To determine the enthalpy of formation of methane (CH4), we will use Hess's Law and the given enthalpies of combustion for various compounds. The relevant enthalpic transformations involved are:
- Combustion of methane:
CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)
The standard enthalpy change for this reaction is provided as -890.3 \ \text{kJ mol}^{-1}.
- Combustion of graphite (carbon to CO2):
C(graphite) + O_2(g) \rightarrow CO_2(g)
The standard enthalpy change for this reaction is -393.5 \ \text{kJ mol}^{-1}.
- Combustion of dihydrogen (H2 to H2O):
H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)
The standard enthalpy change for this reaction is -285.8 \ \text{kJ mol}^{-1}.
Now, the formation reactions are described by:
C(graphite) + 2H_2(g) \rightarrow CH_4(g)
According to Hess's Law, the enthalpy change for the formation of methane can be calculated as:
\Delta H^\circ_f [CH_4(g)] = \Delta H^\circ_2[C(graphite) \rightarrow CO_2(g)] + 2 \times \Delta H^\circ_2[H_2(g) \rightarrow H_2O(l)] - \Delta H^\circ_2[CH_4(g) \rightarrow CO_2(g) + 2H_2O(l)]
Substituting the given values, we have:
\Delta H^\circ_f [CH_4(g)] = (-393.5) + 2 \times (-285.8) - (-890.3)
Simplifying, we find:
\Delta H^\circ_f [CH_4(g)] = -393.5 - 571.6 + 890.3
\Delta H^\circ_f [CH_4(g)] = -74.8 \ \text{kJ mol}^{-1}
Therefore, the enthalpy of formation of methane (CH4(g)) is -74.8\ \text{kJ mol}^{-1}, which matches the correct answer.