Equilibrium and using ICE tables

Question

A reaction vessel initially contains only 3.5 atm of N₂O₄. When the system reaches equilibrium according to the reaction below, there is 1.7 atm of NO₂ in the vessel. What is Kp for this reaction? 2 NO₂ (g) ⇌ N₂O₄ (g)

Answer 1

2NO_{2 (g)} ⇔ N₂O_{4 (g)}

The equilibrium expression for this reaction is K_p = [ N₂O₄] / [NO₂]²

( you should know how to write equilibrium expressions from chemical equations)

For reactions involving only gases, we can use the pressure relationship to calculate the K_p

We need to set up an ICE table ( initial , change, equilibrium)

2NO_{2 (g)} ⇔ N₂O_{4 (g)}

I 0 3.5

C +2x -x

E +2x 3.5 -x

Based on the expression above; K_p = 3.5 -x / (2x)²

We were told that the pressure of NO₂ at equilibrium was 1.7 atm; 2x = 1.7

x = 0.85 .

Plug these numbers into the equation above

K_p= (3.5-0.85) / (2*0.85)² = 2.65 atm / (1.7 atm)² = 0.92 atm^-1