Equilibrium and Concentration
Answer Key
Vocabulary: chemical equilibrium, concentration, equilibrium, equilibrium constant, reaction
quotient, reversible reaction
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
[Note: The purpose of these questions is to activate prior knowledge and get students thinking.
Students are not expected to know the answers to the Prior Knowledge Questions.]
Gary has $5,000 in his bank account and earns a modest salary. Every month he pays for rent,
food, utilities, and entertainment.
A. How will Gary’s account change if he saves more than he spends? It will increase.
B. How will Gary’s account change if he spends more than he saves? It will decrease.
C. What happens if Gary spends exactly as much as he saves? It will stay the same.
Gizmo Warm-up
If Gary spends exactly as much as he earns, his savings will
be in equilibrium. Equilibrium occurs when two opposing
processes occur at the same rate, leading to no net change.
In the Equilibrium and Concentration Gizmo™, you will
investigate how equilibrium can occur in chemical reactions.
To begin, check that Reaction 1 is selected. Set Moles NO2
to 8 and Moles N2O4 to 0.
1. Click Play (
) and observe the colliding molecules. What do you notice?
Sample answer: The molecules move and collide. Occasionally there is a blue or red flash.
In the Gizmo, a blue flash appears every time two reactants combine to form a product. A
red flash appears every time a product dissociates into reactants.
2. Click Reset (
), and set Moles NO2 to 0 and Moles N2O4 to 8. Click Play.
What do you notice now? Sample answer: At first, there are more red flashes than blue
flashes.
3. When a reaction can proceed in either direction, it is a reversible reaction. Based on what
you have observed, is the synthesis of NO2 into N2O4 a reversible reaction? Explain.
I know the synthesis of NO2 into N2O4 is a reversible reaction because both the forward
reaction (blue flashes) and the reverse reaction (red flashes) are observed.
Activity A:
Reversible
reactions
Get the Gizmo ready:
• Click Reset. Reaction 1 should be selected.
• Set Moles NO2 to 8 and Moles N2O4 to 0.
• Move the Sim. speed slider all the way to the right.
Question: What are the characteristics of reversible reactions?
1. Predict: Suppose you began with 8 moles of NO2 in the chamber. What do you think will
happen if you let the reaction go for a long time? Predictions will vary.
2. Test: Click Play. Select the BAR CHART tab and check that Moles is selected. Observe the
bar chart for about 30 seconds. As time goes by, what do you notice about the bars
representing moles NO2 and moles N2O4?
After a while, the bars don’t move as much as at first.
3. Observe: Click Pause ( ). Select the GRAPH tab. Click the (–) zoom control on the
horizontal axis until you can see the whole graph. What do you notice?
At first, the amounts of NO2 and N2O4 change rapidly. After a while the amounts do not
change as much.
This situation, in which the overall amounts of reactants and products does not change
significantly over time, is called a chemical equilibrium.
4. Record: On the BAR CHART tab, turn on Show data values. How many moles of NO2 and
N2O4 are there right now?
Moles NO2 ~4.0
Moles N2O4 ~2.0
[Exact values will vary.]
5. Calculate: Suppose all the NO2 molecules were synthesized into N2O4. Given the equation
2NO2 ⇌ N2O4, how many moles of N2O4 would be produced? 4 moles of N2O4
6. Experiment: Click Reset. On the INITIAL SETTINGS tab, set Moles NO2 to 0 and Moles
N2O4 to 4. Click Play. Click Pause when the bars of the bar chart stop moving very much.
A. List the current amounts of each substance: Moles NO2 ~4.0
Moles N2O4 ~2.0
B. How do these results compare to starting with 8 moles of NO2?
The resulting amounts of NO2 and N2O4 in this experiment are very similar to the
results obtained when starting with 8 moles NO2 and 0 moles N2O4.
(Activity A continued on next page)
Activity A (continued from previous page)
7. Summarize: In each trial, you started with the same amounts of nitrogen and oxygen. In this
situation, did the equilibrium amounts change depending on the direction of the reaction?
The equilibrium amounts were about the same in both situations.
8. Set up the Gizmo: Click Reset and select the EXPERIMENT tab on the left. On the INITIAL
SETTINGS tab on the right, select Reaction 2. Set Moles NO to 5, Moles NO2 to 5, and
Moles N2O3 to 0. What are the reactants and product of this reaction?
Reactants: NO
NO2
Product: N2O3
(Note: In this reaction, some of the NO2 reactants combine to form N2O4, as in reaction 1.)
9. Observe: Recall that a blue flash appears every time two reactants combine to form a
product. A red flash appears every time a product dissociates into reactants. Click Play.
A. At first, do you notice more blue flashes or red flashes? Blue flashes
B. What do you notice about the frequency of blue and red flashes as time goes by?
After a while, the frequency of blue and red flashes is about the same.
C. Click Reset. This time, start the experiment with 0 moles of NO and NO2 and 5
moles of N2O3. Click Play. What do you notice about the red and blue flashes now?
At first, there are more red flashes than blue flashes. Later, the frequencies are even.
10. Explain: Think about how the numbers of blue and red flashes reflect the rates of the
forward (reactants  products) and reverse (products  reactants) reactions.
A. What happens to the rate of the forward reaction as the reactants are consumed?
The rate of the forward reaction decreases as reactants are consumed.
B. What happens to the rate of the reverse reaction as the products are produced?
The rate of the reverse reaction increases as products are produced.
C. Why do reversible reactions always result in chemical equilibria?
Sample answer: Chemical equilibrium occurs when the rates of the forward and
reverse reactions are equal. If there is an excess of reactants, the forward reaction
will occur more quickly than the reverse reaction until there is no longer an excess of
reactants. At this point the rates of the forward and reverse reactions are equal and
equilibrium is reached. The same thing occurs when there is an excess of products.
Activity B:
Get the Gizmo ready:
The equilibrium
constant
• Click Reset. Select Reaction 1.
• Set Moles NO2 to 2 and Moles N2O4 to 7.
Introduction: When investigating the rates of reactions, it often is useful to consider the
concentrations of reactants rather than the total number of moles. Concentrations are often
expressed in moles per liter, or mol/L. Brackets are used to signify concentration. For example,
“[H2] = 5.0 M” means the concentration of hydrogen gas in a chamber is 5.0 moles per liter.
Question: What are the characteristics of reactions in equilibrium?
1. Record: On the BAR CHART tab, select Concentration. Check that Show data values is
on. If necessary, use the arrows to adjust the scale of the chart.
A. What are the current concentrations of each compound?
[NO2] 4.00 M
[N2O4] 14.00 M
Note: “M” is equivalent to “mol/L”
B. Click Play and wait for equilibrium to become established. Click Pause. What are the
approximate equilibrium concentrations?
[NO2] ~12 M
[N2O4] ~6 M
2. Calculate: The value Kc represents the ratio of products to reactants in a reaction at
equilibrium. The greater the amount of products relative to reactants, the higher the resulting
value of Kc. For a general reaction between gases: aA(g) + bB(g) ⇌ cC(g) + dD(g), Kc is
calculated as follows:
[C] c [D] d
Kc =
[ A ] a [B] b
For the current reaction, 2NO2 ⇌ N2O4, we have:
Kc =
[N 2 O 4 ]
[NO 2 ] 2
Based on the current concentrations of NO2 and N2O4, what is Kc? ~0.042
Show your work here:
Kc =
[N2O4 ]
6.0
=
= 0.042
2
[NO2 ]
12.02
[Note: The value of Kc may vary from 0.025 to 0.065. This variation occurs because the
Gizmo model uses a limited number of molecules.]
(Activity B continued on next page)
Activity B (continued from previous page)
3. Gather data: Experiment with a variety of initial concentrations of NO2 and N2O4. For each
set of initial concentrations, use the Gizmo to determine the equilibrium concentrations of
each substance. In the last column, find Kc for that trial. Run three trials for each set of initial
conditions.
Initial
[NO2]
Initial
[N2O4]
Equilibrium
[NO2]
Equilibrium
[N2O4]
Kc
Chosen by
student
Chosen by
student
Values will vary
Values will vary
~0.042
Chosen by
student
Chosen by
student
Values will vary
Values will vary
~0.042
Chosen by
student
Chosen by
student
Values will vary
Values will vary
~0.042
4. Calculate: Find the average value of Kc for each set of three trials. Check calculations.
Trials 1-3: __________
Trials 4-6: __________
Trials 7-9: __________
5. Analyze: What do you notice about the values of Kc?
No matter what initial concentrations are chosen, the value of Kc stays relatively constant.
In general, the value of Kc will be constant for a given reaction at a constant temperature, no
matter the starting concentrations. That is why Kc is known as the equilibrium constant. In
this Gizmo, the values of Kc will vary somewhat because there is a very limited number of
molecules in the chamber.
6. On your own: Use the Gizmo to find Kc for Reaction 4: H2 + I2 ⇌ 2HI. Collect data at least
10 times and average your results to get the best approximation of Kc. Show your data and
work on a separate sheet of paper.
(Hint: Because of the coefficient “2” in front of HI, you will have to square the concentration
of HI to find Kc.)
Kc = ~40 [Check student work. For this reaction, you can expect values of Kc to vary
between 20 and 60.]
7.
Get the Gizmo ready:
Activity C:
Reaction direction
• Click Reset. Check that Reaction 4 is selected.
• Set Moles H2 to 5, Moles I2 to 5, and Moles HI
to 3.
Introduction: For a reversible reaction with equilibrium constant Kc, it often is useful to know in
which direction the reaction will proceed given the starting amounts of reactants A and B and
products C and D. This is done by calculating the reaction quotient, Qc:
Qc =
[C] c [D] d
[ A ] a [B] b
Question: How can you predict the direction of a reversible reaction?
1. List: Select the BAR CHART tab. What are the initial concentrations of each substance?
[H2] 9.23 M
[I2] 9.23 M
[HI] 5.54 M
2. Calculate: Use the equation above to find Qc for the current reaction.
A. What is the current value of Qc? 0.36
B. In activity B, what value of Kc did you arrive at for this reaction? ~40 [Check student
results]
C. How does Qc compare to Kc? Qc is less than Kc.
3. Analyze: Recall that Qc is equal to the ratio of product concentrations to reactant
concentrations.
A. If there is an excess of products, will Qc be greater than or less than Kc? Greater
B. If there is an excess of reactants, will Qc be greater than or less than Kc? Less
C. In the current situation, is there an excess of products or reactants? Reactants
Explain: Qc is less than Kc, so there is an excess of reactants.
D. When the reaction begins, do you expect [HI] to increase or decrease? Increase
Explain: There is an excess of reactants. HI is a product, so it will increase as the
reaction proceeds.
4. Test: Click Play. What happens to [HI]?
[HI] increases. [Note: [HI] may decrease in the first instant but will eventually increase.]
Extension:
Equilibrium
calculations
Get the Gizmo ready:
• Click Reset. Select Reaction 1.
• Set Moles NO2 to 0 and Moles N2O4 to 6.
Goal: Given Kc and initial concentrations, calculate equilibrium concentrations.
1. List: Select the BAR CHART. What is the initial concentration N2O4?
[N2O4]initial = 18.00 M
2. Experiment: Click Play and wait for a few seconds. Click Pause before equilibrium is
reached.
Note: Student results will vary. Sample answers are shown to illustrate the relationship.
A. What is the current concentration of N2O4?
[N2O4] = 11.93 M
B. How much has the concentration of N2O4 gone down? 6.07 M
C. What is the current concentration of NO2?
[NO2] = 6.07 M
D. In general, if [N2O4] is reduced by x, how much does [NO2] increase? x
This result may be surprising. It is true because at constant pressure, the overall
density of particles in the container remains constant. So, if the concentration of one
substance is reduced by x, the concentration of the other substance increases by x.
3. Manipulate: Begin with the general equation for Kc: K c =
[C] c [D] d
.
[ A ] a [B] b
A. What is the equation for Kc for the reaction 2NO2 ⇌ N2O4?
Kc =
[N2O4 ]
[NO2 ]2
B. In this experiment, the initial concentration of NO2 is zero. If the concentration of
N2O4 is reduced by x at equilibrium, the equilibrium concentration of NO2 is equal to
x. Substitute the following values into the equation you wrote in step A:
[N2O4] = ([N2O4]initial – x)
[NO2] = x
Kc =
[N2O4 ]initial − x
x2
C. In activity A, you discovered that Kc for this reaction was close to 0.042. Substitute
this value and the initial concentration of N2O4 into your equation.
0.042 =
18.00 − x
x2
D. Rearrange the terms of your equation to form a quadratic equation in the form
ax2 + bx + c = 0.
0.042 x2 + x – 18.00 = 0
(Extension continued on next page)
Extension (continued from previous page)
4. Solve: Because the equation is in the form ax2 + bx + c = 0, you can use the quadratic
formula (shown below) to solve for x. Ignore negative solutions because the concentrations
cannot be negative. Show your work.
x=
x=
− b ± b 2 − 4ac
2a
a = 0.042, b = 1, c = -18.00
−1 ± 12 − 4 ⋅ 0.042 ⋅ ( −18.0) −1 ± 1 + 3.024 −1 ± 4.024 −1 + 2.006
=
=
=
= 11.98
2 ⋅ 0.042
0.084
0.084
0.084
5. Predict: Based on the value for x, what do you expect the equilibrium concentrations of NO2
and N2O4 to be?
[NO2] 11.98 M
[N2O4] 6.02 M
Check your work by solving for Kc using K c =
[N 2 O 4 ]
[NO 2 ] 2
Kc = 0.042
If you don’t get the correct value of Kc, recheck your work.
6. Test: Click Play and wait for equilibrium to be established. What are the actual equilibrium
values of each substance?
Sample results:
[NO2] 12.18 M
[N2O4] 5.82 M
How close were these results to your predicted results? Answers will vary.
[Results should be relatively close to predicted results given the variation in the Gizmo.]
7. Challenge: Suppose you begin with 6 moles of NO2 and 5 moles of N2O4. Assuming a value
for Kc of 0.042, predict the equilibrium concentrations of NO2 and N2O4. (Use the Gizmo to
determine the initial concentrations.) Show your work on a separate sheet of paper. After
you have made your predictions, click Play and record the experimental results.
Predicted: [NO2] 11.98 M
[N2O4] 6.02 M
Experimental: [NO2] Answers will vary
[Solution is shown on next page.]
[N2O4] Answers will vary
Initial concentrations: [NO2] = 9.82 M, [N2O4] = 8.18 M
K c = 0.042 =
[N2O4 ]initial − x
([NO2 ]initial + x )
2
=
8.18 − x
( 9.82 + x )
2
=
8.18 − x
96.4324 + 19.64 x + x 2
0.042(96.4324 + 19.64 x + x 2 ) = 8.18 − x
4.050 + 0.825 x + 0.042 x 2 = 8.18 − x
0.042 x 2 + 1.825 x − 4.13 = 0
a = 0.042, b = 1.825, c = -4.13
Using the quadratic formula, we have:
x=
=
−1.825 ± 1.8252 − 4 ⋅ 0.042 ⋅ ( −4.13) −1.825 ± 3.331 + 0.694
=
2 ⋅ 0.042
0.084
−1.825 ± 4.025 −1.825 + 2.006
=
= 2.16
0.084
0.084
Recall the initial concentrations: [NO2] = 9.82 M, [N2O4] = 8.18 M
If x = 2.16, then:
Equilibrium [NO2] = 9.82 + 2.16 = 11.98 M
Equilibrium [N2O4] = 8.18 – 2.16 = 6.02 M