Question

The kp for the reaction n2o4 ↔ 2no2 is 640 mm at 775k. calculate the percentage dissociation of n2o4 at equilibrium pressure of 160 mm. at what pressure, the dissociation will be 50%

a. 460 mm

b. 490 atm

c. 480 mm

d. 470 atm

Answer 1

Let there be initially 1 mole  and x moles of reactant reacts before equilibrium.

N2O4→ 2NO2
 1-x           2x

Total No. of moles at equilibrium = 1-x+2x
                                                       = 1+x

         ((2x/1+x) P)2    
Kp=    - - - - - - - -                          P= Total pressure
           (1-x/1+x)P

Substitute Kp and P  as given in the question
and you will get x.

To find when the dissociation will be 50%. Substitute x as 0.5(as 50% dissociation in mole is 0.5) and substitute Kp  that you found in the last step.