The bond enthalpies of $H_{2}, X_{2}$ an HX are in the ratio of 2 : 1 : 2. If enthalpy of formation of HX is $- 50 k J m o l^{- 1}$ , the bond enthalpy of $X_{2}$ is

$100 k J m o l^{- 1}$

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$200 k J m o l^{- 1}$

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$300 k J m o l^{- 1}$

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$400 k J m o l^{- 1}$

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Solution

The correct option is A
$100 k J m o l^{- 1}$

$\frac{1}{2} H_{2} + \frac{1}{2} X_{2} \to H X, L e t x = e n t h a l p y o f X - X b o n d Δ H_{f} (H X) = - 50 = \frac{1}{2} x - x = - \frac{x}{2}$
or, $x = 50 \times 2 = 100 k J m o l^{- 1}$

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Similar questions

Q. The bond enthalpies of

H_{2}, X_{2}

and HX are in the ratio of 2: 1: 2. If the enthalpy for formation of

H X

- 50 k J m o l^{- 1}

, the bond enthalpy of

H_{2}

is:

The standard enthalpy of formation of $N H_{3}$ is $46.0 k J m o l^{- 1}$ . If the enthalpy of formation of $H_{2}$ from its atom is $- 436 k J m o l^{- 1}$ and that of $N_{2}$ is $- 712 k J m o l^{- 1}$ , the average bond enthalpy of $N - H$ bond in $N H_{3}$ is