Question

At –80°C, K for the reaction N2O4(g) 2NO2(g) is 4.66 × 10–8. We introduce 0.049 mole of N2O4 into a 1.0-L vessel at –80°C and let equilibrium be established. The total pressure in the system at equilibrium will be: (0.78 atm)

Answer 1

Calculation of pressure exerted by N2O4 :

We know that PV = nRT

where

P = pressure = ?

V = volume = 1.0 L

n = number of moles = 0.049 mol

R = gas constant = 0.0821 L-atm/(mol-L)

T = temperature = -80 ^oC = -80+273 = 193 K

Plug the values we get P = (nRT) / V

                                      = ( 0.049 x 0.0821x193) / 1.0

                                      = 0.776 atm

                                N₂O₄(g) 2NO₂(g)

initial pressure            0.776              0

change                        -p                +2p

Equb pressure          0.776-p             2p

Equilibrium constant , K = [NO₂]² / [N₂O₄] = 4.66x10^-8

                                       (2p)2 / (0.776-p) = 4.66x10^-8

                solving for p we get , p = 4.0x10^-3 atm

So Equilibrium pressure of N₂O₄(g) = 0.776-p = 0.776-(4.0x10^-3 ) = 0.772 atm

Equilibrium pressure of NO₂ is = 2p = 2x4.0x10^-3 = 8.0x10^-3 atm

So total pressure , P = equilibrium pressure of N₂O₄ + equilibrium pressure of NO₂

                               = 0.772 + 8.0x10^-3

                               = 0.780 atm