Question

Calculate the standard enthalpy of combustion of methane, if the standard enthalpy of formation of methane, carbon dioxide, and water are \( -74.8 \), \( -393.5 \), and \( -285.8 \) kJ mol\(^{-1} \) respectively.

Hint:

The standard enthalpy of combustion is always negative, as combustion reactions are exothermic.

Answer 1

The standard enthalpy of combustion \( (\Delta H_c^\circ) \) of methane is determined using the enthalpy of formation values and the balanced combustion equation: \[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \] Using Hess’s Law: \[ \Delta H_c^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} \] Substituting given values: \[ \Delta H_c^\circ = \left[ \Delta H_f^\circ (CO_2) + 2 \times \Delta H_f^\circ (H_2O) \right] - \left[ \Delta H_f^\circ (CH_4) + 2 \times \Delta H_f^\circ (O_2) \right] \] Since \( \Delta H_f^\circ (O_2) = 0 \): \[ \Delta H_c^\circ = \left[ (-393.5) + 2 \times (-285.8) \right] - (-74.8) \] \[ = (-393.5 - 571.6) + 74.8 \] \[ = -890.3 \, \text{kJ mol}^{-1} \]