Question

Calculate ΔH° for each reaction.

1. $\mathbf{HO}\mathbf{·}\mathbf{+}{\mathbf{CH}}_{\mathbf{4}}\mathbf{\to }\mathbf{.}{\mathbf{CH}}_{\mathbf{3}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$
2. ${\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}\mathbf{+}\mathbf{HBr}\mathbf{\to }{\mathbf{CH}}_{\mathbf{3}}\mathbf{Br}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$
   

Short answer

Answer

1. $∆\mathrm{H}°=-63\mathrm{kJ}/\mathrm{mol}$
2. $∆\mathrm{H}°=-34\mathrm{kJ}/\mathrm{mol}$
   

Step-by-step solution

Step-by-Step SolutionStep 1: ∆H°

$\mathbf{∆}\mathbf{H}\mathbf{°}$ is the enthalpy change between the products and the reactants.

Calculation of Enthalpy change

• The Bond dissociation energies (B.D.E.) for the breaking of bonds (positive values) are added together.
• The Bond dissociation energies for the formation of new bonds (negative values) are added together.

$∆\mathrm{H}°$is calculated by adding all the values of Bond dissociation energies. Thus,

$∆\mathrm{H}°=\mathrm{positive}\mathrm{B}.\mathrm{D}.\mathrm{E}.+\mathrm{negative}\mathrm{B}.\mathrm{D}.\mathrm{E}$

Enthalpy change of the given compounds

a. The bond dissociation energy for the breaking of $+435\mathrm{kJ}/\mathrm{mol}$is .

The bond dissociation energy for the formation of $-498\mathrm{kJ}/\mathrm{mol}$is .

Hence,

$\begin{array}{rcl}∆\mathrm{H}°& =& \mathrm{positive}\mathrm{B}.\mathrm{D}.\mathrm{E}.+\mathrm{negative}\mathrm{B}.\mathrm{D}.\mathrm{E}.\\ & =& +435\mathrm{kJ}/\mathrm{mol}-498\mathrm{kJ}/\mathrm{mol}\\ & =& 63\mathrm{kJ}/\mathrm{mol}\end{array}$

b.The bond dissociation energies for the breaking of ${\mathrm{CH}}_3-\mathrm{OH}\mathrm{and}\mathrm{H}-\mathrm{Br}$ bonds are$+398\mathrm{kJ}/\mathrm{mol}\mathrm{and}+368\mathrm{kJ}/\mathrm{mol}$, respectively.

The bond dissociation energies for the formation of ${\mathrm{CH}}_3-\mathrm{Br}\mathrm{and}\mathrm{H}-\mathrm{OH}$ bonds are $-293\mathrm{kJ}/\mathrm{mol}\mathrm{and}-498\mathrm{kJ}/\mathrm{mol}$, respectively.

Hence,

$\begin{array}{rcl}∆\mathrm{H}°& =& \mathrm{positive}\mathrm{B}.\mathrm{D}.\mathrm{E}+\mathrm{negative}\mathrm{B}.\mathrm{D}.\mathrm{E}.\\ & =& +757\mathrm{kJ}/\mathrm{mol}-791\mathrm{kJ}/\mathrm{mol}\\ & =& -34\mathrm{kJ}/\mathrm{mol}\end{array}$