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 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 ΔS∘ (in kJ) for the reaction:2CH3COOH(g)⇌(CH3COOH)2(g) (Round upto 2 digits after decimal and ln(1.096×103)≈7 and R=8.3 JK−1mol−1)

Acetic acid CH3COOH can form a dimer (CH3COOH)2 in the gas phase. The

1.	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 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 ΔS∘ (in kJ) for the reaction:2CH3COOH(g)⇌(CH3COOH)2(g) (Round upto 2 digits after decimal and ln(1.096×103)≈7 and R=8.3 JK−1mol−1)
Answer» 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 $ln (1.096 \times 10^{3}) \approx 7 a n d R = 8.3 J K^{- 1} m o l^{- 1}$ )