1. SCH4U – Grade 12 Chemistry, University Preparation Mr. Dvorsky
7.3 The Equilibrium Constant: Measuring Equilibrium Concentrations Equilibrium Calculations (ICE Charts) Qualitatively Interpreting the Equilibrium Constant Meaning of Small Equilibrium Constant
SCH4U – Grade 12 Chemistry, University Preparation
Mr. Dvorsky

2. SELF QUIZ: Equilibrium Expressions
Write the equilibrium expression for the following chemical equations:
CO(g) + 1/2 O2(g) CO2(g)
2 HBr(g) H2(g) + Br2(g)
4 NH3(g) + 5 O2(g) NO(g) + 6 H2O(g)

3. We learned in previous lessons that…
LAW OF CHEMICAL EQUILIBRIUM: At equilibrium, there is a constant ratio between the concentration of reactants and products in any change.
Equilibrium constant (Keq or Kc) is the ratio of the forward rate constant and reverse rate constant.
kf = Keq kr
The equilibrium expression is
aA + bB cC+ dD
so: Kc = [C]c[D]d
[A]a[B]b
Equilibrium constant is effected by temperature.

4. What if you don’t know the equilibrium constant?

5. Measuring Equilibrium Concentrations
You can determine some equilibrium concentrations if you know:
the initial concentration(s) of the reactant(s) &
the concentration of one product at equilibrium [which will tell you the change in concentration(s)].
Once you have the equilibrium concentrations, you can find the equilibrium constant!!
Use an ICE table to perform these calculations:
I = initial concentration
C = change in concentration
E = equilibrium concentration

6. Fe3+(aq) + SCN-(aq) Fe(SCN)2+(aq)
colourless red
Fe3+(aq) + SCN-(aq) are both colourless.
Fe(SCN)2+(aq) is red.
Because this reaction involves a colour change, you can determine the equilibrium concentration of Fe(SCN)2+(aq) by measuring the change in the intensity of colour (absorbance).
Knowing the initial concentrations of Fe3+(aq) and SCN-(aq), you can calculate the equilibrium concentration of each ion using the chemical equation.
Using the equilibrium concentrations, you can calculate the equilibrium constant!!

7. Fe3+(aq) + SCN-(aq) Fe(SCN)2+(aq)
Concentration (mol/L)
Fe3+(aq)
SCN-(aq)
Fe(SCN)2+(aq)
Initial
0.0064
0.0010
Change
-4.5 x 10-4
4.5 x 10-4
Equilibrium
0.006
5.5 x 10-4
Kc = [Fe(SCN)2+] = [4.5x10-4] = ≈ 140
[Fe3+][SCN-] [0.006][5.5x10-4]

8. What if you don’t know the equilibrium concentrations?

9. Measuring Equilibrium Concentrations
You can determine some equilibrium concentrations if you know:
the initial concentration(s) of the reactant(s) &
the equilibrium constant.
Using this information and an ICE table, you can calculate the equilibrium concentrations of all the chemical compounds.
Use an ICE table to perform these calculations:
I = initial concentration
C = change in concentration
E = equilibrium concentration

10. Sample Problem
At 700 K, the equilibrium constant is Suppose you start with 1.0 mol of CO(g) and 1.0 mol of H2O(g) in a 5.0 L container. What amount of each substance will be present in the container when the gases are at equilibrium at 700 K?

11. Reactants Products CO(g) H2O(g) H2(g) CO2(g) 1/5 = 0.20 -x +x 0.20 - x
Concentration (mol/L)
CO(g)
H2O(g)
H2(g)
CO2(g)
Initial
1/5 = 0.20
Change -x +x
Equilibrium x
0.20 -x x

12. Doesn’t make sense

13. So… Reactants Products CO(g) H2O(g) H2(g) CO2(g) 1/5 = 0.20 -0.15
Concentration (mol/L)
CO(g)
H2O(g)
H2(g)
CO2(g)
Initial
1/5 = 0.20
Change
-0.15
+0.15
Equilibrium
0.05
0.15
So, in a 5.0 L container:
Amount (n) at Equilibrium
0.05*5 =
0.25 mol
0.15*5 =
0.75 mol

14. The previous problem had perfect squares
The previous problem had perfect squares. It was easy to find the square root both sides and simplify for x.
Many problems do not involve perfect squares.
You will be required to use the quadratic equation to find the value of x.

15. Sample Problem
The above reaction has an equilibrium constant of 25.0 at K. In a 1.00 L reaction vessel, 2.00 mol H2(g) and mol I2(g) react. What is the equilibrium concentration of each gas?
You need to find [H2], [I2], and [HI].
You need to ensure you have a balanced equation.
Set up an ICE table, and calculate the initial concentrations of the gases.
Let x equal all the change in the concentration of all chemical species.
Write the equilibrium expression. Substitute and rearrange to solve for x.

16. Reactants Products H2(g) I2(g) HI(g) 2.00 3.00 -x +2x 2.00 - x 3.00 -x
Concentration (mol/L)
H2(g)
I2(g)
HI(g)
Initial
2.00
3.00
Change -x
+2x
Equilibrium x
3.00 -x 2x

17. Doesn’t make sense

18. So… Reactants Products H2(g) I2(g) HI(g) 2.00 3.00 -1.7 +2(1.7)
Concentration (mol/L)
H2(g)
I2(g)
HI(g)
Initial
2.00
3.00
Change
-1.7
+2(1.7)
Equilibrium
2.00 – 1.7
= 0.3
= 1.3
3.4

19. So what does K mean??

20. Why is the equilibrium constant important?
K describes the extent of a reaction!
A large K >1 means: [products] > [reactants]
 position of equilibrium lies to the right
 equilibrium favours the products
A Kc = 1 means: [products] = [reactants]
A small Kc <1 means: [products] < [reactants]
 position of equilibrium lies to the left
 equilibrium favours the reactants

22. Sample Problem
Consider the reaction of carbon monoxide and chlorine gas to make phosgene. At 870K, the value of Kc is 0.20 and at 370 K, the value of Kc is 4.6 x Based only on the values of Kc, is the production of COCl2(g) more favourable at the higher or lower temperature?
Kc = 0.20 at 870 K
Kc = 4.6 x 107 at 370 K
Is COCl2(g) a product or reactant?
Which Kc favours COCl2(g )?

23. So what does a really small Kc mean??
When Kc is really small compared to the initial concentration, the initial value minus x is approximately equal to the initial value, so you can ignore x!
How do you know when you can do this?
Kc = 4.2 x 1o-8 & [NO] = mol/L
Divide the initial concentration by the value of Kc.
If the answer is > 500, the approximation is justified.
If the answer is < 500 but > 100, it may be justified.
If the answer is < 100, you can’t ignore x and need to solve the equilibrium expression in full.