Types of Equilibrium Constants Homework Help, Tutoring
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Types of Equilibrium Constants
In previous session we have seen what equilibrium constant is. In this session we will learn
how it is represented in terms of concentration and partial pressure.
Kc : if the active masses of the reactants and products is expressed in terms of molar concentration then the equilibrium constant is called as equilibrium constant in terms of concentration and is represented as Kc
Kp: if the reactants and products are in gaseous state then the concentrations are expressed in terms of partial pressure instead of the molar concentrations then the equilibrium constant is termed as equilibrium constant in terms of partial pressure and is represented by Kp
Now, let us see how to represent a general reaction in terms of Kp and Kc
aA + bB ↔ xX + yY
In terms of Kc the above reaction can be written as
[X]x [Y]y
Kc = -----------
[A]a [B]b
Kc = -----------
[A]a [B]b
C Xx . C Yy
Kc = ---------------
CA a . CB b
Kc = ---------------
CA a . CB b
Where C X, C Y, CA, and CB represent the molar concentration of X, Y, A, and B respectively at equilibrium.
In terms of partial pressure the above equation is given as
P Xx . p Yy
Kp = ---------------
PA a . PB b
Kp = ---------------
PA a . PB b
Where P X, P Y, PA, and PB represent the molar concentration of X, Y, A, and B respectively at equilibrium.
Relationship between Kp and Kc:
Again consider the general reaction
aA + bB ↔ xX + yY
as given above the equilibrium constant of this reaction in terms of concentration can be given as
[X]x [Y]y
Kc = -------------- (1)
[A]a [B]b
Kc = -------------- (1)
[A]a [B]b
If A, B, X and Y are gaseous then the equilibrium constant for the above reaction can be given in terms of pressure as below.
P Xx . p Yy
Kp = -------------- (2)
PA a . PB b
Kp = -------------- (2)
PA a . PB b
If the gases are considered as ideal, then we can apply the ideal gas equation
for the above equation.
Ideal gas equation is PV= nRT where n represents the number of moles.
P = n/V RT --------------------------------(3)
P = CRT. --------------------------------(4)
Ideal gas equation is PV= nRT where n represents the number of moles.
P = n/V RT --------------------------------(3)
P = CRT. --------------------------------(4)
Where C= n/V is the number of moles per liter i.e. the molar concentration.
Now let us substitute the value P = CRT in the above given equation 2.
(Cx RT)x (Cy RT)y
Kp = --------------------
(CART)a (CB RT)b
Kp = --------------------
(CART)a (CB RT)b
C Xx . C Yy
Rx Tx . Ry Yy
= ---------------- × ----------------
C Aa . C Bb Ra Ta . Rb Yb
= ---------------- × ----------------
C Aa . C Bb Ra Ta . Rb Yb
C Xx . C Yy
= ---------------- × (RT) (x+y) - (a+b) -------------------------(4)
C Aa . C Bb
= ---------------- × (RT) (x+y) - (a+b) -------------------------(4)
C Aa . C Bb
But,
C Xx . CYy
Kc = ------------
CA a . CB b
Kc = ------------
CA a . CB b
Now, by substituting Kc in the above equation we get
Kp = Kc x (RT) (x+y) - (a+b)-----------------(5)
We know that Δ n = Total no of moles of products -Total no of moles of reactants
Now, by substituting Δ n in the above equation we get Kp = Kc x (RT) Δ n
When Δ n=0 we have
Kp = Kc
Units of Kp and Kc:
As the partial pressure is expressed in atmospheres the units of Kp is Atmosphere.
Similarly, the concentrations are expressed in moles per liter and the dimensions for Kc is given as
(Moles/Liters) Δ n
Example for deriving quantitative expressions for equilibrium constants for a typical reaction
Reaction between hydrogen and iodine:
We know that hydrogen and iodine combine to form hydrogen iodide.
The expression for Kc for this reaction can be given as follow.
Consider a moles of hydrogen and b moles of iodine are allowed to react in a closed vessel of
V liters capacity. Suppose at the stage of equilibrium x moles of hydrogen and x moles of iodine
have reacted to form 2x moles of hydrogen iodide. Now, the reaction can be represented as
Initial moles
a
b 0
H2 + I2 ↔ 2HI
At equilibrium (a-x) (b-x) 2x
Molar concentrations at equilibrium (a-x) (b-x) 2x
----- ----- -----
V V V
H2 + I2 ↔ 2HI
At equilibrium (a-x) (b-x) 2x
Molar concentrations at equilibrium (a-x) (b-x) 2x
----- ----- -----
V V V
By applying the law of chemical equilibrium we get
[HI]2
Kc = ------------
[H2] [I2]
Kc = ------------
[H2] [I2]
[2X/V]2
= -----------------
[a-X/V] [b-x/V]
= -----------------
[a-X/V] [b-x/V]
4X2
= ------------
(a-X) (b-X)
= ------------
(a-X) (b-X)
In the above expression for Kc, Volume and pressure are not used. Hence, at the state of
equilibrium the above reaction is not affected by the change in volume and thus the pressure
of the reaction mixture. Thus, we can say that equilibrium is independent of volume and pressure.
Deriving the expression for Kp
Now, let us derive the equilibrium constant for the formation of hydrogen iodide in terms of
partial pressure of the products and reactants.
Consider the total pressure of the reaction mixture at equilibrium as P atmospheres.
Initial no of gram moles
a
b 0
H2 + I2 ↔ 2HI
No of gram moles at equilibrium (a-x) (b-x) 2x
H2 + I2 ↔ 2HI
No of gram moles at equilibrium (a-x) (b-x) 2x
The total number of gram moles at equilibrium = (a-x) + (b-x) + 2x
= a + b
As the total pressure of the system is P atmospheres.
The partial pressure of H2 at equilibrium is give as
(a-x)
(p H2) = -------- x P atm.
a + b
(p H2) = -------- x P atm.
a + b
The partial pressure of I2 at equilibrium is give as
(b-x)
(p I2) = --------- x P atm.
a + b
(p I2) = --------- x P atm.
a + b
The partial pressure of HI at equilibrium is given as
2x
(pHI) = ------------ x P atm.
a + b
(pHI) = ------------ x P atm.
a + b
By applying the law of equilibrium we have
[pHI]2
Kp = ------------
[p H2] [p I2]
Kp = ------------
[p H2] [p I2]
(2xP/ a+b)2
= -----------------
P(a-x) P(b-x)
-------- ---------
(a+b) (a+b)
= -----------------
P(a-x) P(b-x)
-------- ---------
(a+b) (a+b)
P2 4x2
a+b a+b
= -------- -------- --------
a+b P(a-x) P(b-x)
= -------- -------- --------
a+b P(a-x) P(b-x)
4x2
= ------------
(a-x) (b-x)
= ------------
(a-x) (b-x)
As can be seen above, the final expression does not have P , indicating that the state
of equilibrium is independent of total pressure and hence the total volume.
You can also see that the final expressions obtained for Kp and Kc are the same hence
You can also see that the final expressions obtained for Kp and Kc are the same hence
Kp = Kc
Kp and Kc are identical for all gaseous state reactions in which the total number of moles of
the reactants and products are equal.
Solved problems
1. At 500 degrees centigrade Kp for synthesis of ammonia is 1.42x10-5 atm. Calculate the Kc
Ans: the equation for the above reaction is
Δ n= 2-(1+3)
=-2
Kp = 1.42x10-5
T= 273 + 500
=773 K
R = 0.0821 liter atm.
By substituting the values in the following expression, we have
Kc = Kp/ (RT) Δ n
1.42x10-5/ (0.0821 x 773)-2
=0.057192
Ans: the equation for the above reaction is
N2+3H2 ↔ 2NH3
Δ n= 2-(1+3)
=-2
Kp = 1.42x10-5
T= 273 + 500
=773 K
R = 0.0821 liter atm.
By substituting the values in the following expression, we have
Kc = Kp/ (RT) Δ n
1.42x10-5/ (0.0821 x 773)-2
=0.057192
2. 13.4 c.c. of hydrogen iodide were produced in the interaction of 8.1 c.c. of hydrogen and 9.1 c.c.
of iodine vapour at 443 degrees centigrade. Calculate the equilibrium constant at this temperature of
the reaction.
H2 + I2 ↔ 2HI
Ans: at equilibrium x c.c. of hydrogen reacts with x c.c. of iodine to form 2x c.c. of hydrogen iodide.
It is given that 2x=13.4
Therefore x= 6.7
Ans: at equilibrium x c.c. of hydrogen reacts with x c.c. of iodine to form 2x c.c. of hydrogen iodide.
It is given that 2x=13.4
Therefore x= 6.7
Now let us write the given reaction.
Initial vol
8.1
9.1 0
Initial moles a b 0
H2 + I2 ↔ 2HI
At equilibrium (a-x) (b-x) 2x
(8.1-6.7) (9.1-6.7) 13.4
Initial moles a b 0
H2 + I2 ↔ 2HI
At equilibrium (a-x) (b-x) 2x
(8.1-6.7) (9.1-6.7) 13.4
The equilibrium constant of the reaction is given as
[HI]2
Kc = ------------
(a-x) (b-x)
Kc = ------------
(a-x) (b-x)
(13.4)2
= ------------
(1.4) (2.4)
= ------------
(1.4) (2.4)
= 51.84
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