Question

$AB$, $A_2$ and $B_2$ are diatomic molecules. If the bond enthalpies of $A_2,AB$ and $B_2$ are in the ratio 1 : 1 : 0.5 respectively and enthalpy of formation of $AB$ from $A_2$ and $B_2$ is $-100{\text{kJ/mol}}^{-1}$. What is the bond enthalpy of $A_2$ ?

• A. $400\text{kJ/mol}$
• B. $200\text{kJ/mol}$
• C. $100\text{kJ/mol}$
• D. $300\text{kJ/mol}$

Answer 1

The correct option is A $400\text{kJ/mol}$

$\frac{1}{2}A-A+\frac{1}{2}B-B\to AB$ $\Delta H=-100\text{kJ/mol}$
$\frac{1}{2}x+\frac{1}{2}\left(0.5x\right)-x=-100\Rightarrow \frac{x}{2}+0.25x-x=-100$
$\Rightarrow -0.25x=-100$ $x=400\text{kJ/mol}$
Bond enthalpy $=400\text{kJ/mol}$
Hence, the correct answer is option (a).