Acetic acid, CH 3 COOH can form a dimer ( CH 3 COOH ) 2 in the gas phase 2 CH 3 COOH ( g ) ⇌ ( CH 3 COOH ) 2 ( g ) ; Δ H 298 K = − 66.5 kJ if at 2 5 ∘ C , the equilibrium constant for th

Question

Acetic acid, CH 3 ​ COOH can form a dimer ( CH 3 ​ COOH ) 2 ​ in the gas phase 2 CH 3 ​ COOH ( g ) ⇌ ( CH 3 ​ COOH ) 2 ​ ( g ) ; Δ H 298 K ​ = − 66.5 kJ if at 2 5 ∘ C , the equilibrium constant for this reaction is 1 0 3 then calculate Δ S ∘ for this reaction :-

TestHub · Accepted Answer

Δ G ∘ = Δ H ∘ − T Δ S ∘ − RT ln K eq ​ = − 66.5 × 1 0 + 3 − 298Δ S ∘ − 8.314 × 298 ln ( 1 0 3 ) = − 66.5 × 1 0 3 − 298Δ S ∘ − 298 2.303 × 8.314 × 298 × 3 − 66.5 × 1 0 3 ​ = Δ S ∘ Δ S ∘ = − 0.165 k J m o l − 1 k − 1

Chemistry - Thermodynamics - II Question with Solution | TestHub

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