2CH3CO2H (CH3CO2H)2 Kc 1.51×10^2 at 25℃ 0.0300 mol acetic acid benzene 250.0 mL

2CH3CO2H (CH3CO2H)2 Kc 1.51×10^2 at 25℃ 0.0300 mol acetic acid benzene 250.0 mL

 

 

The equilibrium constant for the dimerization of acetic acid in benzene solution is 1.51×10^2 at 25℃.

2CH3CO2H ⇌ (CH3CO2H)2 ... Kc = 1.51×10^2 at 25℃

(a) What are the equilibrium concentrations of monomer and dimer at 25℃ in a solution prepared by dissolving 0.0300 mol of pure acetic acid in enough benzene to make 250.0 mL of solution?

(b) What is the osmotic pressure of the solution at 25℃?

 

 

2CH3CO2H ⇌ (CH3CO2H)2 ... Kc = 1.51×10^2 (25℃에서)

(a) 25℃에서 순수한 아세트산 0.0300 mol을 충분한 양의 벤젠에 넣어 250.0 mL의 용액을 준비하였다. 단위체와 이합체의 평형에서의 농도는 얼마인가?

(b) 25℃에서 용액의 삼투 압력은 얼마인가?

 

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[CH3CO2H] = 0.0300 mol / (250.0/1000 L)

= 0.120 M

 

 

 

.............2CH3CO2H ⇌ (CH3CO2H)2

평형에서, (0.120–2x) ..... x

 

 

 

Kc = [(CH3CO2H)2] / [CH3CO2H]^2

= x / (0.120 – 2x)^2 = 1.51×10^2 = 151

 

x = [0.120^2 – 2(0.120)(2x) + 4x^2] (151)

 

x = (0.120^2)(151) – 2(0.120)(2x)(151) + (4x^2)(151)

 

x = 2.1744 – 72.48x + 604x^2

 

604x^2 – 73.48x + 2.1744 = 0

 

 

 

근의 공식으로 x를 계산하면,

( 참고: 근의 공식 계산기 https://ywpop.tistory.com/3302 )

 

x = 0.0708 또는 0.0508

 

 

 

[CH3CO2H] = 0.120 – 2(0.0708) = –0.0216

 

[CH3CO2H] = 0.120 – 2(0.0508) = 0.0184 M

---> x = 0.0508 M = [(CH3CO2H)2]

 

 

 

> monomer, [CH3CO2H] = 0.0184 M

> dimer, [(CH3CO2H)2] = 0.0508 M

 

 

 

π = MRT

( 참고 https://ywpop.tistory.com/1921 )

 

 

 

 

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