25℃ K = 0.090 H2O(g) + Cl2O(g) ⇌ 2HOCl(g) 1.0 mol HOCl 2.0 L

25℃ K = 0.090 H2O(g) + Cl2O(g) ⇌ 2HOCl(g) 1.0 mol HOCl 2.0 L

 

 

At 25℃, K = 0.090 for the reaction:

H2O(g) + Cl2O(g) ⇌ 2HOCl(g)

Calculate the concentrations of all species at equilibrium

(b) 1.0 mol pure HOCl is placed in a 2.0 L flask.

 

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▶ 참고: 반응물의 농도가 0인 평형

[ https://ywpop.tistory.com/6299 ]

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[HOCl] = 1.0 mol / 2.0 L = 0.50 M

 

 

 

ICE 도표를 작성하면,

 

 

......... H2O(g) + Cl2O(g) ⇌ 2HOCl(g)

초기(M) . 0 ...... 0 ........ 0.50

변화(M) . +x ..... +x ....... -2x

평형(M) . x ...... x ........ 0.50-2x

 

 

 

K = [HOCl]^2 / [H2O][Cl2O] = 0.090

(0.50-2x)^2 / (x)^2 = 0.090

0.50^2 - (2)(0.50)(2x) + 4x^2 = (0.090)(x^2)

3.91x^2 – 2x + 0.25 = 0

 

 

 

근의 공식으로 x를 계산하면,

x = 0.2941 또는 0.2174

0.50 – 2(0.2941) = -0.0882 이므로,

x = 0.2174 = 0.22 M

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

 

 

 

답:

[HOCl] = 0.50 – 2(0.22) = 0.060 M

[H2O] = [Cl2O] = 0.22 M

 

또는

[HOCl] = 0.50 – 2(0.2174) = 0.0652 M

[H2O] = [Cl2O] = 0.2174 M