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

Options

A. `Delta S^(@)` for dimeriztion is negative
B. `Delta S^(@)` for dimerization is positive
C. `Delta S^(@)` for dimerization
D. `Delta S^(@)` for dimerization is `+1.04J//mol`

Answer

C

Explanation

`{:(2CH_(3)COOH(g),hArr,(CH_(3)COOH)_(2)(g)),(" "1,," "0),(" "1-0.982,," " (0.982)/(2)):}`
`K^(@) = ((CH_(3)COOH)_(2))/((CH_(3)COOH)^(2)) = (0.982)/(2 xx (0.018)^(2)) = 1515.4`
Now, `Delta H^(@)` for dimerization
`= -2 xx 33 kJ = -66kJ`
Thus, `Delta G^(@) = Delta H^(@) - T Delta S^(@)`
`- 2.303 RT log K^(@) = Delta H^(@) - T Delta S^(@)`
`-2.303 xx 8.314 xx 400 log (1515.4) = -66 xx 10^(3) - 400 xx Delta S^(@)`
`-24359.2 = -66000 - 400 Delta S^(@)`
`Delta S^(@) = -(41640.8)/(400) = -140.102J//mol`