Question

use the reaction to answer the question.
$\ce{ch_{4} + 2o_{2} \
ightarrow co_{2} + 2h_{2}o}$
$\ce{h-c}$: 412 kj/mol
$\ce{o=o}$: 496 kj/mol
$\ce{c=o}$: 743 kj/mol
$\ce{o-h}$: 463 kj/mol

Explanation:

To determine the enthalpy change (\(\Delta H\)) of the reaction, we use the formula:
\(\Delta H = \sum (\text{bond energies of reactants broken}) - \sum (\text{bond energies of products formed})\)

Step 1: Analyze bonds in reactants

• \( \text{CH}_4 \): 4 \( \text{C–H} \) bonds.

Energy to break 4 \( \text{C–H} \): \( 4 \times 412 \, \text{kJ/mol} = 1648 \, \text{kJ/mol} \)

• \( \text{O}_2 \): 2 \( \text{O=O} \) bonds (since \( 2 \, \text{O}_2 \) molecules).

Energy to break 2 \( \text{O=O} \): \( 2 \times 496 \, \text{kJ/mol} = 992 \, \text{kJ/mol} \)

Total energy to break reactant bonds:
\( 1648 + 992 = 2640 \, \text{kJ/mol} \)

Step 2: Analyze bonds in products

• \( \text{CO}_2 \): 2 \( \text{C=O} \) bonds.

Energy released forming 2 \( \text{C=O} \): \( 2 \times 743 \, \text{kJ/mol} = 1486 \, \text{kJ/mol} \)

• \( \text{H}_2\text{O} \): 2 molecules, each with 2 \( \text{O–H} \) bonds (total 4 \( \text{O–H} \) bonds).

Energy released forming 4 \( \text{O–H} \): \( 4 \times 463 \, \text{kJ/mol} = 1852 \, \text{kJ/mol} \)

Total energy released forming product bonds:
\( 1486 + 1852 = 3338 \, \text{kJ/mol} \)

Step 3: Calculate \(\Delta H\)

\(\Delta H = (\text{Reactant bond energy}) - (\text{Product bond energy})\)
\(\Delta H = 2640 - 3338 = -698 \, \text{kJ/mol}\)

Answer:

The enthalpy change of the reaction is \(\boldsymbol{-698 \, \text{kJ/mol}}\) (exothermic, as \(\Delta H\) is negative).