Question

A 1 liter vessel is filled with 1 mol of bromine gas and 2 moles of hydrogen gas to form hydrogen bromide gas at 448 degrees celsius. the reaction has an equilibrium constant of 53.5. what is the equilibrium molar concentration of all species?

Answer 1

The reaction is:

H2 + Br2 <—> 2 HBr

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

Let's prepare the ICE table

[H2] [Br2] [HBr]

initial 2.0 1.0 0

change -1x -1x +2x

equilibrium 2.0-1x 1.0-1x +2x

Equilibrium constant expression is

Kc = [HBr]^2/[H2]*[Br2]

53.5 = (4*x^2)/((2-1*x)(1-1*x))

53.5 = (4*x^2)/(2-3*x1*x^2)

107-160.5*x53.5*x^2 = 4*x^2

107-160.5*x49.5*x^2 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 49.5

b = -1.605*10^2

c = 1.07*10^2

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 4.574*10^3

putting value of d, solution can be written as:

x = {1.605*10^2 + √(4.574*10^3)}/99

x = {1.605*10^2 - √(4.574*10^3)}/99

solutions are :

x = 2.304 and x = 0.938

x can't be 2.304 as this will make the concentration negative.so,

x = 0.938

At equilibrium:

[H2] = 2.0- x = 2.0-0.93805 = 1.06195 M

[Br2] = 1.0-x = 1.0-0.93805 = 0.06195 M

[HBr] = 2x = 2*0.93805 = 1.8761 M

[H2] = 1.06 M

[Br2] = 0.062 M

[HBr] = 1.88 M