Question

Bond enthalpies of $H_2, X_2$ and $HX$ are in the ratio $2 : 1 : 2$. If enthalpy of formation of $HX$ is $-50 \,kJ \,mol^{-1}$, the bond enthalpy of $H_2$ is

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

Answer 1

Answer: (A)

$\frac{1}{2} H _{2}+\frac{1}{2} X _{2} \rightarrow HX$
Let bond enthalpy of $X-X$ bond be $x$, then
$\Rightarrow \frac{1}{2} \times 2 x +\frac{1}{2} \times x -2 x =-50$
$\Rightarrow x +0.5 x -2 x =-50 \,KJ / mol $
$\Rightarrow -0.5 x =-50$
$\Rightarrow x =\frac{-50}{-0.5}=100\, KJ / mol$