Determination of Rate Law Lab Report

Determination of a Rate Law Lab Report Essay
Determination of a Rate Law
Megan Gilleland
10.11.2012
Dr. Charles J. Horn
Abstract: This two part experiment is designed to determine the rate law of the following reaction,
2I–(aq) + H2O2(aq) + 2H+I2(aq) + 2H2O(L), and to then determine if a change in temperature has
an effect on that rate of this reaction. It was found that the reaction rate=k[I–]^1[H2O2+]^1, and the
experimental activation energy is 60.62 KJ/mol.
Introduction
The rate of a chemical reaction often depends on reactant concentrations, temperature, and if there's
presence of a catalyst. The rate of reaction for this experiment can be determined by analyzing the
amount of iodine (I2) formed. Two chemical reactions are useful to determining ... Show more
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Find the Ln of [I–]0
Ln(0.015)=–4.19970508
7.Find [H2O2]0
Take (0.10 M H2O2)*(6.00mL)/ ( final volume)=0.015 M
8. Ln of [H2O2]0
Ln(0.015)= –4.19970508
9. Find the Ln of rate:
Ln(2.13675x10–5)=–10.753638
10. The last step for week one calculations is to calculate the average value of k.
Rate= k [I–]1[H2O2]. (2.13675*10–5 ) = k [0.015] [0.015] then solve for k. For this trial,
k=0.09497.
This is then done for all trials. Then, once all five values of k are found, the average is taken by
adding all five values of k and dividing by 5. The experimental k average is 0.105894M/s.
Table 2: Calculations Week 1 | | | | | | | | | | | | solution# | mol s2O3–2 | mol I2 | I2 | (rate)
changeI2/change in temp | [I–]o | ln[I–]o | [H2O2]0 | ln[H2O2]o | ln rate | k | | 1 | 0.001 | 0.0005 |
0.0125 | 2.13675E–05 | 0.015 | –4.19970 | 0.015 | –4.19971 | –10.753 | 0.0949 | | 2 | 0.001 | 0.0005 |
0.0125 | 4.3554E–05 | 0.030 | –3.50655 | 0.015 | –4.19971 | –10.041 | 0.0967 | | 3 | 0.001 | 0.0005 |
0.0125 | 9.54198E–05 | 0.045 | –3.10109 | 0.015 | –4.19971 | –9.2572 | 0.1413 | | 4 | 0.001 | 0.0005 |
0.0125 | 0.000109649 | 0.045 | –3.10109 | 0.025 | –3.68888 | –9.1182 | 0.0974 | | 5 | 0.001 | 0.0005 |
0.0125 | 0.00015625 | 0.045 | –3.09776 | 0.035 | –3.35241 | –8.7640 | 0.0988 | | | | | | | | | | | k avg |
0.1059 | | | | | | | | | | | | | |

Data Week 2
Table 3:
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Relationship Between Surface Area and Rate Of Reaction Essay
An Experiment To Investigate The Relationship Between Surface Area and Rate Of Reaction
Introduction
The rate of reaction (reaction velocity) may be defined as the rate of change of concentration of a
stated reactant or product. The rate of a reaction is found by measuring the amount of a reactant
used up per unit of time or the amount of a product produced per unit of time. A reaction can be
made to go faster or slower by changing a number of factors. In order for a reaction to occur certain
things are necessary: particles must collide with each other and the collision must have enough
energy for the reaction to occur. If this happens the original bonds are broken and new bonds are
formed – so that new products are formed. Successful ... Show more content on Helpwriting.net ...
Also when a collision occurs they have more energy and so are more likely to be successful in
breaking and reforming bonds than when they have less energy and are moving more slowly.
Therefore the rate of the reaction will increase. As a rule of thumb when the temperature increases
by 10C, the rate of the reaction roughly doubles.
The concentration of the acid– The more concentrated the solution, the more particles of reactant are
present in a given volume and the greater the chance of a collision occurring. In dilute acids, with
fewer particles the chance of collisions occurring is reduced.
Therefore if concentration is increased, the reaction rate also increases. This also applies in reactions
between two gases when increasing the pressure has the same effect as increasing the concentration.
The presence of a catalyst – A catalyst increases the rate of a chemical reaction without itself being
used up. Catalysts are usually transition metals. In the presence of a catalyst, less energy is needed
by a collision in order to be successful. Therefore there are relatively more successful collisions and
so the reaction rate increases. I have decided to investigate how a change in surface area will affect
the rate of the reaction between calcium carbonate (marble chips) and hydrochloric acid. The word
equation for this reaction is:
Calcium + Hydrochloric –> Carbon + Water + Calcium
Carbonate Acid Dioxide Chloride
CaCO3(s) + 2HCL(aq) = CO2(G) + H2O(l) +

Kinetics
Section 1:
Kinetics is the study of the rate of chemical processes. The kinetics of the reaction between crystal
violet and NaOH was studied. In order to monitor crystal violet concentration as a function of time,
a spectroscopic colorimeter was used. What is the rate law for decolorization of crystal violet? In
order to figure this out, the rate of the reaction of crystal violet and sodium hydroxide must be
found. In this experiment, the initial goals were to determine the overall rate law for the rate of
decolorization of crystal violet in basic solutions as a function of time and to determine the rate law
for the reaction including the actual value of k; Rate = k[A]x[B]y. The rate of a reaction was
expected to depend on the concentrations ... Show more content on Helpwriting.net ...
Time) because it had a correlation closest to 1. All three orders were graphed and a linear regression
was used to see which graphed order was closest to 1. The order was determined by comparing the
concentration and time to the mathematical predictions made using the integrated rate laws.
Analyzing each graph and finding each correlation helped determine which graph was closest to 1.
The more concentrated a solution is, the higher the absorbance of that solution. This is due to Beer's
Law. The law measures the absorbance of a solution by determining how much light passes through
a solution. As the concentration of a solution increases, fewer wavelengths of light are able to pass
through the concentrated solution. The absorbance at 60 seconds was 0.573 (Figure 1: Table1). To
calculate the concentration (molarity), the Beer's Law equation was used, Abs = slope(m)+b.
Plugging in what is known into the Beer's Law equation resulted in 0.573 = 3.172e+004 + 0, where
the concentration is determined by M = 0.573–0/ 3.172e+004. So, the concentration at 60 seconds
using the equation (M = 0.573–0 / 3.172e+004) was 1.824e–5 M. The 1st order graph resulted in
k=0.006152 (Figure 1: Graph 1). Other groups also resulted in their decolorization of CV to be the
1st rate

Integrated Rate Law Lab Report
The integrated rate law equations all include the same variables:〖[A]〗_t, representing the
concentration of crystal violet at a given time, k, as the true rate constant of the reaction, t,
indicating the given time, and 〖⁡
[A]〗_o, the initial concentration of crystal violet.
By studying the rates of reactions, essentially expressed as how many molecules of reactants are
converted to products in a given time period, the determination of the rate law can found. In this
experiment, the reaction rate for the reaction, shown in eq. 4, was used to find rate law shown in eq.
5; with k being the rate constant, x being the order for crystal violet, and y being the order for
hydroxide.
[〖(CH〗_3 )_2 NC_6 H_4 ]_3 C^+ + OH^– → 〖(CH〗_3 )_2 NC_6 H_4 ]_3 ... Show more
A linear plot for equation 1 indicates a first order reaction with k, the true rate constant, equally the
negated slope of the graph. A mixed second order rate law reaction is when two components, A and
B, crystal violet and hydroxide, are reactants, making it difficult to determine the change in
concentration. Due to the complications of mixed second order rate law, the use of first order
condition, flooding, was used. Since the concentration of hydroxide was exceedingly larger than the
concentration of crystal violet, and does not change significantly throughout the experiment, the
concentration of hydroxide is made constant. An observed rate constant,k_obs, becomes equal to the
product of these constants, as shown in equation 8.
k_obs= k〖[OH]〗^– 1^1
The observed rate constant is the rate constant for a first order reaction that will be observed when
the hydroxide is used excessively in comparison to the crystal violet. Consequentially, the rate law
in equation in equation 9 is solely dependent on the concentration of one reactant, crystal violet,
since the observed rate constant is equal to the concentration of hydroxide (eq. 10).
Rate=k[C〖V]〗^1 [〖NaOH]〗^1

Rate Equation And Order Of Reactant
Rate equation and order of reactant:
The rate of reaction is defined as how fast the reactant is converted to the product and it is measured
in moldm–3s–1
Rate=(change in concentration(〖moldm〗^(–3)))/(time taken (s)) , the unit of rate = moldm–3s–1
Rate = k [A]x [B]y [C]y
Orders of the reaction must be taken in to account when writing a rate equation. Where A, B and C
are reactant, x, y and z are the orders of the reaction with respect to A, B and C.
In my investigation I will vary the concentration of each reactant and I will measure time taken for
the reaction to turn colourless. I will use rate=1/time formula to calculate the rate of the reaction.
This will enable me to draw rate against concentration graph which I will use to ... Show more
Collision theory:
Reactions only occur when particle of the reactant collide with a certain minimum kinetic energy.
This energy is also known as activation enthalpy this energy must be supplied to the reactant to
enable the bonds to stretch and break and the new bonds to form in the product. Most collisions do
not result in a reaction, only the collision with enough activation enthalpy will react to produce the
product. (1)(2) (10)
Concentration: The collision theory states that for any reaction to take place the two reactant
molecule must collide first. The rate of a reaction depends on the frequency of the collision between
the particles. Increasing the concentration of reactants increases the number of particles available to
react in the same volume of solution. More successful frequent collisions occur in a given time, with
sufficient activation energy to break the chemical bond. Therefore the rate of reaction will increase.
(1)(2)(6)
Surface area: Increasing the surface area of a solid reactant means more particles are exposed to the
other reactant, the number of reaction sites increase therefore it will react more quickly increasing
the reaction time.
Catalyst: A catalyst will increase the rate of reaction by lowering the activation enthalpy without
being chemically unchanged. They do this by providing an alternative pathway, forming an
intermediate compound with the reactant and then it forms the product from

Investigating The Rate Law And Activation Energy For 2hcl
The purpose of the experiment is to determine the rate law and activation energy for 2HCl(aq) +
Mg(s) → MgCl2(aq) + H2(g) as the form of magnesium used are magnesium strips. Inserting
magnesium strips to hydrochloric acid and evaluating the pressure of the reaction established the
rate law. The rate law was determined to be 4.77[Mg].767[HCl].864, which was established by the
data collected. By heating and cooling the hydrochloric acid and inserting the magnesium strip
allowed the evaluation of the pressure of the reaction and determined the activation energy. The
activation energy was determined to be 20 KJ/mol, indicating the reaction rate's sensitivity to
temperature. The reaction took longer due to magnesium strips surface area being larger than the
magnesium powder and magnesium shots. The experiments concluded that magnesium strips,
although surface area caused in increase in error, could produce accurate rate law and activation
energy in contrast to magnesium shots and magnesium powder.
Introduction
The purpose of the experiment is to evaluate the rate law result and the activation energy between
magnesium strips and hydrochloric acid. Many experiments and studies have been conducted in the
past to comprehend and use kinetics to evaluate the possibilities of understanding reactions. Kinetics
has the ability to keep human bodies functioning through reactions as well as contributing to the rate
of baking cookies to the rate of food spoilage in the bottom of

Lab Report On Kinetics Of Alcohol Oxidation
Lab Instructor's Name: Zhen Qiao
Student's Name: Nhu Duong
Section # CHEM 102 – 110
Experiment #6
Date of the experiment: 01/29/2016
Title: KINETICS OF ALCOHOL OXIDATION
Drexel University Winter 2016
Introduction:
This report concerns the experiment of determining the order of reaction and the kinetic rate
constant of alcohol oxidation. This experiment relates to the knowledge of chemical kinetics, the
application of Beer's Law, and other calculations.
Chemical kinetics involves the examination of reaction rates, which are the speeds of chemical
reactions. There are chemical reactions which proceed in long periods of time as well as chemical
reactions proceeding in short periods of time. Regarding reaction rates, the reaction order and
kinetic rate constant are considered.
Take a general reaction in solution for instance: aA+bB⇌Product(s) The reaction rate is calculated
using the following function: rate=k〖[A]〗^x 〖[B]〗^y
Where [A] and [B] are the concentrations of A and B in the solution, unit: mol/L or M; x and y is the
reaction order of A and B respectively; k is the kinetic rate constant, its unit depends on the order of
the reaction.
The values of x and y are the partial order corresponding to A and B. The sum of x and y is the
overall reaction order. The values of x and y can be either negative or positive. They can also be
either integer or fractional. The reaction rate can be understood as how fast the reactants are
consumed or how fast the

Hydrazone Reaction Lab Report
The lines in Figures 6 (A) to 6 (F) indicate predicted values, using fitted rate constants for the
reaction. The rate constant valueswere found to increase as expected as the temperature was
increased. The activation energies were determined from the temperature dependency of the rate
constants to be 831, 2494, 3076 and 2660 J/molfor hydrazone, tryptophol, cinnoline derivative, and
polyindole formation, respectively(Table 2). The values indicate the amount of energy required to
cross the energy barrier so that,the reaction can takes place. The activation energy for hydrazone
formation is low(Table 2) indicating that, the reaction takes place at lower temperatures
buttheactivation energy barrier is higherfor tryptophol formation. This indicate that the temperature
of the reaction mixture should be higher for the sigma–tropic rearrangement reaction, but further
increasing temperature also leads to side product formation such as polyindole and cinnoline
derivative, which are still higher activation energy reactions. Thus, by conducting the reaction at
higher temperatures allows the ... Show more content on Helpwriting.net ...
In the current work, the surface tovolume ratio was commissionedat1mm–1. As a result of the
enhanced surfacetovolume ratio, the temperature difference between the reaction mixture passing
through the reactor and reactor wall was insignificant.17Thisbetter heat transfer characteristicresults
in effective cyclization of hydrazone to tryptophol in shorter reaction time. The structures of indole
moiety present in many drug molecules that are widely synthesized in the pharmaceutical industry
are more complex, and one could expect a vast number of intermediates and corresponding reaction
products. However, the typical approachshouldremain similar and can beused in
developingproperkinetic models for other indoles as

Chemical Reaction Lab
The rate law of a chemical reaction is a very useful equation that relates the concentration of
reactants to time. The law was formed from the results of experiments by various European
chemists throughout the second half of the 1800s, when they caught on that there was a correlation
between the concentration of reactants and time. The most basic formula of the equation is
Rate=k[A]n[B]m..., where Rate is the rate of the reaction in concentration per second (M/s), k is a
constant, [A] and [B] are the concentrations of the reactants (M), and n and m are values that rate to
the order of each reactant. All the values in the equation can be determined experimentally by taking
the concentrations of reactants and the corresponding rate at multiple ... Show more content on
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The first solution step is to set the rate of one trial over another to solve for one of the exponents in
the rate formula. To do this trial one and two will be used: 4.69x10–4/9.38x10–
4=k[0.185]n[0.133]m/k[0.185]n[0.266]m. First cancel out k, then multiply both sides of the
equation by 2m to cancel out the entire right side. Next isolate 2m on the left to get 2m=2, which is
equivalent to log(2)/log(2) and equals 1.00. This is the value of m. Repeat the same process for
using trial one and two with m=1 and the result should be n=2.00. The values of m and n are the
orders, so the reaction is second order in A and first order in B. The rate law is now
Rate=k[A]2.00[B]1.00. To solve for the average value of k, the rate constant, isolate k by dividing
rate by [A]2.00[B]1.00 and plug in the values for each trial. From doing this one gets four answers
of k all around 0.103 (1/M2s). Add these four values and then divide by four to get the average of
k=0.103 (1/M2s). Three significant figures are always used because there were three in the data and
only multiplication and division were used. The final rate law is

Investigating the Kinetics of the Reaction Between Iodide...
PLANNING
Investigating the Kinetics of the reaction between Iodide ions and Peroxodisulphate (VI) ions
By the use of an Iodine clock reaction I hope to obtain the length of time taken for Iodine ions (in
potassium iodide) to react fully with Peroxodisulphate ions (in potassium Peroxodisulphate). I will
do three sets of experiments changing first the concentration of iodide ions, then the concentration
of Peroxodisulphate ions and finally the temperature of the solution in which the reaction is taking
place. From these results, I hope to draw conclusions as to the effects of these changes to the
environment of the reaction on the rate and also determine the order of the reaction and the
activation enthalpy.
Background information ... Show more content on Helpwriting.net ...
In terms of log to the base 10 this is:
log k = log A –Ea/ 2.303 RT
Reaction between Iodine ions and peroxodisulphate ions
S2O82–(aq) + 2I–(aq)  2SO42–(aq) + I2(aq)
In order to make the reaction clearer, during my experiment I will add starch and a small known
amount of sodium thiosulphate (to act as a queching agent). The thiosulphate ions turn iodine back
to iodine ions:
2S2O32–(aq) + I2(aq)  S4O62–(aq) + 2I–(aq)
Which means that no starch–iodine colour will appear until all the thiosulphate has been used up.
The amount of time taken for this occur (and the reaction to suddenly turn blue) is the same amount
of time for the reaction to produce the equvilant amount of Iodine.
Apparatus
(For making up solutions) weighing boats scales Beaker (150cm3)
3 Volumetric flasks (250cm3)

Distilled water
Glass rod
(for concentraion and temperature change experiments)
4 thermometers (0–110ºC)
A large number of boiling tubes (roughly 50 depending on repeats)
5 Burettes with funnels for filling
5 Clamp stands (for burrettes)
Stopwatch
(for temperature change only)
Two large beakers (400cm3)
Chemicals
Freshly made starch solution
Pottasium Iodide (made to solution with conc. 1.00 mol dm–3)
Pottasium peroxodisulphate (made to solution with conc.

Ethanol Oxidation Lab Report
Introduction: In this lab we will observe the rate of ethanol oxidation and determine the rate
constant that correspond to it. The rate law for this experiment can be written as: rate=k
〖[〖〖Cr〗_2 O_7〗^(–2)]〗^x 〖[C_2 H_5 OH]〗^y where x and y are both the partial order of
the reactant in a reaction and k is the kinetic rate constant. Since the concentration for ethanol, C_2
H_5 OH, is much higher than the concentration of K_2 〖Cr〗_2 O_7 , we can apply the treatment
of pseudo–order kinetic. The treatment said, when there is a reactant with a high concentration we
can remove it from the rate law because when the reactant with the low concentration is used up
there will be a very small change to the reactant with the high concentration that we can ignore it
like the following: rate=k' 〖[〖〖Cr〗_2 O_7〗^(–2)]〗^x In this experiment, to able to find k we
must first find the pseudo–order constant, k', using experimental data. The first step in finding the
pseudo–order constant is to find the wavelength that can be used to develop good data. This can be
done by using a colorimeter that four different wavelengths for us to test. We have to find the ...
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You will import the data into excel to help you treat the large amount of data. In one of the blank
column you will use formula similar to the one shown above which can help you find the
concentration from the absorbance. You will use also use excel to create a column that has ln [K_2
〖Cr〗_2 O_7] and one that has the formula is 1/K_2 〖Cr〗_2 O_7. When you have all your data
you will create three graphs. The graphs will be 〖time vs [K〗_2 〖Cr〗_2 O_7], 〖time vs
ln[K〗_2 〖Cr〗_2 O_7], 〖time vs 1/[K〗_2 〖Cr〗_2 O_7] just like Figure 2,3,4. You will use
linear regression and the R squared value for all three of the graphs to determine the rate order of
this reaction. You will analysis each graph and determine which

Imperature And The Rate Of Interaction In A Chemical Reaction
Chemistry IA
Title : The relationship between the temperature and the rate of reaction in a chemical reaction.
Introduction : Last academic year, towards the end, in IB chemistry hl class, my class had the
chance to learn about the concentration, temperature, catalysts, etc and their effects on the rate of
reaction in a chemical reaction. I knew that the rate of reaction had a relationship with the factors
but I wanted to know how it changes witness it for myself. I wanted to perform an experiment for
myself to see what it actually is like. This IA is potentially an opportunity for me to test my
understanding on the topic and
Research Question :
How does one determine the activation energy through the relationship between the temperature and
the rate of a said reaction?
Background Information : Rate of reaction in its most basic definition is the speed at which the
reaction takes place. The rate of reaction can be calculated by using
Rate of Reaction = Molestime of reaction
Collision Theory : In order for atoms/molecules/ions to react, they must collide with each other, the
rate of reaction is directly proportional to the collision frequency. The greater the number of
collisions, the faster the reaction will take place.
Activation Energy : For these collisions to take place and the reactions to happen, a minimum
amount of kinetic energy must be possessed by the particles that are going to collide with each
other. This minimum amount of energy required to initiate a

The Importance Of Chemical Kinetics
Chemical kinetics studies and determines the rate or speed at which chemical or physical processes
occur (Oliver,n.d.)(Jircitano, n.d.). The rate of reaction is the speed at which the reactant in a
reaction transforms into products or the change in concentration of a chemical species over the time
taken for that change to occur (Oliver,n.d.)(Jircitano, n.d.)(Mack, n.d.)(Blackburn,n.d.). Chemical
reactions occur at many different rates and in aqueous or equilibrium systems this rate is dependent
on the variables such as the reactivity of reagents, initial concentrations, temperature induced
fluctuations and any means of catalysis. (Oliver,n.d.)(Jircitano, n.d.)(Blackburn,n.d.). In order to
measure rate the change in concentration in a particular reaction must be determined as well as the
time of which this occurred. However studying the concentration at any time during a reaction is
problematic since the reaction must be stopped and a sample must be removed both which
negatively affect accuracy. One class of reactions that enables the change in concentration to be
observed at a particular time is clock reactions. Ina clock reaction an initial induction phase
precedes a significant change in concentration of a particular species. (Preece, King, Billingham,
1999). The rapid increase in concentration results in significant effects such as dramatic color
change. There are two types of clock reaction: 1. Induction where as small concentration grows till
it results in

In this laboratory the objectives were to understand the effects that temperature has on a reaction, be
able to calculate the reaction's rate constant, and to calculate the activations energy from the
recorded rate constant. Activation energy is the minimal amount of energy that is necessary to result
in a chemical reaction. With an insufficient amount of energy, a reaction won't occur, too much and
it will react just fine. A great example of this can be lighting a match. The friction created when
striking the tip covered in Potassium Chlorate against the match pad is enough energy for it to
ignite. Although, if there isn't enough kinetic energy from the swipe, than it won't ignite. The
activation energy in this lab was derived from the equation found from graphing the ln(k) and 1/T:
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The rate law is meant to display the relationship between the reaction rate and the concentration of
the various materials being used. Below is the equation that the was derived from in the reaction and
the rate law:
Balanced Equation: 6I–(aq) + BrO3–(aq) +6H+(aq) → 3I2 (aq) + Br–(aq) +3H2O
Rate Law: rate = k [I–][BrO3–][H+]2 The Arrhenius equation is used to graph the value found into a
relationship in order to determine the activation energy of the reaction. This equation can be easily
related to that of finding the slope of a line. The equation is given below and each components
definition: ln (k) = –(Ea/R) (1/T) + ln (A) y = mx + b k = Rate constant
T = Absolute temperature

Kinetics of Hydrogen Peroxide
February 22, 2007
Chem. 1130
TA: Ms. Babcock
Room 1830 Chemistry Annex
PURPOSE OF THE EXPERIMENT
The major purpose of this experiment is to determine the rate law constant for the reaction of
hydrogen peroxide and potassium iodide. In this experiment, the goal will be to try to measure the
rate law constant at low acidity, since at low acidity, anything less than 1.0 x 10–3M, the effect of
the hydrogen ion is negligible. To calculate the rate, the experiment will have to utilize the rate
equation, which is expressed as Rate = k[H2O2]a[I–]b. At low acidity, the rate of the Hydrogen ion
will not change, from our equation:
H2O2 (aq) + 2I– (aq) + 2H+ (aq) → I2 (aq) + 2H2O ... Show more content on
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Once the timer starts, both group members are to watch for any reaction that occurs. The reaction
should yield a slight change in color. The appearing color will be light blue. Once a reaction is
noticeable, the timer is to stop the stopwatch and the group will record the time down in the
following data table. If the color you observe is a dark blue, then you have passed the reaction point
and you must start the experiment over.
The following data is the data that was collected in the lab. If you wish to see the calculations,
please refer to the calculation Page that follows this section.
[H2O2]: 0.1 M
Table: Collected Data for Rate of Reaction of H2O2
Solution Number 1 2 3 4 5
Time of Color Change 2280 sec 1260 sec 800 sec 390 sec 185 sec
Start Time 0.0 sec 0.0 sec 0.0 sec 0.0 sec 0.0 sec
∆t in seconds 2280 sec 1260 sec 800 sec 390 sec 185 sec
Rate = ∆[ H2O2] / ∆t 2.193E–07 M/s 3.968E–07 M/s 6.250E–07 M/s 1.282E–06 M/s

2.703E–06 M/s
Initial [H2O2] 0.005 M 0.005 M 0.005 M 0.01 M 0.02 M
Initial [I–] 0.0045 M 0.0090 M 0.0150 M 0.0150 M 0.0150 M
The calculations performed in this lab included calculating the amount of sodium thiosulfate in each
of the solutions, calculating the [H2O2] used up during the reaction in each solution, calculating the
initial [H2O2] in each of the five solutions, and calculating the initial [I–] in

Rate Law Lab
The goal of this experiment was to determine the rate law for the reaction between magnesium
metal and hydrochloric acid. In order to find the overall rate law of this reaction, the total pressure
change of the reaction is recorded which allows the rate to be determined by evaluating the slope of
the line pressure versus time before the reaction reaches completion. This technique is using
pressure as a function of time. The equation that will be used to calculate the pressure of the
hydrogen gas that is produced in the reaction is the ideal gas law or PV=nRT. The reaction that is
used in this study between magnesium shot and hydrochloric acid is: Mg(s) + 2H+(aq) + Mg2+(aq)
+ H2(g). The theoretical rate law for this reaction that was expected ... Show more content on
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This means that the concentration of magnesium has no affect on the overall rate of the reaction.
This data also shows that the rate law of HCl is second order, meaning that if the concentration of
HCl were to be doubled, then the rate would increase by a factor of 4 (22) or if the concentration
were to be tripled, the rate would go up by a factor of 9 (32) and so on. This data was found to be
incorrect when comparing this rate law to the theoretical rate law: rate= k(surface area of
Mg)1[HCl]2
This inaccuracy can be contributed to the lack of time that was provided to complete the experiment
and carry out a sufficient number of trials. Because of this inaccuracy, these trials were repeated on
day 2, however with a different pressure probe and Erlenmeyer flask to ensure that this wasn't the
problem for our inaccurate data.
On day 2, the experimentally determined rate law for this reaction was found to be: rate= k(surface
area of

Reaction Kinetics Essay
Reaction Kinetics: Rate of Reaction
Of Tertiary–Butyl Bromide
Purpose: The purpose of this experiment is to find the order of t–BB graphically, to find the k (rate
constant) at 0&#730; C and at room temperature, also to find the Ea (activation energy).
Principles: Several different chemical kinetic principles were used in this experiment. The reaction
rates of this chemical equation were determined experimentally. This then allowed the reaction
mechanisms (i.e. orders of each component, rate constant, etc.). These mechanisms were ultimately
determined to be compiled to form a rate law.
Rate = k[A]m[B]n
Integrated rate laws are used to determine concentrations of reactants at certain times. However, ...
Data: Vo : 102mL Start time: 9:22:25 binf : 21.61mL
A Number Time to Colorless Total Sec bt (mL) (binf –bt) 1 9:25:05 160 2.00mL 19.61
2 9:28:04 339 4.00mL 17.61
3 9:35:34 789 6.00mL 15.61

The Effect Of Temperature And Reactant Concentration On...
This experiment aimed to investigate the relationship between temperature and reactant
concentration on the rate of reaction for the hydrolysis of tert–Butyl Chloride, and to determine the
validity of the proposed mechanism for the reaction. It was hypothesised that measuring the kinetics
of this unimolecular substitution reaction would demonstrate the reaction rate for the hydrolysis of
tert–Butyl Chloride utilising acetone as the solvent which should increase with both its increased
concentration and temperature, to produce the product tert–Butyl alcohol.
When the solution of tert–butyl chloride in acetone is added to water in the presence of a base and a
universal indicator, its reaction to form tert–butyl alcohol is indicated by dramatic color change.
Bromothymol blue is an acid–base indicator which appears blue in an alkaline medium and yellow
in an acidic solution (Artxy, 2011). The solvolysis of tert–butyl chloride is revealed by the indicator
change from blue to yellow as hydrogen halide is produced during the reaction (Mostafa, 2009). The
solution is initially blue because of its Sodium Hydroxide (NaOH) content (alkaline), and the colour
change results as aqueous chlorine removes a proton from the transition state, while the generated
hydrogen ions neutralise the hydroxide ions of the NaOH, causing the change in the pH of the
system (Riley, 1977). This qualitatively determines that the rate of reaction is dependent on the
concentration of tert–butyl chloride.

Essay about Saponifacation of Ethyl Acetate and Soldium...
Ethyl Acetate – NaOH Reaction Kinetics Experiment Martin Novick Group 14, Chemical
Engineering Laboratory Submitted to Prof. David B. Henthorn September 25, 2012 Summary The
goal of this project was to determine the pre–exponential factor, k o , the activation energy, E, and
the reaction rate constants, k, of the saponification process of ethyl acetate using sodium hydroxide
(NaOH) at 5 temperature between 15 and 25 degrees Celsius. Two trails were performed at
temperatures 16, 18, 20, 22, and 24 degrees Celsius. The main equipment of the project were the
jacketed beaker batch reactor and the LabPro conductivity probe. The solution's conductivity
throughout the reaction was collected and plotted in a linearized plot against time to ... Show more
To find the value of the pre–exponential factor, �� , and activation energy,
��, would require linearizing the Arrhenius equation given as: k = k o ��
−�� , where R is the gas constant and T is the temperature the given k is
at in degrees Kelvin. Equation 8 is linearized by taking the natural log of both sides: ln(k) = ln(k o )
− E . RT 1 �� (8) (9) to retrieve a linear plot with Equation 9 shows a
linear relationship between ln(k) and 4 the y–intercept being ln(�� )
and the slope being ��. Hence: k o = ey−intercept . �� =
�� × �� (10) (11)
�� Equipment, Materials, and Method The equipment used were a jacketed batch
reactor beaker, cooling water circulation system, computer, LabPro temperature probe and
conductivity probe, mixing stand and magnetic stir bar. The materials used for this reaction were a
0.08M NaOH solution and a 0.1M ethyl acetate solution. A 20% excess Ethyl acetate was used to
ensure NaOH was the limiting reactant.[1] NaOH was chosen for the limiting reactant because of its
high conductivity relative to Ethyl acetate. The extent of the reaction was monitored by measuring
the conductivity throughout the reaction. With NaOH being the limiting reactant, the change in
conductivity is more visible, and the termination of the reaction can

Formal Report- Kinetics of Reaction: the Iodine Clock...
factors affecting the kinetics of reaction between peroxodisulfate (vi) and iodide
d. del prado1 and j. belano2
1 department of food science and nutrition, college of home economics
2 department of food science and nutrition, college of home economics university of the philppines,
diliman, quezon city 1101, philippines date submitted: january 7, 2013
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ABSTRACT
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In everyday life, several reactions are encountered, but still knowledge on how fast these occur and
the factors affecting it were still insufficient. This study ... Show more content on Helpwriting.net ...
On the first part, 2 beakers with different contents were prepared. 5 runs were prepared, and each
run had different volumes of the substances. The different runs for the effect of Peroxodisulfate (VI)
and iodide ion concentrations on reaction rate were presented on Appendix A. Each run had
different concentrations of substances to test the effect of the change in amount of concentration to
the rate of the reaction. Contents of beaker A and B were mixed; the timer was started immediately,
and was stopped once a blue color appeared in the mixture. [7] Once Peroxodisulfate (VI) ions and
iodide ions combine they produce I2 molecules which in turn reacted with the starch added to form
a blue color. I2 reacted with the starch as fast as it was produced. Hence, it was very difficult to
measure the rate of its reaction. So as to address the problem, S2O3 2– ion was added. This ion
destroyed the I2 by reducing it back to I– as fast as it was produced. The amount of S2O32– ion
added was just small so that it would not consume so much time. If the initial concentration of
S2O32– ion was kept very small, Δ[S2O32–] would be small and Δ[S2O82–] would be even
smaller. Consequently, there would be little change in the concentration of the reactants during the
elapsed time Δt. This was a necessary condition for the initial rates method. The rate of the reaction
was

Reaction Of Ketone Lab Report
Reactions of amines with ketone and aldehydes have been the subject of considerable study for over
half a century. In 1960, Jencks recognized that imine formation involves two steps and second–order
overall.1 Both reaction steps can be the rate–determining step and it depends on the solution pH.
The reaction being studied in the experiment is shown in scheme 1. Scheme 1: Reaction of
benzyldehyde and semicarbazide The first step of the reaction is nucleophile attack and second step
is dehydration. The propose of the experiment is to measure the pseudo–first order rate constant of
different meta– or para– substituted Benzaldehyde and calculate reaction constant ρ to determine the
rate–determining step using the Hammett plot. The initial benzaldehyde ... Show more content on
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With the electron–donating group, positive charge develops at reaction center in transition state
during rate–determining step, so the rate–determining step is till the step 1, and the second
dehydration step is faster (scheme 2). With electron–withdrawing group, negative charge develops
at reaction center in transition state of the rate–determining step, and the dehydration (step 2) builds
up a positive charge near the aromatic ring, because of the resonance stabilized positive iminium
ion, the dehydration step (step 2)) slows down and becomes the rate–determining step with EWG at
pH

Landolt Iod Clock Experiment
Our society today consists of many natural processes of reactions that involve rates. The rate of a
reaction is the speed at which a reaction happens. If a reaction has a low rate, that means the
molecules combine at a slower speed than a reaction with a higher rate. Some reactions take
hundreds, maybe even thousands, of years whilst others can happen in less than one second. The
rate of reaction also depends on the type of molecules that are combining. If there are low
concentrations of an essential element or compound, the reaction will be slower.
The collision theory says that as more collisions in a system occur, there will be more combinations
of molecules bouncing into each other. If you have more possible combinations there is a higher ...
In contrast to this, a relatively stronger molarity of 0.25M NaHSO3 managed to react and change
colour within an average of 6.52 seconds. Similarly, a weaker concentration of KIO3 managed to
react within an average of 47.40 seconds, containing a less concentrated molarity of 0.0125M. A
much higher concentration of 0.1M was able to react at a speed of 6.52 seconds, all kept constant by
a molarity of 0.25M NaHSO3.
We can also evaluate that the temperature of a solution will, in fact, affect the rate of reaction of the
process. By increasing the temperature of a 0.25M solution of NaHSO3, a reaction with a 0.05M
solution of KIO3 displayed an increase in results from the original, unaltered outcome. Previously
this reaction occurred at a speed of 13.05 seconds. However, by increasing the temperature to 30°C
from SLC, this managed to speed up the process and react within 9.27

Crystal Violet Lab
The purpose of this lab experiment was to investigate how fast it would take crystal violet to
decolorize. The concept of kinetics was applied by using rate to figure out the answer to the guiding
question. Rate laws display the mathematical expression of the rate of a chemical reaction and the
concentration of its reactants. Rate laws are expressed in regards to the three types of reaction
orders: 0, 1st, and 2nd. In rate laws, reaction orders are expressed as exponents and showcase the
effect of reactants' concentrations on the rate of a chemical reaction. The rate constant is displayed
as k and represents a certain reaction. Furthermore, the three rate laws are as written, rate= k for
zero order, rate= k[A] for first order, and rate= k[A]2 ... Show more content on Helpwriting.net ...
The actual rate law was rate= k[CV+]1[OH–] 1 with an overall order of 2. This was concluded by
plotting all three reaction order types in separate graphs and determining the straightest graph; then
calculating concentrations in order to determine the rate constants, pseudo k, and rate laws. For
example, M1V1=M2V2 led to [OH–] by calculating (10 ml) (0.10 M) = (20 ml) to equal 0.05 M of
OH. The actual rate constant was found by taking the negative slope of the first order graph (0.0055
s–1) over the concentration of OH (0.05 M), which equated to 0.11 M–1s–1. With this being said,
CV decolorizes at a first order rate law whose graph yields the straightest line containing the highest
R2 value (0.9973) closest to 1. Lastly, comparing the findings of the data from figures 1–4 to other
groups' date, resulted in discrepancies. Other groups claimed that CV would decolorize at a reaction
order of zero because it's graph yielded the highest R2 value. In relation to this, other groups'
findings could have been a result of certain error such as: not having the colorimeter calibrated once
the reaction mixture was prepared, using inaccurate equations to graph data, and not using graduated
cylinders to obtain precise

Landolt Iodine Clock Lab Report
The importance of conducting this experiment is to discover the rate reaction of the Landolt Iodine
Clock. This reaction is used to display the chemical kinetics in action, it was discovered by Hans
Heinrich Landolt in 1886. It is where two colourless solutions are combined and no instant change
appears but over a certain time delay depending on the factors it will instantly change to a dark blue.
The Chemical kinetics of the reaction refers to the rate of the reaction. Different reactions occur at
different rates, for example if it is a proton transfer reaction which is an acid–base reaction it will
often occur at a faster rate. When the molecules collide in the reaction they must have a sufficient
amount of kinetic energy so that the reaction can be initiated. The amount of kinetic energy is
generally dependant on the temperature of the reaction, at higher temperatures there is a higher rate
of reaction because of the increase of kinetic energy in the reactant molecules. The arrhenius
equation was formed in 1899, by chemist Svante Arrhenius it was from a combination of activation
energy and the distribution law. The arrhenius equation is used in science to calculate the
relationship between the rate of a reaction and its increase or decrease in temperature. If a reaction
has higher temperatures with low activation energy it tends to favour increased rate constants. The
equation is in relation to k, which refers to the rate constant in any given equation. With the

Chemical Kinetcs – the Hydrolysis of Pna Ester
"Chemical Kinetcs – The hydrolysis of PNA Ester"
Introduction: Reaction of a compound with water can result in a splitting, or lysis, of the compound
into two parts. Organic molecules containing a group of atoms called an ester can be hydrolyzed by
water to form a –COOH group (carboxylic acid) and an HO–– group (alcohol) as follows:
RCOOR' + H2O ( RCOOH + HOR'
This reaction is spontaneous for almost all esters but can be very slow under typical conditions of
temperature and pressure. The reaction occurs at a much faster rate if there is a significant amount
of base (OH–) in the solution. In this lab experiment, the rate of this reaction will be studied using
an ester called para–nitrophenyl acetate (PNA), which produces an alcohol, ... Show more content
on Helpwriting.net ...
|Table 1 |
|Varying Exp |Test Tube|PNA [45mg/0.5L] |(PO4)3– Buffer| |H2O |Catalyst | | | |
|1.5x[PNA] |2 |3 |7 |2 |0 |– |– |– |– |
|1/2 [PNA] |3 |1 |7 |2 |2 |– |– |– |– |
|Imidazole |4 |2 |7 |2 |0 |Imidazole |0.025 |1 |0.005 |
|Super (10x) Imidazole|5 |2 |7 |2 |0 |Super (10x) Imidazole|0.25 |1 |0.05 |
|2–Me. Imid. |6 |2 |7 |2 |0 |2–Me. Imid. |0.025 |1 |0.005 |
|4–Me. Imid. |7 |2 |7 |2 |0 |4–Me. Imid. |0.025 |1 |0.005 |
|Acidic pH 6.5 |8 |2 |6.5 |2 |1 |– |– |– |– |
|Basic pH 7.5 |9 |2 |7.5 |2 |1 |– |– |– |– |
|Basic pH 8.0 |10

Assignment: Cornerstone 4 Task D.
Cornerstone 4 Task D – Sub Zero Name:________________________
Attachment 4. Data Sheet D
Context
The Alaskan Bering Sea is known for producing one of the world's most prized types of seafood, the
Alaskan king crab. Fishermen often endure harsh seas and bone–chilling sub–zero conditions while
fishing for these creatures. Staying warm in these difficult conditions is crucial, both for them to
survive and to maintain their livelihood. External warming devices, such as hand–warmers, can be
very helpful to fishermen and others who face extreme cold temperatures on a regular basis. You
will use what you know about electron states, chemical reactions, periodic trends and bond energy
to plan a device that uses a chemical reaction to help keep

The Theory And Factors Affecting Reaction Rates
Chemical Background Hydrogen peroxide is a by–product of many reactions that occur within the
body – however, it is toxic so needs to be broken down. The equation for this decomposition is as
follows: 〖2H〗_2 O_2 → 〖2H〗_2 O+ O_2 In the body, this reaction can be catalysed by the
enzyme catalase. Catalase is not removed or used up in this reaction, and speeds up the rate of
reaction. It is acting as a catalyst. The decomposition can be catalysed by other catalysts, however,
and this is the basis of my investigation. By using different catalysts, I can investigate how the rate
of reaction changes with each one and find out if catalase is the best catalyst for this decomposition.
The collision theory and factors affecting reaction rates For a reaction to take place, two particles
must collide with each other so they come in to contact. However, just colliding with each other
does not initiate a reaction. Something called the activation enthalpy must be overcome. The
activation enthalpy is the minimum (kinetic) energy required by a pair of molecules that are
colliding before a reaction can occur. So, for a reaction to take place, pairs of molecules must collide
with enough energy to equal or overcome the activation enthalpy. There are several factors that
affect the rate of reaction (three of which I am investigating). The factors are as follows:
Concentration of reactants – This factor is explored in more depth in a later section. Temperature –
this factor is explored in

Rate Law Lab
The objective of this experiment was to determine the rate law for a chemical reaction between
crystal violet and hydroxide. A rate law is a part of kinetics, which is the study of how fast reactions
occur and how to control the rate of a reaction (4). The rate law is be determined by measuring and
graphing the absorbance of reactants during the reaction. The reaction was first order with respect to
crystal violet (CV+) and hydroxide (OH–). Since crystal violet is in much smaller concentration
than hydroxide, the experiment captured the reaction rate and order of crystal violet while the order
of OH was calculated post–lab using the pseudo first order method (eqn 1,2,3). The rate law for
CV++OH– CVOHis Rate = 0.1644m–1s–1[CV+][OH–].
Introduction
Kinetics is the study of how fast reactions occur and how to control that rate. The rate of a reaction
is defined by the change in concentration of reactants and products over time (4). The rate law for
this reaction is Rate = k[CV+]m[OH–]n, where m is the order with respect to CV+, n is the order
with respect to OH–, and k is the rate constant. Because the concentration of OH is so much larger
than the concentration of CV+, only [CV+] will change when significantly, which is why the
experiment only focuses on the change of concentration in crystal violet while the concentration ...
A rate law is a mathematical equation that describe the rate of a reaction. A reaction rate is the
change in concentration of reactants or products over time (3). A Vernier colorimeter was used to
measure the absorbance (or concentration) of CV+ when mixed with OH. The concentration of CV+
generally decreased over time at a similar rate when mixed with 0.1M and 0.2M hydroxide. The
data was copied into Excel then made into

Valdez 1
Ariana Valdez
Mrs Mcnamara
Chemistry 1st hr
May 31, 2017
Reaction Rate with Concentration and Temperature
According to The Kinetic Molecular Theory
The reaction rate is defined as, the speed at which a chemical reaction proceeds. It is often
expressed in terms of either the concentration (amount per unit volume) of a product that is formed
in a unit of time or the concentration of a reactant that is consumed in a unit of time (according to
Britannica Online Encyclopedia). But not only concentration affects the reaction rate, temperature
also affects the rate. The Kinetic Molecular Theory is described as a gas that has a large number of
submicroscopic particles (atoms or molecules), all of which are in constant rapid motion that has
randomness raising the collisions with each other and with the walls of the container. All chemical
reactions proceed in different ways according to their reacting substances, types of chemical
transformation, and temperature. In this lab, we tested the effects of ... Show more content on
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Changing the concentration affects how many particles of the substance that are needed in the
chemical reaction. The reaction rate can be increased if the concentration of reactants is raised
(according to www.2.chem,uic.edu). When you increase the concentration you are also increasing
the number of particles, this causes for there to be more chances of collisions between molecules
when you mix the two substances. When lowering the concentration and adding something like
water to your solution A, you decrease the number of particles of substance A and therefore there is
a less likely chance that the particles in substance A will collide with substance B because there is an
increase of water molecules that will collide with substance

Essay on Computerized Data Acquisition of a Second Order...
Computerized Data Acquisition of a Second Order Reaction Abstract Introduction The rates at
which reactions occur depend on the composition and the temperature of the reaction mixture.
Usually the rate of reaction is found to be proportional to the concentrations of the reactants raised
to a power.1 There are many reactions that have a rate law in the form of: (1) v = k[A]a[B]b
According to reference1 the power to which the concentration of a species (product or reactant) is
raised in a rate law of this nature is the order of the reaction with respect to that species. In equation
(1) first order with respect to [A] and first order with respect to [B]; however, the overall reaction is
the sum of the individual orders. Thus ... Show more content on Helpwriting.net ...
The Erlenmeyer flask was swirled for 2–3 seconds before pouring the reacting mixture into a 1–cm
cuvette. The cuvette was conditioned with the reacting solution 4 times before being placed into the
sample holder of the spectrophotometer. An absorbance reading was taken at 30 seconds and every
30 seconds thereafter for a total of 6 minutes. The same process was implemented with the Cary 50
Bio except that each sample was analyzed by the computer for 7 minutes and 53 seconds.
Data/Results 0.025 M NaNO3 0.05 M NaNO3 Time (sec) Asb [K3Fe(CN)6] [C6H8O6] Asb
[K3Fe(CN)6] [C6H8O6] 30 0.622 0.0006146 0.0003573 0.653 0.00065 0.0003726 60 0.617
0.0006097 0.0003548 0.640 0.00063 0.0003662 90 0.606 0.0005988 0.0003494 0.628 0.00062
0.0003603 120 0.600 0.0005929 0.0003464 0.619 0.00061 0.0003558 150 0.593 0.0005860
0.0003430 0.609 0.00060 0.0003509 180 0.584 0.0005771 0.0003385 0.600 0.00059 0.0003464 210
0.578 0.0005711 0.0003356 0.591 0.00058 0.0003420 240 0.571 0.0005642 0.0003321 0.583
0.00058 0.0003380 270 0.564 0.0005573 0.0003287 0.575 0.00057 0.0003341 300 0.559 0.0005524
0.0003262 0.567 0.00056 0.0003301 330 0.552 0.0005455 0.0003227 0.560 0.00055 0.0003267 360
0.546 0.0005395 0.0003198 0.553 0.00055 0.0003232 0.01 M NaNO3 0.2 M NaNO3 Time (sec)
Asb [K3Fe(CN)6] [C6H8O6] Asb

Chemical Reaction On Chemical Kinetics
Olivia Isaacs
C127
15 November 2014
Chemical Kinetics
Objective:
This experiment runs many reactions varying the concentrations of the reactants in order to
determine the order for each component and the rate constant.
Introduction:
Chemical kinetics is the study of how fast a chemical reaction occurs and the factors that affect the
speed of reaction.1 Reaction rates are the measure of how much the concentration of reactants
change during a given reaction.1 The rate of change of the reactants, Rate = – Δ [X]/Δt, is related to
the slope of the concentration vs. time graph.1 From observing reaction rates, the overall order of
the reaction and the rate constant can be calculated by using the integrated rate laws. For a zero–
order reaction, the rate law can be written as [A]t = –kt + [A]0, where [A]t is the concentration at a
given time, k is the negative slope, t is the time, and [A]0 is the initial concentration.2 Using the
same variables, a first order reaction can be written as ln[A]t = –kt + ln[A]0 and a second order can
be written as 1/[A]t = kt + 1/[A]0.2 On a graph, these concentrations are plotted vs. time, allowing
the R2 value and equation of the line to be calculated. The R2 value is used in determining the order
of the reaction. The closer the R2 value is to 1, the more likely that the graph displays the correct
reaction order. The y=mx+b equation provides information about the slope and y–intercept, essential
when determining the order and rate constant.

Kinetics Of Acid Catalysed Propanone / Iodine Reaction
Alexander Unsworth – Tomlinson
Candidate Number: 9133
Kinetics of Acid–Catalysed Propanone/Iodine Reaction
Equation for the reaction :
CH3COCH3(aq) + I2(aq) ––> CH3COCH2I(aq) + H+(aq) + I–(aq)
Iodine + Propanone –> Iodopropanone + Hydrogen (cation) + Iodine (anion)
Introduction: Aims:
To vary the concentrations of each reactant along with the sulphuric acid in order to observe and
measure its effect on the overall rate of reaction in absorbance using colourimetry.
2) Calculate the a mean rate constant using orders of reactions and the rate equation allowing for the
overall order or reaction to be found.
3) Calculate the activation energy of the reaction using different versions of the Arrhenius equation.
4) Try and propose a mechanism for the reaction using the orders of reaction taking into account the
iodine, propanone and sulphuric acid.
Chemical Background:
Iodopropanone is formed from a redox reaction between iodine and propanone which is irreversible.
Hydrogen ions (H+) are used as a catalyst for the reactants and are disassociated from Sulphuric
acid. During the reaction the iodine solution turns from a dark brown colour into a colourless
solution this is because the iodine is reacting to produce the iodopropanone and the the colourless
iodine ions. The way I will be observing the progress of the reaction is by using a digital
colourimeter to measure the absorbance value of the change in colour.
Chemical Theory:

The Collision

Batch Reactor Essay
GROUP 2
Kinetics of De–esterification for Synthesis of Benzoic Acid
BATCH REACTOR
Shane Bulk
Chris Crosley
David McGuire
Max Skula
Yunjing Song
Shriram Sundarraj
Nelson Zhou
155:416 Process Laboratory II
Professor Jerry Sheinbeim
January 28 – February 28, 2014
ABSTRACT
The observed reaction that took place in this experiment was the de–esterification of ethyl benzoate
to form benzoic acid. This experiment was used to determine the rate constant k of the synthesis of
benzoic acid at different temperatures and ethanol concentrations. The reaction was carried out in a
batch reactor, where ethyl benzoate was added to a mixture of water, ethanol, and sodium
hydroxide. The agitation speed of the batch reactor was kept ... Show more content on
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Another objective was to calculate the pre–exponential factor based on the activation energy from
the experimental values. These values were compared to literature and previous group's results in
order to draw conclusions and suggest alternatives.
2.2 Theory
The design equation is based on an isothermal Batch reactor. Given the constant volume used in the
system, the rate of reaction can be explained

Magnesium's Reaction Lab
c. In analysis of the graphs created from the magnesium's reaction pressure released over time, it
appears that we may have had a possible source of error. According to our T.A., the powder should
have a faster rate of reaction due to its surface area. However, in our experiment, the reaction rates
were found to be similar due to the closeness of the slopes. For the first reaction (trial) of each
magnesium type, we had used 0.2 grams due to a communication error. For the second reaction
(trial) of each magnesium type, we used 0.02 grams. Despite the mass differences, the slopes were
very close to one another but with the ribbon reactions appearing to be faster due to their slightly
larger slopes. The ribbon reactions were shown to have a slope in ... Show more content on
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In–Lab Questions
1. The parameter (time frame) that's determined from the slope of the graph at "short times" is
typically where the rate is bound to be the same over a similar time period. Usually this is within a
time interval where the graph is increasing (or decreasing) at a similar rate without the initial burst
or the slowing down of the rate towards the end.
2. Slope can be used to determine the effect of the magnesium on the reaction rate by how close it is
to the number one, when it is showing an increase in reaction rate. As hypothesized, the magnesium
with the greatest surface area would have the fastest reaction rate. Hence, it would have the greater
slope value as well.
3. Please refer to the graphs at the beginning of the post–lab and question 2a. The effect of
magnesium on a reaction rate depends on the magnesium's shape and surface area, greater the
surface area, the faster the reaction rate.
4. If HNO3 was used in place of HCl, I'd expect the rate of change to be about the same as both are
strong acids. However, if acetic acid (CH3COOH) were to be used instead of HCl, I'd except the
reaction rate to be slower as acetic acid is a weak acid and it will dissociate in the water unlike the
stronger

Rate Law Lab
The rate law of a reaction relates the concentration of the molecules involved in the reaction to time,
allowing us to determine how fast the reaction proceeds and what the reaction mechanism is. Based
on the data given on the reaction rate of A and B, we can find the order of the reactants and the
average rate constant for the experimental rate law. The order of A and B are determined by
comparing the trials where the other's concentration is held constant. This is done by creating a ratio
between two trials' rate=k[A]m[B]n. For A, I looked at trials 1 and 3. In the ratio, the k and the
concentration of B cancels out. This leaves the ratio of the instantaneous initial rate equaling to the
ratio of the concentration of A ((1.88x10–3/4.64x10–4)=(0.370/0.185)m). Reducing these ratios
yielded the equation 4.05=2m. Taking the natural log of both sides yields m=2. This can be double
checked by also doing the calculation with trials 2 and 4, where the concentration of B is also held
constant. It also resulted in m=2. Thus, A is second order. Because of this, a linear relationship will
be seen in the plot of 1/[A] versus time. This makes sense because when the concentration of A is
doubled, the rate is quadrupled. ... Show more content on Helpwriting.net ...
In the ratio, the k and the concentration of A cancels out. The ratio of the instantaneous initial rate
then equals the ratio of the concentration of B ((9.38x10–4/4.64x10–4)=(0.266/0.133)n).
Simplifying the equation yields 2.02=2.00m. Solving for n gives n=1. This was also repeated for
trials 3 and 4, where the concentration of A is, again, held constant. The calculation also gave n=1.
This also makes sense because when the concentration of B was doubled, the rate doubled as well.
Thus, B is first order and a linear relationship will be seen in the plot of ln[B] versus time. With A
being second order and B being first order, the overall reaction order for the rate law is then

Chemical Kinetics Lab Report
Review 3: Text Chemical kinetics is the study of rates and mechanisms of chemical reactions. In our
study of chemical kinetics, experimental data identifying the initial concentrations of reactants and
the instantaneous initial rates of multiple trials is used to determine the rate law for the reaction, the
order of the reactants, the overall reaction order, and the average rate constant. By comparing the
instantaneous initial rates and the initial concentrations of the reactants for two trials, it is possible
to deduce the order of each reactant. In order to determine the order of A, the two trials must be
selected such that the concentration of A changes while the concentration of B is held constant. In
this case, trial 1 and trial 3 or trial 2 and ... Show more content on Helpwriting.net ...
k=(0.103M–2s–1+0.103M–2s–1+0.103M–2s–1+0.103M–2s–1)/4=0.103 M–2s–1 The average rate
constant is 0.103M–2s–1. Combining everything, the rate law for the reaction is Rate=(0.103M–2s–
1)[A]2[B]1. We know that the reaction is 2A+2B→C+D. Based on the orders we calculated for A
and B, we know that this reaction is not an elementary reaction because both of the coefficients of A
and B are 2, which do not match the calculated orders of A and B, which are 2 and 1 respectively.
Also, if this were an elementary reaction, we would expect 3 molecules to perfectly collide with
each other, which is highly unlikely. As a result, it is more likely that there was an intermediate and
that multiple steps were involved. Through experimental data, we can not only determine the order
of reactants, but also the rate law, the average rate constant, and the overall order of the reaction.
Using the orders calculated, we can also determine the integrated rate plot that best represents the
reactants and the type of

In this lab, the study of the kinetics of a chemical reaction will be investigated. One of these
reactions involved the oxidation of iodide ions by bromate ions in the presence of an acid.
6I–(aq) + BrO3–(aq) + 6H+(aq) → 3I2(aq) + Br–(aq) + 3H2O(l)
The reaction rate for this is slow at room temperature. The definition of reaction rate is the rate of
change in concentration of either reactants or products. It is dependent on the concentration of the
reactants and on the temperature. In a rate law, a mathematical expression, the relationship between
the reaction rate and the concentrations of reactants can be demonstrated. For example, the rate law
for the rate of decrease in concentration of bromate ion would be:
Rate=(–∆[BrO_3^–])/∆t=k[I^–

Relative Reactivity Of Alkyl Halides
Relative Reactivity of Alkyl Halides in Nucleophilic Substitution Reactions
Charlie Doyle
Madison McGough
Annie Chang
Introduction
Both Sn1 and Sn2 reactions are nucleophilic substitution reactions, though they are slightly
different. Sn2 reactions have bimolecular displacement and are also concerted, meaning the bond
making and the bond breaking processes happen in one step.1 Sn1 reactions require two steps and
have unimolecular displacement. This difference can be seen when comparing Figure 1 and Figure 2
below. The strength of the nucleophile does not effect Sn1 reactions, while the strongest nucleophile
is required for Sn2 substitution reactions.2 Other important considerations include the effect of ...
For Part A: five drops of 2–bromo–2–methylpropane were pipetted into test tube 1; five drops of 2–
bromobutane were pipetted into test tube two; five drops of 1–bromobutane were pipetted into test
tube three; and five drops of 1–chlorobutane were pipetted into test tube four and each were labeled
accordingly. Then, twenty drops of a 15% solution of sodium iodide (NaI) in acetone (for Sn2
reactions) was added to all four test tubes, shaking each once to mix contents, and the time from
when the first drop hit to when cloudiness or a precipitant formed was recorded. The solutions were
then disposed of in the appropriate waste container. For Part B, the test tubes were cleaned and
allowed to dry. The process of pipetting the alkyl halides into the individual four test tubes was
repeated. 20 drops of a 1% solution of silver nitrate (AgNO3) in ethanol (for Sn1 reactions) was
added to each tube, shaking each once to mix contents, and again recording the

Determination of Rate Law Lab Report

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Determination of Rate Law Lab Report