Question

Consider the following reaction: ${\mathbf{CH}}_{\mathbf{4}}\mathbf{+}\mathbf{Cl}\mathbf{·}\underset{}{\mathbf{\to }}\mathbf{·}{\mathbf{CH}}_{\mathbf{3}}\mathbf{+}\mathbf{HCl}\mathbf{.}$

a. Use curved arrows to show the movement of electrons.

b. Calculate $\mathbf{∆}\mathbf{H}\mathbf{°}$using the bond dissociation energies in Table 6.2.

c. Draw an energy diagram assuming that ${\mathbf{E}}_{\mathbf{a}}\mathbf{=}\mathbf{16}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{.}$

d. What is ${\mathbf{E}}_{\mathbf{a}}$for the reverse reaction $\left(·{\mathrm{CH}}_3+\mathrm{HCl}\stackrel{}{\to }{\mathrm{CH}}_4+\mathrm{Cl}·\right)$?

Short answer

Answer

a.

Curved arrows showing the movement of electrons

b. $∆\mathrm{H}°=+4\mathrm{kJ}/\mathrm{mol}$

c. Energy diagram of c.

d.

Energy Diagram of d.

Step-by-step solution

Step-by-Step SolutionStep 1: Radicals

Radicals are short-lived and highly reactive. Radicals are chemical species that contain an unpaired free electron.

Arrow representation

Half-headed curved arrows are used to represent the homolytic cleavage of bonds in radical formation and propagation reactions.

Explanation

a. The free radical Chlorine tends to break one C-H bond in ${\mathrm{CH}}_4$homolytically, leading to the formation of HCL and $·{\mathrm{CH}}_3$radical. The movement of electrons by using curved arrows is shown as follows.

Curved arrows showing the movement of electrons

b. Since energy is given to break off the old bond and some energy is released during the formation of new bonds, the net energy gives the enthalpy change of the reaction. So, $∆\mathrm{H}°$is given by:

For the breaking of ${\mathrm{CH}}_3-\mathrm{H}∆{\mathrm{H}}^{°}=+435\mathrm{kJ}/\mathrm{mol}$

For the formation of $\mathrm{H}-\mathrm{Cl}∆{\mathrm{H}}^{°}=-431\mathrm{kJ}/\mathrm{mol}$

Hence,

$\begin{array}{rcl}∆\mathrm{H}& =& +435\mathrm{kJ}/\mathrm{mol}-431\mathrm{kJ}/\mathrm{mol}\\ & =& +4\mathrm{kJ}/\mathrm{mol}\end{array}$

c. The energy diagram with $∆{\mathrm{H}}^{°}=4\mathrm{kJ}/\mathrm{mol}$and ${\mathrm{E}}_{\mathrm{a}}=16\mathrm{kJ}/\mathrm{mol}$is shown as follows:

Energy Diagram of c.

d. The ${\mathrm{E}}_{\mathrm{a}}$for the reverse reaction is calculated by subtracting the energy of the products from that of the Transition state.

Energy Diagram of d.