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One mole of CH(3)COOH undergo dimerizati...

One mole of `CH_(3)COOH` undergo dimerization in vapour state at `127^(@)C` as `2CH_(3)COOH (g) hArr (CH_(3)COOH)_(2)` if dimer formation is due to two H-bonds involved in dimer, each of 33kJ strenght and the degree of dimerization of acetic acid `98.2%` which is correct

Text Solution

Verified by Experts

The correct Answer is:
`-104`
`2underset(1)(2CH_(3)COOH)(g) `K=((CH_(3)COOH)_(2))/(CH_(3)COOH)^(2) = 0.982/(2 xx (0.018)^(2)) = 1515.4`
Now, `DeltaH^(@)` for dimerization `=-2 xx 33 kJ = -66 kJ`
Thus, `DeltaG^(@) = DeltaH^(@) - TDeltaS^(@)`
`DeltaS^(@) =-(41640.8)/400 = -104 JK^(-1) mol^(-1)`
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