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:
<em>(1) </em>C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ
<em>(2) </em>CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ
<em>(3) </em>H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ
<em>(4) </em>C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ
<em>(5) </em>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 = <em>-747.54kJ</em>
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Thus <em>ΔH° of the reaction is -747.54kJ</em>
