Which one of the following reactions would you expect to have the lowest ∆S°? C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(g) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(g) C2H4(g) + O2(g) → 2CO2(g) + 2H2O(g)

Explanation:

The reaction undergoing least change in volume will have lowest change in entropy .

1 ) C₂H₂(g) + 5/2O₂(g) → 2CO₂(g) + H₂O(g)

change in volume = 2.5 +1 - ( 2 + 1 ) = 0.5

2 ) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

change in volume = 2 +1 - ( 2 + 1 ) = zero

3 ) C₂H₆(g) + 7/2O₂(g) → 2CO₂(g) + 3H₂O(g)

change in volume = 3 +2 - ( 3.5  + 1 ) = 0.5

4 ) C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

change in volume = 2 +2 - ( 3  + 1 ) = zero

Hence least change in volume is in case no - 2 and 4 .

In reaction

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g) and

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

∆S° or change in entropy is least .

the answer would be a