Question

A flask is charged with 1.800 atm of N₂O₄(g) and 1.00 atm of NO₂(g) at 25 ^∘C , and the following equilibrium is achieved:

N₂O₄(g)⇌2NO₂(g)

After equilibrium is reached, the partial pressure of NO₂ is 0.519 atm .

Part A:

What is the partial pressure of N₂O₄ at equilibrium?

Part B:

Calculate the value of K_p for the reaction.​

Part C:

Calculate the value of K_c for the reaction.​

Answer 1

A)

ICE Table:

p(N2O4) p(NO2)

initial 1.8 1.0

change -1x +2x

equilibrium 1.8-1x 1.0+2x

Given at equilibrium,

p(NO2) = 0.519

1.0+2x = 0.519

x = -0.2405

p(N2O4) = 1.8-x

= 1.8 + 0.2405

= 2.04 atm

Answer: 2.04 atm

B)

Equilibrium constant expression is

Kp = p(NO2)^2/p(N2O4)

= (0.519)^2/(2.04)

= 0.132

Answer: 0.132

C)

T= 25.0 oC

= (25.0+273) K

= 298 K

Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant

Δ n = 1

use:

Kp= Kc (RT)^Δn

0.132 = Kc *(0.08206*298.0)^(1)

Kc = 5.398*10^-3

Answer: 5.398*10^-3