For the equilibrium Br2(g) + Cl2(g) ⇌ 2BrCl(g) at 400 K, Kc = 7.0

For the equilibrium Br2(g) + Cl2(g) ⇌ 2BrCl(g) at 400 K, Kc = 7.0

 

 

For the equilibrium Br2(g) + Cl2(g) ⇌ 2BrCl(g) at 400 K, Kc = 7.0.

If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0 L container at 400 K, what will be the equilibrium concentration of Br2, Cl2, and BrCl?

what will be the equilibrium concentration of each of the components?

 

 

온도 400 K에서 다음 반응의 평형 상수 Kc = 7.0이다.

Br2(g) + Cl2(g) ⇌ 2BrCl(g)

400 K의 3.0 L 용기에 Br2 0.25 mol과 Cl2 0.55 mol을 넣으면

Br2, Cl2, BrCl의 평형 농도는 어떻게 되겠는가?

 

 

 

초기 반응물의 몰농도를 계산하면,

[Br2] = 0.25 mol / 3.0 L = 0.08333 M

 

[Cl2] = 0.55 mol / 3.0 L = 0.1833 M

 

 

 

ICE 도표를 작성하면,

.......... Br2(g) + Cl2(g) ⇌ 2BrCl(g)

초기(M): 0.08333 ... 0.1833 ... 0

변화(M): –x ... –x ... +2x

평형(M): 0.08333–x ... 0.1833–x ... 2x

 

 

 

Kc = [BrCl]^2 / [Br2] [Cl2]

= (2x)^2 / (0.08333–x) (0.1833–x) = 7.0

 

 

 

4x^2 = (7.0) (0.08333–x) (0.1833–x)

 

4x^2 = (7.0) (0.08333×0.1833 – 0.26663x + x^2)

 

4x^2 = (7.0×0.08333×0.1833 – 7.0×0.26663x + 7x^2)

 

3x^2 – 1.86641x + 0.106920723 = 0

 

 

300x250

 

 

근의 공식으로 x를 계산하면,

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

 

x = 0.063837

 

 

 

평형에서,

[Br2] = 0.08333–x

= 0.08333 – 0.063837 = 0.019493 M

 

[Cl2] = 0.1833–x

= 0.1833 – 0.063837 = 0.119463 M

 

[BrCl] = 2x

= 2 × 0.063837 = 0.127674 M

 

 

 

답: [Br2] = 0.020 M, [Cl2] = 0.12 M, [BrCl] = 0.13 M

 

 

 

 

[참고] 검산

Kc = [BrCl]^2 / [Br2] [Cl2]

= 0.127674^2 / (0.019493 × 0.119463)

= 6.9999 ≒ 7.0

 

 

 

 

[키워드] Br2(g) + Cl2(g) ⇌ 2BrCl(g) 평형 기준