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 $1:1:0.5$ and enthalpy of formation of $AB$ from $A_2$ and $B_2$ is $-100kJmo{l}^{-1}$. What is the bond energy of $A_2$?

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

Answer 1

The correct option is D $400kJmo{l}^{-1}$

Let bond energy of $A_2$ be $X$ then bond energy of $AB$ is also $X$ and bond energy of $B_2$ is $X/2$.

Enthalpy of formation of $AB$ is $-100kJ/mole$:

$A_2+B_2\to 2AB$;

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

or $-100=(\frac{1x}{2}+\frac{x}{4})-x$

$\therefore -100=\frac{2x+x-4x}{4}\therefore x=400kJ$

Hence, option D is correct