N2O4 <—> 2 NO2 0 1.3120 (initial) x 1.3120-2x (at equilibrium) Kc = [NO2]^2/[N2O

Question

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

Explanation / Answer

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?

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.