The vapor pressure of water at 27°C is 0.2463 atm .Calculate the values of Kc and Kp at 27°C for the equilibrium H2O(l)H2O(g)

Dear Student
 In the above reaction 
H2O(l)  =  H2O(g)
Vapour pressure of water at 27 degree centigrade is 0.2463 atm 
Kp = 1/PH2O
 So     Kp = 1/0.2463 = 4.06 /atm
Again we know Kp = Kc (RT)n
Where n = number of mole change
So     4.06 = Kc/(0.082*300)-1
So Kc = 4.06*0.082*300 = 99.8 as T=27 degree centigrade = 300 K

Regards

  • -27
What are you looking for?