91Ó°ÊÓ

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).

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

${{\mathrm{Δ\pm á}}^{∘}}_{\mathrm{overall}}={\underset{}{∑}{\mathrm{Δ\pm á}}^{∘}}_{\mathrm{bonds}\mathrm{broken}}-\underset{}{\overset{}{∑}}{{\mathrm{Δ\pm á}}^{∘}}_{\mathrm{bonds}\mathrm{formed}}$

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

Bonds broken:

role="math" localid="1648192060232" ${\mathrm{CH}}_3-\mathrm{H}\left({\mathrm{Δ\pm á}}^{∘}=4×435\mathrm{kJ}/\mathrm{mol}\right)$and role="math" localid="1648192297307" $O-O\left({\mathrm{Δ\pm á}}^{∘}=2×497\mathrm{kJ}/\mathrm{mol}\right).$

The bonds formed are:

role="math" localid="1648192315890" $\mathrm{OC}-\mathrm{O}\left({\mathrm{Δ\pm á}}^{∘}=2×535\mathrm{kJ}/\mathrm{mol}\right)$and role="math" localid="1648192388083" $\mathrm{HO}-\mathrm{H}\left({\mathrm{Δ\pm á}}^{∘}=4×498\mathrm{kJ}/\mathrm{mol}\right).$

role="math" localid="1648193161702" $$\begin{gathered} {{\mathrm{Δ\pm á}}^{∘}}_{\mathrm{overall}}={\underset{}{∑}{\mathrm{Δ\pm á}}^{∘}}_{\mathrm{bonds}\mathrm{broken}}-\underset{}{\overset{}{∑}}{{\mathrm{Δ\pm á}}^{∘}}_{\mathrm{bonds}\mathrm{formed}} \\ =\left(4x435+2×497\right)\mathrm{kJ}/\mathrm{mol}-\left(2×535+4×498\right)\mathrm{kJ}/\mathrm{mol} \\ =2784\mathrm{kJ}/\mathrm{mol}-3062\mathrm{kJ}/\mathrm{mol} \\ =-328\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

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

Bonds broken:

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

Bonds formed:

$\mathrm{OC}-\mathrm{O}({\mathrm{Δ\pm á}}^{∘}=8×535\mathrm{kJ}/\mathrm{mol})$and$\mathrm{O}-\mathrm{O}\left(\right({\mathrm{Δ\pm á}}^{∘}=12×498\mathrm{kJ}/\mathrm{mol})$