Question

Suppose a 500. mL flask is filled with 1.7 mol of O_2 and 1.1 mol of NO. The following reaction becomes possible: N_2(g) + O_2(g) 2NO(g) The equilibrium constant K for this reaction is 3.80 at the temperature of the flask. Calculate the equilibrium molarity of NO.

Answer 1

initial concentration of O2 = number of mol/volume in L
= 1.7/0.5
= 3.4 M

initial concentration of NO = number of mol/volume in L
= 1.1/0.5
= 2.2 M

N2 + O2 <------> 2NO
0 3.4 2.2 (initial)
x 3.4+x 2.2-2x (at equilibrium)

Kc = [NO]^2/ {[N2][O2]}
3.80 = (2.2-2x)^2 / {(x)(3.4+x)}
3.80*(x)(3.4+x) = (2.2-2x)^2
12.92*x + 3.80*x^2 = (4.84 + 4x^2 -8.8*x)
0.2*x^2 - 21.72*x + 4.84 = 0
solving for x we get,
x = 108 and x = 0.223 M
x can't be greater than 2.2. so x = 0.223 M

[NO] = 2.2 -2x
= 2.2-2*0.223
=1.8
Answer: 1.8 M