Question

The equilibrium constant, Kc, for the following reaction is 1.80×10-2 at 698 K.

2HI(g) ---> H2(g) + I2(g)

Calculate the equilibrium concentrations of reactant and products when 0.384 moles of HI are introduced into a 1.00 L vessel at 698 K.

[HI] = M

[H2] = M

[I2] = M

Answer 1

Kc = 1.8*10^-2 = 0.018

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

initially:

[HI] = mol/V = 0.384/1 = 0.384

[I2] = 0

[H2] = 0

in equilibrium

[HI] = 0.384 -2x

[I2] = 0 + x

[H2] = 0 + x

substitute in Kc expression

Kc = 0.018

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

Kc = (x*x)/(0.384 -2x)^2

0.018 = (x*x)/(0.384 -2x)^2

0.018 = (x^2)/(0.384 -2x)^2

sqrt(0.018 ) = sqrt( (x^2)/(0.384 -2x)^2

0.13416 = x/(0.384 -2x)

solve for x

0.13416*(0.384 -2x) = x

7.4537x = (0.384 -2x)

9.4537x = 0.384

x = 0.384 / 9.4537

x = 0.04061

in equilibrium:

[HI] = 0.384 -2x = 0.384-2*0.04061 = 0.30278

[I2] = 0 + x = 0.04061

[H2] = 0 + x = 0.04061

proof:

Q = (0.04061^2)/0.30278^2 = 0.017989

which is pretty near to 0.018, so this must be correct