Consider the following chemical equilibrium reaction 2 mol o

Consider the following chemical equilibrium reaction. 2 mol of H2 is placed (along with 2 mol of I2) in a 1 L vessel. Calculate the concentration of HI when equilibrium is reached. The value of the equilibirum constant is 49.7.

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

Solution

initial concentration of H2 = 2 mol / 1 L = 2 mol / L = 2 M

initial concentration of I2 = 2 mol / 1 L = 2 M

now

H2 + I2 ---> 2 HI

using ICE table

initial concentration of H2 , I2 , HI are 2 , 2 , 0

change in concentration of H2,I2,HI are -x , -x , +2x

equilibrium concentration of H2,I2,HI are 2-x ,2-x , 2x

now

K = [HI]^2 / [H2]eq [I2]eq

49.7 = [2x]^2 / [2-x][2-x]

49.7 = ( [2x] /[2-x])^2

sqrt(49.7) = (2x) / (2-x)

7.0498 (2-x) = 2x

x = 1.558

so

[HI]eq = 2x = 2 (1.558)

[HI]eq = 3.116 M