Butane (C4H10) (g), (ΔHf = -125.7 kJ/mol), combusts in the presence of oxygen to form CO2 (g) (ΔHf = -393.5 kJ/mol), and H2O (g) (ΔHf = -241.82 kJ/mol) in the reaction: 2C4H10 (g) + 13O2 (g) → 8CO2 + 10H2O (g). What is the enthalpy of combustion, per mole, of butane
The Correct Answer and Explanation is:
Correct answer:
The enthalpy of combustion per mole of butane is -2657.3 kJ/mol.
Explanation:
To find the enthalpy of combustion of butane, we use the standard enthalpies of formation (ΔHf) and apply Hess’s Law. The combustion reaction is:
2C4H10 (g) + 13O2 (g) → 8CO2 (g) + 10H2O (g)
The standard enthalpy of formation (ΔHf) values are:
- C4H10 (g): -125.7 kJ/mol
- CO2 (g): -393.5 kJ/mol
- H2O (g): -241.82 kJ/mol
- O2 (g): 0 kJ/mol (because it is an element in its standard state)
We use this formula:
ΔHrxn = [ΣΔHf (products)] – [ΣΔHf (reactants)]
Step 1: Calculate ΔHf of products
- 8 mol CO2 × (-393.5 kJ/mol) = -3148.0 kJ
- 10 mol H2O × (-241.82 kJ/mol) = -2418.2 kJ
Total products enthalpy = -3148.0 + (-2418.2) = -5566.2 kJ
Step 2: Calculate ΔHf of reactants
- 2 mol C4H10 × (-125.7 kJ/mol) = -251.4 kJ
- 13 mol O2 × 0 = 0
Total reactants enthalpy = -251.4 kJ
Step 3: Find ΔHrxn
ΔHrxn = -5566.2 – (-251.4) = -5314.8 kJ for 2 moles of butane
Step 4: Per mole of butane
-5314.8 kJ ÷ 2 mol = -2657.4 kJ/mol
Rounded to one decimal place: -2657.3 kJ/mol
This negative value means the reaction is exothermic, releasing a large amount of energy. Combustion reactions of hydrocarbons like butane are commonly used in fuels due to this high energy output. The energy is mainly released as heat, which makes butane a good candidate for heating and cooking purposes.
