5.00 L 1.00 mol H2 1.00 mol H2 2.50 mol HI 500 K Kc 129

A 5.00 L reaction vessel is filled with

1.00 mol of H2, 1.00 mol of I2 and 2.50 mol of HI.

Kc (at 500 K) is 129.

Calculate the equilibrium concentrations of

H2, I2 and HI at 500 K given the reaction:

H2(g) + I2(g) ⇌ 2HI(g)

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[H2] = 1.00 mol / 5.00 L

= 0.200 M = [I2]

[HI] = 2.50 mol / 5.00 L

= 0.500 M

H2(g) + I2(g) ⇌ 2HI(g)

Kc = [HI]^2 / [H2] [I2] = 129

[그림] H2(g) + I2(g) ⇌ 2HI(g) 반응의 일반적인 ICE 도표와 평형상수식.

위 그림에서,

초기(M): C C 0 대신,

초기(M): 0.200 0.200 0.500 이므로,

(0.500 + 2x)^2 / (0.200 – x)^2 = 129

[(0.500 + 2x) / (0.200 – x)]^2 = 129

(0.500 + 2x) / (0.200 – x) = 129^(1/2) = 11.3578167

0.500 + 2x = (0.200)(11.3578167) – 11.3578167x

13.3578167x = (0.200)(11.3578167) – 0.500

x = [(0.200)(11.3578167) – 0.500] / 13.3578167

= 0.1326237

= 0.133

답: 평형에서,

[H2] = [I2] = 0.200 – x

= 0.200 – 0.133 = 0.067 M

[HI] = 0.500 + 2x

= 0.500 + 2(0.133) = 0.766 M

[검산]

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

= (0.500 + 2(0.1326237))^2 / (0.200 – 0.1326237)^2

= 128.9999

≒ 129

[키워드] H2(g) + I2(g) ⇌ 2HI(g) 평형 기준