Question

Calculate the enthalpy change for the process

$CC{l}_4\left(g\right)\to C\left(g\right)+4Cl\left(g\right)$

and calculate bond enthalpy of C-Cl in $CC{l}_4\left(g\right)$

${\Delta }_{vap}{H}^{\ominus }\left(CC{l}_4\right)=30.5$ kJ mol${}^{-1}$.

${\Delta }_f{H}^{\ominus }\left(CC{l}_4\right)=-135.5$ kJ mol${}^{-1}$.

${\Delta }_a{H}^{\ominus }\left(C\right)=715.0$ kJ mol${}^{-1}$.

where ${\Delta }_a{H}^{\ominus }$ is enthalpy of atomisation

${\Delta }_a{H}^{\ominus }\left(C{l}_2\right)=242$ kJ mol${}^{-1}$.

Answer 1

The given chemical reactions are:

$CC{l}_4\left(l\right)\to CC{l}_4\left(g\right){\Delta }_{vap}{H}^{\ominus }=30.5$ kJ mol${}^{-1}$.

$C\left(g\right)+2C{l}_2\left(g\right)\to CC{l}_4\left(g\right){\Delta }_f{H}^{\ominus }=-135.5$ kJ mol${}^{-1}$.
$C\left(s\right)\to C\left(g\right){\Delta }_a{H}^{\ominus }=715.0$ kJ mol${}^{-1}$.
$C{l}_2\left(g\right)\to 2Cl\left(g\right){\Delta }_a{H}^{\ominus }=242$ kJ mol${}^{-1}$.

Now for calculating enthalpy change for the process $CC{l}_4\left(g\right)\to C\left(g\right)+4Cl\left(g\right)$, we have to follow following algebric process

Equation$\left(2\right)+2\times$ Equation$\left(3\right)-$Equation$\left(1\right)-$Equation$\left(4\right)$

$\Delta H={\Delta }_a{H}^{\ominus }+2{\Delta }_a{H}^{\ominus }C{l}_2-{\Delta }_{vap}{H}^{\ominus }-{\Delta }_f{H}^{\ominus }$

$=715+2\times 242-30.5-(-135.5)$

$\Delta H=1304$ kJ mol${}^{-1}$.

Now,

The bond enthalpy of $C-Cl$ in $CC{l}_4=\frac{1304}{4}=326$ kJ mol${}^{-1}$.