Question

The value of Kc for the reaction of dinitrogen tetroxide to make nitrogen dioxide is 1.9636....

The value of Kc for the reaction of dinitrogen tetroxide to make nitrogen dioxide is 1.9636. The concentration of nitrogen dioxide 1.3120 M with no dinitrogen tetroxide. What is the equilibrium concentration (in M) of nitrogen dioxide?

Homework Answers

Answer #1

N2O4    <—>    2 NO2
0       1.3120   (initial)
x       1.3120-2x   (at equilibrium)

Kc = [NO2]^2/[N2O4]
1.9636 = (1.3120-2x)^2 / (x)
1.9636*x = 1.7213 + 4*x^2 - 5.248*x
4*x^2 - 7.212*x + 1.7213 = 0

This is quadratic equation (ax^2+bx+c=0)
a = 4
b = -7.212
c = 1.721

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 = 24.47

roots are :
x = 1.52 and x = 0.2831

x can't be 1.52 as this will make the concentration negative.so,
x = 0.2831

At equilibrium:
[NO2] = 1.3120-2x
= 1.3120-2*0.2831
= 0.7458 M

Answer: 0.7458 M

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