A_2+B_2 arrow 2 AB . \Delta H_f^0=-200 ~kJ ~mol^-1 AB, A_2 and B_2 are diatomic molecules. If the bond enthalpies of A_2, ~B_2 and AB are in the ratio 1: 0.5: 1 , then the bond enthalpy of A_2 is \underline\hspace2cm kJ ~mol^-1 (Nearest integer)

Thermodynamics PYQ — JEE Main 2023 · 13 Apr · Shift 1 · Q21 with Solution

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Correct answer: 400

Given ratio for bond enthalpies of $\mathrm{A_2} : \mathrm{B_2} : \mathrm{AB}$ is $1 : 0.5 : 1$

Let $D(\mathrm{A_2}) = x$ $D(\mathrm{B_2}) = 0.5x$ $D(\mathrm{AB}) = x$

Reaction: $\mathrm{A_2 + B_2 \to 2AB}$

Standard enthalpy change is $\Delta H = \text{bond enthalpies of bonds broken} - \text{bond enthalpies of bonds formed}$

Here,

So, $\Delta H = D(\mathrm{A_2}) + D(\mathrm{B_2}) - 2D(\mathrm{AB})$

Substitute values: $-200 = x + 0.5x - 2x$

$-200 = 1.5x - 2x = -0.5x$

Thus, $x = 400$

Since $x = D(\mathrm{A_2})$ , $D(\mathrm{A_2}) = 400\ \text{kJ mol}^{-1}$

Stored correct answer = $400$

This matches our derived answer.