Question

Suppose a 500. mL flask is filled with 0.20 mol of NO3 and 2.0 of NO the following reaction becomes possible NO3(g) + NO (g) =2NO2(g) the equilibrium constant K for this reaction is 4.05 at the themperature of the flask .calculate the equilibrium Molarity of NO round to two decimal places final answer in Molarity .

Answer 1

initial concentration of NO3 = mol of NO3 / volume

= 0.20 mol / 0.500 L

= 0.40 M

initial concentration of NO = mol of NO / volume

= 2.0 mol / 0.500 L

= 4.0 M

Let's prepare the ICE table

[NO3] [NO] [NO2]

initial 0.4 4.0 0

change -1x -1x +2x

equilibrium 0.4-1x 4.0-1x +2x

Equilibrium constant expression is

Kc = [NO2]^2/[NO3]*[NO]

4.05 = (4*x^2)/((0.4-1*x)(4-1*x))

4.05 = (4*x^2)/(1.6-4.4*x + 1*x^2)

6.48-17.82*x + 4.05*x^2 = 4*x^2

6.48-17.82*x + 0.05*x^2 = 0

Let's solve this quadratic equation

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

a = 5*10^-2

b = -17.82

c = 6.48

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 = 3.163*10^2

putting value of d, solution can be written as:

x = {17.82 + √(3.163*10^2)}/0.1

x = {17.82 - √(3.163*10^2)}/0.1

solutions are :

x = 3.56*10^2 and x = 0.364

x can't be 3.56*10^2 as this will make the concentration negative.so,

x = 0.364

At equilibrium:

[NO] = 4.0-1x = 4.0-1* 0.364 = 3.64 M

Answer: 3.64 M