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 $-100$ $kJ$ mol${}^{-1}$. What is the bond energy of $A_2$?

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

Answer 1

The correct option is D $400$ $kJ$ mol${}^{-1}$

According to given ratios,

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 $-100$ $kJ/$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, the correct option is $\text{D}$