Question

Suppose a 250.mL flask is filled with 0.20mol of N2 and 1.4mol of O2 . The following reaction becomes possible: N2(g)+O2(g)= 2NO(g )The equilibrium constant K for this reaction is 4.94 at the temperature of the flask. Calculate the equilibrium molarity of O2 . Round your answer to two decimal places.

Answer 1

initial concentration of N2 = mol of N2 / volume in L

= 0.20 mol / 0.250 L

= 0.80 M

initial concentration of O2 = mol of O2 / volume in L

= 1.4 mol / 0.250 L

= 5.6 M

Let's prepare the ICE table

[N2] [O2] [NO]

initial 0.8 5.6 0

change -1x -1x +2x

equilibrium 0.8-1x 5.6-1x +2x

Equilibrium constant expression is

Kc = [NO]^2/[N2]*[O2]

4.94 = (4*x^2)/((0.8-1*x)(5.6-1*x))

4.94 = (4*x^2)/(4.48-6.4*x + 1*x^2)

22.1312-31.616*x + 4.94*x^2 = 4*x^2

22.1312-31.616*x + 0.94*x^2 = 0

Let's solve this quadratic equation

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

a = 0.94

b = -31.62

c = 22.13

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 = 9.164*10^2

putting value of d, solution can be written as:

x = {31.62 + √(9.164*10^2)}/1.88

x = {31.62 - √(9.164*10^2)}/1.88

solutions are :

x = 32.92 and x = 0.7152

x can't be 32.92 as this will make the concentration negative.so,

x = 0.7152

At equilibrium:

[O2] = 5.6-1x = 5.6-1*0.71521 = 4.88 M

Answer: 4.88 M