Allan Paul M. Villa
September 12, 2019
CHEM 18.1 24-L
September 19, 2019
Group 4
CHEMICAL KINETICS: DETERMINATION OF THE REACTION ORDER
I.
Introduction
The branch of chemistry that concerns with reaction rates is called Chemical Kinetics. It
plays a vital role in chemical reactions since it does not only determine the factors that affect the rates
and the reaction mechanisms by which reactions happen but also shows whether the reaction will
proceed. It also displays quantitative data which determines the amount of energy needed and how
rapid reactions will take place which is referred as the reaction rate [1]. This reaction rate is the
amount of a chemical change occurring per unit of time and depends on the reactants, the
concentrations of the reactants, the temperature at which the reaction takes place, and any catalysts or
inhibitors that affect the reaction [2].
Reaction rate increases as the concentrations of the reactants are increased. Generally, it also
increases rapidly along with increasing the temperature. Same thing happens for reactions that occur
on a surface rather than in solution, the rate increases as the surface area is increased. While catalysts
speed up reactions and inhibitors slow them down [3].
In this experiment the reaction rates were observed as samples were subjected to variation of
concentrations of reactants.
In order to examine this, this experiment determined the order of the reaction of acetone with
iodine in an acidic medium and the order was obtained through the use of Method of Initial Rates
which is a process that involved a series of experiments in which the initial concentration of one of
the reactants was held constant and others varied in convenient multiples in order to determine the
rate law for the reaction. Meanwhile, the order of the reaction dictated how a reactant affected the
speed of conversion of reactants to products
O
C
H3
C
I-I
CH
3
O
HCl
HI
C
H3
C
CH2
I
In determining the rate law, numerous steps were done. The experiment was performed in a
systematic manner and information were gathered and presented in tabular forms.
The concentrations of the reactants in reaction mixtures were calculated first followed by the
determination of the reaction rates.
These reaction rates were summarized into rate laws which described the relation between the
velocity of a reaction and the concentration of chemical reactants.
For the general reaction aA+bB→C with no intermediate steps in its reaction mechanism,
meaning that it is an elementary reaction, the rate law is given by:
Rate = k [A]x [B]y [C]z
In this equation [A], [B] and [C] expressed the concentrations of acetone, iodine and
hydrochloric acid, respectively, in units of moles per liter. The exponents x, y and z are the individual
orders, and k is the rate constant [4]. Using the same formula, the ratio of rate law was determined to
calculate the values of x, y and z or the orders of the reaction with respect to the reagents which are
considered to be the foundation of the rate law expression [5]. After obtaining the values, these were
plugged in all the trials to get the rate constant. These rates constants were then averaged and the
calculated value together with the calculated orders was used in the writing of the final rate law of the
reaction.
Similar study (Iodination reaction) began in 19th century when it was noticed that the color
of iodine slowly fades when the solution also contains acetone. Since then, reactions involving a
reactant containing a central carbonyl group have become some of the popular organic chemical
reactions. The iodination of acetone is also catalyzed by hydrogen ions. The effects of varying the
concentrations of acetone, iodine and hydrogen ions have been studied earlier and it has been found
that the reaction is zero order with respect to iodine. [6]
I.
Materials
A. Reagents
4 M Acetone
0.0012 M I2
1 M HCl
starch solution
B. Apparatus and Equipment
Test Tube Rack
5 Test Tubes
5 125-mL Erlenmeyer Flasks
10-mL Graduated Cylinder
Rectangular Container
II.
Procedure
To show the dependence of the rate of reaction upon the concentration of reactants, certain
amounts of HCl, I2, starch and water with a final volume of 25mL (as summarized in Table 1) were
placed in the Erlenmeyer flasks. Each flask was labelled trial 1 to 5 in order to avoid confusion and
errors as the experiment wore on. Subsequently, needed volume of acetone were put in the test tubes
and were labelled trial 1 to 5 too. After this, the flasks and test tubes were soaked in the water bath
and allowed the temperature to equilibrate for 2 minutes.
Trial
1
2
3
4
5
4 M Acetone
2.50
5.00
2.50
2.50
3.75
0.0012 M I2
2.50
2.50
2.50
5.00
3.75
Volume, mL
1 M HCl
2.50
2.50
5.00
2.50
3.75
Starch Solution
2.50
2.50
2.50
2.50
1.25
H 2O
15
12.50
12.50
12.50
12.50
Table 1. Volume of Reagents
After stabilizing the temperature of the samples, the test tubes and flasks were removed in the
water bath and the content of the test tubes were poured into the respective Erlenmeyer flasks. The
timer was then started as soon as the acetone was transferred to the flasks. The mixtures were swirled
in a constant direction and in the same manner until the dark blue color no longer persisted. To ensure
this, the mixture was observed in a white background. The time that the color disappeared in each
flask and became clear was recorded on Table 2.1.
Finally the group cleaned up the equipment and solutions used in the experiment. The
solutions were disposed of in the waste container labelled E501. Then the glasswares were cleaned in
the sinks.
III.
Data/Observations
After performing the experiment, the following sets of information were gathered.
Run
1
2
3
4
5
[I2] in mixture,
M
1.2x10-4
1.2x10-4
1.2x10-4
2.4x10-4
1.8x10-4
Time
started, s
0
0
0
0
0
Time finished,
s
327
305
315
1048
180.05
Time
elapsed, s
327
305
315
1048
180.05
-d[I2]/dt, M/s
3.7x10-7
3.9x10-7
3.8x10-7
2.3x10-7
1.0x10-6
Table 2.1. Determination of the Various Reaction Rates
Table 2.1 shows the initial concentration of Iodine in each trial calculated based upon the
stock concentration and the volumes listed in Table 1 under the formula C 1V1=C2V2. Where C1 is the
Molarity of the given substance, V1 is the volume used of that substance, C 2 is the total Molarity used
and V2 is the total volume used. The reaction time were recorded on this table in seconds and the
reaction rates were calculated as the change in concentration of iodine over reaction time or through
the equation, Rate of Reaction =
Run
1
2
3
4
5
[C3H6O]
0.40
0.80
0.40
0.40
0.60
−∆[ I ]2
.
∆t
Concentration in the Mixture, M
[I2]
1.2x10-4
1.2x10-4
1.2x10-4
2.4x10-4
1.8x10-4
[HCl]
0.10
0.10
0.20
0.10
0.15
Reaction Rate,
M/s
3.7x10-7
3.9x10-7
3.8x10-7
2.3x10-7
1.0x10-6
Table 2.2. Determination of the Reaction Order
While Table 2.2 shows the initial concentration of the other reagents (acetone and HCl) per
run calculated based upon the values in Table 1 under the same formula C 1V1=C2V2 used in finding
the initial concentration of iodine. The reaction rate can also be seen on this table.
Reactant
Reaction Order with Respect
to Reactant
I2
C3H6O
HCl
-1
0
0
Effect of Doubling
Concentration to the reaction
time (increase/decrease/no
effect)
Increase
Decrease
Decrease
Lastly, Table 3 shows the reaction orders with respect to the reactants which were obtained by
finding the values of x, y and z through utilizing the rate law equation Rate = k [A]x [B]y [C]z. In
order to get these values, ratio of the rate laws for each trial where in specific reactants were set
constant was used.
To find the order of Iodine (y), Trials 1 and 4 were treated. Following the formula
r1
r4
=
k [ A]1x [B] 1y [C ]1z
, where r1 and r4 is the reaction rate on runs 1 and 4, respectively; [A],
k [ A ]4x [B] 4y [C ]z4
[B] and [C] is the molarity of acetone, iodine and HCl, respectively the obtained value is -0.69,
rounding this off gives us a value of -1.
On the other hand, using the same treatment, the calculated order of acetone (x) using trials 1
and 2 is 0.076 or 0 while the order of HCl (z) using trials 1 and 3 is 0.038 or 0 when rounded in the
nearest whole number.
Also, the effect of doubling the concentration of either of the reactants to the reaction time is
transparent on this table. In trial 2, the amount of acetone was doubled (see table 1) and it can be seen
in table 2.1 that the reaction time decreased. The same manner happened to HCl when its volume was
doubled in trial 4 (see table 1). Meanwhile, when the volume of iodine was doubled, there was an
increase in reaction time.
After x, y and z had been determined, the rate constant or k was calculated. This was done by
using each reaction and determining the individual k, then averaging all of them to determine the
overall rate constant. Using the equation Rate = k [A]x [B]y [C]z, k is determined from a simple
algebraic equation of dividing the rate by the concentration of A, B and C all taken to its order.
Simply written, it is k =
R
. Upon plugging the values, this gave a rate constant of
[ A ] [ B]y [C] z
x
8.9x10-10 M-1 sec-1 on runs 1 to 4 and 3.0x10-9 M-1 sec-1 on run 5. Hence, the computed average rate
constant is 1.3x10-9 M-1 sec-1.
IV.
Discussion
Based on the results, when we qualitatively compare the standard run (run 1) to the doubled
acetone run (run 2) the rate of the reaction barely changed; therefore we can infer that the reaction is
zero order with respect to acetone. Same thing can be drawn when we compare the standard run to the
doubled acid run (run 3). There was no change in reaction rate at all, thus, the reaction is zero order
with respect to HCl. Similar conclusions can be inferred using the calculated value (0.08 and 0.04 or
roughly 0) of each ratio of the runs. While, upon comparing the standard run and the doubled iodine
run (run 4) the reaction rate decreased and quantitatively it gave us -0.68 or -1 value.
With these values and the aforementioned average rate constant, we can now write the final
rate law as:
Rate = 1.3x10-9 M-1 sec-1 [C3H6O]0[I2]-1[HCl]0.
V.
Conclusion
A series of steps is required to determine the rate law of a chemical reaction. By varying the
concentrations of reactants in the iodination of acetone it was determined that the reaction is -1 order
with respect to the iodine concentration and zero order with respect to both acetone and hydrochloric
acid concentration. Through the reaction rate we can also determine which reaction is the fastest and
slowest. In this experiment the 5 th run is the fastest and the 4 th run is the slowest in accordance with
varying concentrations of the reactants.
Through analyzing the calculated reaction order the reaction rates were independent of that of
the concentrations of acetone and hydrochloric acid or the rates did not changed even if the
concentrations were doubled. Consequently, the negative 1 order value with respect to iodine
indicates that the iodine concentration is inversely proportional to the rate of the reaction.
Unfortunately, these obtained results did not matched up with the expected ones. The reaction
should have been 1st order with respect to acetone and HCl which means that these two should have
been directly proportional to the increase of their concentrations. Meaning, as their concentrations
were doubled, the reaction rate should be doubled too. On the other hand the reaction should have
been zero order with respect to iodine. There should have been no relative change in the reaction rate
when the iodine concentration was doubled [6].
With this, we can say that there could be errors committed while performing the experiment.
The experiment done might have been subjected to different factors. There could have been wrong
measurements of the reactants and/or there was a relative difference while swirling the mixture.
Impurities in the reactants could also be one factor that affected the results as well as the temperature.
The allotted 2 minutes could have not been enough to well-equilibrate the samples and cancel this
variable on the determination of the reaction rates.
VI.
Literature Cited/Bibliography
[1] Retrieved June 6, 2019 (September 17, 2019) from the LibreTexts: http://www.libretexts.org
[2] Retrieved February 2, 2010 (September 17,2019): cms.montgomerycollege.edu
[3] HILL JW, PETRUCCI RH, MC CREARY TW, PERRY SS. 2010. Rates and Mechanisms of
Chemical Reactions. In: General Chemistry Fourth Edition. USA: Pearson Education Ed. Chapter 13.
[4] Retrieved April 1, 2014 (September 17, 2019) from the Odinity: www.odinity.com
[5] Retrieved February 2, 2010 (September 17, 2019) from the MNState: web.mnstate.edu
[6] SHUKLA ST, KULKARNI S. 2015. Kinetics of iodination of acetone, catalyzed by HCl and
H2SO4. India. 7(8):226-229
Sample Calculations
Calculating the initial concentration of I2 in the mixture
Runs 1, 2 & 3
I
¿
volume of I 2 ¿ stock
¿
[I2]initial = volume of solution¿
=
total
¿
¿
¿
¿
(0.0012 M )(2.50 mL)
25 mL
= 1.2x10-4M
(0.0012 M )(5.00 mL)
25 mL
= 2.4x10-4M
(0.0012 M )(3.75 mL)
25 mL
= 1.8x10-4M
Run 4
I
¿
volume of I 2 ¿ stock
¿
[I2]initial = volume of solution¿
=
total
¿
¿
¿
¿
Run 5
I
¿
volume of I 2 ¿ stock
¿
[I2]initial = volume of solution¿
=
total
¿
¿
¿
¿
Calculating the initial concentration of C3H6O in the mixture
Runs 1, 3 & 4
volume of acetone ¿stock
¿
volume of solution ¿total
[acetone]initial =
=
¿
acetone¿ stock ¿
¿
¿
Run 2
(4 M )(2.50 mL)
25 mL
= 0.40M
volume of acetone ¿stock
¿
volume of solution ¿total
[acetone]initial =
=
¿
acetone¿ stock ¿
¿
¿
(4 M )(5.00 mL)
25 mL
= 0.80M
(4 M )(3.75 mL)
25 mL
= 0.60M
Run 5
volu me of acetone ¿stock
¿
volume of solution¿total
[acetone]initial =
¿
acetone¿ stock ¿
¿
¿
=
Calculating the initial concentration of HCl in the mixture
Runs 1, 2 & 4
volume of HCl ¿ stock
¿
volume of solution ¿ total
[HCl]initial =
=
¿
HCl ¿stock ¿
¿
¿
(1 M )(2.50 mL)
25 mL
= 0.10M
(1 M )(5.00 m L)
25 mL
= 0.20M
(1 M )(3.75 mL)
25 mL
= 0.15M
Run 3
volume of HCl ¿ stock
¿
volume of solution ¿ total
[HCl]initial =
=
¿
HCl ¿stock ¿
¿
¿
Run 5
volume of HCl ¿ stock
¿
volume of solution ¿ total
[HCl]initial =
=
¿
HCl ¿stock ¿
¿
¿
Calculating the rate of reaction
Run 1
Rate of Reaction = –
I
[¿¿ 2]
∆
∆t
¿
= –
I 2 ¿initial
¿
I 2 ¿ final−¿
¿
¿
=
0 M −1.2 x 1 0−4 M
327 sec
= –
I 2 ¿initial
¿
I 2 ¿ final−¿
¿
¿
=
0 M −1.2 x 1 0
305 sec
= –
I 2 ¿initial
¿
I 2 ¿ final−¿
¿
¿
=
0 M −1.2 x 1 0
315 sec
= –
I 2 ¿initial
¿
I 2 ¿ final−¿
¿
¿
=
0 M −2.4 x 1 0− 4 M
1048 sec
= –
I 2 ¿initial
¿
I 2 ¿ final −¿
¿
¿
=
0 M −1.8 x 10 M
180.05 sec
= 3.7x10-7
M
sec
Run 2
Rate of Reaction = –
I
[¿¿ 2]
∆
∆t
¿
−4
M
= 3.9x10-7
M
sec
Run 3
Rate of Reaction = –
I
[¿¿ 2]
∆
∆t
¿
−4
M
= 3.8x10-7
M
sec
Run 4
Rate of Reaction = –
I
[¿¿ 2]
∆
∆t
¿
= 2.3x10-7
M
sec
Run 5
Rate of Reaction = –
M
sec
I
[¿¿ 2]
∆
∆t
¿
−4
= 1.0x10-6
Comparing run 1 (standard run) and run 4 (doubled iodine)
x
r1
r4
y
=
M
sec
M
2.3 x 10−7
sec
=
=
M
sec
M
3.9 x 10−7
sec
=
=
M
sec
M
3.8 x 10−7
sec
=
−7
3.7 x 10
z
k [ A]1 [B] 1 [C ]1
x
y
z
k [ A ]4 [B] 4 [C ]4
=
k [0.40 M ]1x [1.2 x 1 0−4 M ]1y [0.10 M ]1z
k [0.40 M ]x4 [2.4 x 1 0−4 M ]4y [0.10 M ]4z
M
sec
ln
M
2.3 x 1 0−7
sec
−4
y
[1.2 x 1 0 M ]
ln
[2.4 x 1 0−4 M ]❑y
3.7 x 1 0
−7
=
y = – 0.69
Comparing run 1 (standard run) and run 2 (doubled acetone)
r1
r2
k [ A ]1x [B]1y [C ]z1
k [ A ]2x [B]2y [C ]z2
=
x
−4
y
3.7 x 10−7
z
k [0.40 M ]1 [1.2 x 10 M ]1 [0.10 M ]1
x
−4
y
z
k [0.80 M ]2 [1.2 x 1 0 M ]2 [0.10 M ]2
M
sec
ln
M
3.9 x 1 0−7
sec
x
[ 0.40 M ]
ln
x
[0.80 M ]❑
3.7 x 1 0
−7
=
x = 0.076
Comparing run 1 (standard run) and run 3 (doubled HCL)
r1
r3
=
x
1
x
3
y
1
y
3
z
1
z
3
k [ A ] [B] [ C ]
k [ A ] [B] [ C ]
k [0.40 M ]x1 [1.2 x 10−4 M ]1y [0.10 M ]1z
k [0.40 M ]x3 [1.2 x 1 0−4 M ]3y [0.20 M ]3z
3.7 x 10−7
M
sec
ln
M
3.8 x 1 0−7
sec
[0.10 M ] z
ln
[0.20 M ]❑z
3.7 x 1 0−7
=
z = 0.038
Calculating the rate constant (k)
Run 1
Rate = k [C3H6O]0[I2]-1[HCl]0
Rate = k (0.40)0(1.2x10-4)-1(0.10)0
M
0.40 ¿
¿
0.10 M ¿ 0
k = 1.2 x 10−4 M ¿−1 ¿ =
¿
−7 M
3.7 x 10
sec
¿
8.9x10-10 M-1 sec-1
Run 2
Rate = k [C3H6O]0[I2]-1[HCl]0
Rate = k (0.40)0(1.2x10-4)-1(0.10)0
M
0.80 ¿
¿
0.10 M ¿ 0
k = 1.2 x 10−4 M ¿−1 ¿ =
¿
−7 M
3.9 x 1 0
sec
¿
8.9x10-10 M-1 sec-1
Run 3
Rate = k [C3H6O]0[I2]-1[HCl]0
Rate = k (0.40)0(1.2x10-4)-1(0.10)0
M
0.40 ¿
¿
0
0.20 M ¿
k = 1.2 x 10−4 M ¿−1 ¿ =
¿
M
3.8 x 10−7
sec
¿
8.9x10-10 M-1 sec-1
Run 4
Rate = k [C3H6O]0[I2]-1[HCl]0
Rate = k (0.40)0(1.2x10-4)-1(0.10)0
M
0.40¿
¿
0.10 M ¿0
k = 2.4 x 1 0−4 M ¿−1 ¿ =
¿
−7 M
2.3 x 1 0
sec
¿
8.9x10-10 M-1 sec-1
Run 5
Rate = k [C3H6O]0[I2]-1[HCl]0
Rate = k (0.40)0(1.2x10-4)-1(0.10)0
M
0.60 ¿
¿
0.15 M ¿ 0
k = 1.8 x 1 0−4 M ¿−1 ¿ =
¿
−6 M
1.0 x 10
sec
¿
3.0x10-9 M-1 sec-1
Calculating the average rate constant
kave =
∑k
n
−10
kave =
4( 8.9 x 1 0
−1
−1
−9
−1
−1
M se c )+3.0 x 10 M se c
5
kave = 1.3x10-9 M-1 sec-1