Kinetic Equilibrium, Reaction Rate and Free Energy
In studying chemistry we concern ourselves with a wide variety of physical and chemical changes.
Physical change: No change in chemical composition. No new substances are produced.
Examples: ice melting, water boiling, water freezing, sugar dissolving in tea, etc.
Chemical Reaction: Change in chemical composition. New substances created. Reactions:
bonds broken, new bonds made. Examples: synthesizing drugs, batteries generating a voltage,
twinkies being converted to fat, etc.
All of these processes (both physical and chemical) have the following in common:
 They occur at a certain rate.
 They proceed until kinetic equilibrium is reached.
 The process involves a transfer of energy into or out of the system.
Understanding kinetic equilibrium, reaction rate and free energy changes is extremely important if
we are to make any headway in understanding chemistry. This simulated lab is designed to improve
your understanding and appreciation of the relationship between rate, equilibrium and energy.
Understanding these concepts will provide you with a framework in which to organize your growing
chemical knowledge.
Reactants
Products
Reaction rate: The study of rates of reaction or kinetics applies to the speed of reaction. The rate of a
reaction is determined by measuring the concentration of reactant (or formation of product) over time.
Rate of Reaction = rate of disappearance of reactant
= -kforward [reactant]/Δtime
or
Rate of Reaction = rate of appearance of product
= +kreverse [product]/Δtime
kforward and kreverse are constants
Equilibrium: Equilibrium applies to the extent of reaction. How much/little product forms. At
equilibrium, the concentrations of reactants and products no longer change with time. This gives the
perception that the reaction has ceased. However, equilibrium is a dynamic balance between forward
and reverse reactions. The reaction does not stop when equilibrium has been reached. At equilibrium,
the rates of forward and reverse reactions are such that the concentrations of reactants and products
remain constant. The equilibrium constant (K) is an indication of the extent of reaction.
K = [Products]/ [Reactants]
K>1


the product concentration is greater than the reactant concentration at equilibrium.
The right side (products) of the reaction is favored


the product concentration is less than the reactant concentration at equilibrium.
the left side (reactants) of the reaction is favored.
K< 1
PART 1: Equilibrium : K = 1
Assign jobs to the members of your group.
 Reactant (the loud pushy person)
 Product (that quiet shy guy, what’s his name again?)
 Checker(the smart person in the group)
 Recorder( the organized person in the group)
For groups of three, one person can act as recorder and checker.
Start with:
 16 pennies
 and kfor = 1/2 krev = 1/2
Place 16 pennies face up on the grid. This represents the reaction at time=0. In this simulation, face
up pennies will represent reactants [R] and face down pennies will represent products [P]. The
reaction will proceed in the forward direction and [R] will be transformed into [P]. This
transformation will be simulated by lab partner one (Reactant) who will flip over some fraction of the
face up pennies. As soon as product forms (face down pennies) the reaction will proceed in the reverse
direction to form more [R]. The reverse reaction will be simulated by lab partner two (Product) who
will flip over some face down pennies. After repeating this procedure for some time you will reach a
point (if you didn’t screw up) where the number of face up and face down pennies will remain
constant.
Time one:
The number of pennies flipped over each round depends on the rate of the forward and reverse
reactions.
Lab partner 1(Reactant):
Calculate the forward reaction rate
rate forward = -kforward[R]
Flip that many pennies over.
Round down any fractions to the nearest
whole number.
The smart person should check your
answer. The recorder should record the
results in the table.
Time two:
Lab partner 1(Reactant):
Repeat the procedure you followed in round one.
Lab partner 2 (Product):
Calculate the forward reaction rate:
rate reverse = kreverse[P]
Round down any fractions to the nearest
whole number.
Flip that many pennies over.
The smart person should check your
answer. The recorder should recode the
results in the table.
Time 3 +: Repeat this procedure, taking turns, until you reach a point when the amount of [R] and [P]
remains constant. Continue the process for two or three more cycles.
PART 1a: Equilibrium: K = 1
 Complete the table.
Q*
Time [P]/[R]
1
0
2
3
4
5
[R]
16
Rate forward Rate Back Net Rate
kf = 1/2
kr = 1/2
forward
kf[R]
kr[P]
kf[R]-kr[P]
8
0
8
[P]
0
*Q = reaction quotient [P]/[R]. Q = K at equilibrium.
 Plot the results using excel. For chart type choose: scatter with data points connected
o Series 1: [R] vs. Time
 Use Time for the x values and [R] for the y values
o Series 2: [P] vs. Time
 Use Time for the x values and [P] for the y values
Questions:
1. Calculate the equilibrium constant using K = [P]/[R].
2. What happens to the rate of the forward reaction as you approach equilibrium?
3. What happens to the rate of the reverse reaction as you approach equilibrium?
4. What observation can you make about the rate of both forward and reverse reactions at equilibrium?
5. If the rate of the forward and reverse reactions were doubled, what would happen to the equilibrium
constant?
Part 1b: Repeat the Simulation with the reaction rates halved (kfor and krev = 1/4)
Rate forward Rate Back Net Rate
Q*
kf = 1/4
kr = 1/4
forward
Time [P]/[R]
[R]
kf[P]
kr[R]
kf[R]-kr[P]
[P]
1
0.00
16
4
0
4
0
2
0.33
12
4
3
4
5
6
7
8
 Complete the table.
 Plot the results as before.
Questions
6. Calculate the equilibrium constant for the reaction using K=[P]/[R].
7. Which reaction reached equilibrium the fastest?
8. Is it is possible to predict the rate of a reaction if you know its equilibrium constant?
9. What is the relationship between K (the equilibrium constant) and kfor and krev?
Use the following relationships and your mastery of algebra to come up with an equation for K
expressed in terms of kfor and krev.
 K = [P]/[R]
 At equilibrium:
o rate forward = rate back
o kfor[R] = krev[P]
10. Calculate K for both reactions in Part A using your formula from question 9.
Reaction 1:
K=
Reaction 2:
K=
PART 2: Equilibrium: K < 1
Repeat the procedure you used in Part A for reaction 3 using the following values:
Start with:
 18 pennies
 and kfor = 1/4 krev = 1/2
TIME
1
2
3
4
5
6
7
Q
[P]/[R]
0
[R]
18
 Plot the results using as before.
Rate forward
kf = 1/4
kf[R]
4
Rate Back
Kr=1/2
kr[P]
0
Net Rate
forward
kf[R]-kr[P]
4
[P]
0
Questions
11. Calculate the equilibrium constant for the reaction using K=[P]/[R].
12. Calculate the equilibrium constant using K=kfor/ krev
PART 3a: Equilibrium: K > 1
Repeat the procedure you used in Part A for reaction 3 using the following values:
Start with:
 18 pennies
 and kfor = 1/3 krev = 1/6
 Plot the results as before.
Time
1
2
3
4
5
6
7
8
Q
[P]/[R]
0.00
[R]
18
rate forward
kf = 1/3
kf[R]
rate back
kr = 1/6
kr[P]
Net Rate
forward
kf[R]-kr[P]
Questions
13. Calculate the equilibrium constant for the reaction using K=[P]/[R].
14. Calculate the equilibrium constant using K=kfor/ krev
[P]
PART 3b: Repeat the exercixe BUT this time start with 18P and zero R.
Time
1
2
3
4
5
6
7
8
9
Q
[P]/[R]
na
5.00
[R]
0
3
rate forwardrate back Net Rate
kf = 1/3
kr = 1/6 forward
kf[R]
kr[P]
kf[R]-kr[P]
0
3
-3
[P]
18
15
15. What effect did starting the reaction with Product present but no Reactant present have on the
equilibrium?
Le Chateliers Principle and Metabolism
Metabolism refers to the chemical processes occurring within a living cell or organism that are
necessary for the maintenance of life. In metabolism some substances are broken down to yield energy
for vital processes while other substances, necessary for life, are synthesized.
Enzymes are proteins that catalyze the millions of metabolic reactions occurring in your body.
Most enzymes are located in cell membranes (cell wall, endoplasmic reticulum, golgi apparatus…).
Enzymes that are involved in the same metabolic pathway are often found clustered together in these
cell structures. This allows them to couple their reactions and employ Le chatelier’s principle to
increase yields. The product produced by one enzyme serves as a reactant for a second enzyme
nearby. In other words, enzyme one produces a high local concentration of starting material for
enzyme two. This increases the yield of the second metabolic process. Also, the second enzyme, in
consuming the product of the first reaction increases the yield of the first process.
The fluid Mosaic model of a cell membrane. Enzymes involved in the
same metabolic pathway are located near each other within the lipid
bilayer.
Active transport is another way biological systems take advantage of Le Chateliers principle.
Cell functions are compartmentalized into organelles. An organelle is a discrete structure of a cell
having specialized functions. An organelle is to the cell what an organ is to the body
The chemical processes that occur in the various organelles are isolated from the rest of the
cellular matrix by lipid membranes. Proteins in the membrane walls of the organelles can transport
molecules into or out of the organelle. In doing this, the concentration of reactants or products can be
increased or decreased to regulate the yield of metabolic reactions.
PART 4: Le Chateliers Principle
Repeat the procedure you used in Part A for reaction 3 using the following values:
Start with:
 18 pennies
 and kfor = 1/4 krev = 1/2
 You don’t have to Plot the results
Follow any additional instructions listed in the table.
Net Rate
kf = 1/4
kr = 1/2
forward
Q
kf[R]
kr[P]
d[P]/dt
time
[P]/[R]
[R]
Rate forward
Rate Back
kf[R]-kr[P]
1
0.00
18
4
0
4
2
3
4
5
6
remove [P]
7
8
9
10
11
add 6 [R]
12
14
13
14
15
16
remove [P]
17
18
19
20
21
add 6[R]
22
14
23
24
25
26
[P]
0
0
0
Questions.
16. What effect does removing product have on the equilibrium constant?
17. What effect does adding more reactant to the reaction have on the equilibrium constant?
18. How could you improve the yield on a reaction that has an equilibrium constant < 1.
k1 [R]
k-1 [P]