the vapor pressure of water at 27degree calcius is 0.2463atm. calculate the values of KP and KC at 27degree Celsius for the equilibrium water liquid equilibrium water gas
Answers
Given: Vapour pressure of H2O
= 0.2463 atm
Temperature = 27°C
To Find: Kp =?
Kc = ?
Solution:
- In the terms of partial pressure the concentration ratios of products and reactants at equilibrium is described by Kc
- The reaction is given as :-
H2O(l) ---> H2O(g)
As suggested by equation Vapour phase is start forming from the liquid molecules.
- So, The Kp is just equal to vapour pressure of water.
Kp = P(water) = 0.2463 atm (given)
- By the formula given below:-
Kp = Kc (RT)^Δn(g)
Kp is related to Kc to the equilibrium constant in terms of molar concentration .
- Here, value if R = 0.0821 atm
( gas constant)
T = 27°c = 273 + 27
= 300 Kelvin
Δn(g) = 1
- By putting value into equation (i)
0.2463 = Kc ( 0.0821/300)
Kc = (0.0821/300)/0.2463
Kc = 1.11 × 10^-3
- So, The values of Kp and Kc are 0.2463 and 1.11 × 10^-3