Question

If enthalpy of a reaction is $1482kJ/mol$ and bond dissociation enthalpy of $(C-H)$ bond is $415kJ/mol$ than bond dissociation enthalpy of $(C-Cl)$ bond is:
$C{H}_{2}C{l}_{2\left(g\right)}\to C_{\left(g\right)}+2H_{\left(g\right)}+2C{l}_{\left(g\right)}$

• A. $326kJ/mol$
• B. $1067kJ/mol$
• C. $652kJ/mol$
• D. $dataisinsufficient$

Answer 1

Answer: (A)

$326kJ/mol$

$326kJ/mol$

The above reaction involves breakage of two types of bonds, i.e., $2C-H$ bonds and $2C-Cl$ bonds. The enthalpy of the reaction is $1482kJ/mol$ (given).

The bond dissociation enthalpy of $C-H$ bond is $415kJ/mol$ (given).

Enthalpy of the $=(BonddissociationenthalpyofC-Hbond)+2(BonddissociationenthalpyofC-Clbond.)$

$1482=2\left(415\right)+2\left(x\right)$

$1482=830+2x$

$\Rightarrow 2x=1482-830$

$x=\frac{652}{2}$

$x=326kJ/mol$

$\therefore$ Bond dissociation enthalpy of $C-Cl$ bond is $326kJ/mol$.