The Rate Law
background:
For most chemical reactions, changing the concentrations of
the reactants will change the rate of the reaction. The rate
of a reaction is the change of concentration taking place
per change in time.
You will study the reaction (net ionic reaction shown):
6 I–(aq) + BrO3–(aq) + 6 H+(aq) →
3 I2(aq) + Br–(aq) + 3 H2O
You will determine the rate law for the reaction, and in a following experiment, you will study the effect of temperature
on the rate constant.
For the generic reaction:
2A+B→C
we would measure the rate by measuring either the increase
in the concentration of the product, ∆[C], with a change in
time, ∆t, or by measuring the decrease in the concentration of one of the reactants. In this example, there are two
moles of substance A reacting with one mole of substance
Look carefully at the reaction. You see that iodide ions
on the left are oxidized to iodine molecules on the right.
Also notice that the bromine in BrO3–, the bromate ion, is
reduced to Br– during the reaction. This reaction could be
described in the following way: Bromine atoms in a plus
five oxidation state pick electrons off of six iodide ions in an
acidic solution. The bromines end up with a minus 1 state,
the oxygens that were attached to them attach to hydrogen
ions, ending up in water molecules, and the iodines end up
as aqueous iodine molecules.
B. To account for the different number of moles of A and
− ∆[A]
B reacting, we would use
to show the decrease in
2
the concentration of substance A, and –∆[B] to show the
decrease in the concentration of substance B. So, the rate
of our generic reaction could be shown in any one of the
following equivalent forms:
− ∆[A] − ∆[B] ∆[C] .
rate =
=
=
2∆t
∆t
∆t
Rate has the units: moles per liter per time.
How do you suppose this all takes place? Do 13 particles
(six I-’s, one BrO3–, six H+’s) all somehow collide, hold
on to each other, and react? Or does the reaction take place
in a series of collisions, each one involving two or three
particles, each one resulting in part of the reaction taking
place? This is one of the purposes behind the study of rate
laws—establishing the sequence of events occurring during
the transformation of reactants into products. This sequence
of events is called a reaction mechanism.
It might seem intuitively obvious to you that if we increase the
amount of substance A or substance B present, the reaction
will go at a higher rate. That is to say, the more of substance
A or substance B that is present, the faster substance C gets
produced. Indeed, this is usually the case.
The reaction you are studying has the rate law:
−
rate = k[I − ] x [BrO 3 ] y [H + ] z .
(eqn 1)
You will experimentally determine the values of x, y, and z.
Some reactions may be zero order with respect to a particular
reactant, which means that the reactant will not appear in the
rate law expression. Prior experimental work has already
been done to present the rate law in this form.
The quantitative relationship between the concentrations of
substances in a reaction, and the rate of a reaction, is called
the rate law for a reaction. If the generic reaction used here
had a typical and simple rate law, it would be:
rate = k[A]x [B]y
k is called the rate constant. Its value depends on the nature
of the substances and the temperature. Each concentration
is raised to a power. The value of the power is called the
order of the reaction for each reactant. For example, if x
and y were found to each equal 1, then the rate law for the
reaction would be: rate = k[A][B]. Both A and B would be
first order reactants. The overall rate order of the reaction
would be 2 (the sum of the reactant orders).
The order of a reaction can only be determined by experiment. A balanced equation for a reaction gives no clues to
the rate law for the reaction.
The initial rate method will be used to determine the order
of each reactant (the values of x, y, and z). In this method,
the reaction is clocked as it proceeds for only a small portion of the total time it would take to reach completion.
You will repeatedly time the reaction during the first 5%
or so of the reactants being used up. Each time you run the
reaction, you will double one, then another, then the third
reactant’s concentration compared to the original set of
concentrations.
1
Given the quantity of S2O32– used (0.000200M), equation
1 on page 1 becomes:
3.33x10 −5 M
−
= k[I − ]x [BrO 3 ]y [H + ]z
∆t
(eqn 2)
The change in the concentration of S2O32– is a minus value,
since it goes down from a 0.000200M to 0M. That is why
the 3.33 x 10-5 is a positive rather than a negative value.
If you divide the rate of the reaction with the original set of
concentrations into the rate of the reaction with the concentration of one of the reactants doubled, you can calculate
how the change in concentration of that one reactant affected
the rate of the reaction. That is to say, you can determine
the order for that one reactant. If, for example, you ran the
reaction twice, and everything was the same in the two
runs except that you doubled the concentration of I–, and
the rate doubled after you doubled the [I–], then you could
conclude that the value of x in equation 1 is 1. If the rate
quadrupled after you doubled the [I–], then the value of x
is 2. And if the rate went up eight-fold, the value of x is
3 (21 = 2, 22 = 4, 23 = 8). Orders do not have to be whole
numbers, although in the reaction you are running, they are
small, whole numbers.
The first run you do will be compared to each of the next
three runs, so take extra care on run one.
After the first four runs, you will do a fifth, but only after
calculating the values of x, y, and z, and then calculating the
average value of k for the reaction. The fifth run will test
your measuring and calculating abilities. For the quantities
of materials mixed, you are asked to predict the time the
reaction should take. The closeness of your prediction and
the actual time of reaction may result in bonus points added
to this lab grade, depending on the instructor.
The timing method you will use is based on the properties of one of the products of the reaction, the I2 molecule.
Two reactions that the I2 molecule can undergo in this
experiment are:
The experiment will be run at room temperature, which
we will assume stays constant. Record the temperature at
the beginning and at the end of the experiment to see how
good an assumption this is.
A fast reaction:
1) I2(aq) + starch → deep blue complex;
And a very fast reaction:
2) I2(aq) + 2S2O32–(aq) → 2I–(aq) + S4O62–(aq)
If there is any S2O32– (thiosulfate ion) present, I2 will react
with it rather than react with starch. Every time you run the
reaction in this experiment, you will add some starch and
enough Na2S2O3 solution to give a [S2O32–] of 0.000200M.
The presence of starch and of S2O32– does not interfere
with any of the reactants used, only with the I2.
The rate of the chemical reaction you are studying could be
given in any of the following equivalent ways, by measuring
either the rate of disappearance of reactants or the rate of
appearance of products:
−∆[I − ] −∆[BrO−3 ] −∆[H + ]
rate =
=
=
6∆t
∆t
6∆t
or
∆[I 2 ] ∆[Br − ] ∆[H 2 O]
rate =
=
=
3∆t
∆t
3∆t
Reaction rate experiments depend upon the ability to measure
concentrations. The measuring technique employed in this
experiment depends on iodine, so:
∆[I 2 ]
rate =
3∆t .
Because the iodine will be reacting with S2O32– ions:
− ∆[S 2O 2−
3 ]
rate =
6∆t
2
experiment
Work in pairs.
Supplies:
• 5-10 ml graduated cylinders (3 from cart)
• 1 stopwatch (from cart)
• 5 transfer pipets
• 1-125ml, 1-250 ml flask
• 1 thermometer
• 80 ml of 0.0100M KI,
• 70 ml of 0.00100M Na2S2O3,
• 80 ml of 0.0400M KBrO3
• 85 ml of 0.100M HCl
(Use the smaller beakers from the lab drawers. Estimate volumes from the markings on the beakers. Do
not waste reagents.)
General mixing information: You will run the reaction
five times, four at the beginning of the experiment and one
after calculations.
You will use 10.0 ml of the Na2S2O3 solution for each
run.
Use the quantities of each of the other reagents given in
the table below. Take care in measuring the volumes. After
pouring each solution into the graduated cylinder, use the
transfer pipet to adjust the bottom of the meniscus right to
the line.
Place the KI, the Na2S2O3, and any water into the 250 ml
flask. Place the KBrO3, the HCl, and 3 to 4 drops of starch
solution into the 125 ml flask.
Bring the reagents to your work area in clean, labeled
containers. Put about 50 ml of deionized water in another
beaker. One of the graduated cylinders and one of the
transfer pipets will be used for each reagent. Label these
as well. Make sure they are clean and dry to start. If necessary, rinse the cylinder and pipet with a small amount of
its reagent. The amount of reagent the directions give is
enough for an extra run, if necessary, as well as a few extra
milliliters for rinsing.
Pour the contents of the 125 ml flask into the 250 ml flask,
starting the stopwatch at the moment of transfer. Swirl the
mixture for 10 seconds or so. Set the flask down on a piece
of white paper and watch for the first appearance of a blue
color. When the color first appears, stop the stopwatch
and record the time. Pour the mix into the sink. Rinse the
flasks well with deionized water, allow them to drip-drain,
then reload them for the next run. Do not dry the inside of
Initial Room Temperature_____
R eaction
Mixture
250 ml flask contains 10.0 ml Na2S 2O 3, plus:
0.0100M K I (ml)
H 2O (ml)
125 ml flask ( + 3-4 drops starch)
0.0400M K B rO 3 (ml)
0.100M H C l (ml)
1
10.0
10.0
10.0
10.0
2
20.0
0
10.0
10.0
3
10.0
0
20.0
10.0
4
10.0
0
10.0
20.0
10.0
6.0
15.0
and after calculations are finished,
5
9.0
Room Temperature at End of Experiment______
Each mix has a total volume of 50.0 ml (including the Na2S2O3). This is purposely done to ease the calculations. When
the reagents are poured together, each one dilutes the other. The concentrations given at the head of each column are the
pre-mix concentrations. You will use the post-mix concentrations in the calculations. To calculate concentrations after a
dilution, use the following equation:
original volume
final conc = original conc x
final volume
In the table on the following page, record the time (in seconds) for each reaction and calculate the concentrations of each
substance in each mix. Also, calculate the rate for each reaction, by dividing the time into 3.33 x 10–5M (see equation 2 on
the previous page). Write out the rates using scientific notation. If you write each rate with the same exponent, it will be
easier to visualize what is happening. That is, if one rate is 3.50 x 10‑7, and some other one shows up on your calculator as
1.45 x 10-6, re-write the second number as 14.5 x 10-7, so the two numbers have the same “index”. It’s easier to notice the
1:4 ratio of these numbers in this form.
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data and calculations
-5
(3.33 x 10 M /∆t)
R eaction M ix
C on cen tr ation s after M ixin g
–
R ate (M · sec-1 )
T ime (∆t) (sec)
–
[I ]
+
[B r O 3 ]
[H ]
1
M
M
M
2
M
M
M
3
M
M
M
4
M
M
M
M
M
M
F or the next m ix, just calculate the concentrations. Do not m ix until later.
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3.33 x 10 −5 M
−
= k[I − ]x [BrO 3 ]y [H + ] z
∆t
Equation 2 on page 34 gives:
. You will use the data in runs 1 and 2 to calculate
x, the data in runs 1 and 3 to calculate y, and the data in runs 1 and 4 to calculate z.
rate =
Calculation of x: Write out equation 2 using the numbers in the above data table for runs 2 and 1:
rate
[I–]x
[BrO3-]y
[H+]z
=
k
2)
________ =
k
________ ________ ________
1)
________ =
k
________ ________ ________
Calculation
Here is how to do it:
Divide, term by term, the data from run 1 into the data from run 2. Notice how all the units cancel, how the k’s
cancel, and how the [BrO3–] and [H+] terms cancel.
x
 0.00400M 
The [I–] terms should look like this:  0.00200M  , which works out to (2.00)x.
Example: Suppose the value obtained when rate 1 is divided into rate 2 is 2.25. You would have the following
x
equation to solve: 2. 25 = ( 2. 00 ) .This can be rewritten as: log 2.25 = x log 2.00 . To solve, you would take:
log 2. 25
x=
log 2.00 , which gives x = 1.17. This reaction has whole number orders, so you would round x to 1.
Your value for x: _____, rounded = __
Calculation of y: Write out equation 2 using the numbers in the above data table for runs 3 and 1:
rate
[I–]x
[BrO3-]y
[H+]z
=
k
3)
________ =
k
________ ________ ________
1)
________ =
k
________ ________ ________
Calculation
Your value for y: _____, rounded = __
Calculation of z: Write out equation 2 using the numbers in the above data table for runs 4 and 1:
rate
[I–]x
[BrO3-]y
[H+]z
=
k
4)
________ =
k
________ ________ ________
1)
________ =
k
________ ________ ________
4
Calculation
Your value for z: _____, rounded = __
Calculation of the value of k: Use the values of x, y, z, and the concentrations and rates from the table on page 36 to
calculate the value of k for each run.
rate
k= − x
− x
− y
+ z
[I ] [BrO − ] y [H + ] z . Remember to substitute the actual rounded‑off values of x, y, and
rate = k[I ] [BrO ] [H ] , so
z. Show your setup for each calculation, then give the value for each k. Also calculate the average value of k. If one value
of k is out of line with the other three, check with the instructor.
R eaction 1 setup:
R eaction 2 setup:
k1 =
k2 =
R eaction 3 setup:
R eaction 4 setup:
k3 =
k4 =
Average value of k =
L 3 ·m ol -3 ·sec-1
Run number 5: Now, finally, you can run the reaction with mix number 5. First, however, calculate how many seconds
− x
− y
+ z
the reaction should take. Use rate = k[I ] [BrO ] [H ] . Use the average value of k and the concentrations for the reactants in mix 5 from the table on page 36. Apply the values of x, y, and z correctly. Solve the equation for rate. Since
3.33 x 10 −5 M
rate =
∆t
, you can now solve for ∆t.
Show the instructor the calculated value for ∆t before you begin. Call the instructor to your place to observe the color
change and the time of the reaction. If the reaction takes place within 5 seconds of your predicted time, you may receive
bonus points.
Calculation:
Predicted value for ∆t
__________sec
Observed value for ∆t__________sec
Instructor’s comment_______________________________________________________________________________ Care in all measurements is important in this experiment. However, the uncertainties in measuring require a bit of good
luck to come within one or two seconds of the predicted time. So if you are a good experimenter, but you missed out on
any bonus points in this experiment, it might be disappointing, but it is not unexpected.
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Name_________________________________________ Grade___________ Date ___________
Questions
1. How much did the temperature change during the course of the experiment? Increasing temperature speeds up a
reaction. Decreasing temperature slows down a reaction. If there were a temperature change of over 2 degrees, try to
interpret if variations in your data are consistent with the temperature change.
2. On page 33, three lines from the bottom of the left-hand column, the rate law for the generic reaction could be verbally described as: The reaction is first order in substance A, first order in substance B, and second order overall. Use
this style to describe the rate law for the reaction you used in this experiment.
3. If the order of a reaction is the same as its stoichiometry, the order of each reactant is the same number as the coefficient of that reactant in the balanced equation. When this is the case, it is possible (but not necessary) that the reaction takes place in one step. If the order of a reaction is different than its stoichiometry, then the reaction must take
place in steps. For the reaction used in this experiment, show how the rate law (equation 1 on page 33) would look if
the reaction took place in one step:
4. Since the observed rate law is different than the one in question 3, the reaction must take place in steps. Remember, particles are in constant motion, and at room temperature, they reach fairly high velocities. Think of a series
of collisions, each one resulting in a partial change for the colliding species. For example, the first step might be
the collision of a BrO3– with an I– to form BrO2– and IO‑. The reaction in each step of a series of steps is called an
elementary process. Since neither BrO2– nor IO– show up as final products in the reaction, they would be called
intermediate products. They must be consumed in subsequent steps in the series of reactions that make up the reaction mechanism. Working out reaction mechanisms is an important part of chemical kinetics. Working out reaction
mechanisms in biochemical processes is at the heart of modern medical research. Studies on AIDs and other diseases
operate using this technique. They look to see what molecules a disease organism uses as it attacks a host, and how
these molecules fit in with the host molecules. If a researcher discovers all the molecular steps involved in a disease,
what do you think her next step is?
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