Instructions for Grading
Grade only the 13 questions at the end of the exercise.
Each question is worth 3 points. You decide what your answer is worth.
To get full credit for a question you must include the parts of the key that are in bold and underlined. Your answer does not have to be identical to the key but it should provide the same information. Again, you decide what your answer is worth.
Next to each question, indicate the number of points that you gave yourself (from 0 to 3).
At the top of the first page, write the total number of points (X) as X/39.
An Introduction to Equilibrium
As you can see from the list below, most of SC112 deals with the concept of equilibrium
Chapter 12: Solution equilibrium and colligative properties
Chapter 13: Rates of reactions (Kinetics). This is intimately related to equilibrium.
Chapter 14: Basic concepts about chemical equilibrium
Chapter 15: Acid/Base equilibrium
Chapter 16: More about equilibrium of reactions in aqueous solutions.
Chapter 18: Thermodynamics (What determines the size of an equilibrium constant?)
Chapter 19: Redox equilibrium (Electrochemistry)
Learning objectives:
1. Understand the meaning of “dynamic equilibrium”
2. Understand the meaning of “equilibrium constant”
3. Understand how equilibrium can be “disturbed” and how equilibrium is reestablished.
4. Understand the effects of concentration, catalysts, and temperature on the rate of a reaction.
5. Understand the effects of catalysts and temperature on the equilibrium constant.
6. Understand the meaning of “closed system” vs. “open system”
7. Understand one way to drive a reaction to completion.
Procedure
We will model the simple chemical reaction, A = B (“=” means a double arrow) by transferring water from one container to another with various size beakers.
Analogies
Transfer of water from left to right Forward reaction (A → B)
Transfer of water from right to left Reverse reaction (A ← B)
Volume (or height) of water in the container on the left Concentration of reactants
Volume (or height) of water in the container on the right Concentration of products
Size of the beaker in your left hand Rate constant for the forward reaction
Size of the beaker in your right hand Rate constant for the reverse reaction
Volume of water in a right hand beaker Rate of the forward reaction
Volume of water in the left-hand beaker Rate of the reverse reaction
Number of times water has been transferred Passage of time
Value of Q at equilibrium = Keq
I. Dynamic equilibrium and the equilibrium constant. (Note: This is my data.)
Volume of beaker in left hand (rate constant of forward reaction): 250 mL
Volume of beaker in right hand (rate constant of reverse reaction): 100 mL
Transfer # (time) / Volume of water in 250 mL beaker(rate of the forward reaction) / Volume of water in 100 mL beaker
(rate of the reverse reaction) / Height of water in left-hand box
(concentration of reactants) / Height of water in right-hand box
(concentration of products) / Ratio of heights (products/reactants)
(Reaction quotient, Q)
1 / 130 mL / 0 mL / 3.4 cm / 0 cm / 0.0
5 / 90 mL / 20 mL / 2.1 cm / 1.4 cm / 0.7
10 / 60 mL / 35 mL / 1.6 cm / 1.9 cm / 1.2
15 / 50 mL / 50 mL / 1.5 cm / 2.0 cm / 1.3
20 / 50 mL / 50 mL / 1.5 cm / 2.0 cm / 1.3
When the ratio stops changing, the reaction is “at equilibrium” and Q = Keq (equilibrium constant)
Keq = ______1.3______(Q and Keq have no units)
What happened to the rates of the forward and reverse reactions as the reaction proceeded?
The forward reaction slowed down (due to decreasing water level) and the reverse reaction sped up (due to increasing water level).
How did the rates of the forward and reverse reactions compare when the reaction reached equilibrium?
The rates of the forward and reverse reactions were the same at equilibrium. At this point the water levels stopped changing so the value of Q stopped changing.
Ratef = Rater
II. Effect of initial concentrations of reactants and products on Keq
Add some water to the left-hand (reactant) box.
Height of water in left-hand box (“initial concentration of reactants”) = ______2.5 cm______
Height of water in right-hand box (“initial concentration of products”) = ______2.0 cm______
Q = ______0.8 (notice that Q is now less than Keq so no longer at equilibrium)______
Re-establish equilibrium by performing the reaction as before.
Height of water in left-hand box (“equilibrium concentration of reactants”) = ______1.8 cm______
Height of water in right-hand box (“equilibrium concentration of products”) = ______2.4 cm______
Keq = ___1.3 (same as previous experiment even though the amount of water in each box is different!)__
Pour all the water into the right-hand (product) box and establish equilibrium
Height of water in left-hand box (“equilibrium concentration of reactants”) = ______1.8 cm______
Height of water in right-hand box (“equilibrium concentration of products”) = ______2.4 cm______
Keq = _____1.3 (same again!)______
Do the initial concentrations of reactants and products affect the value of the equilibrium constant?
No
Do the initial concentrations affect the equilibrium concentrations of reactants and products?
Yes
Keq DOES NOT DEPEND ON INITIAL CONCENTRATIONS OF REACTANTS AND PRODUCTS (ALTHOUGH ABSOLUTE AMOUNT OF REACTANTS OR PRODUCTS AT EQUILIBRIUM DOES).
III. Effect of a catalyst on Keq
Pour all the water into the left-hand (reactant) box. Use a 1000 mL beaker (left hand) and a 400 mL beaker (right hand) to establish equilibrium. Count the number of transfers required to reach equilibrium.
How will the new beaker sizes affect the rates of the forward and reverse reactions?
Both rates should be 4 times faster than before. (This should not change the equilibrium constant!)
Number of transfers required to reach equilibrium ______5 (4 times faster!)_____
Keq = _____1.3 (same as before!)______
IV. Effect of temperature on the equilibrium constant.
Pour all of the water into the left-hand (reactant) box. Do the reaction with two 400 mL beakers. (This is analogous to raising the temperature.) Count the number of transfers required to reach equilibrium.
How will the rates of the forward and reverse reactions compare to their values in experiment I?
The forward reaction will be 400/250 = 1.6 times faster and the reverse reaction will be 400/100 = 4 times faster. This should result in a smaller equilibrium constant (less product, more reactant than before). In fact, since the two beakers are the same size, we’d expect an equilibrium constant of 1!
Number of transfers required to reach equilibrium = _____Fewer than in expt. I as expected______
Keq = _____1.0 (as predicted!)______
Raising the temperature or adding a catalyst has a similar effect on a reaction, what is it?
They both increase the reaction rate so equilibrium is reached more quickly.
Why does the equilibrium constant change when you raise (or lower) the temperature but does not change when you add a catalyst?
Adding a catalyst increases the rate constants (beaker sizes) for the forward and reverse reactions by the same factor (in this case both increased by a factor of 4).
Raising the temperature also increases both rate constants but by different factors (1.6 and 4 in our experiment).
V. Effect of removing product on equilibrium (Effect of an “open system”)
Starting at equilibrium, remove some of the water from the “product” side. While one partner attempts to reestablish equilibrium, have the other partner continue to remove product. What happens?
All of the reactant is consumed!
What is the definition of a “closed system”?
One in which reactants and products can not escape and remain in contact with each other. This doesn’t mean that the reaction vessel must be sealed. If there are no gases, then the reactants and products will not escape.
Questions:
1. What is meant by “dynamic equilibrium”?
For a chemical reaction, equilibrium is the point in the reaction at which there is no net change in the amounts of reactants or products. The reaction has not stopped at this point. Both the forward and reverse reactions are occurring but at the exact same rate. Thus, it is a “dynamic” equilibrium (as opposed to a “static” equilibrium).
2. Explain why a reaction in a closed system eventually reaches equilibrium.
If a reaction is started with only reactants present (no products), then only the forward reaction can occur and the rate of the reverse reaction is zero. As reactants are consumed and products are formed, the rate of the forward reaction decreases and the rate of the reverse reaction increases. Eventually, the two rates must become equal. (Regardless of the starting concentrations of reactants and products, a reaction will always reach equilibrium. See below.)
3. Define Q (reaction quotient) and Keq (equilibrium constant) and explain the relationship between these two values.
For the simple reaction that we modeled in this exercise: A = B
(note: You can give the equation above or you can describe Q in words. The formula for Q is more complicated for a more complicated reaction.)
Starting with only A, Q is initially equal to zero and increases as the reaction proceeds. Q reaches its maximum value when the reaction reaches equilibrium. Keq is the value of Q when the reaction reaches equilibrium.
For gases, partial pressures can be substituted for molar concentrations. (Do you see why?) The numerical value of the equilibrium constant will change but it will still be a “constant”.
Later you will see that equilibrium constant expression simply says that the rate of the forward reaction is equal to the rate of the reverse reaction!! The size of the equilibrium constant says nothing about the net rate of the reaction. A reaction with a very large equilibrium constant could take years or milliseconds to reach equilibrium.
4. How does changing the initial concentrations of reactants and/or products affect Keq?
The initial concentrations have no effect on Keq. Although the equilibrium concentrations of reactants and products depend on the initial concentrations, the value of Keq does not. Regardless of the initial concentrations, you will always end up in the same equilibrium state (same Keq).
5. Does changing the initial concentrations of reactants change the equilibrium concentration of products?
Yes. If you start with more reactants, you end up with more products. However, the equilibrium constant does NOT change!
6. What is a rate constant? What happens to rate constants when you add a catalyst or increase the temperature?
A rate constant is a measure of how easily a reaction occurs when reactants collide with each other. The rate of a reaction depends on two things, the concentrations of reactants (depth of water) and the rate constant (size of beaker). As you will learn, a rate constant is not a rate but it is proportional to the rate. If a rate constant doubles, then the rate doubles. Numerically, a rate constant is the same as the rate if all of the reactant concentrations are 1 M.
Every reaction has a characteristic rate constant that depends on temperature and on the presence of a catalyst. When you add a catalyst or increase the temperature, rate constants increase which is why a reaction goes faster.
7. Why does the addition of a catalyst NOT affect the value of Keq?
Adding a catalyst does not change Keq because a catalyst increases the forward AND reverse rate constants by the same factor. In part III, we increased both rate constants by a factor of 4 so both forward and reverse reactions went 4 times faster and equilibrium was not disturbed.
8. Why does changing the temperature change the value of Keq?
When temperature increases, the forward and reverse rate constants both increase but one increases more than the other. Thus, if a reaction is at equilibrium, a temperature shift will disturb the equilibrium because now the two rates are no longer equal. To re-establish equilibrium, either more product or more reactant is produced which, in turn, changes the equilibrium constant. We modeled temperature change by changing the ratio of the sizes of the beakers. Both beakers got larger but one size increased more than the other size. (125 mL and 250 mL beakers were both increased to 400 mL.)
9. Why do different reactions have different values of Keq?
By now you should see that Keq depends on the ratio of the rate constants for the forward and reverse reactions. Every reaction has a characteristic rate constant so every reaction has a different value of Keq that depends only on temperature. (Changing the temperature changes the ratio of the rate constants but adding a catalyst does not.
10. Name 3 ways to change the rate of a reaction (in one direction).
1. Change the concentration of reactants. Rate increases with increasing concentration (USUALLY)
2. Add a catalyst. Rate increases.
3. Change the temperature. Rate increases with increasing temperature.
11. Must a “closed system” be inside a sealed container?
No. The container does not have to be sealed as long as nothing can escape. For example, a reaction that occurs entirely in aqueous solution in an open beaker is a closed system because nothing escapes from the solution. However, if one of the reactants or products is a gas, then the container must be sealed or the gas will escape.
12. Why can’t equilibrium be established in an “open system”?
Imagine a reaction at equilibrium. Now open the system by allowing reactants or products to escape. The equilibrium is disturbed. The reaction will attempt to reestablish equilibrium but it is a hopeless attempt because stuff is always escaping. To establish equilibrium, reactants and products must be kept in constant contact so the reaction can occur in both directions. See what happened in part V!
13. Describe one way to drive a reaction to completion (consume all of the limiting reactant).
A reaction can be driven to completion by constantly removing one or more products as they are formed. This causes the forward reaction to continue in a futile attempt to reach equilibrium.
Here is another way. If there is more than one reactant, starting with a great excess of one of them will cause the limiting reactant to be completely consumed. (Do you see why?)
Some reactions have such large equilibrium constants that they naturally go to completion.
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