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"; Reaction Kinetics Experiment. - Chem Homework Help
Reaction Kinetics experiment.

 1. Determine which of the following statements are true by comparing the rate (1/t) in Experiment 1 with the rates in Experiments 2 and 3 where [I–(aq)] and [S2O8 2–(aq)] are halved. Explain your choices. (use attatched)2. Use your answers to the previous question to write an experimentally determined rate law for the oxidation of iodide ions with persulfate ions. 3. i) Describe how the rate of oxidation of iodide ions with persulfate ions depends on the initial concentration of the reactants.ii) Would you expect the rate of all reactions to be related to reactant concentrations in the exact same way as you saw here?Consider the following proposed mechanisms for the reaction of iodide ions with persulfate ions:4. i) Place a tick√next to the rate determining step (RDS) in each of the mechanisms above.ii) Which mechanism(s) support/s your experimental results?State your reasoning clearly, using the RDS to support your answer.iii) Which mechanism(s) do/does NOT support/s your experimental results?State your reasoning clearly, using the RDS to support your answer. Consider your results for Experiments 1, 4, 5 and 6. 5. Did the reaction rate increase or decrease with increasing temperature? Explain briefly why this should be so. 6. Showing your calculations, use your data from Experiments 4 and 5 to justify or reject the following general statement: ‘The rates of most simple reactions approximately double for each 10K rise in temperature.’ 7. Why is it not as accurate to use your data for Experiments 1 and 4 to answer the above question? Compare your results for Experiments 1 and 7. 8.Describe the effect of copper ions on your reaction rate.

EXPERIMENT 9F

Reaction Kinetics

This experiment is done in pairs.

 

Useful background reading (this is not compulsory but may be helpful):

“Chemistry Core Concepts”. 1st Edition. John Wiley & Sons Australia Ltd.

Section 13.1, 13.2, 13.3 (pages 564-565), 13.4, 13.5 (pages 575-577); Worked examples 13.1,

13.2; Practice exercises 13.1, 13.2

“Chemistry Core Concepts”. 2nd Edition. John Wiley & Sons Australia Ltd.

Section 13.1, 13.2, 13.3 (pages 759-760), 13.4, 13.5 (pages 774-776); Worked examples 13.1,

13.2; Practice exercises 13.1, 13.2

 

What is the relevance of this prac?

This experiment relates to the reaction kinetics section of your lectures. You will see how reactant concentration, temperature and addition of a catalyst affect the rate of a reaction.

Learning Objectives (remember these are different to the scientific objectives):

On completion of this practical, you will have:

  • Learned how an increase in reactant concentration affects reaction rate.
  • Learned how an increase or decrease in temperature affects reaction rate.
  • Learned how the addition of a catalyst affects reaction rate.
  • Learned how to use experimentally-derived data to deduce a rate law for a reaction.

In this experiment you will study factors affecting the rate of the oxidation of iodide ions, I(aq), with persulfate ions, S2O28(aq):

S2O28(aq) + 2I(aq) → 2SO24(aq) + I2(aq)

Peroxydisulfate (persulfate – note that the two central oxygen atoms are peroxide-like and hence have an oxidation state of -1 instead of the usual -2):

O O

O

O

O

O

S

S

O O

You will thereby have the opportunity to develop some understanding of this reaction and also the principles of reaction kinetics.

Introduction (see pages 16-18 of this prac script for further background information)

Why should the rate of a reaction be important? Indeed, what is the rate of a chemical reaction? To answer the last question first: the rate of a chemical reaction is a measure of how fast products are being formed by that chemical reaction. Conversely, it is also a measure of how fast reactants are being consumed. This will be discussed in more detail later on.

The rates of chemical reactions are of great importance in industrial and biological processes. The economic viability of many industrial processes is largely affected by the rate at which the products can be formed. Of more vital importance, in the true sense of the word, every chemical reaction taking place in your body is occurring at a rate carefully controlled by the most complex of catalysts—enzymes. Life would be impossible without the rates of countless, complicated chemical processes being controlled with exceptional precision by exquisitely formed enzyme catalysts.

The thermodynamic favourability of a process (ie the work needed to complete that process or the energy released by that process) can be determined by applying the Laws of Thermodynamics and simply measuring the initial and final energy of the system. This tells us nothing about the rate at which the process can occur. Indeed, many highly favourable processes will not proceed at a measurable rate until something ‘triggers’ or catalyses them eg: A litre of petrol will burn vigourously in air to produce a great deal of energy. However, no obvious reaction will be observed until the vapour above the petrol is ignited (carefully!).[1]

Reaction kinetics is the study of the rate and mechanism of chemical reactions. Our understanding of chemistry is greatly increased by studying the details of chemical processes.

MECHANISM

Refer to Brown et al, “Chemistry: The Central Science”, 2nd Edition, section 12.6.

As chemists, we often write equations describing chemical reactions. Consider the reaction you will study in this experiment:

S2O28(aq) + 2I(aq) → 2SO24(aq) + I2(aq)

When we write such an equation we are making a statement about the starting materials and the end products of the reaction. A chemical equation does not imply that the process takes place in one step. To state that this is the reaction mechanism is to imply that the three reactant molecules will collide absolutely simultaneously and then instantly be converted to products. This is extremely unlikely.

When studying reaction kinetics an attempt is made to describe a chemical reaction as a series of single steps or elementary steps. The elementary steps must satisfy two requirements:

  1. The sum of the elementary steps must give the overall balanced equation for the reaction.
  2. The rate determining step (RDS), which is the slowest step in the sequence of steps leading to product formation, should predict the same rate law as is determined experimentally.

If it is possible to write a series of chemical equations that describe each of the elementary steps in a chemical process then that series of steps is a reaction mechanism.[2] Thus, the sequence of steps in the study of a reaction mechanism are:

measuring the rate of a reaction

 

formulating the rate law

 

postulating a

reasonable reaction mechanism

In the experiment you will do today, you will measure the rate of the reaction of the oxidation of iodide ions with persulfate ions under various conditions and use your results to formulate an experimentally determined rate law.

Several possible hypothetical mechanisms for the reaction being studied and the rate law or rate expression which can be derived from them will be provided. Your experimental rate law will either support or contradict the mechanisms provided. The mechanism which supports your experimentally derived rate law will be the likely mechanism for the reaction.

For example, one of the proposed mechanisms for the oxidation of iodide ions with persulfate ions is:

I(aq)+ I(aq) slow *[I….I]2–(aq)

RDS

*[I….I]2–(aq) + S2O28(aq) fast I2(aq) + 2SO24(aq)

These two equations are elementary steps in the overall chemical reaction. They state that two iodide molecules collide to form an intermediate di-iodide species which lasts long enough to react, at some time later, with one molecule of persulfate to form the final products. This is much more plausible than a simultaneous and instantaneous reaction, however it is only one of several possible mechanisms which you will consider.

RATE DETERMINING STEP AND RATE LAW

Refer to Brown et al, “Chemistry: The Central Science”, 2nd edition, section 12.6.

As already stated it is often the case that one of the elementary steps in a reaction mechanism is much slower than the rest. If this is so, the overall reaction rate is equal to the rate of this slowest step and this step is called the rate determining step (RDS).

Some reactants may be not be involved in the RDS.

Consider the hypothetical mechanism for the oxidation of iodide ions with persulfate ions above. The first step is the rate determining step so the rate of the reaction can be determined from the first step alone:

rate = k [I] [I]

 rate = k [I]2

This is the rate law or rate expression for the reaction.

If this rate law is the same as the experimentally determined rate law formulated from your experimental results, the mechanism is the likely one for the reaction. If the rate law contradicts your results, other mechanisms must be considered.

RATE

The rate of a chemical reaction is defined as being the absolute value of the rate at which the concentration of reactants or products are changing — divided by the stoichiometric ratio if necessary eg:

aA + bB → cC + dD

The rate of disappearance of A, eg. is (by definition):

d[A] 1

Rate = –  (ie the rate of disappearance of A)

dt a

RATE LAWS AND ELEMENTARY STEPS

Refer to Brown et al, “Chemistry: The Central Science”, 2nd edition, sections 12.4 and 12.6.

Once the mechanism of a reaction is determined, the observed rate of the reaction may be rationalised and, if necessary, the reaction conditions can be adjusted to control the rate of the process.

While it is not possible generally to predict the rate of a chemical reaction from the equation that describes it, it is possible to do so for an elementary step in a reaction mechanism.

For two atoms, ions or molecules to react they must first collide. The rate of the reaction is therefore proportional to the rate of collision. The rate of collision is proportional to the concentration of the species involved in the elementary step.

The rate of an elementary step is proportional to the concentration of reactants involved in that step.

Consider three simple examples of elementary steps:

Example One

 

Example Two

 

Example Three

A →

B

C + D →

E

2F → G

Rate 

[A]

Rate 

[C] × [D]

Rate  [F]2

 

Example One

In Example One above, the process is only affected by the concentration of species A and this process is described as a First-Order Process.

Rate = k × [A]

k = rate constant

For a First-Order process the concentration of the reactant and products approach their equilibrium values exponentially.[3]

Diagram

9

.1

 

First

Order Process

 

 

An important property of First-Order processes is that the time taken for the concentration of reactants to halve (from any arbitrary starting concentration) is always constant at constant temperature. This period of time is characteristic of the rate constant and is called the half-life of the reaction (t½ ).

Example Two

In Example Two, the rate of the reaction is affected by the concentration of the two species involved. Such a process is called a Second-Order Process.

Rate = k × [C] × [D]

In contrast to example one, the half-life of a Second-Order process increases as the reaction proceeds.

Diagram 9.2 Second-Order Process

word image 13

Example Three

In Example Three on page 5 the rate of the reaction is only affected by the concentration of F but in this case the rate will be proportional to the square of the concentration of F. Such a process is also called a Second-Order Process.

The mathematics of second-order processes is much more complicated than that of firstorder ones and will not be examined here.

EFFECT OF TEMPERATURE ON RATE

Refer to Brown et al, “Chemistry: The Central Science”, 2nd edition, section 12.5.

Earlier it was argued that the rate of an elementary step is proportional to the concentration of species involved in that step because the rate is proportional to the rate of collisions between the species involved. Increasing the concentration of species is not the only way to increase the rate of reaction. The rate of collision can also be increased by increasing the temperature of the reaction.

Not all collisions will result in the desired reaction to occur. Collision theory states that colliding molecules must possess sufficient energy between them so that a certain minimum activation energy (Ea) is exceeded.[4] As the temperature of a reaction mixture is raised, the thermal energy of the reacting species is raised and so the number of ‘successful’ collisions is increased dramatically.

The Arrhenius Equation describes the relationship between rate constant and temperature:

k

=

A × e–Ea/RT

k

=

rate constant

Ea

=

activation energy (Joules)

R

=

gas constant (8.314 J K-1 mol-1)

T

=

temperature (Kelvin)

A

 

=

collision frequency

CATALYSTS

Refer to Brown et al, “Chemistry: The Central Science”, 2nd edition, section 12.7.

In the previous section it was stated that in order for collisions between reactants to be ‘successful’ the molecules had to possess sufficient energy to overcome an activation energy characteristic of the particular reaction. A consequence of the Arrhenius Equation is that as the temperature increases so too does the rate constant.

The rate constant could also be increased if the activation energy can be reduced. Catalysts are species that are not consumed in a reaction but provide an alternative pathway for a reaction. This alternative pathway has a lower activation energy and therefore the reaction will proceed faster.

THIS EXPERIMENT

It is often possible to plot the changing concentration of reactants or products of a chemical reaction as a function of time. The rate of the reaction at any time during the measurement period can then be determined by measuring the slope of the tangent to the curve at that point.

The experiment that you will perform is designed so that it is reasonable to assume that the rate is constant throughout the measurement period.[5] Therefore, it is reasonable to assume that the average rate of the reaction is equal to the initial rate of the reaction.

Under these conditions the time taken to produce a certain quantity of product can be used to calculate the initial rate of the reaction.

d[Product] [Product]

Rate = [dt] t (for small Δt)

1

 Dt (if Δ[Product] is constant)

The following diagram indicates the following important points regarding the experiment you will perform:

  • The rate of a reaction is approximately constant if the observation period is much less than the half-life of the reaction.
  • The time taken to form a certain, fixed quantity of product (Δt) is inversely proportional to the rate constant (or rate).

Diagram 9.3 First-Order Processes

word image 14

INDICATOR

How can the progress of the reaction being studied be conveniently followed?

The reaction we are studying produces iodine (I2):

S2O28(aq) + 2I(aq) → 2SO24(aq) + I2(aq)

It is convenient to use the formation of iodine to follow the reaction and to do so we will add another reagent: sodium thiosulfate (Na2S2O3). Iodine is reduced rapidly by thiosulfate (S2O23(aq)) to form iodide ions and sulfate ions:

S2O23(aq) + 4I2(aq) + 5H2O → SO24(aq) + 8I(aq) + 8H+(aq)

By including a precisely measured quantity of thiosulfate in the reaction mixture we can ensure that no free iodine will be present in the mixture until all the thiosulfate has been consumed. As soon as the thiosulfate ions supply is exhausted the iodine concentration will rise.[6]

The appearance of free iodine in the reaction mixture will be more easily noted by adding a small quantity starch. Starch forms a strongly-coloured, dark blue complex with iodine. The solution will suddenly turn dark blue when the thiosulfate is exhausted.

The time (Δt) taken for the solution to turn blue after the reagents are mixed is the time required to produce enough free iodine to react with all of the thiosulfate. If the amount of thiosulfate is kept constant, the amount of iodine needed to produce a colour change will be constant. So, the initial rate of the reaction will be proportional to 1/Δt.

Diagram 9.4

word image 15

One of the objectives in this practical class is to determine the effect of temperature on reaction rate. In order to do this you must use the thermostatted water baths to adjust and control the reaction temperature. You should organise your time so that solutions are placed in the correct water bath at least ten minutes before you wish to use them. This portion of Table 9.1 is included to help you prepare for the practical class. The entire table with room for your experimental data may be found in your report book.

 

Hazardous substances

  

Ammonium persulfate

(NH4)2S2O8

harmful

Ammonium sulfate

(NH4)2SO4

non hazardous

Copper sulfate

CuSO4

harmful

Sodium thiosulfate

Na2S2O3

non hazardous

Potassium chloride

KCl

non hazardous

Potassium iodide

KI

non hazardous

 

 

 

 

Experiment 9F equipment in your basket

3 x 100mL glass beaker

 

3 * 10mL measuring cylinder

 

1 * 500mL plastic beaker for ice bath

Additional equipment and location

On the bench: Test tubes (14) in rack, potassium chloride, copper sulfate, starch, ammonium sulfate, 10mL pipette with filler

Side trolley: Ammonium persulfate, potassium iodide, sodium thiosulfate

 

On sink: ice, water baths

 

Please ensure this is correct and your area is clean and tidy before leaving the laboratory!

 

 

PROCEDURE

  1. Use Table 9.1 in your report book to prepare test tubes containing the correct volumes of each of the solutions provided.
  2. Place test tubes into water baths as soon as possible and allow at least ten minutes for thermal equilibrium to be established. Measure and record the temperature of the water bath.
  3. Start the stop-watch as you carefully pour the contents of the test tube B into test tube A. Pour the contents of test tube A back into test tube B to mix the solutions and then replace test tube B into the water bath.
  4. Stop the stop-watch as soon as a blue colour appears in your test tube. Record (in seconds) the time taken for the blue colour to appear.
  5. Record the final temperature of the solution in test tube B.

NOTE

    • All volumes are in millilitres (mL) unless otherwise specified
    • While measuring cylinders are provided for measuring most solutions, the volume of sodium thiosulfate (Na2S2O3) must be measured more precisely than the others—a pipette is provided to measure volumes of this solution.

 

Table 9.1 – All volumes are in millilitres unless otherwise specified

Reagent 1 2 3 4 5 6 7

(concentration)

iodide (0.20 M) chloride (0.20 M) thiosulfate (0.005 M) starch

10.0

5.0

10.0

10.0

10.0

10.0

10.0

0.0

5.0

0.0

0.0

0.0

0.0

0.0

10.00

10.00

10.00

10.00

10.00

10.00

10.00

1

1

1

1

1

1

1

persulfate (0.10 M) sulfate (0.10 M) copper sulfate

10.0

10.0

5.0

10.0

10.0

10.0

10.0

0.0

0.0

5.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

2 drops

target temperature

room

room

room

room

+10 K

room

+20 K

273 K

room

Test tube

A

 

 

 

Test tube

B

 

 

 

 

REPORT BOOK QUESTIONS

These questions from the Report Book are included in the manual to help you prepare for the Experiment and the Computer Preliminary Exercises.

  1. Determine which of the following statements are true by comparing the rate (1/t) in Experiment 1 with the rates in Experiments 2 and 3 where [I(aq)] and [S2O28(aq)] are halved. Explain your choices.

Rate  [I (aq)] Rate  [I(aq)]

2

1

[

I

(

aq

)

]

 

2

8

2

(

aq

)]

 

[

S

2

O

8

2

(

aq

)

]

 

 

Rate  [I(aq)]

Rate 

Rate  [S O Rate  [S2O28– 2 (aq)]

 

Rate  Rate 21–(aq)]

 [S2O8

 

  1. Use your answers to the previous question to write an experimentally determined rate law for the oxidation of iodide ions with persulfate ions.
  2. i) Describe how the rate of oxidation of iodide ions with persulfate ions depends on the initial concentration of the reactants. ii) Would you expect the rate of all reactions to be related to reactant concentrations in the exact same way as you saw here?

Consider the following proposed mechanisms for the reaction of iodide ions with persulfate ions:

Proposed Mechanism 1

I(aq) + I(aq)

slow



*[I….I]2–(aq)

*[I….I]2–(aq) + S2O28(aq) rate law : rate

fast



=

I2(aq) + 2SO24(aq) k[I]2

Proposed Mechanism 2

I(aq) + S2O28(aq) fastÛ *[I….S2O8]3–(aq)

equilibrium

*[I…..S2O8]3–(aq) + I(aq) slow I2(aq) + 2SO24(aq)

rate = k[[I…..S2O8]3–][I] therefore,

rate law : rate = Ke.k[I]2[S2O28]

The rate law is derived by writing an expression for the equilibrium constant, Ke, for the first step. It is not necessary for you to know how to derive it but your demonstrator will help you if you are interested.

Proposed Mechanism 3

I(aq) + S2O28(aq)

slow



*[I….S2O8]3–(aq)

*[I….S2O8]3–(aq) + I(aq) rate law : rate

fast



=

I2(aq) + 2SO24(aq) k[I–][S2O82–]

Proposed Mechanism 4

2I(aq) + S2O28(aq) rate law : rate

slow



=

I2(aq) + 2SO24(aq) k[I–]2[S2O82–]

  1. i) Place a tick √ next to the rate determining step (RDS) in each of the mechanisms above.
    1. Which mechanism(s) support/s your experimental results? State your reasoning clearly, using the RDS to support your answer.
    2. Which mechanism(s) do/does NOT support/s your experimental results? State your reasoning clearly, using the RDS to support your answer.

Consider your results for Experiments 1, 4, 5 and 6.

  1. Did the reaction rate increase or decrease with increasing temperature? Explain briefly why this should be so.
  2. Showing your calculations, use your data from Experiments 4 and 5 to justify or reject the following general statement:

‘The rates of most simple reactions approximately double for each 10K rise in temperature.’

  1. Why is it not as accurate to use your data for Experiments 1 and 4 to answer the above question?

Compare your results for Experiments 1 and 7.

  1. Describe the effect of copper ions on your reaction rate.

 

Information Sheet

HARMFUL

(can affect health if exposed to large doses or to low doses over a long period of time)

OXIDISING AGENT

(may cause an oxidation process or liberate oxygen which can start a fire in other material or increase the violence of a fire)

AMMONIUM PERSULFATE

IDENTIFICATION

Name Ammonium persulfate

Structure

O

S

O

O

S

O

O

O

O

O

NH

4

NH

4

PHYSICAL DESCRIPTION AND PROPERTIES

Description

 

colourless, odourless solid

Boiling Point

 

not available

Melting Point

 

120° C

Vapour Pressure

 

not relevant

Flammability

 

Contact with combustible material may cause a fire.

Density

 

1.98 g/cm3 at 20° C

Solubility

 

Soluble: water

Reactivity

 

Contact with combustible material should be avoided

 

 

as ammonium persulphate is a fire promoting

 

 

material and can cause a fire in combustible material.

HEALTH HAZARD INFORMATION

Major hazards

Inhalation or skin contact can cause sensitisation. Harmful if swallowed.

Toxicity

Inhalation: May cause sensitisation (development of an allergy to even minute quantities of the substance which becomes evident on re-exposure). Inhalation may irritate the respiratory tract and result in delayed oedema.

Eye contact: May irritate the eye.

Skin contact: May cause sensitisation (development of an allergy to even minute quantities of the substance which becomes evident on re-exposure). May irritate the skin.

Swallowing: Harmful if swallowed. May irritate the digestive tract, causing nausea and vomiting.

FIRST AID INFORMATION

Eyes: Hold the eyelid wide open, wash the eye for at least 10 minutes with flowing water.

Lungs: Remove patient to fresh air. If unconscious, do not give anything to drink. If no pulse, apply artificial respiration or cardiopulmonary respiration. If there is a pulse, place in coma position. If conscious, make the casualty lie or sit quietly, give oxygen if available.

Mouth: If unconscious, treat as described above. If conscious, drink lots of water, induce vomiting.

Skin: Remove contaminated clothing immediately. Wash off skin with running water for at least 10 minutes.

DISPOSAL OF SMALL AMOUNTS/SPILLAGES BY DEMONSTRATORS

Sweep up material. Place in appropriate container for further disposal. Do not store with combustible material.

Other Information

The full Material Safety Data Sheet for this chemical is available from Chemwatch Gold, on-line at:

http://www.adelaide.edu.au/hr/ohs/legislation/chemwatch

 

Information Sheet

HARMFUL

(can affect health if exposed to large doses or to low doses over a long period of time) COPPER SULFATE

IDENTIFICATION

Name Copper sulfate

Structure CuSO4

PHYSICAL DESCRIPTION AND PROPERTIES

Description Pentahydrate: blue crystalline solid

Aqueous solution: blue liquid

Boiling Point Not relevant

Melting Point Not relevant

Vapour Pressure Not relevant

Flammability Not relevant

Density Not relevant

Solubility Highly soluble: water

Reactivity

HEALTH HAZARD INFORMATION

Major hazards

Harmful if swallowed. Can cause irritation of skin, eyes and mucous membranes.

Toxicity

Inhalation: May irritate the mucous membranes, cause coughing, and dyspnoea (difficulty in breathing).

Eye contact: May irritate the eye with risk of corneal clouding.

Skin contact: May irritate the skin on contact.

Swallowing: Can cause pain, vomiting, diarrhoea, a drop in blood pressure, tachycardia (increase in heart rate), collapse and acidosis (reduced alkalinity of the blood). After a latency period death may result.

FIRST AID INFORMATION

Eyes: Hold the eyelid wide open, wash the eye for at least 10 minutes with flowing water. Summon medical assistance immediately.

Lungs: Remove patient to fresh air. If unconscious, do not give anything to drink. If no pulse, apply artificial respiration or cardiopulmonary respiration. If there is a pulse, place in coma position. If conscious, make the casualty lie or sit quietly, give oxygen if available.

Mouth: If unconscious, treat as described above. If conscious, drink lots of water. Induce vomiting.

Skin: Remove contaminated clothing immediately. Wash off skin with running water for at least 10 minutes.

DISPOSAL OF SMALL AMOUNTS/SPILLAGES BY DEMONSTRATORS

Mop up solution and place in container for further disposal.

Other Information

The full Material Safety Data Sheet for this chemical is available from Chemwatch Gold, on-line at:

http://www.adelaide.edu.au/hr/ohs/legislation/chemwatch

 

EXPERIMENT 6

REACTION KINETICS

Extra Background

 

INFORMATION

What are oxidation numbers and how do you use them?

Oxidation numbers are a useful way to keep track of electrons during oxidation and reduction reactions. The oxidation number or oxidation state is the charge an atom would have in a molecule (or ionic compound) if the electrons had been transferred completely.

For H-Cl the Cl is more electronegative (that is it attracts the electrons in the bond more than the hydrogen) and so would be 1- if the electron was completely transferred (noble gas configuration with 8 valence electrons). The Cl therefore has an oxidation number or state of -1. The hydrogen then has on oxidation state or number of +1.

For P4O10 the O is more electronegative (that is it attracts the electrons in the bond more than the phosphorus) and so would be 2- if the electrons were completely transferred (noble gas configuration with 8 valence electrons). The O therefore has an oxidation number or state of -2. Since the molecule has 10 oxygens this would mean a total of 20 negative charges which must be balanced by the 4 phosphorus atoms. The P then has on oxidation state or number of +5.

The following rules apply to assigning oxidation numbers

Free elements have an oxidation number of zero. Br2, Be, S (as in S8) and O2

Ions consisting of the single atom have an oxidation number equal to their charge. Li+ (+1), A13+ (+3). O2- (-2)

Oxygen normally has the oxidation number -2, except in H2O2 and O22- where it is -1.

Hydrogen normally has the oxidation number +1

If the molecule is neutral, then the sum of the oxidation numbers of all the elements must be zero. In an Ionic species the sum of the oxidation numbers of all the elements must be equal to the net charge on the ion.

What is kinetics and what do you do with it?

We need to know how fast a reaction is going to be so that we do not have explosions or wait for weeks for a reaction to go to completion. The area of chemistry dealing with how fast a reaction occurs is called chemical kinetics. The reaction rate is the change in the concentration of reactant (what you started with) or product (what you end up with) with time.

 

For the simple reaction A  B a graph of the reaction rate could look like this:

 

10

 

word image 35

A molecules

 

B molecules

 

 

8

 

 

number of 6 molecules

 

4

 

 

2

 

1 2 3 4 5 6 7 8 9 10 11 12

time

To determine the rate of a reaction we measure a change in the concentration of one of the reactants or products with time. If the rate is dependent on the concentration of one of the reactants (say A) then the rate is proportional to the [A] or rate a [A]. If the rate was only dependent on the concentration of A then the term k (known as the rate constant) can be introduced as a constant of proportionality between the reaction rate and the concentration of A.

rate

k 

[A] or rate = k[A]

The value of k is constant and is not affected by the concentration of A. The rate of the reaction will be greater when the concentration of A is higher, but the ratio of the rate divided by [A] will be constant.

Some reaction rates are dependent on the concentration of a number of reactants. For example if the rate is proportional to each of [A] and [B] then the rate = k [A][B]. This is known as the rate law, which is the expression relating the rate of a reaction to the rate constant (k) and the concentrations of the reactants. It is important to realise that the rate law cannot be determined by simply inspecting the stoichiometry of a reaction. The rate law must be determined by experiment.

If the rate law for the reaction takes the form rate = k [A]x[B]y then the reaction order is the sum of the powers to which the [] are raised, eg x + y. A first order reaction is a reaction in which the rate depends on the reactant concentration raised to the power one. A second order reaction is a reaction in which the rate depends on the reactant concentration raised to the power two or on the concentration of two different reactants each raised to the power one.

The Arrhenius Equation relates the rate of a reaction to the temperature, k = Ae-Ea/RT Ea is the activation energy for the reaction (the energy to reach the activated complex), R is the gas constant and T is the absolute temperature. The term A is the frequency factor and represents the collision frequency (molecules must collide with each other to react).

The rate constant decreases with increasing activation energy and increases with increasing temperature. As an approximation, the rate of a reaction doubles with every 10 K increase in temperature.

One of the reasons for studying the rate of a reaction is to propose a mechanism for the reaction; that is, what happens to the molecules during the reaction. The overall reaction is the sum of the individual reactions which take place. The sequence of simple steps that eventually leads to the product is the reaction mechanism. A number of intermediates can be formed during the reaction and will be consumed during subsequent steps. These intermediates appear in the reaction mechanism but not in the overall reaction.

What are we actually measuring when we measure the rate of a reaction if there are a number of steps and intermediates? The slowest step in the reaction mechanism, called the rate determining step, is what we are measuring.

Is there a way we can make reactions proceed more rapidly? One of them is to increase the temperature. The other is to add a catalyst. A catalyst will lower the activation energy (increase the rate) by providing an alternative pathway for the reaction. A catalyst will be involved in an intermediate step but will be regenerated in a subsequent step and so not be consumed in the overall reaction.

 

 

WATER BATHS (+10 K, +20 K)

 

word image 36

  1. It is not recommended that you try this experiment!

  2. It is unfortunate that the symbols used to describe elementary steps in a reaction mechanism are identical to those used in normal chemical equations. You need to learn to take note of the context of the equations.

  3. Because the rate of a reaction varies as the concentration of at least one of the reactants varies, it is not convenient to compare the rates of different reactions directly. However, the rate constant, which is characteristic of a particular reaction under certain conditions, does not vary with concentration and so is a useful measure for comparing the rates of different reactions.

  4. Brown et al, “Chemistry: The Central Science”, 2nd edition, section 12.5.

  5. For this approximation to be reasonable, Δt, the period of observation, must be much less than the half-life of the reaction ie: Δt << t ½.

  6. If necessary, the number of moles of iodide oxidised by persulfate before the solution turns blue can be calculated from the number of moles of thiosulfate added and the stoichiometry of the previous equation.

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