Question

The standard enthalpy of formation of CH₄(g), CO₂(g) and H₂O(I) are -75 kJ mol^-1, -393 kJ mol^-1 and -286 kJ mol^-1 respectively. The amount of heat liberated (in kJ) when 3.2g of methane gas is burnt under standard conditions is

• A. 89
• B. 278
• C. 890
• D. 965
• E. 178

Answer 1

Answer: (E)

The combustion of methane (CH$_4$) can be represented by the following balanced chemical equation: \[ \text{CH}_4(g) + 2 \text{O}_2(g) \to \text{CO}_2(g) + 2 \text{H}_2\text{O}(l) \] The heat liberated during the combustion of methane can be calculated using the standard enthalpy of formation values for the reactants and products. The standard enthalpy change of the reaction \( \Delta H_{\text{reaction}} \) is given by: \[ \Delta H_{\text{reaction}} = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants}) \] Substituting the values for the standard enthalpy of formation: \[ \Delta H_{\text{reaction}} = \left[ (-393 \, \text{kJ/mol}) + 2(-286 \, \text{kJ/mol}) \right] - \left[ (-75 \, \text{kJ/mol}) + 0 \right] \] \[ \Delta H_{\text{reaction}} = \left[ -393 - 572 \right] - (-75) \] \[ \Delta H_{\text{reaction}} = -965 + 75 = -890 \, \text{kJ/mol} \] Thus, the enthalpy change for the combustion of one mole of methane is -890 kJ/mol. Now, we need to calculate the heat released when 3.2 g of methane is burnt. The molar mass of methane (CH$_4$) is: \[ \text{Molar mass of CH}_4 = 12 + 4 = 16 \, \text{g/mol} \] The number of moles of methane in 3.2 g is: \[ \text{moles of CH}_4 = \frac{3.2 \, \text{g}}{16 \, \text{g/mol}} = 0.2 \, \text{mol} \] The heat released is: \[ \text{Heat released} = 0.2 \, \text{mol} \times (-890 \, \text{kJ/mol}) = -178 \, \text{kJ} \]

The correct option is (E) : \(178\)