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Chemical Reaction Rates II
Chemical Reaction Rates II
Solving Kinetics Problems Involving Integrated Rate Law
OBJECTIVE
Students will use a graphing calculator and concentration-time data to determine the order of a reactant
using the graphical methods of integrated rate law.
LEVEL
Chemistry
T E A C H E R
P A G E S
NATIONAL STANDARDS
UCP.1, UCP.2, UCP.3, B.3, B.6
CONNECTIONS TO AP
AP Chemistry:
III. Reaction types D. Kinetics 1. Concept of rate of reaction 2. Use of experimental data and
graphical analysis to determine reactant order, rate constants, and reaction rate laws 3. Effect of
temperature changes on rates 4. Energy of activation; the role of catalysts
TIME FRAME
two 45 minute class periods
MATERIALS
graphing calculator
paper and pencil
TEACHER NOTES
This activity is designed to provide an introduction to the types of problems students will face on the
AP* Chemistry exam. This lesson should follow Chemical Reaction Rates I, a lesson on differential rate
law found in this guide. The mathematical skills and graphing calculator skills are those students should
have mastered in their mathematics class since they are required of Algebra I. Do not be intimidated by
the use of the graphing calculator, your students use it regularly in their math classes. Students should
complete the conclusion questions for homework. The kinetics topics involving mechanisms and the
Arrhenius equation for calculating the activation energy are not addressed in this lesson.
Prior to this lesson, students should have an understanding of basic collision theory, especially reaction
coordinate diagrams (a.k.a. potential energy diagrams). From these diagrams they should be able to
compare the energy of the reactants, the energy of the products, and the activation energy. Emphasize
that collisions between molecules must occur for chemical reactions to take place. Also emphasize that
a catalyst lowers the activation energy by providing an alternate collision pathway, one that requires less
energy. Remind students that a catalyst is neither a reactant nor a product, so it can participate in
chemical reactions over and over again. Be sure students also understand the role of a reaction
intermediate, a highly unstable species that is formed at the peak of the potential energy diagram. It is
sometimes called the activated complex.
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ANSWERS TO THE CONCLUSION QUESTIONS
1. Data for the decomposition of N2O5 in a solution at 55°C are as follows:
[N2O5]
(mol/L)
2.080
1.670
1.360
0.720
Time
(min)
3.07
8.77
14.45
31.28
Use graphical methods of the integrated rate law to determine the following:
b. What is the rate constant for the reaction?
0.0375
• The | slope | = k, so k =
min
d. What is the concentration of N2O5 at 2.00
minutes?
• Press ψ ρ ⊆ and enter 2 as the x value.
• When x = 2.00 min, y = 0.771 which is the
natural log of the correct concentration.
• ln[N 2O5 ] = 0.771 = 2.16 M
P A G E S
c. Write the rate law that is consistent with the data?
• Rate = k [N2O5]
ln[N 2 O5 ]
e. At what time is the concentration of N2O5
time (min)
equal to 0.55 M?
• First, write an equation in Y2 (Press ο until you arrive at Y2=) that is the ln of 0.55, then
press σ. If you do not see the horizontal line representing ln(0.55), then press π and change
the Ymin. Press σ to view the graph and see if your horizontal line appears. The ln(0.55) =
−0.598, so we chose −0.8. Students do this in math class all the time. This is not a new skill
for them.
• Now the horizontal line appears, but it still does not intersect our regression equation. Press
π again and adjust the Xmax. This may take some trial and error, we settled on 50 as an
Xmax. Press σ. The intersection should now be on your screen.
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T E A C H E R
a. What is the order for the reaction?
• The statistics displayed next to the data above are from the linear regression of L1,L3 which
translates to ln[N 2 O5 ] vs. time. This regression had the highest r value ∴ the reaction is first
order with respect to N2O5.
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•
Chemical Reaction Rates II
Next calculate the x and y values at the intersection. Press ψ ρ • ⊆ ⊆ ⊆. The x value is the
time in minutes which is equal to 38.5 minutes.
P A G E S
2. Sucrose, C12H22O11, decomposes in dilute acid solution to form the two isomers glucose and fructose
that both have the chemical formula of C6H12O6. The rate of this reaction has been studied in acid
solution, and the data in the table were obtained.
T E A C H E R
Time (min) [C12H22O11] (mol/L)
0
0.316
25
0.289
50
0.264
150
0.183
200
0.153
Use the graphical method of the integrated rate law to determine the following:
a. What is the order for the reaction?
• The statistics displayed next to the data above are from the linear regression of L1,L3 which
translates to ln[C12H22O11] vs. time. This regression had the highest r value ∴ the reaction is
first order with respect to C12H22O11.
b. What is the rate constant for the reaction?
•
−3
The slope = k, so k = 3.64 × 10
min
c. Write the rate law that is consistent with the data?
• Rate = k [C12H22O11]
d. What is the concentration of [C12H22O11] at 75.0 minutes?
• When x = 75.0 min, y = −1.42 which is the natural log of the correct concentration.
• e−1.42 = 0.242 M
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e. At what time is the concentration of [C12H22O11] equal to 0.115 M?
•
•
•
First, write an equation in Y2 that is the natural log of 0.115 M, then press σ. Adjust your
window until the horizontal line appears.
Next calculate the x and y values at the intersection.
The x value is the time in minutes which is equal to 278 minutes.
T E A C H E R
3. Ammonia decomposes when heated according to the equation
NH3(g) → NH2(g) + H(g)
The data in the table for this reaction were collected at 2500K.
Time (h)
0
15
30
60
P A G E S
[NH3] (mol/L)
8.00 × 10−7
7.20 × 10−7
6.55 × 10−7
5.54 × 10−7
Use graphical methods of the integrated rate law to determine the following:
a. What is the order for the reaction?
• The statistics displayed next to the data above are from the linear regression of L1,L4 which
1
vs. time. This regression had the highest r value ∴ the reaction is
translates to
[NH 3 ]
second order with respect to NH3.
b. What is the rate constant for the reaction?
9250
• The slope = k, so k =
M •h
c. Write the rate law that is consistent with the data?
• Rate = k [NH3]2
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Chemical Reaction Rates II
d. What is the concentration of NH3 at 90.0 hours?
• Adjust your window as necessary.
• When x = 90.0 min, y = 2,082,266 which is the reciprocal of the correct concentration.
•
1
1
=
= 4.80 × 10−7 M
[NH3 ] [2,082,266 ]
T E A C H E R
P A G E S
e. At what time is the concentration of NH3 equal to 7.00 × 10−7 M?
•
•
First, write an equation in Y2 that is the reciprocal of 7.00 × 10−7, then press σ. Adjust your
window until you see the intersection.
Next calculate the x and y values at the intersection. Press ψ ρ • ⊆ ⊆ ⊆. The x value is the
time in minutes which is equal to 19.3 minutes.
4. The reaction 2 HOF(g) → 2 HF(g) + O2(g) occurs at 45°C. The following data was collected:
Time (min)
0
5
10
15
20
[HOF] (mol/L)
0.850
0.754
0.666
0.587
0.526
a. What is the order for the reaction?
• The statistics displayed next to the data above are from the linear regression of L1,L3 which
translates to ln[HOF] vs. time. This regression had the highest r value ∴ the reaction is first
order with respect to HOF.
b. What is the rate constant for the reaction?
0.0242
• The slope = k, so k =
min
c. Write the rate law that is consistent with the data?
• Rate = k [HOF]
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d. What is the concentration of [HOF] at 60.0 minutes?
• Adjust your window as necessary.
• When x = 60.0 min, y = −1.62 which is the natural log of the correct concentration.
• e−1.62 = 0.198 M
e. At what time is the concentration of [HOF] equal to 0.250 M?
•
•
T E A C H E R
•
First, write an equation in Y2 that is the natural log of 0.250 M, then press σ. Adjust your
window until the horizontal line appears. This one was tricky since it was the Ymin that
needed adjusting since the natural logs are all negative.
Next calculate the x and y values at the intersection.
The x value is the time in minutes which is equal to 50.5 minutes.
Time (min)
15
30
80
120
P A G E S
5. Data for the decomposition of dinitrogen oxide into its elements at 900°C is as follows:
[N2O] (mol/L)
0.0830
0.0680
0.0360
0.0220
a. What is the order for the reaction?
• The statistics displayed next to the data above are from the linear regression of L1,L3 which
translates to ln[N2O] vs. time. This regression had the highest r value ∴ the reaction is first
order with respect to N2O.
b. What is the rate constant for the reaction?
0.0126
• The slope = k, so k =
min
c. Write the rate law that is consistent with the data?
• Rate = k [N2O]
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Chemical Reaction Rates II
d. What is the concentration of [N2O] at 240.0 minutes?
• Adjust your window as necessary.
• When x = 240.0 min, y = −5.34 which is the natural log of the correct concentration.
• e−5.36 = 0.0048 M = 4.8 ×10−3 M
P A G E S
e. At what time is the concentration of [N2O] equal to 0.0500 M?
•
T E A C H E R
•
•
First, write an equation in Y2 that is the natural log of 0.0500 M, then press σ. You should
not have to adjust your window.
Next calculate the x and y values at the intersection.
The x value is the time in minutes which is equal to 54.2 minutes.
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Chemical Reaction Rates II
Solving Kinetics Problems Involving Integrated Rate
Chemical kinetics is the study of the speed or rate of a chemical reaction under various conditions.
Collisions must occur in order for chemical reactions to take place. These collisions must be of
sufficient energy to make and break bonds. The collisions must also be effective which means they not
only have sufficient energy during the collision but that the molecules also collide in the proper
orientation. The speed of a reaction is expressed in terms of its rate — some measurable quantity is
changing with time.
Rate =
Change in concentration of a reactant or a product
Change in time
Most commonly, we discuss four different types of rate when discussing chemical reactions: relative
rate, instantaneous rate, differential rate, and integrated rate. Although these sound complicated, they
are actually quite simple. This activity will focus primarily on the last type, integrated rate law.
First, we need to discuss factors that affect the rate of a chemical reaction:
1. Nature of the reactants—some reactant molecules react quickly, others react very slowly.
2. Concentration of reactants—more molecules in a given volume means more collisions which
means more bonds are broken and more bonds are formed.
3. Temperature—“heat ‘em up & speed ‘em up”; the faster the molecules move, the more likely they
are to collide and the more energetic those collisions will be.
• An increase in temperature produces more successful collisions because more collisions will
meet the required activation energy. There is a general increase in reaction rate with increasing
temperature.
• In fact, a general rule of thumb is that increasing the temperature of a reaction by 10°C doubles
the reaction rate.
4. Catalysts—accelerate chemical reactions by allowing for more effective collisions, thus lowering
the activation energy. The forward and reverse reactions are both accelerated to the same degree.
Catalysts are not themselves transformed in the reaction, so they may be used over and over again.
5. Surface area of reactants—exposed surfaces affect speed.
• Except for substances in the gaseous state or in solution, reactions occur at the boundary, or
interface, between two phases.
• The greater the surface area exposed, the greater the chance of collisions between particles,
hence, the reaction should proceed at a much faster rate. Example: coal dust is very explosive as
opposed to a piece of charcoal. Solutions provide the ultimate surface area exposure!
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Chemical Reaction Rates II
INTEGRATED RATE LAW
Differential rate law expresses how the rate depends on concentration. This is the most common type
of rate problem and the subject of Chemical Reaction Rates I. Integrated rate law expresses how the
concentrations depend on time. If the data you have been given contains “Rate” and “Concentration”
data, use the differential rate law methods. If the data you have been given contains “Concentration”
and “Time” data, use the integrated rate law methods. Why on Earth do we need two methods? It is a
result of experimental convenience.
When we wish to know how long a reaction must proceed to reach a predetermined concentration of
some reagent, we can construct curves or derive an equation that relates concentration and time. This
sounds scary, but is quite simple as long as you appreciate some elegant patterns. First the graphs…
When graphing the concentration of a reactant vs. time, one of the following two shapes is observed:
Zero order, always a straight line.
No further analysis is needed—just
interpret the graph.
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If the graph is a curve, it could be 1st or 2nd
order. Further analysis is needed to determine
which one. This involves creating two
additional graphs in search of a linear
relationship.
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GRAPHICAL METHODS FOR DISTINGUISHING FIRST AND SECOND ORDER
REACTIONS
Why are we in search of a linear relationship? y = mx + b is the friendly little equation format for a
straight line that you already know and love. It allows you to quickly solve for either x or y values. The
next section shows you HOW to determine zero, first, and second order relationships from
concentration vs. time graphs.
Question: How do I get a linear relationship?
Answer: Set up your axes so that time is always on the x-axis. Next, sketch your two new graphs so that
you can determine whether the reactant is first or second order. Plot the natural log (ln [A], NOT log) of
the concentration on the y-axis of the first graph and the reciprocal concentration on the y-axis of the
second graph. You are in search of linear data! Here comes the elegant part… If you do the set of
graphs in this order with the y-axes being “concentration”, “natural log of concentration” and
“reciprocal concentration”, the alphabetical order of the y-axis variables leads to 0, 1, 2 orders
respectively for that reactant.
First order
k = negative slope
Zero order
k = negative slope
Second order
k = the slope
You can now easily solve for either time or concentration once you know the order of the reactant. Just
remember y = mx + b. Pick up the variables that gave you the straight line and insert them in place of
x and y in the equation. “A” is reactant A and Ao is the initial concentration of reactant A at time zero
[the y-intercept].
zero order
first order
second order
y
[A]
ln[A]
1/[A]
=
=
=
=
mx
−kt
−k t
kt
+
+
+
+
b
[Ao]
ln [Ao]
1/[Ao]
Also recognize that slope = k, since the rate constant is NEVER negative. If you are asked to write
the rate expression [or rate law] it is simply Rate = k[A].order you determined from analyzing the graphs
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Chemical Reaction Rates II
PURPOSE
In this activity you will use a graphing calculator to determine the order of a reactant from
concentration-time data using the method of integrated rate law.
MATERIALS
calculator
paper and pencil
PROCEDURE
Solve the problems found on your student answer page. Be sure to show all work paying attention to the
proper use of significant digits and units.
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Chemical Reaction Rates II
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Name _____________________________________
Period _____________________________________
Chemical Reaction Rates II
Solving Kinetics Problems Involving Integrated Rate
ANALYSIS
The following TI-83 or TI-83+ calculator steps may prove useful when applying the graphical methods
of the integrated rate law.
To determine the order of a reactant:
• Press ℘ ∂ ⊆ to clear all lists.
• Press ψ ⊇ and scroll down to DIAGNOSTICS ON and press ⊆ ⊆ so that your calculator will display
the linear correlation coefficient, r, for each linear regression you perform.
• Press ζ and use your arrow keys to select FLOAT, press ⊆ ψ ζ to quit.
• Press to clear your screen.
• To enter data, press and ⊆ to select EDIT. Use the summary below to prepare your analysis of the
data:
L1:
L2:
L3:
L4:
•
•
•
•
•
•
•
time (x variable throughout)
concentration
ln concentration
reciprocal concentration
[A] straight line = zero order
ln [A] straight line = first order
1/[A] straight line = second order
Perform 3 linear regressions. Begin by pressing ∼ ∂, LinReg(ax + b) is now displayed.
Press ψ ℵ ′ ψ ℑ ⊆ to select your x-values from L1 (time) and your y-values from L2. Make note of
the r value. You seek the best linear fit for your data. The best fit will have an r value closest to ±1.
Press ψ ⊆ to re-display the LinReg command, press | ψ ℜ to replace L2 with L3, press ⊆ to execute
the command and make note of the r value.
Press ψ ⊆ to re-display the LinReg command, press | ψ ∂ to replace L3 with L4, press ⊆ to execute
the command and make note of the r value.
Examine your r values and decide which set of data gives the best linear fit.
Paste the regression equation into Y= by pressing ψ ⊆ until you return to the LinReg command that
yielded the best fit.
Once the L1, Lwhichever you chose combination is properly displayed, press ′ ∼ to Y-VARS then ⊆ ⊆.
If you were successful, you’ll see LinReg(ax +b) L1, Lwhichever you chose, Y1 displayed on your screen.
Press ⊆ to execute the command and paste the regression equation into Y1.
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Chemical Reaction Rates II
More helpful hints:
Recall that the order of the reaction is 0; 1; 2 respectively for each L1, L2; L1,L3; L1,L4 combination
respectively and slope = k and Rate = k[reactant]order
Next, since you have created a line, never forget: y = mx + b (TI uses an “a” instead of an “m”)
If L1,L3 was your best “r” then, the reaction is first order and
y = mx + b
becomes
ln [conc.] = −kt + ln [conc.]o
*Notice that you must respect the sign of k when substituting into the y = mx + b format.
Do the same substitutions into y = mx + b for the other relationships.
To determine the concentration at a given time:
• Set up your STAT PLOT by pressing ψ ο. Make sure only one plot is on and that you choose ∀ and
select the list combination that generated the regression line you pasted into Y1.
• Press θ → to display the graph.
• Press ψ ρ ⊆ to select CALCULATE and VALUE. An X= is displayed in the lower left-hand corner
of the screen. Enter the time you were given in the problem and press ⊆. Record the y-value
displayed. Recall what you placed on your y-axis. It is most likely not L2 or the concentration since
zero order reactions are rare. It is more likely either the natural log value of the concentration or the
reciprocal of the concentration. Solve for the actual concentration before recording your answer.
• If you get an ERR:INVALID message, press π and re-set your Xmax or Xmin value. The calculator
can only calculate what it displays. Choose a value for Xmax or Xmin that includes your desired
time value.
To determine the time at a given concentration:
• Press ο then to Y2. Recall the command that generated your regression equation in Y1. Write an
equation that is either the concentration, ln concentration or reciprocal concentration based on
whether your y-value was L2, L3, or L4 respectively.
• Press ψ ρ • to select CALCULATE and INTERSECT. Press ⊆. Record the x-value displayed since
it is the time value.
• If you get an ERR:INVALID message, press and re-set your Ymax or Ymin values.
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CONCLUSION QUESTIONS
1. Data for the decomposition of N2O5 in a solution at 55°C are as follows:
[N2O5]
(mol/L)
2.080
1.670
1.360
0.720
Time
(min)
3.07
8.77
14.45
31.28
Use the graphical method of the integrated rate law to determine the following:
a. What is the order for the reaction?
b. What is the rate constant for the reaction?
c. Write the rate law that is consistent with the data?
d. What is the concentration of N2O5 at 2.00 minutes?
e. At what time is the concentration of N2O5 equal to 0.55 M?
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Chemical Reaction Rates II
2. Sucrose, C12H22O11, decomposes in dilute acid solution to form the two isomers glucose and fructose
that have the chemical formula of C6H12O6. The rate of this reaction has been studied in acid
solution, and the data in the table were obtained.
Time (min)
0
25
50
150
200
[C12H22O11] (mol/L)
0.316
0.289
0.264
0.183
0.153
Use the graphical method of the integrated rate law to determine the following:
a. What is the order for the reaction?
b. What is the rate constant for the reaction?
c. Write the rate law that is consistent with the data?
d. What is the concentration of [C12H22O11] at 75.0 minutes?
e. At what time is the concentration of [C12H22O11] equal to 0.115 M?
3. Ammonia decomposes when heated according to the equation
NH3(g) → NH2(g) + H(g)
[NH3] (mol/L)
8.00 × 10−7
7.20 × 10−7
6.55 × 10−7
5.54 × 10−7
Time (h)
0
15
30
60
The data in the table for this reaction were collected at 2500K.
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Chemical Reaction Rates II
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Use the graphical method of the integrated rate law to determine the following:
a. What is the order for the reaction?
b. What is the rate constant for the reaction?
c. Write the rate law that is consistent with the data?
d. What is the concentration of NH3 at 90.0 hours?
e. At what time is the concentration of NH3 equal to 7.00 × 10–7 M?
4. The reaction 2 HOF(g) → 2 HF(g) + O2(g) occurs at 45°C.
The following data was collected:
Time (min)
0
5
10
15
20
[HOF] (mol/L)
0.850
0.754
0.666
0.587
0.526
a. What is the order for the reaction?
b. What is the rate constant for the reaction?
c. Write the rate law that is consistent with the data?
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Chemical Reaction Rates II
d. What is the concentration of [HOF] at 60.0 minutes?
e. At what time is the concentration of [HOF] equal to 0.250 M?
5. Data for the decomposition of dinitrogen oxide into its elements at 900°C is as follows:
Time (min)
15
30
80
120
[N2O] (mol/L)
0.0830
0.0680
0.0360
0.0220
a. What is the order for the reaction?
b. What is the rate constant for the reaction?
c. Write the rate law that is consistent with the data?
d. What is the concentration of [N2O] at 240.0 minutes?
e. At what time is the concentration of [N2O] equal to 0.0500 M?
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Chemical Reaction Rates II
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