Question

# Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial...

Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.802 M, what is the equilibrium concentration (in M to 4 decimal places) of NO?

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Keq = [NO]^2[Br2]/[NOBr]^2

initially

[NOBr] = 0.802

[NO] = 0

[Br2] = 0

in equilibrium

[NOBr] = 0.802 -2x

[NO] = 0 +2x

[Br2] = 0 + x

substitute

Keq = [NO]^2[Br2]/[NOBr]^2

(3.07*10^-4)= (2x)^2(x) / (0.802 -2x )

0.802 - 2x = 4/0.000307 * x^3

13029.31x^3 +0x^2 +2x - 0.802 =0

x = 0.038188

[NOBr] = 0.802 -2*0.038188 = 0.725624

[NO] = 0 +2*0.038188 = 0.076376

[Br2] = 0 + 0.038188

Proof

Q= [NO]^2[Br2]/[NOBr]^2 = (0.076376^2)(0.038188) / (0.725624^2) = 4.2*10^-4 approx, near to 3*10^-4