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Enthalpy

AB, A2, and ...

Question

$A B, A_{2}$ , and $B_{2}$ are diatomic molecules. If the bond enthalpies of $A_{2}, A B$ , and $B_{2}$ are in the ratio $2 : 2 : 1$ and ethalpy of formation $A B$ from $A_{2}$ and $B_{2}$ is $- 100 k J m o l^{- 1}$ . What is the bond energy of $A_{2}$ ?

A

200 k J m o l^{- 1}

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B

100 k J m o l^{- 1}

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C

300 k J m o l^{- 1}

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D

400 k J m o l^{- 1}

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Solution

The correct option is D $400 k J m o l^{- 1}$
Solution:- (D) $400 K J / m o l$
Given:-
${B . E .}_{A_{2}} : {B . E .}_{A B} : {B . E .}_{B_{2}} = 2 : 2 : 1$
Let bond enthalpy of $B_{2}$ be $x K J / m o l$
Therefore,
${B . E .}_{A_{2}} = {B . E .}_{A B} = 2 x$
Given that ${Δ H_{A B}}_{(f o r m a t i o n)} = - 100 K J / m o l$
$\frac{1}{2} A_{2} + \frac{1}{2} B_{2} ⟶ A B; Δ H = - 100 K J / m o l$
$A_{2} + B_{2} ⟶ 2 A B; Δ H = - 200 K J / m o l$
As we know that,
$Δ H =$ Bond enthalpy of reactant $-$ Bond enthalpy of product
$∴ Δ H = ({B . E .}_{A_{2}} + {B . E .}_{B_{2}}) - (2 \times {B . E .}_{A B})$
$- 200 = (2 x + x) - (2 \times 2 x)$
$- 200 = 3 x - 4 x$
$\Rightarrow x = 200$
Therefore,
${B . E .}_{A_{2}} = 2 x = 2 (200) = 400 K J / m o l$
Hence the bond energy of $A_{2}$ is $400 K J / m o l$ .

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