Question

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

• A. 400 kJ/mole
• B. 300 kJ/mole
• C. 500 kJ/mole
• D. 900 kJ/mole

Answer 1

The correct option is A 400 kJ/mole

$\frac{1}{2}{A}_2+\frac{1}{2}{B}_2\to AB\Delta H=-100J$

Let the bond enthalpies of $A_2:B_2:AB=1:0.5:1$

From reaction : $0.5x+0.25x-1x=100\Rightarrow x=\frac{100}{0.25}=400kJ$

Bond enthalpy of $A_2=1x=400kJmo{l}^{-1}$