Question

Calculate ΔH° for each oxidation reaction. Each equation is balanced as written; remember to take into account the coefficients in determining the number of bonds broken or formed.

[ ${\mathbf{\Delta H}}^{\mathbf{\circ }}$for ${\mathbf{O}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{497}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$ ; ${\mathbf{\Delta H}}^{\mathbf{\circ }}$for one C=O in ${\mathbf{CO}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{535}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$]

a.role="math" localid="1648191068323" ${\mathbf{CH}}_{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{O}}_{\mathbf{2}\mathbf{}}\stackrel{}{\mathbf{\to }}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

b.$\mathbf{2}{\mathbf{CH}}_{\mathbf{4}}\mathbf{+}\mathbf{7}{\mathbf{O}}_{\mathbf{2}\mathbf{}}\stackrel{}{\mathbf{\to }}\mathbf{4}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

Short answer

Answer

a. ${\mathrm{\Delta H}}^{\circ }$ for the reaction is width="98">-328kJ/mol.

b. ${\mathrm{\Delta H}}^{\circ }$ for the reaction is localid="1648191467506" $-1121\mathrm{kJ}/\mathrm{mol}$.

Step-by-step solution

Step-by-Step SolutionStep 1: Enthalpy of a reaction

The enthalpy of a reaction is determined by taking the difference of the enthalpy of reactants (sum of enthalpies of a bond broken) and the enthalpy of products (sum of enthalpies of a bond formed).

Formula used to calculate the enthalpy of a reaction

The formula used to calculate the enthalpy of a given reaction is:

${{\mathrm{\Delta H}}^{\circ }}_{\mathrm{overall}}={\sum _{}{\mathrm{\Delta H}}^{\circ }}_{\mathrm{bonds}\mathrm{broken}}-\sum _{}^{}{{\mathrm{\Delta H}}^{\circ }}_{\mathrm{bonds}\mathrm{formed}}$

Calculating enthalpy of a reaction

a. ${\mathrm{CH}}_4+2{\mathrm{O}}_2\stackrel{}{\to }{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$

Bonds broken:

role="math" localid="1648192060232" ${\mathrm{CH}}_3-\mathrm{H}\left({\mathrm{\Delta H}}^{\circ }=4\times 435\mathrm{kJ}/\mathrm{mol}\right)$and role="math" localid="1648192297307" $O-O\left({\mathrm{\Delta H}}^{\circ }=2\times 497\mathrm{kJ}/\mathrm{mol}\right).$

The bonds formed are:

role="math" localid="1648192315890" $\mathrm{OC}-\mathrm{O}\left({\mathrm{\Delta H}}^{\circ }=2\times 535\mathrm{kJ}/\mathrm{mol}\right)$and role="math" localid="1648192388083" $\mathrm{HO}-\mathrm{H}\left({\mathrm{\Delta H}}^{\circ }=4\times 498\mathrm{kJ}/\mathrm{mol}\right).$

role="math" localid="1648193161702" $$\begin{gathered} {{\mathrm{\Delta H}}^{\circ }}_{\mathrm{overall}}={\sum _{}{\mathrm{\Delta H}}^{\circ }}_{\mathrm{bonds}\mathrm{broken}}-\sum _{}^{}{{\mathrm{\Delta H}}^{\circ }}_{\mathrm{bonds}\mathrm{formed}} \\ =\left(4x435+2\times 497\right)\mathrm{kJ}/\mathrm{mol}-\left(2\times 535+4\times 498\right)\mathrm{kJ}/\mathrm{mol} \\ =2784\mathrm{kJ}/\mathrm{mol}-3062\mathrm{kJ}/\mathrm{mol} \\ =-328\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

b. $2{\mathrm{CH}}_4+7{\mathrm{O}}_2\stackrel{}{\to }4{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}$

Bonds broken:

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{CH}}_2-\mathrm{H}({\mathrm{\Delta H}}^{\circ }=12\times 410\mathrm{kJ}/\mathrm{mol}),\mathrm{O}-\mathrm{O}\left(\right({\mathrm{\Delta H}}^{\circ }=7\times 497\mathrm{kJ}/\mathrm{mol})\mathrm{and}\mathrm{C}-\mathrm{C}( \\ {\mathrm{\Delta H}}^{\circ }=2\times 368\mathrm{kJ}/\mathrm{mol}) \end{gathered}$$

Bonds formed:

$\mathrm{OC}-\mathrm{O}({\mathrm{\Delta H}}^{\circ }=8\times 535\mathrm{kJ}/\mathrm{mol})$and$\mathrm{O}-\mathrm{O}\left(\right({\mathrm{\Delta H}}^{\circ }=12\times 498\mathrm{kJ}/\mathrm{mol})$

$$\begin{gathered} {{\mathrm{\Delta H}}^{\circ }}_{\mathrm{overall}}={\sum _{}{\mathrm{\Delta H}}^{\circ }}_{\mathrm{bonds}\mathrm{broken}}-\sum _{}^{}{{\mathrm{\Delta H}}^{\circ }}_{\mathrm{bonds}\mathrm{formed}} \\ =\left(12\times 410+2\times 368\right)\mathrm{kJ}/\mathrm{mol}-\left(8\times 535+12\times 498\right)\mathrm{kJ}/\mathrm{mol} \\ =9135\mathrm{kJ}/\mathrm{mol}-10256\mathrm{kJ}/\mathrm{mol} \\ =-1121\mathrm{kJ}/\mathrm{mol} \end{gathered}$$