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Heat of Formation

AB,A2 and B2...

Question

$A B, A_{2}$ and $B_{2}$ are in the ratio $1 : 1 : 0.5$ and the enthalpy of formation of $A B$ from $A_{2}$ and $B_{2}$ is $200 k J {m o l}^{- 1}$ . What is the bond enthalpy of $A_{2}$ ?

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is A $800 k J {m o l}^{- 1}$
Let energy of formation of $A_{2}$ be $E$
then reaction is,
$\frac{1}{2} A_{2} + \frac{1}{2} B_{2} ⟶ A B$
Energy of 0.5 mol $A_{2}$ bond $= 0.5 E$
Energy of 0.5 mol $B_{2}$ bond $= \frac{0.5 E}{2}$ (As ratio is 0.5)
Energy of 1 mol $A B = E$ (As ratio is 1)
Total energy $= E - 0.75 E = 0.25 E = 200$
$E = 800 k J / m o l$

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