The atomisation enthalpy of $\mathrm{CH}_4$ is $1660 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The…
The atomisation enthalpy of $\mathrm{CH}_4$ is $1660 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The $\mathrm{C}-\mathrm{H}$ bond enthalpy of each successive step in
$\mathrm{CH}_4 ightarrow \mathrm{CH}_3 ightarrow \mathrm{CH}_2 ightarrow \mathrm{CH}$ are $+15,+30$ and $+45 \mathrm{~kJ} \mathrm{~mol}^{-1}$ higher than the mean bond enthalpy of $\mathrm{CH}$ bonds, respectively. The bond enthalpy of the last $\mathrm{C}-\mathrm{H}$ unit is
$400 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$325 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$475 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$385 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
The atomisation enthalpy of $\mathrm{CH}_4$ is $1660 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
$\mathrm{CH}_4(g) \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g) ; \quad \Delta_a H=1660 \mathrm{kJmol}^{-1}$
$\mathrm{CH}_4 \xrightarrow{15} \mathrm{CH}_3 \xrightarrow{30} \mathrm{CH}_2 \xrightarrow{45} \mathrm{CH}$
Mean $\mathrm{C}-\mathrm{H}$ bond enthalpy $=\frac{1660}{4}$
$\therefore \quad 415=15+30+45+x$
$\Rightarrow \quad x=415-90=325 \mathrm{~kJ} \mathrm{~mol}^{-1}$
^