1. Chemical Systems & Equilibrium
Chapter 7

2. Factors affecting an equilibrium
Temperature, pressure and alterations to concentration adjust equilibria.
These changes result in a favour to either the forward or reverse reaction and cause the equilibrium concentrations to “shift” to establish a new stable state.
This principle was studied in 1884 by Henry Louis Le Châtelier.

3. Le Châtelier’s Principle
When a chemical equilibrium is disturbed by a change in property, (temperature, pressure, concentration), the system adjusts in a manner that relieves the change.
An equilibrium shift is an adjustment that results in changes to the concentrations of the equilibrium reactants and products.

4. Change in concentration of reactants
Shift a
2 CO2(g) + energy D 2 CO(g) + O2(g)
[O2(g)]
[CO2(g)]
[CO(g)]
[CO2(g)] added
Original equilibrium established
“Equilibrium Shift” to new equilibrium concentrations
Time (s)
Concentration (mol/L)

5. Change in concentration of reactants
An increase in the concentration of reactants results in an increase in the concentration of the products.
Notice that the [CO2] is reduced and the [CO] and [O2] is increased during the adjustment.
The reaction shifts to “counter” the increase in CO2.

6. Change in concentration & Kinetic Theory
The addition of a substance increases the likelihood of a collision and thereby increases the rate of reaction in that direction.
The opposing reaction lags the initial response but eventually “catches up” and establishes an new equilibrium.

7. Reactants + energy  products
Change in temperature
Adding heat favours the endothermic reaction – it is a “reactant”.
Reactants + energy  products
Consider energy as a “product” or a “reactant” in predicting the shift.
The equilibrium shifts in the direction that ABSORBS the heat.

8. Concentration (mol/L)
Change in temperature
Shift a
2 CO2(g) + energy D 2 CO(g) + O2(g)
Original equilibrium established
“Equilibrium Shift” to new equilibrium concentrations
heat added
Time (s)
Concentration (mol/L)
[CO2(g)]
[CO(g)]
[O2(g)]

9. Change in temperature & Kinetic Theory
Adding heat increases the rate of reaction due to the more frequent and energetic collisions.
The reaction that uses heat as a reactant (endothermic) will be affected quickest and thereby shift the equilibrium in that direction.
The increased rate produces more products, which are reactants for the reverse reaction.
The increased heat and reactants result in the reverse reaction “catching up” and establishing new equilibrium concentrations.

10. Change in pressure/volume
According to Boyle’s law, the concentration of a gas is directly proportional to the pressure of the gas.
An increase in pressure reduces the volume and increases the concentration of the gas.
Therefore, changing the pressure or volume of an equilibrium involving gases may affect the equilibrium.

11. Change in Pressure or Volume
Shift \
2 CO2(g) + energy D 2 CO(g) + O2(g)
[O2(g)]
[CO2(g)]
[CO(g)]
Increased pressure or reduced volume
Original equilibrium established
“Equilibrium Shift” to new equilibrium concentrations
Time (s)
Concentration (mol/L)

12. Change in pressure/volume
An increase in pressure, or a decrease in volume favours the reaction that produces fewer moles of gas.
They occupy less volume and thereby relieve the stress of the added pressure.
An increase in pressure adds energy to the system and therefore, reacts similar to the addition of heat.

13. Change in pressure/volume & Kinetic Theory
Reducing the volume (increasing pressure) increases the likelihood of collision, thereby increasing the rate of reaction.
The reaction that has more particles (increased number of moles) will have more collisions and increase at a quicker rate.
As the shift occurs the products of the quicker reaction will increase and improve the rate of the reverse reaction.

14. Quantitative analysis of equilibria
Using Le Châtelier’s Principle & the Equilibrium Law one can quantitatively assess;
The equilibrium constant (Keq)
The position or progress of the equilibrium system with the reaction quotient (Q)
The equilibrium concentrations

15. Determining Keq To determine the value for Keq, one must;
Use the balanced chemical equation and the Equilibrium Law to produce and equilibrium expression.
Substitute given values for equilibrium concentrations into the equilibrium expression to solve for Keq.

16. The equilibrium concentrations for the chemicals are; [N2] = 0
The equilibrium concentrations for the chemicals are; [N2] = 0.40 mol/L, [H2] = 0.15 mol/L & [NH3]=0.20 mol/L. Determine the value for the equilibrium constant (Keq) under these conditions.
Balanced equation for the reaction
N2 (g) H2 (g)  2 NH3 (g)
Generate the equilibrium expression
K eq = [ NH 3 ] 2 [ N 2 ] [ H 2 ] 3
Substitute equilibrium concentrations into the expression to solve for Keq.
K eq = [0.20] 2 [0.40] [0.15] 3 =29.6

17. Determining the position or status of an equilibrium system.
Calculate the reaction quotient using the equilibrium concentrations.
Compare the reaction quotient value to the equilibrium constant value.
Determine whether the reaction is reactant or product rich and adjust the equilibrium to establish the equilibrium.

18. At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium?
Balanced equation for the reaction
N2 (g) H2 (g)  2 NH3 (g)
Generate the equilibrium expression and reaction quotient expression
Q = [ NH 3 ] 2 [ N 2 ] [ H 2 ] 3

19. At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium?
Substitute the supplied concentrations into the expression to solve for Q.
Q = [0.20] 2 [0.10] [0.30] 3 =14.8
Compare the Q value to the known Keq value.
Q < Keq < 25.0

20. At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium?
Use the Q and Keq values to determine the position of the equilibrium system.
𝑄 ≈ [Products] [Reactants]
Since Q < Keq, the system has fewer products than the equilibrium state. Therefore the reaction needs to progress toward the products to attain equilibrium. (i.e.- Shift to the right.)

21. Calculating equilibrium concentrations
To determine the equilibrium concentrations, one must use:
A balanced equation for the equilibrium reaction
An equilibrium expression and constant, Keq
An Initial concentration, Change in concentration, and Equilibrium concentration (ICE) table.

22. Calculating equilibrium concentrations
In cases where the Keq value is very small the addition and subtraction of the change value (x) may be insignificant and thereby omitted.
Use the ratio of the smallest initial concentration and the Keq to assess the affect of the change.

23. Calculating equilibrium concentrations
If clue > 500, the addition or subtraction of “x” is insignificant and may be ignored.
If 100<clue<500, the addition or subtraction of “x” is probably insignificant, but should be checked.
If clue<100, the addition or subtraction of “x” is significant and must be included in the calculations.

24. Calculating equilibrium concentrations
In cases where the Keq expression results in the resolution of a quadratic equation, the use of the quadratic formula my prove helpful.
When the quadratic formula produces two answers select the one that is viable. Remembering that you can not have a negative concentration!

25. Example – At high temperatures, as with lightning, nitrogen and oxygen will react to produce nitrogen monoxide. A chemist puts moles of N2(g) and mol of O2(g) in a 1.0 L flask at high temperature, where the Keq= 4.2 x What is the concentration of the NO(g) in the mixture at equilibrium?
Strategy
Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.
Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.
Set up an ICE table letting “x” represent the change in concentrations.
Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.
Calculate the required value(s).

26. Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.
The clue is far greater than 500 so we can ignore the changes in N2(g) and O2(g).
0.850−𝑥≅0.850

27. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.
N2(g) + O2(g)  2 NO(g)
N2(g) + O2(g)  NO(g)

28. Set up an ICE table letting “x” represent the change in concentrations.
Concentration (mol/L)
N2(g) +
O2(g) 
2 NO(g)
Initial concentration
0.085
0.038
Change in concentration -x 2x
Equilibrium concentration
0.085 – x ≈0.085
0.038 – x ≈0.038

29. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.
The negative value is impossible as one can not have a negative concentration.

30. Calculate the required value(s).
What is the NO(g) concentration at equilibrium?
The NO(g) concentration at equilibrium is 1.2 x mol/L.

31. Example – In a 1. 00 L flask, 2. 00 mol of H2(g) is combined with 3
Example – In a 1.00 L flask, 2.00 mol of H2(g) is combined with 3.00 mol of I2(g) which produces HI(g). The Keq for the reaction is 25 at 1100 K. What is the What is the concentration of each gas in the mixture at equilibrium?
Strategy
Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.
Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.
Set up an ICE table letting “x” represent the change in concentrations.
Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.
Calculate the required value(s).

32. Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored.
The clue is much less than 500 so we can not ignore the changes in H2(g), I2(g) and HI(g).

33. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression.
H2(g) + I2(g)  HI(g)
H2(g) + I2(g)  2 HI(g)

34. Set up an ICE table letting “x” represent the change in concentrations.
Concentration (mol/L)
H2(g) +
I2(g) 
2 HI(g)
Initial concentration
2.00
3.00
Change in concentration -x 2x
Equilibrium concentration
2.00 – x
3.00 – x

35. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.

36. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.

37. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.
4.3 and 1.7 are determined through the quadratic formula calculation, but only one is correct! – 4.3 = negative value so 1.7 mol/L is the correct value. (A negative concentration is impossible!) Therefore the final equilibrium concentrations are: [H2]eq = 2.00 – 1.7 = 0.3 mol/L [I2]eq = 3.00 – 1.7 = 1.3 mol/L [HI]eq = 2( 1.7) = 3.4 mol/L

38. The Solubility Equilibrium
Remember from SPH3U:
Solubility is the amount of solute that dissolves in a given amount of solvent (usually water) at a particular temperature.
Measured in g/100 mL or mol/L (molar concentration)
Very soluble - > 10 g/100 mL Soluble – 1 – 10 g/100 mL Slightly soluble – 0.1 – 1 g/100 mL Insoluble - < 0.1 g/100 mL

39. The Solubility Equilibrium
Remember from SPH3U:
Produced solubility curves
Most solids increased in solubility as the temperature increased. While, most gases decreased in solubility as the temperature increased.
Terminology – Saturated, Unsaturated, Supersaturated

40. The Solubility Process
Supersaturated solution
Saturated solution
Solubility (gm/100 ml)
Unsaturated solution
Temperature (oC)

41. The Solubility Equilibrium
Remember from SPH3U:
Dissociation of ionic compounds into ions
Ionic equations, net ionic equations, spectator ions
Electrolytes, non-electrolytes
Solubility charts
Predict precipitates in ionic solution combinations (Pg 801 & Pg 487 Table 4)

42. The Solubility Equilibrium
A solution equilibrium can be treated as a chemical equilibrium.
The solubility equilibrium must be saturated (has excess salt (solute)). Therefore, they occur with low solubility compounds.
Remember, solids have a constant concentration and are not used in the equilibrium law/expression. The equilibrium constant is simplified to the product of the ion equilibrium concentrations raised to their proper exponents.
The equilibrium constant (Keq) can be considered a solubility product constant (Ksp) in a solubility equilibrium.

43. The Solubility Product Constant
A solution equilibrium can be treated as a chemical equilibrium.
The solubility equilibrium must be saturated and have excess solute (salt), therefore they occur with low solubility compounds.
The balanced chemical reaction (dissociation equation) can be used to generate an equilibrium expression.
Remember, solids have a constant concentration and are not used in the equilibrium law expression. The equilibrium constant is simplified to the product of the ion concentration raised to their proper exponents.
The equilibrium constant (Kc) can be considered a solubility product constant (Ksp) in an solubility equilibrium.

44. The Solubility Product Constant
A solution equilibrium can be treated as a chemical equilibrium.
PbI2 (s)  Pb 2+(aq) I 1-(aq) (saturated sol’n)
PbI2 is a solid so its concentration does not alter and is therefore removed from the equilibrium expression.

45. The Solubility Product Constant
A general application of the solubility equilibria results in the following Ksp expression.
BC(s) D b B+(aq) + c C-(aq)
Where BC(s) is a slightly soluble salt, and B+(aq) and C-(aq) are dissociated aqueous ions.
Ksp values are reported in reference tables such as those in the appendix. (Pg 802)

46. The Solubility Product Constant usage
Calculating solubility constants from solubility data. (i.e.- solubility of g/100 25oC)
Predicting whether a precipitate will form in a solubility equilibrium.
Trial ion product, Q– the reaction quotient applied to the ion concentrations.
Q > Ksp (supersaturated) precipitate will form Q = Ksp (saturated) precipitate will not form Q < Ksp (unsaturated) precipitate will not form

47. The Solubility Product Constant usage
The common ion effect
The reduction in the solubility of a salt caused by the addition of a second salt with a common ion.
Le Chatelier’s principle is applied and results in a shift of the solubility equilibrium.
Use equilibrium (ICE) table to solve for new concentrations.

48. Calculating solubility constants from solubility data
Determine the solubility of AgCl (s) at SATP given that its Ksp is 1.8X
Strategy
Balanced chemical equation for the system.
Equilibrium Law expression.
Initial, Change in & Equilibrium Concentration table.
Sub-n-solve.

49. Calculating solubility constants from solubility data
Determine the solubility of AgCl (s) at SATP given that its Ksp is 1.8X
Strategy
Balanced chemical equation for the system.
AgCl(s) D Ag +1(aq) + Cl-1(aq)

50. Calculating solubility constants from solubility data
Determine the solubility of AgCl (s) at SATP given that its Ksp is 1.8X
Equilibrium Law expression.
Ksp = [Ag+1] [Cl-1] = 1.8 x 10-10

51. Calculating solubility constants from solubility data
Determine the solubility of AgCl (s) at SATP given that its Ksp is 1.8X
Initial, Change in & Equilibrium Concentration table.
AgCl (s) D
Ag+1(aq)
+ Cl+1(aq)
[Initial]
[constant]
[change in] +x
[Equilibrium]
0+x = x

52. Calculating solubility constants from solubility data
Determine the solubility of AgCl (s) at SATP given that its Ksp is 1.8X
Sub-n-solve.

53. Calculating solubility constants from solubility data
Determine the solubility of AgCl (s) at SATP given that its Ksp is 1.8X
Check assumptions

54. Determining if a precipitate will form
Strategy
Balanced chemical equation for the system.
Identify the concentrations to be considered.
Use the equilibrium law expression to generate the “Trial Ion Product (Qsp)” expression.
Use given concentration and the expression to solve for Qsp.

55. Determining if a precipitate will form
Strategy
Compare the Qsp value to the known Ksp value to determine if the situation is at equilibrium.
Qsp = Ksp  the solution is saturated & at equilibrium so no precipitate will form.
Qsp < Ksp  the solution is unsaturated so no precipitate will form.
Qsp > Ksp  the solution is “supersaturated” so a precipitate will form.

56. Determining if a precipitate will form
If 50 mL of mol/L NiCl2 (aq) and 50 mL of mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x 10-4.
NiCl2 (aq) + Na2CO3 (aq) → NiCO3 (s) + 2 NaCl (aq)
(Use the solubility rules to identify which of the products is most likely to form a solid.)
NiCO3 (s) ⇌ Ni 2+(aq) + 2 Cl-(aq)

57. Determining if a precipitate will form
If 50 mL of mol/L NiCl2 (aq) and 50 mL of mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x 10-4.
NiCl2 (aq) → Ni 2+(aq) + 2 Cl- (aq)
[Ni2+]= mol/L(50 mL)/(100 mL) = mol/L
Na2CO3 (aq) → 2 Na+(aq) + CO32- (aq)
[CO32-]= mol/L(50 mL)/(100 mL) = mol/L

58. Determining if a precipitate will form
If 50 mL of mol/L NiCl2 (aq) and 50 mL of mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x NiCO3 (s) ⇌ Ni 2+(aq) + CO32- (aq)

59. Determining if a precipitate will form
If 50 mL of mol/L NiCl2 (aq) and 50 mL of mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x Qsp > Ksp ( > 8.2 x 10-4 ) In this case the solution is either supersaturated or a precipitate will be formed.