Question

Data: ΔHf values: CH₄(g), -74.8 kJ; CO₂(g), -393.5 kJ; H₂O(l), -285.8 kJ. Using the ΔHf data above, calculate ΔHrxn for the reaction below. Reaction: CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(l)

A. -604.2 kJ
B. 890.3 kJ
C. -997.7 kJ
D. -890.3 kJ
E. None of the above

Answer 1

To calculate the enthalpy change for the combustion of methane (CH₄), the standard enthalpies of formation are substituted into the reaction enthalpy equation. The calculation leads to an enthalpy change (ΔHrxn) of -890.3 kJ, which is answer option D.

Step-by-step explanation:

The question is about calculating the enthalpy change of a reaction (ΔHrxn) using the standard enthalpies of formation (ΔHf) for the reactants and products. To find the ΔHrxn for the combustion of methane (CH₄(g)), we can use the following equation:

ΔHrxn = [ΔHf CO₂(g) + 2ΔHf H₂O(l)] - [ΔHf CH₄(g) + 2ΔHf O₂(g)]

We substitute the provided ΔHf values into this equation:

ΔHrxn = [-393.5 kJ + 2(-285.8 kJ)] - [-74.8 kJ + 2(0 kJ)]

Calculate the sums:

ΔHrxn = [-393.5 kJ - 571.6 kJ] - [-74.8 kJ]

Combine the terms:

ΔHrxn = -965.1 kJ + 74.8 kJ

ΔHrxn = -890.3 kJ

Therefore, the correct answer is D. -890.3 kJ, which represents the enthalpy change for the combustion of methane.