Question

The following reaction was performed in a sealed vessel at 737 ∘C :

H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.65M and [I2]=2.90M. The equilibrium concentration of I2 is 0.0900 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Answer 1

Solution :-

Lets make the ICE table for the given reaction

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

3.65 M              2.90 M                                   0

-x                           -x                                         +2x

3.65-x               2.90-x =0.0900                       2x

Using the equilibrium concentration of the I2 lets calculate the value of the X

2.90-x = 0.0900

x= 2.90 – 0.0900 = 2.81 M

now lets find equilibrium concentration of the H2 and HI

H2 = 3.65 M – x = 3.65 M - 2.81 M = 0.84 M

HI = 2x = 2*2.81 M = 5.62 M

Now lets write the Kc equation

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

    = [5.62]^2/[0.84][0.0900]

    = 418

Kc= 418

So the equilibrium constant Kc = 418