Question

Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production of SNG is
4CO(g) + 8H2(g) → 3CH4(g) + CO2(g) + 2H2O(l) ΔHo = ?

Use appropriate data from the following list to determine ΔHo for this SNG reaction.
C(graphite) + 1/2O2(g) → CO(g) ΔHo = -110.5 kJ
CO(g) + 1/2O2(g) → CO2(g) ΔHo = -283.0 kJ
H2(g) + 1/2O2(g) → H2O(l) ΔHo = -285.8 kJ
C(graphite) + 2H2(g) → CH4(g) ΔHo = -74.81 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔHo = -890.3 kJ

Answer 1

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

(1) C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

(2) CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

(4) C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

(5) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = -747.54kJ

Thus ΔH° of the reaction is -747.54kJ