Acetic acid CH3COOH can form a dimer CH3COOH2 in the gas phase- The dimer is held together by two H - bonds with a total strength of 77-43 kJ per mol of dimer-If at 27-C- the equilibrium constant for the dimerization is 1-096 - 103- then calculate the magnitude of - S- in kJ for the reaction-2CH3COOH g - CH3COOH2 g Round upto 2 digits after decimal and ln1-096- 103 - 7 and R- 8-3 JK-1mol-1

Δ G^∘ = RT ln K = 8.3 × 300×ln (1.096 × 10^3) = 17430 J = 17.43 kJ Δ G^∘ nbsp;= Δ H^∘ T Δ S^∘ ⇒ 17.43 = 77.43 300 ×Δ S^∘ ⇒Δ S^∘ = 77.4 ...

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Byju's Answer

Standard XIII

Chemistry

Entropy for a Reversible Process

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

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Solution

$Δ G^{\circ} = - R T ln K = - 8.3 \times 300 \times ln (1.096 \times 10^{3}) = - 17430 J = - 17.43 k J Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} \Rightarrow - 17.43 = - 77.43 - 300 \times Δ S^{\circ} \Rightarrow Δ S^{\circ} = \frac{- 77.43 + 17.43}{300} = - 0.2 k J$

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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 strength of

66.5 k J

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If at

25^{o} C

, the equilibrium constant for the dimerization is

1.3 \times 10^{3}

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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 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

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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

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ΔG∘=− RTlnK=−8.3×300×ln(1.096×103)=−17430 J=−17.43 kJΔG∘=ΔH∘−TΔS∘⇒−17.43=−77.43−300×ΔS∘⇒ΔS∘=−77.43+17.43300=−0.2 kJ

$Δ G^{\circ} = - R T ln K = - 8.3 \times 300 \times ln (1.096 \times 10^{3}) = - 17430 J = - 17.43 k J Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} \Rightarrow - 17.43 = - 77.43 - 300 \times Δ S^{\circ} \Rightarrow Δ S^{\circ} = \frac{- 77.43 + 17.43}{300} = - 0.2 k J$