Calculate the enthalpy of combustion of butane, C4H10, for the formation of H2O and CO2 The enthalpy of formation of butane is - 126 kJ/mol.

The enthalpy of the combustion of butane is - 2,875.5 kJ/mol

Explanation:

To find the enthalpy of the combustion of butane, first we need to write the balanced equation. A combustion is when a substance reacts with oxygen and, as it says in the task, forms water and carbon dioxide. Since enthalpies are usually calculated at 1 bar and 298 K (standard conditions), formed water is in the liquid state.

The balanced equation is:

C₄H₁₀ (g) + 6.5 O₂ (g) = 4 CO₂ (g) + 5 H₂O (l)

Now we can find the enthalpy of this reaction using the formula:

ΔH°r = Σ n (p).ΔH°f (p) - Σ n (r).ΔH°f (r)

where,

ΔH°r is the standard enthalpy of the reaction (in this case the reaction is the combustion, so this the data we are looking for).

n represents the number of moles of reactants and products (which can be found in the balanced equation).

ΔH°f are the standard enthalpies of formation of reactant and products (and can be found in tables).

To apply this formula we need to search the ΔH°f, which are:

C₄H₁₀ (g) - 126 kJ/mol O₂ (g) 0 kJ/mol (by convention, all elements in its most stable state have enthalpy of formation equal to zero). CO₂ (g) - 393.5 kJ/mol H₂O (l) - 285.5 kJ/mol

Replacing this data in the formula:

ΔH°r = [4mol. (-393.5 kJ/mol) + 5mol. (-285.5kJ/mol) ] - [1mol. (-126 kJ/mol) + 6.5. (0 kJ/mol) ]

ΔH°r = - 2,875.5 kJ

Since this enthalpy corresponds to the combustion of 1 mol of butane (according to the balanced equation), we can say that the enthalpy of the combustion of butane is - 2,875.5 kJ/mol.