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Standard XII

Chemistry

Heat of Formation

Bond energy o...

Question

Bond energy of $A B$ , $A_{2}$ 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 $- 100$ kJ mol $^{- 1}$ . Then the bond enthalpy of $A_{2}$ is:

400 kJ mol

^{- 1}

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200 kJ mol

^{- 1}

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100 kJ mol

^{- 1}

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300 kJ mol

^{- 1}

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Solution

The correct option is B 400 kJ mol $^{- 1}$
$\frac{1}{2} A_{2} + \frac{1}{2} B_{2} \to A B; Δ H = - 100 k J / m o l$

$Δ H =$ B.E. of reactants - B.E. of products

Let B.E. of AB be $x$ .

$∴$ B.E. of $A_{2} = x$ ; B.E. of $B_{2} = \frac{x}{2}$

$∴ Δ H = \frac{x}{2} + \frac{x}{4} - x$

$∴ - 100 = \frac{- x}{4}$

$∴ x = 400$ kJ mol $^{- 1}$

$∴$ Bond enthalpy of A $_{2} = 400 k J m o l^{- 1}$

Correct option is 'A'

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