Question

Consider the equilibrium

N2(g) + O2(g) ⇄ 2 NO(g)

At 2300 K the equilibrium constant K = 1.7 × 10-3. Suppose that 0.0150 mol NO(g), 0.250 mol N2(g), and 0.250 mol O2(g) are placed into a 10.0-L flask and heated to 2300 K. The system is not at equilibrium. Determine the direction the reaction must proceed to reach equilibrium and the final equilibrium concentrations of each species.

to the right

to the left

[N2] =____ mol/L

[O2] = ____mol/L

[NO] =____ mol/L

Answer 1

A)

Direction

Q = [NO]^2/[N2][O2]

[NO2] = 0.015/10 = 0.0015

[N2] = 0.25/10 = 0.025

[O2] = 0.25/10 = 0.025

Q = (0.0015^2)/(0.025*0.025) = 0.0036 = 3.6*10^-3

K = 1.7*10^-3 =

Since Q > K, then expect more NO formation

Then

initially

[NO2] = 0.0015

[N2] = 0.025

[O2] = 0.025

in equilibrium

[NO] = 0.0015 + 2x

[N2] = 0.025 - x

[O2] = 0.025 - x

substitute in Kc

K = [NO]^2/[N2][O2]

1.7*10^-3 = (0.0015 + 2x)^2 / ( 0.025 - x)^2

sqrt((1.7*10^-3)) = (0.0015 + 2x) / ( 0.025 - x)

0.04123*(0.025 - x) = 0.0015 + 2x

0.04123*0.25 - 0.04123x = 0.0015 + 2x

x(2+0.04123) = 0.04123*0.25 -0.0015

x = 0.0088075 / 2.04123

x = 0.004314

so

[NO] = 0.0015 + 2*0.004314 = 0.010128 M

[N2] = 0.025 - 0.004314 = 0.020686 M

[O2] = 0.025 - 0.004314 = 0.020686 M