Calculate the enthalpy variation that occurs in the combustion reaction of butan

Question

Calculate the enthalpy variation that occurs in the combustion reaction of butane under standard conditions. b) How much heat will be released in the complete combustion of the 12 kg of butane contained in a cylinder? Data: standard formation enthalpies: CO2 (g) = -393kJ / mol, H2O (l) = -286kJ / mol; C4H10 (g) = - 125 kJ / mo Calculate the enthalpy variation that occurs in the combustion reaction of butane under standard conditions. b) How much heat will be released in the complete combustion of the 12 kg of butane contained in a cylinder? Data: standard formation enthalpies: CO2 (g) = -393kJ / mol, H2O (l) = -286kJ / mol; C4H10 (g) = - 125 kJ / mo

Explanation / Answer

Combustion of a hydrocarbon will produce CO2 and Water (assuming complete combustion)

1. Lets write the chemical equation

C4H10 + O2 ==== CO2 + H2O, notice that you have 4 carbons on the left so multiply by 4 the CO2 so:

C4H10 +O2 ==== 4 CO2 + H2O, notice that you have 10 hydrogens on the left and 5 on the right so:

C4H10 + O2 ==== 4 CO2 + 5 H2O,

Now count the oxygens, we have 2 on the left and 13 on the left side, the best thing to do is to multiply by 13/2 the oxygen so:

C4H10 + 13/2 O2 ==== 4 CO2 + 5 H2O, now just to have a better equatin multiply by 2 all the equation

2 C4H10 + 13 O2 ==== 8 CO2 + 10 H2O,

Now apply the next equation

H rxn = H products - H reactants

H products = n products * Hformation

H reactants = n reactants * Hformation

H rxn = ( 8 * -393 + 10*-286) - (2 * -125 + 13 *0)

H rxn = -5754 KJ this is for the combustion of 2 moles of butane so

H rxn = -2877 KJ / mol

H is enthalpy

lets find how many moles are 12 Kg or 12 000 grams

molar mass = 58.12 g/gmol

moles = mass / molar mass = 12 000 / 58.12 = 206.469 moles so

H rxn = -2877 KJ / mol * 206.47 moles = -594, 014.2 Kj, the negative sign means release of energy

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