Answer:
- 2977 kJ
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
