Question

The reaction $C{H}_4\left(g\right)+C{l}_2\to C{H}_{3}Cl\left(g\right)+HCl\left(g\right)$ has $\delta H=-25kcal.$ From the data given below, what is the bond enthalpy of $Cl-Cl$ bond?

Bond	Bond Energy (kcal)
$C-C$	$84$
$C-Cl$	$81$
$C-H$	$x$
$Cl-Cl$	$y$
$H-Cl$	$103$

Given: $x:y=9:5$.

• A. $70kcal$
• B. $80kcal$
• C. $67.75kcal$
• D. $56.75kcal$

Answer 1

The correct option is D $56.75kcal$

$C{H}_4\left(g\right)+C{l}_2\to C{H}_{3}Cl\left(g\right)+HCl\left(g\right)$ has $\delta H=-25kcal.$
For this reaction:
$\delta H=4B.E(C-H)+B.E.(Cl-Cl)-3B.E.(C-H)$ $-B.E.(C-Cl)-B.E.(H-Cl)=-25kcal$

$x+y-103-81=-25x=\left(9/5\right)y$.
So, $y=159\times \frac{5}{14}=56.75kcal$