Question

The standard heat of formation of CH\(_4\), CO\(_2\) and H\(_2\)O (l) are \( -76.2 \), \( -394.8 \) and \( -285.82 \) kJ mol\(^{-1} \), respectively. Heat of vaporization of water is 44 kJ mol\(^{-1} \). Calculate the amount of heat evolved when 22.4 L of CH\(_4\), kept under normal conditions, is oxidized into its gaseous products

• A. 802 kJ
• B. 878.4 kJ
• C. 702 kJ
• D. 788.4 kJ
• E. 500 kJ

Hint:

Always check the physical state of the products. If the question asks for "gaseous products," you must subtract the heat of vaporization from the enthalpy of liquid water formation.

Answer 1

The Correct Option is D

Concept: The heat of combustion is calculated using the formula: $\Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$. Since the product specified is gaseous water, we must account for the heat of vaporization.

Step 1: {Write the balanced chemical equation for oxidation.}
The combustion of methane is: $$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$$

Step 2: {Calculate the heat of formation for gaseous water.}
Given $\Delta H_f$ for liquid water is $-285.82 \text{ kJ/mol}$ and heat of vaporization is $44 \text{ kJ/mol}$: $$\Delta H_f(\text{H}_2\text{O}, g) = \Delta H_f(\text{H}_2\text{O}, l) + \Delta H_{vap}$$ $$\Delta H_f(\text{H}_2\text{O}, g) = -285.82 + 44 = -241.82 \text{ kJ/mol}$$

Step 3: {Calculate the standard enthalpy of reaction.}
$$\Delta H_{rxn} = [\Delta H_f(\text{CO}_2) + 2\Delta H_f(\text{H}_2\text{O}, g)] - [\Delta H_f(\text{CH}_4) + 2\Delta H_f(\text{O}_2)]$$ $$\Delta H_{rxn} = [-394.8 + 2(-241.82)] - [-76.2 + 0]$$ $$\Delta H_{rxn} = [-394.8 - 483.64] + 76.2$$ $$\Delta H_{rxn} = -878.44 + 76.2 = -802.24 \text{ kJ/mol}$$

Step 4: {Calculate heat evolved for 22.4 L.}
At Normal Conditions (STP), $22.4 \text{ L}$ corresponds to $1 \text{ mole}$. The total heat evolved for $1 \text{ mole}$ to gaseous products is approximately $802 \text{ kJ}$.