Respuesta :
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 (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.
The enthalpy of combustion of butane from the balanced equation is; ΔH°_rxn = -2878 KJ
Balanced equation for the combustion of butane is;
C₄H₁₀(g) + 6.5 O₂(g) = 4 CO₂(g) + 5 H₂O(l)
Now from the balanced equation, we have;
4 moles of C₄H₁₀
6.5 moles of O₂
4 moles of CO₂
5 moles of H₂O
- Now, the formula for enthalpy of combustion is;
ΔH°_rxn = ΣΔH°_f,products - ΣΔH°_f,reactants
Where;
ΣΔH°_f = (number of moles × enthalpy of formation)
From tables;
Enthalpy of formation of CO₂ = -393.5 kJ/mol
Enthalpy of formation of O₂ = 0 KJ/mol
Enthalpy of formation of H₂O = -286 kJ/mol
Enthalpy of formation of C₄H₁₀ = -126 KJ/mol
Thus;
For CO₂; ΣΔH°_f = (4 × -393.5) = −1,574 KJ
For O₂; ΣΔH°_f = 0 KJ
For H₂O; ΣΔH°_f = (5 × - 286) = -1430 KJ
For C₄H₁₀; ΣΔH°_f = (1 × - 126) = - 126 KJ
Thus;
ΔH°_rxn = (-1574 - 1430) - (-126)
ΔH°_rxn = -2878 KJ
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