Question

Butane (C���H������(g)) reacts with oxygen to produce carbon dioxide (CO���) and water (H���O) according to the equation below:

2C���H������(g) + 13O���(g) ��� 8CO���(g) + 10H���O(g)

What is the enthalpy of combustion (per mole) of C���H������(g)?
Use ��Hrxn = ������Hf(products) - ������Hf(reactants).

Options:
A) -2,657.5 kJ/mol
B) -5,315.0 kJ/mol
C) -509.7 kJ/mol
D) -254.8 kJ/mol
1) -2,657.5 kJ/mol
2) -5,315.0 kJ/mol
3) -509.7 kJ/mol
4) -254.8 kJ/mol

Answer 1

The enthalpy of combustion of butane is calculated using the enthalpies of formation for the products and reactants. For butane, it is determined to be -2657.5 kJ/mol, making Option A the correct answer.

Step-by-step explanation:

To calculate the enthalpy of combustion of butane (C4H10(g)), we use the provided combustion equation and the formula ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants). With the enthalpy of formation of butane being -126 kJ/mol, we need the enthalpies of formation for CO2 and H2O; these are typically -393.5 kJ/mol and -241.8 kJ/mol, respectively. Plugging these into the formula, we calculate the enthalpy of combustion for 2 moles of butane and then divide by 2 to find the value per mole.

For the given reaction, 2 moles of butane combine with 13 moles of oxygen to form 8 moles of carbon dioxide and 10 moles of water. The enthalpy change for the reaction can be calculated as follows:

ΔHrxn = [8(-393.5) + 10(-241.8)] - [2(-126) + 13(0)] kJ = -5315 kJ for 2 moles of butane

So for 1 mole, it would be half of that value: -2657.5 kJ/mol. Therefore, the correct answer is Option A) -2657.5 kJ/mol.