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Standard XII

Chemistry

Entropy

Acetic acid ...

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

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Solution

$Δ G^{o} = - 2.303 R T log K$

$= - 2.303 \times 8.314 \times 298 log (1.3 \times 10^{3}) = - 17767.688 J = - 17.767 k J$

$Δ G^{o} = Δ H^{o} - T Δ S^{o}$

$- 17.767 = - 66.5 - 298 \times Δ S^{o}$

$Δ S^{o} = \frac{- 66.5 + 17.767}{298} = - 0.163 k J$

Thus, $Δ S^{o} = - 0.163 K J$

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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 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

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 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 strength of 66.5kJ per mol of the dimer.If at 25oC, the equilibrium constant for the dimerization is 1.3×103. Calculate ΔSo for the reaction:2CH3COOH(g)⇌(CH3COOH)2(g)

ΔGo=−2.303RTlogK=−2.303×8.314×298log(1.3×103)=−17767.688J=−17.767 kJΔGo=ΔHo−TΔSo−17.767=−66.5−298×ΔSoΔSo=−66.5+17.767298=−0.163 kJThus, ΔSo=−0.163 KJ

$Δ G^{o} = - 2.303 R T log K$

$= - 2.303 \times 8.314 \times 298 log (1.3 \times 10^{3}) = - 17767.688 J = - 17.767 k J$

$Δ G^{o} = Δ H^{o} - T Δ S^{o}$

$- 17.767 = - 66.5 - 298 \times Δ S^{o}$

$Δ S^{o} = \frac{- 66.5 + 17.767}{298} = - 0.163 k J$

Thus, $Δ S^{o} = - 0.163 K J$