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Q.

Bond enthalpies of H_{2 ,}X₂and HX are in the ratio 2: 1:2.If enthalpy of formation of HX is — 50 kJ mol^-1 the bond enthalpy of X₂ is

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a

100kJ mol^-1

b

400kJ mol^-1

c

200kJ mol^-1

d

300kJ mol^-1

answer is A.

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Detailed Solution

$\frac{1}{2} H_{2} + \frac{1}{2} X_{2} ⟶ HX$
Let the bond enthalpy of X — X bond be x.
${ΔH}_{f} (HX) = - 50 = \frac{1}{2} {ΔH}_{H - H} + \frac{1}{2} {ΔH}_{X - X} - {ΔH}_{H - X} = \frac{1}{2} 2 x + \frac{1}{2} x - 2 x = \frac{- x}{2} ∴ x = 50 \times 2 = 100 kJ {mol}^{- 1}$

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