Question

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol^-1, -393.5 kJ mol^-1, and -285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄(g) will be

• A. \(–74.8\ kJ mol^{–1}\)
• B. \(–52.27 \ kJ mol^{–1}\)
• C. \(+74.8 \ kJ mol^{–1}\)
• D. \(+52.26\ kJ mol^{–1}\)

Answer 1

The Correct Option is A

To determine the enthalpy of formation of methane \((\text{CH}_4(g))\), we will use Hess's Law, which states that the change in enthalpy for a chemical reaction is the same, regardless of the path taken. The enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

The chemical equation for the formation of methane from its elements in their standard states is:

C(\text{graphite}) + 2H_2(\text{gas}) \rightarrow CH_4(\text{gas})

We are provided with the following enthalpy changes:

• Enthalpy of combustion of methane: \Delta H = -890.3\ \text{kJ mol}^{-1}
• Enthalpy of combustion of graphite: \Delta H = -393.5\ \text{kJ mol}^{-1}
• Enthalpy of combustion of dihydrogen: \Delta H = -285.8\ \text{kJ mol}^{-1}

The combustion reactions for each species are:

1. For \text{CH}_4\): \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\), \Delta H = -890.3\ \text{kJ mol}^{-1}
2. For C: \(C(\text{graphite}) + O_2(g) \rightarrow CO_2(g)\), \Delta H = -393.5\ \text{kJ mol}^{-1}
3. For H_2: \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)\), \Delta H = -285.8\ \text{kJ mol}^{-1}

According to Hess's Law, the standard enthalpy of formation of methane is calculated using the equation:

\Delta H_f^\circ (\text{CH}_4) = \Delta H^\circ (\text{combustion of methane}) - [ \Delta H^\circ (\text{combustion of carbon}) + 2 \times \Delta H^\circ (\text{combustion of hydrogen}) ]

Substituting the given values:

\Delta H_f^\circ (\text{CH}_4) = -890.3 - [(-393.5) + 2 \times (-285.8)]

\Delta H_f^\circ (\text{CH}_4) = -890.3 + 393.5 + 2 \times 285.8

\Delta H_f^\circ (\text{CH}_4) = -890.3 + 393.5 + 571.6

\Delta H_f^\circ (\text{CH}_4) = 74.8 \ \text{kJ mol}^{-1}\

Therefore, the standard enthalpy of formation of methane \((\text{CH}_4(g))\) is -74.8\ \text{kJ mol}^{-1}, which is the correct option.