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14 January, 20:03

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.

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  1. 14 January, 20:22
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    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.
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