AB, A 2 and B 2 are diatomic molecules. If the bond enthalpies of A 2 , B 2 and AB are in the ratio 1 : 0.5 : 1, then the bond enthalpy of A 2 is ________ kJ mol - 1 (Nearest Integer) [2023]

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Solution

Q.
$A_{2} + B_{2} \to 2 A B, Δ H_{f} ° = - {200 kJ mol}^{- 1}$

AB, $A_{2}$ and $B_{2}$ are diatomic molecules. If the bond enthalpies of $A_{2}$ , $B_{2}$ and AB are in the ratio 1 : 0.5 : 1, then the bond enthalpy of $A_{2}$ is ________ kJ ${mol}^{- 1}$ (Nearest Integer) [2023]

Ans.
(800)

$A_{2} + B_{2} \to 2 A B; Δ H_{f} ° = - {200 kJ mol}^{- 1}$

$\Rightarrow Δ H_{f} ° (A B) = - {200 kJ mol}^{- 1}$

$∴$ $Δ H_{R}^{0} for reaction A_{2} + B_{2} \to 2 A B = - {400 kJ mol}^{- 1}$

$Given: Bond enthalpy ratio of A_{2}, B_{2} and A B is 1 : 0.5 : 1$

$Assume bond enthalpy of A_{2} = x {kJ mol}^{- 1}$

$∴$ $Bond enthalpy of B_{2} = 0.5 x {kJ mol}^{- 1}$

$∴$ $Bond enthalpy of A B = x {kJ mol}^{- 1}$

$- 400 = x + 0.5 x - 2 x$

$- 400 = - 0.5 x$

$∴$ $x = 800 kJ/mol$

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