The equilibrium constant Kc is 54.3 at 430°C for the following reaction: H2(g) + I2(g) ⇌...

The equilibrium constant Kc is 54.3 at 430°C for the following reaction: H2(g) + I2(g) ⇌ 2HI(g) Initially, 0.90 M H2, I2, and HI are introduced into a 5.0-L flask and allowed to come to equilibrium. What are the equilibrium concentrations of H2, I2, and HI in the flask? (10 points) (Does the size of flask matter since Molarity is given?)

Expert Solution

H2(g) + I2(g) ⇌ 2HI(g)

I 0.9 0.9 0.9

C +x +x -2x

E 0.9+x 0.9+x 0.9-2x

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

54.3 = (0.9-2x)^2/(0.9+x)(0.9+x)

54.3 = (0.9-2x)^2/(0.9+x)^2

7.37 = 0.9-2x/0.9+x

7.37*(0.9+x) = (0.9-2x)

x = -0.6

[H2] = 0.9+x = 0.9-0.6 = 0.3M

[I2] = 0.9+x = 0.9-0.6 = 0.3M

[HI] = 0.9-2x = 0.9-2*-0.6 = 2.1M

queen_honey_blossom answered 2 years ago

The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430 ℃

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