Answer:
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 (<em>in this case the reaction is the combustion, so </em><u><em>this the data we are looking for</em></u>).
n represents the number of moles of reactants and products <em>(which can be found in the balanced equation).</em>
ΔH°f are the standard enthalpies of formation of reactant and products (<em>and can be found in tables</em>).
To apply this formula we need to search the ΔH°f, which are:
- 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.