4.00 mol H2(g), 4.00 mol F2(g), 6.00 mol HF(g)

4.00 mol H2(g), 4.00 mol F2(g), 6.00 mol HF(g)

 

 

When hydrogen reacts with fluorine,

hydrogen fluoride is formed

according to the following equation:

H2(g) + F2(g) ⇌ 2HF(g)

 

The equilibrium constant, K, is 1.15×10^2 at SATP.

Calculate the concentrations of all entities at equilibrium

if 4.00 mol H2(g), 4.00 mol F2(g), and 6.00 mol HF(g)

are initially placed into a 2.00 L container.

 

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각 화학종의 몰농도를 계산하면,

 

4.00 mol / 2.00 L

= 2.00 M = [H2] = [F2]

 

 

6.00 mol / 2.00 L

= 3.00 M = [HF]

 

 

 

 

[그림] H2(g) + F2(g) ⇌ 2HF(g) 반응의 일반적인 ICE 도표와 평형 상수식.

 

 

 

ICE 도표를 작성하면,

 

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

초기(M): 2.00 ............ 2.00 ............ 3.00

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

평형(M): 2.00–x ........ 2.00–x ........ 3.00+2x

 

 

 

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

= (3.00+2x)^2 / (2.00–x) (2.00–x)

= (3.00+2x)^2 / (2.00–x)^2

 

1.15×10^2 = [(3.00+2x) / (2.00–x)]^2

 

(3.00+2x) / (2.00–x) = (1.15×10^2)^(1/2) = 10.7238

 

3.00+2x = (2.00)(10.7238) – 10.7238x

 

12.7238x = (2.00)(10.7238) – 3.00

 

x = [(2.00)(10.7238) – 3.00] / 12.7238

= 1.45 M

 

 

 

답:

[H2] = 2.00 – x = 2.00 – 1.45 = 0.55 M = [F2]

 

[HF] = 3.00 + 2x = 3.00 + 2(1.45) = 5.90 M

 

 

 

 

[참고] 검산

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

= (5.90)^2 / (0.55)^2

= 115.07

≒ 1.15×10^2

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/19700 ] If 3.000 mol of each species (H2, F2, and HF) are put in a 1.500 L vessel

 

 

 

[키워드] H2(g) + F2(g) ⇌ 2HF(g) ICE table 기준문서

When hydrogen reacts with fluorine, hydrogen fluoride is formed according to the following equation: H2(g) + F2(g) ⇌ 2HF(g) Ke = 1.15×10^2 at SATP Calculate the concentrations of all entities at equilibrium if 5.00 mol of H2(g) and 5.00 mol of F2(g) are initially placed into a 2.00 L reaction vessel.