Question

I2(g)+Cl2(g)⇌2ICl(g),Kp=81.9 (at 25∘C) The reaction between I2 and Cl2 is carried out at the same temperature, but with the following initial partial pressures: PI2=0.170atm, PCl2=0.170atm, and PICl=0.00 atm. find equibilibrum partial pressure of ICl.

Answer 1

first construct the ICE table and substitute the values in the kp equation calculate for values and finally you can get the euilibrium partial pressure of ICl
I2    +   Cl2 ---------------------> 2 ICl

I 0.170    0.170                              0

C   -X          -X                                +2X

E    (0.170-X) (0.170-X)                  2X

Kp = [ICl]² / [I2] [Cl2]

81.9 = [2X]² / [0.170-X][0.170-X]

81.9 * [0.170-X][0.170-X] = [2X]²

solving for x = 0.147 atm

at equilibrium ICl partial pressure is => 2x = 2*0.147 = 0.294 atm