Question

Give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)+5315 kJ.

Express your answer using four significant figures. If the value is positive, do not include the + sign in your answer.

Answer 1

The ΔH value for the combustion of butane is given as 5315 kJ for the reaction as written, which is for 2 moles of butane. To find the enthalpy of combustion per mole, divide this number by 2, yielding -2658 kJ/mol.

Step-by-step explanation:

The student asked to give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) + 5315 kJ. The ΔH value for the reaction is given as +5315 kJ, indicating the amount of energy released during the combustion of butane.

Now, the enthalpy of combustion is typically expressed on a per mole basis, referring to the amount of heat released when one mole of a substance is burned. Since the reaction above shows the combustion for 2 moles of butane, we can calculate the enthalpy of combustion per mole by dividing the total energy by 2. So, the enthalpy of combustion for one mole of butane would be 5315 kJ / 2 = 2657.5 kJ/mol, which is typically reported in kilojoules per mole (kJ/mol).

Expressed with four significant figures, the enthalpy of combustion per mole of butane would be -2658 kJ/mol (the negative sign indicates that energy is released).