Question

Calculate the heat of combustion of methane using standard heats of formation below: ΔH° CO₂(g) 387 kJ mol-1 ΔH° H₂O(l) = – 268 kJ mol-1 ΔH° CH₄(g) 74.5 kJ mol-1

a) 302.5 kJ/mol
b) 442.5 kJ/mol
c) 161.5 kJ/mol
d) 817 kJ/mol

Answer 1

To calculate the heat of combustion of 1 mole of ethanol, use the enthalpies of formation. The heat of combustion is calculated as the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. In this case, the heat of combustion of ethanol is -1090 kJ/mol.

Step-by-step explanation:

To calculate the heat of combustion of 1 mole of ethanol, we can use the enthalpies of formation provided. The balanced equation for the combustion of ethanol is:

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

Using the enthalpies of formation:

ΔH° C₂H₅OH(l) = -278 kJ/mol

ΔH° CO₂(g) = -394 kJ/mol

ΔH° H₂O(l) = -286 kJ/mol

The heat of combustion can be calculated as the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:

ΔH° combustion = (2 × ΔH° CO₂(g)) + (3 × ΔH° H₂O(l)) - ΔH° C₂H₅OH(l)

ΔH° combustion = (2 × -394 kJ/mol) + (3 × -286 kJ/mol) - (-278 kJ/mol)

ΔH° combustion = -788 kJ/mol - 858 kJ/mol + 278 kJ/mol

ΔH° combustion = -1368 kJ/mol + 278 kJ/mol

ΔH° combustion = -1090 kJ/mol