A 2 + B 2 → 2 AB . Δ H f = - 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 ______

Question

A 2 + B 2 → 2 AB . Δ H f = - 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 _________kJ mol – 1 (Nearest integer)

TestHub · Accepted Answer

First write balanced reaction for the formation of one mole of $AB$ with the help of diatomic molecule:

$\frac{1}{2} A_{2} + \frac{1}{2} B_{2} \to AB {ΔH}_{f} = - 200 kJ / mol$

Now,

$∆ H_{reaction} = \frac{1}{2} ∆ H_{A - A} + \frac{1}{2} ∆ H_{B - B} - ∆ H_{AB} (let, ∆ H_{AB} = x)$

$\frac{1}{2} x + \frac{1}{2} (0.5 x) - x = - 200 \Rightarrow \frac{x}{2} + 0.25 x - x = - 200$

$\Rightarrow - 0.25 x = - 200 \Rightarrow x = 800 kJ / mol$

Bond enthalpy $= 800 kJ / mol .$

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