Question

The air pollutant NO is produced in automobile engines from the high-temperature reaction:

N2(g)+O2(g)⇌2NO(g) . Kc=1.7×10−3.  Calculate the equilibrium concentrations if the initial concentrations are 2.31 M N2 and 0.59 M O2. (This N2/O2 concentration ratio is the ratio found in air.)

Answer 1

[N2] = 2.31

[O2] = 0.59

then

Kc = [NO]^2 / ([N2][O2])

Kc = 1.7*10^-3

then, in equilibrium

due to stoichiomerty coefficients

[N2] = 2.31 - x

[O2]= 0.59 -x

[NO] = 0 + 2x

then

Kc = [NO]^2 / ([N2][O2])

1.7*10^-3 = (2x)^2 / (2.31 - x)( 0.59 -x)

solve for x

( 1.7*10^-3)*(2.31*0.59 +(-2.31-0.59)x + x^2) = 4x^2

(2.31*0.59 +(-2.31-0.59)x + x^2) = (4/( 1.7*10^-3))x^2

1.363 -2.9x + x^2 = 2352.9*x^2

2351.9*x^2 + 2.9x -1.363 = 0

x = 0.02346

then

[N2] = 2.31 - x = 2.31-0.02346 = 2.28654

[O2]= 0.59 -x = 0.59- 0.02346 = 0.56654

[NO] = 0 + 2x = 2*0.02346 = 0.04692