Question

At constant pressure, the combustion of 6.00 g of C₄H1₁₀g) releases 297 kJ of heat. What is AH for the reaction given below?

2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂0(1) AH = ?

Answer 1

The enthalpy change for the complete combustion of 2 moles of butane (C4H10) at constant pressure is -5767 kJ. This is calculated by first determining the per mole enthalpy change from the given data and then scaling up to the stoichiometric amount in the balanced chemical equation.

Step-by-step explanation:

To determine the enthalpy change (ΔH) for the given reaction 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l), we can use the information provided about the combustion of 6.00 g of C4H10(g) releasing 297 kJ of heat. First, we need to calculate the molar mass of C4H10, which is 58.12 g/mol. Then, we can find the number of moles in 6.00 g of C4H10 and scale the ΔH value to a per mole basis.

The calculation is: number of moles = 6.00 g / 58.12 g/mol = 0.103 moles. Since the combustion of this amount of C4H10 releases 297 kJ, the enthalpy change for one mole would be (297 kJ / 0.103 moles) = 2883.5 kJ/mol. However, our balanced equation has 2 moles of C4H10, so we multiply the ΔH value by 2 to get the total enthalpy change for the reaction: ΔH = 2 * 2883.5 kJ/mol = 5767 kJ.

Therefore, the enthalpy change for the reaction is ΔH = -5767 kJ, considering that the combustion process is exothermic and releases heat energy.