H2 F2 HF 1.15×10^2 3.000 mol 1.500 L equilibrium concentration

H2 F2 HF 1.15×10^2 3.000 mol 1.500 L equilibrium concentration

 

 

Given: H2(g) + F2(g) ⇌ 2HF(g) ... Kc = 1.15×10^2

If 3.000 mol of each species (H2, F2, and HF) are put in a 1.500 L vessel,

what would be the equilibrium concentration of each species?

 

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각 화학종의 몰농도

= 3.000 mol / 1.500 L

= 2.000 M

 

 

 

ICE 도표를 작성하면,

 

............. H2(g) ... + ... F2(g) ... ⇌ ... 2HF(g)

초기(M): 2.000 .......... 2.000 .......... 2.000

변화(M): –x ............... –x ............... +2x

평형(M): 2.000–x ...... 2.000–x ...... 2.000+2x

 

 

 

K = [HF]^2 / [H2] [F2]

= (2.000+2x)^2 / (2.000–x) (2.000–x)

= (2.000+2x)^2 / (2.000–x)^2

 

1.15×10^2 = [(2.000+2x) / (2.000–x)]^2

 

(2.000+2x) / (2.000–x) = (1.15×10^2)^(1/2) = 10.7238

 

2.000+2x = (2.000)(10.7238) – 10.7238x

 

12.7238x = (2.000)(10.7238) – 2.000

 

x = [(2.000)(10.7238) – 2.000] / 12.7238 = 1.528 M

 

 

 

답:

[H2] = 2.000 – 1.528 = 0.472 M = [F2]

[HF] = 2.000 + 2(1.528) = 5.056 M

 

 

 

 

[참고] 검산

K = [HF]^2 / [H2] [F2]

= (5.056)^2 / (0.472)^2

= 114.7

≒ 1.15×10^2