Question

Consider the following reaction (assume an ideal gas mixture):

2NOBr(g) <--> 2NO(g) + Br2(g)

A 1.0-liter vessel was initially filled with pure NOBr, at a pressure of 0.685 atm, at 300 K. After equilibrium was established, the partial pressure of NOBr was 0.279 atm. What is Kp for the reaction? Report answer to 4 decimal points.

Answer 1

Iniitial conditions

P-NOBr = 0.685 atm

P-NO = 0

P-Br2 = 0

the extent in equilbirium

P-NOBr = 0.685 -2x

P-NO = 0 +2x

P-Br2 = 0+x

in equilbrium, we know

P-NOBr = 0.685 -2x = 0.279

solve for x

x = (0.279 - 0.685 )/(-2)

x = 0.203

now..

P-NOBr = 0.685 -2*0.203 = 0.279

P-NO = 0 +2*0.203 = 0.406

P-Br2 = 0+0.203 = 0.203

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)

For pressure

Kp = P-NO^2 * P-Br2 / (P-NOBr)^2

Kp = (0.406^2)(0.203)/ (0.279^2)

Kp = 0.4298