TWO TYPES OF RATE LAW
¾ Differential rate law—data table contains concentration and rate data. Use table logic or ugly
algebra to determine the orders of reactants and the value of the rate constant, k.
¾ Integrated rate law—data table contains concentration and time data. Use graphical methods to determine the order of a given reactant. The value of the rate constant k is equal to the absolute value of the slope of the best fit line which was decided by performing 3 linear regressions and analyzing the regression correlation coefficient r. Not nearly as hard as it sounds!
INTEGRATED RATE LAW: CONCENTRATION/TIME RELATIONSHIPS
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.
Set up your axes so that time is always on the x-axis. Plot the concentration of the reactant on the y-axis of the first graph. Plot the natural log of the concentration (ln [A], NOT log[A]) on the y-axis of the second graph and the reciprocal of the concentration on the y-axis of the third 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.
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You can now easily solve for either time or concentration once you know the order of the reactant. Just remember y = mx + b. Choose the set of variables that gave you the best straight line (r value closest to ±1) and insert them in place of x and y in the generalized equation for a straight line. “A” is reactant A and Ao is the initial concentration of reactant A at time zero [the y-intercept].
y = mx + b | |
zero order | [A] = −kt + [Ao] |
first order | ln[A] = −kt + ln [Ao] |
second order | 1/[A] = kt + 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
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 Using the graphing calculator: Set up your calculator so that time is always in L1.
Use L2, L3 and L4 to display the y-variables. Remember the list for what is placed on the y-axis is alphabetical (concentration, natural log of concentration and reciprocal concentration).
L1 = time (x-variable throughout!)
L2 = concentration [A] straight line = zero order
L3 = ln concentration ln [A] straight line = first order
L4 = reciprocal concentration 1/[A] straight line = second order
Use this system to set up the data given in the following exercise.
Exercise The decomposition of N2O5 in the gas phase was studied at constant temperature. 2 N2O5(g) → 4 NO2(g) + O2(g) The following results were collected: [N2O5] Time (s) 0.1000 0 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400 Determine the rate law and calculate the value of k. What is the concentration of N2O5(g) at 600 s? At what time is the concentration of N2O5(g) equal to 0.00150 M ? Ans: rate = [N2O5] ; k = −slope of graph of ln [N2O5] vs. time = 6.93 × 10−3 s−1; 0.0016 M; 606 s |
We are going to perform 3 linear regressions to determine the order of the reactant. They will be L1,L2; L1,L3; L1,L4. Next, we will determine which regression has the best r-value [linear regression correlation coefficient in big people language!] We will also paste the best regression equation Y= so that we can easily do other calculations commonly required on AP Chemistry Exam problems.
1. To begin let’s do some housekeeping. Press ‘ repeatedly until your home screen is cleared.
2. Press yö to paste the ClrAllLists command on your screen. Press Í to execute the command.
3. Press yÊ to access the catalogue. Press †repeatedly to scroll down the alphabetical list of commands until you reach DiagnosticsOn. Press ÍÍ to paste the command on your screen and execute the command. If you skip this step, you will never see the r-value!
4. Press …Í to enter data into your lists. Put times in L1 and concentrations (or absorbances) in L2.
5. Batch transform the concentrations into ln[concentration] and place them into L3. How?
Press ~ and } to get to the “tippy top” of L3. Next, press μyÁ to calculate the natural logs of the concentrations stored in L2 and plop them into L3.
6. Batch transform the concentrations into reciprocal concentrations by pressing ~ and } to get to the “tippy top” of L4. Press yÁ— to calculate the reciprocals of the concentrations stored in L2 and plop them into L4.
7. Press …~ to get to CALC and then press ¶ to choose LinReg. The command will be
pasted on your screen. Press yÀ¢yÁ¢~ÍÍÍto add L1, L2, Y1 to your screen. That translates to “Oh, wise calculator, please run a linear regression picking up the x-values from L1, the y-values from L2 and paste the regression equation (y = mx + bformat) into Y1 so that it will graph and I may use it to quickly solve additional problems.” Your screen now displays the statistics for the regression. You are in search of an r that is closest to ±1. Jot down your r for L1,L2.
8. Press yÍ to get the LinReg L1, L2, Y1 on the screen again. Use your | arrow to position your cursor over L2. Press y to replace L2 with L3 and press Í to calculate the new regression statistics. Jot down your r for L1, L3.
9. Press yÍ to get the LinReg L1, L3, Y1 on the screen again. Use your | arrow to position your cursor over L3. Press y¶ to replace L3 with L4 and press Í to calculate the new regression statistics. Jot down your r for L1, L4.
10. Determine the best r-value.
ƒ IF it was L1, L2 then zero order for that reactant.
ƒ IF it was L1, L3 then first order for that reactant.
ƒ IF it was L1, L4 then second order for that reactant.
11. Press yÍ as many times as it takes you to get back to the LinReg command that generated the best r-value. Calculate the regression again. Why? Because |slope| = k & Rate = k[rxt]order
To answer the rest of the questions in our example:
2. Press q® to fit your data to the graph window and display your regression line. The regression equation is in your Y= if you care to look at it.
Once you have the CORRECT equation for the reaction’s rate law in your calculator so that it can draw the CORRECT linear regression line…
14. Press p to check the max and min x and y-values that the Zoom 9 command assigned to the graph window. You can now solve for any concentration EXACTLY between those max and min values. What if your window doesn’t have the proper time or ln[conc.] or 1/[conc.] range? CHANGE IT!
15. To solve for time, display your graph by pressing s.
16. Press yr to get to calculate then choose À which is “value”. Now your screen has the graph displayed AND in the lower left corner an X= with a flashing cursor. Enter the time you want the concentration for and voila! Trouble is that the x-value may be either ln[conc.] or 1/[conc.], so you are not finished yet. Jot down the x-value. To get the anti natural log, press yμ and the x-value you wrote down. If you need the reciprocal, type in the value and press
— to calculate the concentration. NOTE: If the time is outside your window range, you’ll get an error message which is a reminder to reset your window.
17. To solve for a concentration is only a tad more complicated. Press o and † to get to Y2. Type in the concentration value with the proper function (such as ln or reciprocal) applied to it that corresponds to the time that you seek. Press s. If you see an intersection, peachy. If you do not, then adjust your window.
18. With the graph displaying an intersection, press yr to get to calculate but press · to choose “intersection”. Press ÍÍÍ to display the time value that you seek.
HALF-LIFE AND REACTION RATE FOR FIRST ORDER REACTIONS, t1/2
• the time required for one half of one of the reactants to disappear.
• [A] = 2[A]o or [A] = 2 so... ln [A] = k t2 and... ln 2 = t2
[A]o [A]o/2
• Rearrange , evaluate ln 2 and solve for t2and you get
t2 = 0.693 k
Exercise A certain first-order reaction has a half-life of 20.0 minutes. a. Calculate the rate constant for this reaction. b. How much time is required for this reaction to be 75% complete? 3.47 × 10−2 min−1; 40 minutes |
Exercise
The rate constant for the first order transformation of cyclopropane to propene is 5.40 × 10−2/hr. What is the half-life of this reaction? What fraction of the cyclopropane remains after 51.2 hours? What fraction remains after 18.0 hours?
12.8 hr; 0.063; 0.38
Exercise For the reaction of (CH3)3CBr with OH−, (CH3)3CBr + OH− → (CH3)3COH + Br− The following data were obtained in the laboratory. TIME (s) [(CH3)3CBr] 0 0.100 30 0.074 60 0.055 90 0.041 Plot these data as ln [(CH3)3CBr] versus time. Sketch your graph. Is the reaction first order or second order? What is the value of the rate constant? straight line with a negative slope; 1st order since plot of ln[(CH3)3CBr] vs. time is linear; 9.9 × 10−3 s−1 |
Exercise 12.5 Butadiene reacts to form its dimer according to the equation 2 C4H6 (g) Æ C8H12 (g) The following data were collected for this reaction at a given temperature: [C4H6] Time (± 1 s) 0.01000 0 0.00625 1000 0.00476 1800 0.00370 2800 0.00313 3600 0.00270 4400 0.00241 5200 0.00208 6200 a. What is the order of this reaction? Explain. Sketch your graph as part of your explanation. Write the rate law expression: b. What is the value of the rate constant for this reaction? c. What if the half-life for the reaction under the conditions of this experiment? 2nd since plot of 1/[C4H6] vs. time yields a straight line with a |slope| = k = 6.14 × 10−2 M/s; 1630 s |
HALF-LIFE AND REACTION RATE FOR ZERO ORDER REACTIONS, t1/2
- the time required for one half of one of the reactants to disappear, BUT
ƒ Rate = k[A]0 = k (a big fat 1) = k
ƒ Integrated rate law is [A] = -kt + [A]o
- [A] = 2[A]o or [A] = 2 so...
[A]o
[A]o = − k t1/2 + [A]o
2
k t1/2 = [A]o solve for t1/2, t1/2 = [A]o for a ZERO order rxn.
2k 2k
Zero-order reactions are most often encountered when a substance such as a metal surface or an enzyme is required for the reaction to occur. The enzyme or catalyst may be come saturated and therefore an increase in the [reactant/substrate] has no effect on the rate.
INTEGRATED RATE LAWS FOR REACTIONS WITH MORE THAN ONE REACTANT
• Must [still] be determined by experiment! But we use a technique called “swamping”.
• Flood the reaction vessel with high concentrations of all but one reactant and perform the experiment. The reactants at high concentrations like say, 1.0 M compared to the reactant with a low concentration say, 1.0 x 10-3 M, stay the same.
- “In English”—the rate is now dependent on the concentration of the little guy since the big guy’s aren’t changing, therefore the rate = k’ [little guy]
- We now re-write the rate as a pseudo-rate-law and k’ is a pseudo-rate-constant
This is what is happening in the Crystal Violet lab!
A SUMMARY:
REACTION MECHANISMS
REACTION MECHANISMS
The sequence of bond-making and bond-breaking steps that occurs during the conversion of reactants to products.
• Must be determined by experiment! Must agree with overall stoichiometry AND the experimentally determined rate law!
• ELEMENTARY STEPS
- molecularity--number of molecules that participate in an atomic rearrangement - unimolecular: involves one reactant molecule
- bimolecular: involves a collision between two reactant molecules
- termolecular: simultaneous collision between three reactant molecules [very rare!]*
• RATE EXPRESSIONS FOR ELEMENTARY STEPS--the rate expression cannot be predicted from overall stoichiometry. The rate expression of an elementary step is given by the product of the rate constant and the concentrations of the reactants in the step.
ELEMENTARY STEP | MOLECULARITY | RATE EXPRESSION |
A→ products | unimolecular | rate = k[A] |
A + B → products | bimolecular | rate = k[A][B] |
A + A → products | bimolecular | rate = k[A]2 |
2 A + B → products* | termolecular* | rate = k[A]2[B] |
• THE PHYSICAL SIGNIFICANCE OF RATE EXPRESSIONS FOR ELEMENTARY STEPS
- the more molecules the more collisions, the faster the rate
- the faster the molecules are moving, the more likely they will collide, the faster the rate
• MOLECULARITY AND ORDER
- an elementary step is a reaction whose rate law can be written from its molecularity - NOT true of the overall reaction order!
Exercise Nitrogen oxide is reduced by hydrogen to give water and nitrogen, 2 H2(g) + 2 NO(g) → N2(g) + 2 H2O(g) and one possible mechanism to account for this reaction is 2 NO(g) → N2O2(g) N2O2(g) + H2(g) → N2O(g) + H2O(g) N2O(g) + H2(g) → N2(g) + H2O(g) What is the molecularity of each of the three steps? Show that the sum of these elementary steps is the net reaction. bimolecular; unimolecular; unimolecular |
• REACTION MECHANISMS AND RATE EXPRESSIONS
- determined by experiment
- the rate of the overall reaction is limited by, and is exactly equal to, the combined rates of all elementary steps up to and including the slowest step in the mechanism
- the slowest step is the rate determining step
- reaction intermediate--produced in one step but consumed in another.
- catalyst--goes in, comes out unharmed and DOES NOT show up in the final rxn.
Exercise
The balanced equation for the reaction of the gases nitrogen dioxide and fluorine is
2 NO2 (g) + F2 (g) → 2 NO2F (g)
The experimentally determined rate law is
Rate = k [NO2][F2]
A suggested mechanism for the reaction is
NO2 + F2 → NO2F + F Slow
F + NO2 → NO2F Fast
Is this an acceptable mechanism? That is, does it satisfy the two requirements? Justify.
Yes. It is bimolecular in the first step which is the slow step, which yields a rate law expression that agrees with the experimentally determined rate law that was given.
12.8 CATALYSIS
Alter the mechanism so the activation energy barrier can be lowered.
• Catalysts are not altered during the reaction--they serve to lower the activation energy and speed up the reaction by offering a different pathway for the reaction
• ΔE is NOT changed for the process
• Biological catalysts are enzymes--proteins w/ specific shapes ATP synthetase is the most important enzyme in the human body!
HETEROGENEOUS CATALYST--different phase than reactants,
usually involves gaseous reactants adsorbed on the surface of a solid catalyst
