Question

Consider the following reversible reaction,

N2O4 and 2NO2

Assuming that inital partial presssure of the gases are 0.5 atm and 0.6 atm respectively, predict what happens to the reaction and explain your response. Assume Kp=31

Answer 1

You want to know what happen to the reaction but in what aspect? you want to know the equilibrium pressure? the direction of the reaction, or what happens when it reach the equilibrium or what happens if you lower the value of pressure?.

Either way, I'll calculate the pressures in equilibrium with the data you provide:

r: N₂O₄ <------> 2NO₂

i: 0.5 0.6

e: 0.5-x 0.6+2x

Kp = 31 = (0.6+2x)² / 0.5-x

31(0.5-x) = 0.36+2.4x+4x²

15.5 - 31x = 0.36+2.4x+4x²

4x²+33.4x-15.14 = 0

Solving for X:

x = -33.4 (33.4² + 4*4*15.14)^1/2 / 2*4

x = -33.4 36.85 / 8

x₁ = 0.43 atm

x₂ = -8.78 atm

Then, the pressure in equilibrium are:

PN₂O₄ = 0.5-0.43 = 0.07 atm

PNO₂ = 0.6+2(0.43) = 1,46 atm

Hope this helps