Question

A and B react as follows: A + B = C + D. The equilibrium constant is 2.0 x 10^3. If 0.30 mol of A and 0.80 mol of B are mixed in 1 L, what are the concentrations for A, B, C, and D after reaction?

Answer 1

Since volume is 1 L, number of moles is same as molarity

Let's prepare the ICE table

[A] [B] [C] [D]

initial 0.3 0.8 0 0

change -1x -1x +1x +1x

equilibrium 0.3-1x 0.8-1x +1x +1x

Equilibrium constant expression is

Kc = [C]*[D]/[A]*[B]

2000.0 = (1*x)(1*x)/((0.3-1*x)(0.8-1*x))

2000.0 = (1*x^2)/(0.24-1.1*x + 1*x^2)

480-2200*x + 2000*x^2 = 1*x^2

480-2200*x + 1999*x^2 = 0

Let's solve this quadratic equation

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

a = 1.999*10^3

b = -2.2*10^3

c = 4.8*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 = 1.002*10^6

putting value of d, solution can be written as:

x = {2.2*10^3 + √(1.002*10^6)}/3.998*10^3

x = {2.2*10^3 - √(1.002*10^6)}/3.998*10^3

solutions are :

x = 0.8006 and x = 0.2999

x can't be 0.8006 as this will make the concentration negative.so,

x = 0.2999

At equilibrium:

[A] = 0.3-1x = 0.3-1*0.29991 = 0.00009 M

[B] = 0.8-1x = 0.8-1*0.29991 = 0.50009 M

[C] = 0+1x = 0+1*0.29991 = 0.29991 M

[D] = 0+1x = 0+1*0.29991 = 0.29991 M