Question

0.250 moles of H2 and 0.250 moles of I2 were placed in a 1.00 L flask at 500˚. The equilibrium constant, Kc, for the reaction H2(g)+I2(g)-> 2HI(g) is 54.3. Calculate the equilibrium concentrations of all species.

Answer 1

since volume is 1 L, concentration will be same as number of moles

Let's prepare the ICE table

[H2] [I2] [HI]

initial 0.25 0.25 0

change -1x -1x +2x

equilibrium 0.25-1x 0.25-1x +2x

Equilibrium constant expression is

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

54.3 = (2*x)^2/(0.25-1*x)^2

sqrt(54.3) = (2*x)/(0.25-1*x)

7.369 = (2*x)/(0.25-1*x)

1.84221-7.36885*x = 2*x

1.84221-9.36885*x = 0

x = 0.197

At equilibrium:

[H2] = 0.25-x = 0.25-0.19663 = 0.0534 M

[I2] = 0.25-x = 0.25-0.19663 = 0.0534 M

[HI] = 2x = 2*0.19663 = 0.393 M

[H2] = 0.0534 M

[I2] = 0.0534 M

[HI] = 0.393 M