|
|
kinetics/equilibrium |
||
|
honors |
|||
|
.....aims |
|
||
Notes
MASS-ACTION EXPRESSION (EQUILIBRIUM CONSTANT)
For the reaction aB + bB <---> cC + dD Keq = [C]c [D]d / [A]a [B]b
Note: Solids and liquids have constant concentration therefore they do not appear in Keq
Example
BF3(g) + 3H2O(l) <---> 3HF(aq) + H3BO3(aq) Keq = [HF]3 [H3BO3] / [BF3]
The K eq of a reverse reaction is the reciprocal of the forward reaction.
Example:
Keq of forward reaction is 1 x 10-11 than the reverse reaction Keq is 1 / 1 x 10-11
Solubility Product Constant (Ksp)
Used for salts.
Example:
PbCl2(s) ---> Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+] [Cl-]2
LAWS OF THERMODYNAMICS
The first Law of Thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
∆E = q − w
q = heat added to the system
w = work done by the system on the surroundings
The Second Law of Thermodynamics states that in any spontaneous process there is always an increase in the entropy of the universe.
∆S universe = ∆S system + ∆S surrounding
The Third Law of Thermodynamics states that a crystal at 0 K has 0 entropy.
GIBBS FREE ENERGY
The function G (free energy) is used to find the spontaneity of a reaction.
∆G = ∆H-T∆S T in Kelvin
When ∆G < 0 the forward reaction is spontaneous therefore reactants form products.
When ∆G > 0 the reverse reaction is spontaneous therefore products form reactants.
When ∆G = 0 the reaction is at equilibrium

Summary
|
|
Spontaneity |
∆G |
= |
∆H |
- |
T |
∆S |
|
Case #1 |
forward |
- |
|
- |
|
all |
+ |
|
Case #2 |
forward |
- |
|
+ |
|
high |
+ |
|
Case #3 |
forward |
- |
|
- |
|
low |
- |
|
Case #4 |
reverse |
+ |
|
+ |
|
all |
- |
RATE LAWS
Most chemical reaction rates vary with the concentration of the reactants. The rate law of a reaction is only determined by experiment.
In first order reactions (k[A]1), as concentration changes the rate changes in the same way.
In second order reactions (k[A]2), as concentration changes the change in rate is the square of the concentration change.
In zero order reactions (k[A]0 = k), as concentration changes the rate is constant.
k = a proportionality constant
A = a reactant in a reaction
Example
What is the rate law for the reaction 2NO(g) + 2H2(g) --> N2(g) + 2H2O(g) given the following data?
|
Run |
Reactants Concentration (mol/L) |
Rate of reaction |
|
|
[NO] |
[H2] |
||
|
1 |
0.10 |
0.010 |
0.062 |
|
2 |
0.10 |
0.040 |
0.246 |
|
3 |
0.30 |
0.010 |
0.558 |
First, write the rate law for the reaction
Rate = k[NO]n [H2]m
Second, calculate “n”
Find the runs where [H2] remains constant
Comparing run 1 and 3
|
Δ [NO] |
Δ rate |
n |
Rate law |
|
3 times (0.30/0.10) = 3 |
9 times (0.558/0.062) = 9 |
3n = 9 therefore, n = 2 |
k[NO]2 |
Third, calculate “m”
Find the runs where [NO] remains constant
Comparing run 1 and 2
|
Δ [H2] |
Δ rate |
m |
Rate law |
|
4 times (0.040/0.010) = 4
|
4 times (0.246/0.062) = 4
|
4m = 4 therefore, m = 1 |
k[H2]1 |
Therefore, the rate law for the reaction is
Rate = k[NO]2 [H2]1
Problem
The following data was obtained for the reaction 2A(g) + B2(g) --> 2C(g)
|
Run |
[A] |
[B2] |
Rate (mol/L.s) |
|
1 |
2 |
2 |
4 |
|
2 |
4 |
2 |
16 |
|
3 |
2 |
8 |
16 |
What is the rate law for the reaction?