Question

$AB,A_{2}and{B}_2$ are diatomic molecules. If the bond enthalpies of $A_2,ABand{B}_2$ are in the ratio 1 : 1 : 0.5 and the enthalpy of formation of AB from $A_{2}and{B}_2$ is $-100kJmo{l}^{-1}$, what is the bond enthalpy of $A_2$ ?

• A. $400kJmo{l}^{-1}$
• B. $200kJmo{l}^{-1}$
• C. $100kJmo{l}^{-1}$
• D. $300kJmo{l}^{-1}$

Answer 1

Answer: (A)

$400kJmo{l}^{-1}$

Let the bond enthalpy of $A_2$ be a kJ/mol.
$e_{A--A}=a$
$e_{A--B}=a$
$e_{B--B}=0.5a$

Also, the balanced chemical equation for the formation of AB is $\frac{1}{2}{A}_2+\frac{1}{2}{B}_2⟶AB$;

$\Delta H=-100kJmo{l}^{-1}$

Hence, $\Delta H=-\left[e_{A--B}\right]+\frac{1}{2}[e_{A--A}+e_{B--B}]$

$=-a+\frac{1}{2}[a+0.5a]$

$-100=-0.25a$
$a=400kJmo{l}^{-1}$

Hence, the bond enthalpy of $A_2$ is $a=400kJmo{l}^{-1}$