Question

Suppose a 250.mL flask is filled with 1.9mol of H2 and 1.7mol of I2. The following reaction becomes possible:

H2(g) + I2(g)   2HI(g)

The equilibrium constant K for this reaction is 5.99 at the temperature of the flask.

Calculate the equilibrium molarity of H2. Round your answer to two decimal places.

Answer 1

Initial concentration of H2 = mol of H2 / volume in L
= 1.9 mol / 0.250 L
= 7.6 M

Initial concentration of I2 = mol of I2 / volume in L
= 1.7 mol / 0.250 L
= 6.8 M

ICE Table:

[H2] [12] initial 7.6 6.8 -1x - 1x change equilibrium 7.6-1x 6.8-1x

Equilibrium constant expression is
Kc = [HI]^2/[H2]*[I2]
5.99 = (4*x^2)/((7.6-1*x)(6.8-1*x))
5.99 = (4*x^2)/(51.68-14.4*x + 1*x^2)
3.096*10^2-86.26*x + 5.99*x^2 = 4*x^2
3.096*10^2-86.26*x + 1.99*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1.99
b = -86.26
c = 3.096*10^2

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 4.976*10^3

roots are :
x = 39.4 and x = 3.949

x can't be 39.4 as this will make the concentration negative.so,
x = 3.949

At equilibrium:
[H2] = 7.6-1x = 7.6-1*3.949 = 3.65 M

Answer: 3.65 M