Question

The combustion of methane is represented by the equation: \[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \] Given: $\Delta H_{\text{CH}_4} = -75 \, \text{kJ/mol}$ 
$\Delta H_{\text{CO}_2} = -393.5 \, \text{kJ/mol}$ 
$\Delta H_{\text{H}_2\text{O}} = -285.8 \, \text{kJ/mol}$ 
What is the enthalpy change ($\Delta H$) for the combustion of 1 mole of methane?

• A. -890.1 kJ/mol
• B. -650.1 kJ/mol
• C. -1000.5 kJ/mol
• D. -500.0 kJ/mol

Hint:

Use Hess's law: \(\Delta H = \sum \Delta H_{\text{products}} - \sum \Delta H_{\text{reactants}}\).

Answer 1

The Correct Option is A

To determine the enthalpy change (\(\Delta H\)) for the combustion of 1 mole of methane, we must use the enthalpies of formation for the reactants and products. The overall reaction is:
\[\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)\]
Using the formula for enthalpy change:
\[\Delta H = \Sigma \Delta H_{\text{products}} - \Sigma \Delta H_{\text{reactants}}\]

Given values:

• \(\Delta H_{\text{CH}_4} = -75 \, \text{kJ/mol}\)
• \(\Delta H_{\text{CO}_2} = -393.5 \, \text{kJ/mol}\)
• \(\Delta H_{\text{H}_2\text{O}} = -285.8 \, \text{kJ/mol}\)

The reaction includes 1 mole of \(\text{CO}_2(g)\) and 2 moles of \(\text{H}_2\text{O}(l)\) as products.

• Total \(\Delta H_{\text{products}}\):\[\Delta H_{\text{CO}_2} + 2\times\Delta H_{\text{H}_2\text{O}} = -393.5 + 2(-285.8)\]
• \[-393.5 + (-571.6) = -965.1 \, \text{kJ/mol}\]

\[\Delta H_{\text{reactants}} = \Delta H_{\text{CH}_4} + 2\times\Delta H_{\text{O}_2}\]

Since oxygen is in elemental form, its standard enthalpy of formation is 0:

• \(\Delta H_{\text{reactants}} = -75 \, \text{kJ/mol} + 2(0) = -75 \, \text{kJ/mol}\)

Now, calculate \(\Delta H\):

\[\Delta H = (-965.1) - (-75) = -965.1 + 75 = -890.1 \, \text{kJ/mol} \]

Therefore, the enthalpy change for the combustion of 1 mole of methane is: -890.3 kJ/mol