5 How to calculate equilibrium concentrations?
LO: To calculate equilibrium concentrations from Kc and initial concentrations
To calculate equilibrium concentrations once the equilibrium has been
disturbed.
Calculating equilibrium concentrations from Kc
and initial concentrations
Example 1
CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq)
Assume that you have prepared a solution of acetic acid with
initial concentration of 0.3 mol L-1.
Calculate the concentrations of CH3COOH, CH3COO- , H3O+ at
equilibrium.
Kc = 1.8 x 10-5 for the following reaction at 25oC
a) Calculate the concentrations of the compounds at equilibrium:
Initial concentrations
[CH3C OOH]
[CH3C OO-]
[H3O+]
0.3
0
0
Change in concentrations
Equilibrium concentrations
b) Write the equilibrium constant expression:
a) Insert the equilibrium concentrations from the table
a) Calculate the concentrations of the compounds at equilibrium:
[CH3C OOH]
[CH3C OO-]
[H3O+]
Initial concentrations
0.3
0
0
Change in concentrations
-x
+x
+x
0.3 – x
Equilibrium concentrations
x
b) Write the equilibrium constant expression:
Kc=
!"#$$% ["#$']
!"#$$"
a) Insert the equilibrium concentrations from the table
Kc=
!"#$$% ["#$']
!"#$$"
=
*+
,../*
= 1.8 x 10-5
x
d) Solve the equation
!"
= 1.8 x 10−5
0.3 − !
x 2 = 1.8 x 10—5 (O.3 –x )
x 2 + 1.8 x 10—5x - 5.4 x 10—6 = 0
x 2 = 1.8 x 10—5 x O.3
x = 0.0023
x 2 = 5.4 x 10—6
! " = 5.4 x 10—6
x = 0.0023
[CH3C OOH]
[CH3C OO-]
[H3O+]
Initial concentrations
0.3
0
0
Change in concentrations
-x
+x
+x
0.0023
0.0023
Equilibrium concentrations
0.3 – x
0.2977
x
x
Note
That we found
[H3O+] = 0.0023 mol L-1
pH = - log [H3O+] = 2.6
You have prepared a solution of acetic acid with initial concentration of 0.3
mol L-1.
[H3O+] = 0.3 x 1.8 x 10−5 = 5.4 x 10−6 = 0.0023 mol L-1
Example 2
Let’s revisit the glycolysis process
Assume that you have prepared a 1 L of a solution where the glucose derivative is in
equilibrium with the fructose derivative and that at equilibrium the [glucose-6-P] =
0.07 mol L-1 and [fructose-6-P] = 0.03 mol L-1.
Now we will disturb the equilibrium by adding another 0.07 mol of glucose-6-P to
the 1 L of a solution at equilibrium,(without changing the volume by much).
Calculate the concentrations of of glucose-6-P and fructose-6-P once the new
equilibrium is established.
a) Write the equilibrium constant expression and calculate the
value for K:
b) Calculate the concentrations of the compounds at equilibrium:
[glucose-6-p]
Initial concen.
Change in concen.
Equilibrium concen.
[fructose-6-P]
a) Write the equilibrium constant expression and calculate the
value for K:
Kc=
!"#$%&'()*)+
,-#$&'()*)+
a) Calculate the concentrations of the compounds at equilibrium:
Kc=
!"#$%&'()*)+
,-#$&'()*)+
Initial concen.
Change in concen.
Equilibrium concen.
/./1
= /./2 = 0.43
[glucose-6-p]
[fructose-6-P]
0.14
0.03
-x
+x
0.14 - x
0.03 + x
c) Insert the provided values
d) Solve the equation
c) Insert the provided values
Kc=
*+,-./012324
56,-/012324
=
7.789:
7.;<2:
= 0.43
d) Solve the equation
0.03 + %
= 0.43
0.14 − %
0.03 + x = 0.43 (0.14 – x )
0.03 + x = 0.0602 – 0.43 x
x + 0.043 x = 0.0602 - 0.03
1.43 x = 0.0302
x = 0.021
0.12
0.05