Answer:
ΔH = 15.3kJ
Explanation:
It is possible to obtain standard enthalpy change of a reaction from the sum of half-reactions (Hess's law). Thus:
(1) C(s) + H₂O(g) ⇄ CO(g) + H₂(g) ΔH = 131.3kJ
(2) CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g) ΔH = -41.2kJ
(3) CH₄(g) + H₂O(g) ⇄ 3H₂(g) + CO(g) ΔH = 206.1kJ
The sum of 2×(1) + (2) gives:
2C(s) + 3H₂O(g) ⇄ CO₂(g) + 3H₂(g) + CO(g) ΔH = 131.3kJ×2 + -41.2kJ = 221.4kJ
Now, the sum of this reaction with -(3) gives:
2C(s) + 2H₂O(g) ⇄ CH₄(g) + CO₂(g) ΔH = 221.4kJ - 206.1kJ = 15.3kJ
I hope it helps!