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Standard XI

Chemistry

Le Chatelier's Principle for Del N Greater Than Zero

Acetic acid C...

Question

Acetic acid $C H_{3} C O O H$ can form a dimer $(C H_{3} C O O H)_{2}$ in the gas phase. The dimer is held together by two H-bonds with a total strenght of 66.5 kJ per mol of dimer
If at $25^{o} C,$ the equilibrium constant for the dimerization is $1.3 \times 10^{3} .$ Calculate $△ S^{o}$ for the reaction:
$2 C H_{3} C O O H (g) ⇌ (C H_{3} C O O H)_{2} (g)$

- 0.263 k J

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- 0.463 k J

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- 0.163 k J

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- 0613 k J

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Solution

The correct option is C $- 0.163 k J$
WE know,
$△ G^{o} = - 2.303 R T l o g K$
putting the values
$= - 2.303 \times 8.314 \times 298 l o g (1.3 \times 10^{3})$
$= - 17767.688 J m o l^{- 1} = - 17.767 k J m o l^{- 1}$

We also know that,
$△ G^{o} = △ H^{o} - T △ S^{o}$
putting the values
$- 17.767 = - 66.5 - 298 \times △ S^{o}$
$△ S^{o} = \frac{- 66.5 + 17.767}{298} = - 0.163 k J m o l^{- 1} K^{- 1}$

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Similar questions

Q. Acetic acid

{C H}_{3} C O O H

can form a dimer

{({C H}_{3} C O O H)}_{2}

in the gas phase. The dimer is held together by two H-bonds with a total strength of

66.5 k J

per mol of the dimer.

If at

25^{o} C

, the equilibrium constant for the dimerization is

1.3 \times 10^{3}

. Calculate

Δ S^{o}

for the reaction:

2 {C H}_{3} C O O H (g) ⇌ {({C H}_{3} C O O H)}_{2} (g)

Q. Acetic acid

C H_{3} C O O H

can form a dimer

(C H_{3} C O O H)_{2}

in the gas phase. The dimer is held together by two H-bonds with a total strenght of 66.5 kJ per mol of dimer

If at

25^{o} C,

the equilibrium constant for the dimerization is

1.3 \times 10^{3} .

Calculate

△ S^{o}

for the reaction:

2 C H_{3} C O O H (g) ⇌ (C H_{3} C O O H)_{2} (g)

Q. Acetic acid

C H_{3} C O O H

can form a dimer

(C H_{3} C O O H)_{2}

in the gas phase. The dimer is held together by two H - bonds with a total strength of

77.43 k J p e r m o l

of dimer.

If at

27^{\circ} C

, the equilibrium constant for the dimerization is

1.096 \times 10^{3}

, then calculate the magnitude of

Δ S^{\circ}

(in

k J

) for the reaction:

2 C H_{3} C O O H (g) ⇌ (C H_{3} C O O H)_{2} (g)

(Round upto 2 digits after decimal and

ln (1.096 \times 10^{3}) \approx 7 a n d R = 8.3 J K^{- 1} m o l^{- 1}

)

Q. Acetic acid forms dimer in vapour phase. The dimer is held together by two hydrogen bonds with a total strength of

66.5

kJ per mole of dimer. If at

25^{o}

C, the equilibrium constant for the dimerization is

1.3 \times 10^{3}

Δ S^{o}

for the reaction

2 C H_{3} C O O H ⇌ (C H_{3} C O O H)_{2}

will be (

l o g 1.3 = 0.1139

) :

Q. Acetic acid forms a dimer in the gas phase

The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer. At

25^{o} C,

the equilibrium constant for the dimerisation is

1.3 \times 10^{3}

(pressure in atm). What is magnitude

△ S^{0}

for the reaction? Assume that

△ H

does not vary with temperature.

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Acetic acid CH3COOH can form a dimer (CH3COOH)2 in the gas phase. The dimer is held together by two H-bonds with a total strenght of 66.5 kJ per mol of dimer If at 25oC, the equilibrium constant for the dimerization is 1.3×103. Calculate △So for the reaction: 2CH3COOH(g)⇌(CH3COOH)2(g)