H2(g) + I2(g) ⇌ 2HI(g) 평형상수 54.4 at 355℃

H2(g) + I2(g) ⇌ 2HI(g) 평형상수 54.4 at 355℃

 

 

The equilibrium constant for the reaction

H2(g) + I2(g) ⇌ 2HI(g)

was found to be 54.4 at 355℃.

What are the equilibrium concentrations if 0.200 mol H2 and 0.200 mol I2 are mixed in a 0.500 L flask, and allowed to come to equilibrium?

 

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[H2] = [I2] = 0.200 mol / 0.500 L = 0.400 M

 

 

Kc = [HI]^2 / [H2][I2]

 

 

54.4 = (2x)^2 / (0.400-x)^2

( 식 설명 https://ywpop.tistory.com/6300 )

 

54.4 = (2x / 0.400-x)^2

2x / 0.400-x = 54.4^(1/2)

2x = (54.4^(1/2))(0.400) - (54.4^(1/2))x

(2 + (54.4^(1/2)))x = (54.4^(1/2))(0.400)

 

x = ((54.4^(1/2))(0.400)) / (2 + (54.4^(1/2)))

= 0.31467245

= 0.315

 

 

답:

[H2] = [I2] = 0.400-x = 0.400 – 0.315 = 0.085 M

[HI] = 2x = 2 × 0.315 = 0.630 M

 

 

 

[검산]

[HI]^2 / [H2][I2] = 0.630^2 / 0.085^2 = 54.9 ≒ 54.4

 

또는

(2 × 0.31467245)^2 / (0.400 – 0.31467245)^2 = 54.4

 

 

 

[참고] H2(g) + I2(g) ⇌ 2HI(g) 평형상수 일반식.