Chapter 16 – Kinetics and Equilibrium
Kinetics – study of the rate of a reaction
Factors that affect the rate of a reaction:
1. Concentration - Higher concentrations react faster.
2. Temperature
- Increase in temperature increases the
reaction rate.
3. (For Solids) Particle Size - Smaller particles react faster
4. Catalyst
- A substance that increases the rate of a reaction,
but does not get used up.
Biological catalysts are called enzymes.
Inhibitor - substance that slows a reaction
down, usually by interfering with a catalyst.
Collision Theory – If 2 molecules collide with enough energy, and of proper
orientation, they will react to become something new.
Catalysts often increase reaction rates by causing molecules to orient
properly, or because they cause a reaction to require less energetic
collisions.
Reaction Mechanism – a sequence of reactions that show more details of
how reactants change to products.
Mechanisms involve intermediates – products that form, then become a
reactant.
O3
→
O
O + O3
O3
+
+
O2
→ O2
+
O2
O3 → O2 + O2 + O2
O3 + Cl → OCl + O2
O3 + OCl → O2 + O2 + Cl
O3 + O3 → 3O2
Activation Energy Diagram – graph relating energy of a reaction to time.*
System – the reactants and products
Surroundings – everything else
Ea – activation energy or minimum energy needed to initiate a reaction.
Activated Complex – top of the graph where reactants hit each other hard
enough to break bonds and form products.
Enthalpy - H - Potential energy of a substance.
∆H – Change in enthalpy of a system during a reaction.
Exothermic reaction – energy is released from the system to the surroundings.
Heat is given off from these reactions.
Endothermic reaction – energy from the surroundings is transferred into
the system. Often, the surroundings will get colder.
Equilibrium
When the forward and reverse reaction are occurring at the same rate,
and it appears that the reaction has stopped.
Terms: Equilibrium favors reactants
Equilibrium favors products
Forward reaction
Reverse reaction
Reversable reaction
Static vs. dynamic equilibrium
Physical vs. chemical equilibrium
To attain equilibrium, the system must be in a closed container.
The Equilibrium Expression and Constant (K)
Problem: Write the equilibrium expression for:
1) H2(g) + F2(g) ↔ 2HF(g)
2) N2(g) + 3H2(g) ↔ 2NH3(g)
At equilibrium, the
Calculate K
[N2] = 0.921 M
[H2] = 0.763 M
[NH3] = 0.157 M
Is K really a constant even if the beginning concentrations start out different?
For the reaction 2SO2(g) + O2(g) ↔ 2SO3(g)
Exp. #1
[SO2]
[O2]
[SO3]
Initial
2.00 M
1.50 M
3.00 M
Equilibrium
1.50 M
1.25 M
3.50 M
Exp. #2
[SO2]
[O2]
[SO3]
Initial
0.500 M
0M
0.350 M
Equilibrium
0.590 M
0.045 M
0.260 M
K is always a constant, as long as
the temperature does not change.
If temperature changes, so will K
Homogenous Equilibrium - all reactants and products are in the same phase
Heterogenous Equilibrium - more than 1 phase of matter exists among the
reactants and products.
Solids and pure liquids don’t change in their concentration, therefore,
they are not included in the equilibrium constant.
Problem: Determine the molarity of 1 liter of water.
Determine the molarity of 1/2 liter of water.
Write the equilibrium constant for:
CaCO3(s) ↔ CaO(s) + CO2(g)
2H2O(l) ↔ 2H2(g) + O2(g)
NaCl(s) ↔ Na+(aq) + Cl-(aq)
Le Chateliers Principle - If a change is imposed on a system at
equilibrium, the position of the equilibrium will shift in a
direction that will reduce the effect of that change.
Factors that will shift equilibrium to favor more reactants or products.
1. Concentration
- When a reactant or product is added, the system shifts away from the
added component. The added component will be partly consumed.
- When a reactant or product is removed, the system shifts toward the
removed component. Part of that removed component will be replaced.
c
o
n
c
e
n
t
r
a
t
i
o
n
N2(g) + 3H2(g) ↔ 2NH3(g)
time
Hb(aq) + 4O2(g) ↔ Hb(O2)4(aq)
Which way will equilibrium shift?
NaCl (s) ↔ Na+ (aq)
+ Cl- (aq)
Add NaNO3(aq)
Add CaCl2(aq)
Add NaCl(s)
Add AgNO3(aq)
Universal Indicator :
pH = 3
Red
4
Orange
5
Shrimp
increase in [H+]
6
Yellow
↔
7
Green
8
Blue
9
Violet
decrease in [H+]
If there are 2 or more liquids in an equation, their concentrations can change
HC2H3O2(l) + CH3CH2OH(l) C4H8O2(l) + H2O(l)
Add CH3CH2OH(l)
Add H2O(l)
2. Pressure
- An Increase in pressure shifts equilibrium in the direction of the least
number of moles of gas.
- A Decrease in pressure shifts equilibrium in the direction of the most number
of moles of gas.
C3H8(g) + 5O2(g) ↔ 3CO2(g) + 4H2O(g)
Which direction will equilibrium shift when pressure is increased?
CaCO3(s) ↔ CaO(s) + CO2(g)
Increase pressure is the same as
decreasing the volume of the
container
2NO2(g)
Brown
↔
N2O4(g)
Colorless
N2(g) + 3H2(g) ↔ 2NH3(g)
Would you increase or decrease
pressure to make more NH3?
3. Temperature
- Raising the temperature will shift equilibrium in the endothermic direction.
- Lowering the temperature will shift equilibrium in the exothermic direction.
Heat + Co(H2O)6+2(aq) + 4Cl- (aq) ↔ CoCl4-2(aq) + 6H2O(l)
(pink)
(blue)
What color will this solution become when cooled?
3. Temperature
- Raising the temperature will shift equilibrium in the endothermic direction.
-Lowering the temperature will shift equilibrium in the exothermic direction.
Which side of the equation would we write “energy”?
2NO2(g)
Brown
↔
N2O4(g)
Colorless
Magnitude of K
For
A(g) ↔ B(g)
K=
[B]
[A]
If K = big number ..?
Products are favored, and equilibrium
lies to the right. (reaction goes
to completion.)
If K = small number..?
N2(g) + 3H2(g) ↔ 2NH3(g)
K = 0.0602
At equilibrium, [N2] = 0.921 M
[H2] = 0.763 M
Calculate [NH3]
Reactants are favored,equilibrium
Lies to the left, the reaction tends
not to occur.
Equilibrium Problems from worksheet packet:
1. Calculate K at 350 oC for the following reaction if the final concentration
of each chemical species is
[N2] = 2.36
[H2] = 7.08
[NH3] = 4.72
N2(g) + 3H2(g)
↔ 2NH3(g) + energy
2a) Predict which way equilibrium will shift if the temperature of the
reaction is raised to 450 oC. (left or right)
b) Do you predict K will be larger, smaller, or the same value at this higher
temperature?
4) Diatomic oxygen and triatomic oxygen (ozone) are in equilibrium according
to the following equation:
2O3(g) ↔ 3O2(g)
At a temperature of 1727 oC, the concentration of each species is found to be:
[O2] = 0.050
[O3] = 7.0 x 10-9
a) Based on the concentration of oxygen and ozone at equilibrium, does
equilibrium favor the: reactants or products?
b) What is the value of K?
Initial/Final Problem:
9. Given the reaction at equilibrium: H2(g) + CO2(g) ↔ H2O(g) + CO(g)
Calculate K if 0.70 moles of H2 are added to 0.50 moles of CO2, and when
equilibrium is reached, 0.30 moles of H2O(g) are present. This reaction
takes place in a 1.0 Liter container.
10. Calculate K for the following reaction at equilibrium:
H2(g) + I2(g) + energy ↔ 2HI(g)
The reaction begins with 1.40 moles Hydrogen and 0.60 moles of iodine in a
1.0 liter flask, and at equilibrium, there are 1.10 moles of HI
14. At a certain temperature, 30. % of the original amount of NO2(g) will
convert to products.
a) Determine the molarity of each component at equilibrium if 0.80 moles of
NO2(g) is placed in a 1.0 liter flask.
2NO2(g) ↔ 2NO(g) + O2(g)
b) Calculate K for this reaction.
23. At a given temperature, 40. % of NO2(g) converts to products. Determine
the molarity of each component and the value of K if 4.0 moles of NO2(g) are
placed in a 1.0 liter flask.
2NO2(g) ↔ 2NO(g) + O2(g)
Solubility Equilibrium
Saturated?
Write a dissociation equation and Ksp (solubility product constant) for:
PbCl2 (s)
Ag2CrO4(s)
Two types of solubility problems:
1. Calculate Ksp when given solubility of a substance.
Problem #1: The solubility of CuBr is 2.0 x 10-4 mole/liter. Calculate Ksp
Problem #2: If 2.5 x 10-3 mole PbF2 dissolve/liter, Calculate Ksp.
2. Calculate solubility of a substance given Ksp
Problem #3: The Ksp of AgI = 1.5 x 10-16. Determine how many moles/Liter and
grams per liter will dissolve.
Problem #4: The Ksp of PbCrO4 is 2.8 x 10-13. Determine its solubility. (How
many moles/Liter and grams/Liter will dissolve?
Problem #5: The Ksp of Mg(OH)2 = 4.0 x 10-12. Determine how many moles/Liter
and grams/Liter will dissolve.