Question

What is the standard enthalpy change for the following reaction?

3CH4(g) + 4O3(g) ------> 3CO2(g) + 6H2O(g)
Substance ΔH°f (kJ/mol)
CH4(g) –74.87
O3(g) +142.7
CO2(g) –393.5
H2O(g) –241.8

Answer 1

- 2977 kJ

Step-by-step explanation:

Let's consider the following reaction.

3 CH₄(g) + 4 O₃(g) → 3 CO₂(g) + 6 H₂O(g)

We can find the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = [3 mol × ΔH°f(CO₂(g)) + 6 mol × ΔH°f(H₂O(g))] - [3 mol × ΔH°f(CH₄(g)) + 4 mol × ΔH°f(O₃(g))]

ΔH°r = [3 mol × (-393.5 kJ/mol) + 6 mol × (-241.8 kJ/mol)] - [3 mol × (-74.87 kJ/mol) + 4 mol × (142.7 kJ/mol)]

ΔH°r = - 2977 kJ