$\Delta H_{rxn} = \sum BE(\text{reactants}) - \sum BE(\text{products})$
$-800 = (x + 0.75x) - 2x = 1.75x - 2x = -0.25x$
$x = \dfrac{-800}{-0.25} = 3200$ kJ/mol
$BE_{B_2} = 0.75 \times 3200 = 2400$ kJ/mol
Let $BE_{A_2} = x$, $BE_{B_2} = 0.75x$, $BE_{AB} = x$
$\Delta H_{rxn} = \sum BE(\text{reactants}) - \sum BE(\text{products})$
$-800 = (x + 0.75x) - 2x = 1.75x - 2x = -0.25x$
$x = \dfrac{-800}{-0.25} = 3200$ kJ/mol
$BE_{B_2} = 0.75 \times 3200 = 2400$ kJ/mol
Hence, the correct answer is option: 'C'.