Question

Calculate ?Horxn for the combustion reaction of CH4 shown below given the following:

a.Hof CH4 (g) = -74.8 kJ/mol; ?Hof CO2 (g) = -393.5 kJ/mol; ?Hof H2O (l) = -285.5 kJ/mol.
b.Hof O2 (g) = 0 kJ/mol
c.CH4(g) + 2 O2(g) à CO2(g) + 2 H2O(l) ?Horxn = ?

Answer 1

The enthalpy change for the combustion of 1 mol of methane (CH4) is -965 kJ.

Step-by-step explanation:

The student has asked to calculate the enthalpy change (ΔHrxn) for the combustion reaction of methane (CH4).

The balanced chemical equation for the combustion of methane is:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

Using the enthalpy of formation (ΔHf°) values given:

• ΔHf° CH4 (g) = -74.8 kJ/mol
• ΔHf° O2 (g) = 0 kJ/mol (since O2 is a pure element in its standard state)
• ΔHf° CO2 (g) = -393.5 kJ/mol
• ΔHf° H2O (l) = -285.5 kJ/mol

The equation to calculate ΔHrxn is:

ΔHrxn = Σ(ΔHf° products) - Σ(ΔHf° reactants)

Plugging in the values:

ΔHrxn = [1(-393.5 kJ/mol) + 2(-285.5 kJ/mol)] - [1(-74.8 kJ/mol) + 2(0 kJ/mol)]

ΔHrxn = (-393.5 kJ/mol + 2 × -285.5 kJ/mol) - (-74.8 kJ/mol)

ΔHrxn = -965 kJ/mol

Therefore, the enthalpy change for the combustion of 1 mol of methane is -965 kJ.