Question

In a chemical reaction,4 mol of nitrogen monoxide and 2.5 mol of oxygen are mixed together in a container with a volume of 1000ml to produce nitrogen dioxide.if the kc for this reaction is 0.002.
Determined the equilibrium concentrations for each substance?

Answer 1

The balanced chemical equation for the reaction is:

2NO (g) + O2 (g) -> 2NO2 (g)

Let x be the change in concentration of NO and O2, and 2x be the change in concentration of NO2.

Initial concentrations:
[NO] = 4 mol / 1 L = 4 M
[O2] = 2.5 mol / 1 L = 2.5 M
[NO2] = 0 mol / 1 L = 0 M

At equilibrium:
[NO] = 4 - 2x
[O2] = 2.5 - x
[NO2] = 2x

Using the equilibrium constant expression:

Kc = [NO2]^2 / ([NO]^2 * [O2])
0.002 = (2x)^2 / ((4-2x)^2 * (2.5-x))

Solving for x gives:
0.002 = 4x^2 / (16 - 16x + 4x^2) * (2.5 - x)
0.002 = 4x^2 / (16 - 16x + 4x^2) * (2.5 - x)
0.002 = 4x^2 / (16 - 64x + 64x^2 + 2.5x - 10x^2)
0.002 = 4x^2 / (54 - 61.5x + 54x^2)

Solving for x gives:
0.002 = 4x^2 / (54 - 61.5x + 54x^2)
x = 0.408

Substitute x back into the equilibrium concentrations:
[NO] = 4 - 2(0.408) = 3.184 M
[O2] = 2.5 - 0.408 = 2.092 M
[NO2] = 2(0.408) = 0.816 M

Therefore, the equilibrium concentrations are:
[NO] = 3.184 M
[O2] = 2.092 M
[NO2] = 0.816 M