Question

For the following reaction, Kc = 0.513 at 500 K. N2O4(g)⇌2NO2(g) If a reaction vessel initially contains an N2O4 concentration of 5.50×10−2 M at 500 K, what are the equilibrium concentrations of N2O4 and NO2 at 500 K? [N2O4], [NO2] = ____ M

Answer 1

                    [N2O4]              [NO2]             

initial             0.055               0                 

change              -1x                 +2x               

equilibrium         0.055-1x            +2x               

Equilibrium constant expression is
Kc = [NO2]^2/[N2O4]
0.513 = (4*x^2)/((0.055-1*x))
0.02822-0.513*x = 4*x^2
0.02822-0.513*x-4*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -4
b = -0.513
c = 2.822*10^-2

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 = 0.7146

roots are :
x = -0.1698 and x = 0.04154

since x can't be negative, the possible value of x is
x = 0.04154

At equilibrium:
[N2O4] = 0.055-x = 0.055-1*0.04154 = 0.0135 M
[NO2] = +2x = +2*0.04154 = 0.0831 M