Edexcel A2 Chemistry 4 3 Notes

A2 Chem

Unit 4: Chapter 1 – How Fast? - Rates Specification Reference: 

Demonstrate an understanding of the terms ‘rate of reaction’, ‘rate equation’, ‘order of reaction’, ‘rate constant’, ‘half-life’, ‘rate-determining step’, ‘activation energy’, ‘heterogeneous and homogeneous catalyst’.

Rates – Definitions  

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Rate of reaction – is the change in the amount or concentration of reactants or products per unit time (normally per second). Rate equation – (rate = k[A]m[B]n). This is known as a rate equation. [A] and [B] represent the concentrations of A and B in mol dm-3, and k is called the rate constant. The indices m and n are usually whole numbers (1, 2 …) but they can be fractional or zero. Order of reaction – the order of reaction is the sum of the powers to which the concentrations of the reactants are raised in the experimentally determined rate equation (m + n). Partial order – of one reactant is the power to which the concentration of that reactant is raised in the rate equation. Rate constant – the rate constant, k, is the constant of proportionality that connects the rate of the reaction with the concentration of the reactants. Its value alters with temperature. A reaction with a large activation energy has a low value of k. Half-life – is the time taken for any reactant concentration to fall to half of its initial value. Rate-determining step – is the slowest step that controls how fast the overall reaction occurs. Activation energy – is the minimum energy needed by reactant particles (molecules or ions) before products can form. Heterogeneous catalyst – these are catalysts that are in a different phase from the reactants. Homogeneous catalyst – the catalyst and the reactants are in the same phase, usually the gas phase or in solution.

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Specification Reference: 

Select and describe a suitable experimental technique to obtain rate data for a given reaction, e.g. colorimetry, mass change and volume of gas evolved.

Experimental Methods Rate of Reaction (mol dm-3 s-1) =

𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑇𝑖𝑚𝑒

1) Colorimetry If a reactant or product is coloured, the concentration of the coloured species can be measured using a spectrophotometer. The amount of light of a particular frequency that is absorbed depends on the concentration of the coloured substance. Colorimetry measures the intensity of a colour in the reaction mixture with time, such as the in the oxidation of iodide ions to give brown iodine. In clock reactions the reaction is timed until a sudden colour change happens when a certain amount of product is formed. For example, in the reaction between propanone and iodine, the brown colour fades. CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + H+(aq) + I+(aq) Colourless Brown Colourless 2) Mass Change Mass change is used when a gas is produced. For example, when calcium carbonate reacts with acids to release carbon dioxide the mass of the flask decreases.

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CaCO3(s) + 2HCl

(aq)

→ CaCl2(aq) + H2O(l) + CO2(g)

Recording the loss in mass over a certain period of time at regular intervals using the apparatus shown gives an indication of the rate of reaction. 3) Volume of Gas Evolved If the reaction produces a gas, the volume of gas produced can be measured at regular time intervals. The volume of gas is proportional to the moles of gas and can, therefore, be used to measure the concentration of the product. The rate of the reaction of an acid with a solid carbonate can be studied in this way. The acid is added to the carbonate and the volume of carbon dioxide noted every 30 seconds. CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)

Specification Reference: 

Investigate reactions which produce data that can be used to calculate the rate of the reaction, its half-life from concentration or volume against time graphs, e.g. a clock reaction.

‘Clock’ Reactions In a „clock‟ reaction, the reactants are mixed and the time taken to produce a fixed amount of product is measured. The experiment is then repeated several different times using different starting concentrations.

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1) The Iodine ‘Clock’ The oxidation of iodide ions by hydrogen peroxide in acid solution can be considered a „clock‟ reaction: H2O2(aq) + 2I-(aq) + 2H+(aq) → I2(s) + 2H2O(l)    

25cm3 of hydrogen peroxide solution is mixed in a beaker with 25cm 3 of water and a few drops of starch solution are added. 25cm3 of potassium iodide solution and 5cm3 of a dilute solution of sodium thiosulfate are placed in a second beaker. The contents of the two beakers are mixed and the time taken for the solution to go blue is measured. The experiment is repeated with the same volumes of potassium iodide and sodium thiosulfate but with 20cm3 of hydrogen peroxide and 30cm3 of water, and then with other relative amounts of hydrogen peroxide and water, totalling 50cm3.

The reaction produces iodine, which reacts with the sodium thiosulfate. When all the sodium thiosulfate has been used up, the next iodine that is produced reacts with the starch to give an intense blue-black colour. The amount of iodine produced in the measured time is proportional to the volume of sodium thiosulfate solution taken. Therefore, the average rate of reaction for each experiment is proportional to 1/time.

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2. The Sulfur ‘Clock’ Sodium thiosulfate is decomposed by acid, producing a precipitate of sulfur. S2O32-(aq) + 2H+(aq) → S(s) + SO2(aq) + H2O(l)      

A large X is drawn on a white tile with a marker pen. 2cm3 of sodium thiosulfate solution is mixed with 25cm3 of water in a beaker. 25cm3 of dilute nitric acid is placed in a second beaker. The first beaker is placed on top of the X and the contents of the second one are added. The mixture is stirred and the time (t) taken for sufficient sulfur to be produced to hide the X when looking down through the beaker is measured. The experiment is repeated with different relative amounts of sodium thiosulfate and water, totalling 50cm3.

The number of moles of sulfur produced is the same in all experiments. Therefore, the average rate of reaction for each experiment is proportional to 1/t.

Half-life 

Half-life – is the time taken for any reactant concentration to fall to half of its initial value.

Half-life of First Order Reaction The following equation and graph shows the decomposition of sulfur dichloride oxide:

SO2Cl2(g) → SO2(g) + Cl2(g)

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The time it takes SO2Cl2 to halve is 2300 seconds. Notice that this time is constant – it takes the same time for the concentration to fall from 0.50 mol dm-3 to 0.25 mol dm-3 as it takes to fall from 0.25 mol dm-3 to 0.125 mol dm-3. If the half-life is short, the reaction is rapid; if the half-life is long, the reaction is slow. The half-life is independent of concentration. All first order reactions have constant half-lives at a given temperature.

Half-life and Radioactive Decay Radioactive decay is an example of a first order reaction because its rate is independent of the concentration of the radioactive material. Many radioactive elements have very long half-lives and scientists use these to help to work out the age of rocks in the Earth‟s crust.   

Carbon-14 is radioactive. Radioactivity of carbon-14 halves every 5730 years. The difference between the historical ratio and that measured in the mass spectrometer is the basis of carbon-14 i.e. the percentage abundance of radioactive isotopes is measured.

Half-life in Second Order Reactions The half-life of a first order reaction is independent of the initial concentration of the reactants. However, the half-life of a second order reaction does depend on the initial concentrations of the reactants.

The graph above shows the change in concentration of HI with time for the reaction: 2HI(g) → H2(g) + I2(g) (@508⁰C) Page 6 of 20

The initial HI concentration is 0.10 mol dm-3 and it takes 125 seconds for the concentration to drop to 0.05 mol dm-3 – the half-life for the first part is 125 seconds. Think of 0.05 mol dm-3 as the new “initial concentration”, which takes 250 seconds to fall to 0.0250 mol dm-3. So in second order reactions, halving the initial concentration doubles the half-life.

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Half-life is inversely proportional to the initial concentration of the reactants. T1/2 =

1 𝑘 × 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑖𝑛𝑖𝑡𝑖𝑎𝑙

Specification Reference: 

Present and interpret the results of kinetic measurements in graphical form, including concentration-time and rate-concentration graphs.

Concentration-Time Graphs The simplest type of relationship between variables is the linear one of general form y = mx + c, where m is the slope or gradient and c the intercept on the y-axis.   

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The gradient of a curve is found by drawing a tangent. The gradient of the tangent to the concentration–time graph at a particular time gives the rate at that moment (concentration). The initial rate of reaction is found from the gradient of the tangent to the concentration–time graph at t = 0. It is often difficult to monitor reaction rates or concentration continuously. In practice you carry out a series of reactions where you vary the initial concentration of each reactant in turn. You then plot a graph of initial rate against initial concentration for each reactant. You can find the orders from the shape of each graph, and the value of k and units by substituting your measurements into the rate equation: k=

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𝑅𝑎𝑡𝑒 𝐴𝑥𝐵𝑦

If it is a horizontal line, it means that the reaction is not taking place (rate of reaction = 0). If a straight line sloping downwards is obtained, the slope is constant. This means that the rate is constant. This only occurs when the reaction is zero order. If a downward curve is obtained, with decreasing slope, the rate is decreasing as [A] falls. Therefore, the reaction is first order or greater.

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Rate-Concentration Graphs If the rate of a reaction, or some quantity that is proportional to the rate, is plotted against the concentration of one reactant, the order with respect to that reactant can be found. 1/time for the reaction to proceed to a certain point is often used as a measure of the rate.  If the graph of rate (or 1/time) against [reactant] is a straight line, the reaction is first order with respect to that reactant.  If the graph of rate (or 1/time) against [reactant]2 is a straight line, the reaction is second order with respect to that reactant.

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Specification Reference: 

Investigate the reaction of iodine with propanone in acid to obtain data for the order with respect to the reactants and the hydrogen ion and make predictions about molecules/ions involved in the rate-determining step and possible mechanism (details of the actual mechanism can be discussed at a later stage in this topic).

Deducing the Order of Reaction from Initial Rate Data To do this, you need data from at least three experiments. For example, data from the reaction below could be used: xA + yB → products Experiment 1 2 3     

[A] /mol dm-3 0.1 0.2 0.2

[B] /mol dm-3 0.1 0.1 0.2

Initial Rate / mol dm-3 p q r

Comparing experiments 1 and 2 showed that [A] has doubled but [B] has stayed the same. If the rate is unaltered (q = p), the order with respect to A is 0. If the rate doubles (q = 2p), the order with respect to A is 1. If the rate quadruples (q = 4p), the order with respect to A is 2. The order with respect to B can be determined in a similar way, by analysing experiments 2 and 3.

Worked Example In the table above, suppose p = 0.0024, q = 0.0096 and r = 0.0096. (1) Determine the partial orders of A and B. From experiments 1 and 2: when [A] doubles, the rate quadruples (0/0096 = 4 x 0.0024). Therefore the order with respect to A is 2. From experiments 2 and 3: when [B] doubles, the rate is unaltered. Therefore, the order with respect to B is 0. (2) Determine the total order. Total Order is 2 + 0 = 2. (3) Write the rate equation. The rate equation is: Rate = k[A]2 (4) Calculate the value of the rate constant. k = rate/[A]2 = 0.0024 mol dm-3s-1/(0.1 mol dm-3)2 = 0.24 mol-1 dm3s-1. Page 9 of 20

Iodine-Propanone Reaction In acid solution iodine reacts with propanone to form iodopropanone and hydrogen iodide. H+(aq) CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + HI(aq) This reaction can be monitored by the following method:  Place 25cm3 of iodine solution of known concentration in a flask and add 25cm3 of sulfuric acid solution (an excess) and 50cm3 of water. Add 5cm3 of pure propanone from a burette, start the clock as you do so.  At noted intervals of time remove 10cm3 of the reaction mixture (aliquots) and quench it by adding it to a slight excess of aqueous sodium hydrogencarbonate. (This reacts with the acid catalyst, stopping the reaction – ensuring no further change in iodine concentration occurs during titration.)  Quickly titrate the remaining iodine against standard sodium thiosulfate solution, using starch as an indicator, near the end-point.  If the starch is added too soon, an insoluble starch-iodine complex is formed and the titre will be too low.  Repeat by removing samples at regular intervals. The rate equation is of the form: Rate = k[propanone]x[H+]y[I2]z Because a large excess of propanone was taken, the concentration of propanone, written as [propanone], remains approximately constant. Even though acid is produced in the reaction, the original amount of acid catalyst was large and so [H +] remains effectively constant. This means that the rate equation becomes: Rate = k‟[I2]z where k’ = k[propanone]x[H+]y The volume of sodium thiosulfate is proportional to the amount of iodine and hence the iodine concentration. So a graph of volume of sodium thiosulfate against time will have the same shape as the graph of [I2] against time. The result of this experiment is shown in the graph below: As this graph is a straight line, the slope and hence the rate is constant and so the reaction is zero order with respect to iodine. A conclusion that can be drawn from this is that iodine must enter the mechanism after the rate-determining step.

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Experiment 1 2 3 4 5

[CH3COCH3] /mol dm-3 0.1 0.1 0.1 0.2 0.1

[I2] /mol dm-3 0.1 0.2 0.2 0.1 0.1

[H+] /mol dm-3 0.1 0.1 0.2 0.1 0.2

Relative Rate 2 2 4 4 4

The table above is an example of some results obtained from this experiment. We can conclude that:  Doubling the concentration of iodine has no effect on the relative rate of reaction (experiments 1 and 2)  Doubling the concentration of propanone doubles the relative rate of reaction (experiments 1 and 4)  Doubling the concentration of hydrogen ions doubles the relative rate of the reaction (experiments 1, 3 and 5) Mechanism for the Reaction of Iodine and Propanone in Acid Solution The reaction involves various steps and the slowest step, the rate-determining step, does not involve iodine. This is why iodine does not appear in the rate equation. Hydrogen ions act as a catalyst – they are regenerated during the reaction. A possible mechanism for the reaction of iodine and propanone in acid solution is given below. Step 1 An H+ ion protonates the oxygen atom in propanone: (CH3)2C=O + H3O+ ⇌ (CH3)2C=O+H + H2O This is a reversible reaction involving proton transfer (acid-base reaction). Remember that a protonated water molecule, H3O+, behaves as an H+ ion. Step 2 The electrons in the C=O bond partly shift to form a carbocation – i.e. the positive charge is transferred from the oxygen to the carbon: (CH3)2C=O+H ⇌ (CH3)2C+-OH Step 3 This carbocation loses a proton and slowly changes into the enol form. The enol has both alkene and alcohol functional groups and is isomeric with the original ketone: (CH3)2C+-OH + H2O ⇌ CH3C(OH)=CH2 + H3O+

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This involves breaking a strong C-H bond, hence this step has a high activation energy and slow speed. The positive charge on the adjacent carbon of the carbocation facilitates in „pulling‟ the C-H bond pair to form the C=C bond and releases the proton to form an H3O+ ion. The rate of formation of the enol thus depends on the concentrations of the ketone and the acid. Step 4 The iodine molecule acts as an electrophile and undergoes a quick electrophilic addition reaction (like other alkenes). This produces a protonated iodoketone: CH3C(OH)=CH2 + I2 → CH3C(=O+H)-CH2I + IStep 5 A water molecule then rapidly removes the proton in another acid-base reaction to form the iodoketone: CH3C(=O+H)-CH2I + H2O → CH3COCH2I + H3O+

In this example the low, rate-determining steps 1, 2 and 3 are the first stage in the reaction and need only CH3COCH3 and H+. The remaining steps happen very quickly. This is also the case when oxygen and hydrogen bromide react together at 700K. However, the slowest reaction is not always the first. In the reaction between bromide ions and bromate (V) ions in acid solution the most likely mechanism is that HBr and HBrO3 are made very rapidly before the third, relatively slow, reaction between the two of them takes place. This is the rate-determining step, and is followed by two more rapid steps to complete the reaction.

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Using a Colorimeter A colorimeter measures the absorption of light during the progress of the experiment. First, the colorimeter has to be calibrated using standard iodine solutions. The reaction mixture containing iodine will be light brown in colour and as the reaction proceeds the solution becomes paler and more light is transferred (when using a colorimeter there is no need for starch indicator). The chosen filter should let through only the wavelength to be absorbed by the coloured iodine solution. Since iodine is brown-red, a bluegreen filter is used.

Specification Reference:  Deduce from experimental data for reactions with zero, first and second order kinetics: (i) Half-life (the relationship between half-life and rate constant will be given if required) (ii) Order of Reaction (iii) Rate Equation (iv) Rate-determining step related to reaction mechanisms (v) Activation energy (by graphical methods only; the Arrhenius equation will be given if needed).

Rate-Determining Steps Reaction do not just happen whenever all the relevant molecules collide at once – they happen in steps. The slowest step controls how fast the overall reaction occurs – it is called the rate-determining step. Kinetic measurements establish the order of the reaction for each species. For a reaction between A, B and C the rate equation could be: Rate = k[A][C]2 Since substance B does not occur in the rate equation, any step involving molecule B must be fast. This rate equation demonstrates that:  The reaction is first order with respect to A – so A is involved in the rate determining step.  The reaction is second order with respect to substance C – so two moles of C are involved in the rate-determining step. Page 13 of 20

Activation Energy and Catalysts When reactant molecules collide, it may result in a chemical reaction. There is an energy requirement before this can happen. The activation energy is the minimum energy needed by reactant particles (molecules or ions) before products can form.

The diagram shows that the reaction is: • Exothermic – the products are at a lower energy level than the reactants • Subject to an energy barrier, the activation energy, in route 1 • Able to follow an alternative pathway in route 2, with a lower barrier A catalyst increases the reaction rate by providing an alternative reaction pathway with a lower activation energy. Such catalysed reactions are faster. Catalysts and reactants can be in the same physical state/phase, called homogeneous - for example, all liquids. Or they can be in different states/phases, when they are called heterogeneous. Process Haber Synthesis Catalytic Converter on car Contact Process, Sulfuric Acid manufacture Esterification

Reactants Nitrogen, Hydrogen Exhaust Gases

Catalyst Iron

Type of Catalysis Heterogeneous

Platinum

Heterogeneous

Sulfur Dioxide, Oxygen Gases

Vanadium (V) Oxide

Heterogeneous

Solutions of Acid, Alcohol

Hydrogen Ions

Homogeneous

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Most catalysts are transition metals (or their compounds) because they have variable oxidation states and can alter the number of bonds available to reactants.

Inorganic catalysts are used to catalyse a wide range of different reactions. Although catalysts are not permanently altered during the reactions that they catalyse, they can be poisoned by some impurities and will not work again. Catalytic Converters A catalytic converter removes pollutant gases from the exhaust by oxidising or reducing them. The exhaust gases pass through a converter containing a precious metal catalyst, usually an alloy of platinum or rhodium. Several reaction may take place. NOx and CO may take part in a redox reaction which removes both of them at the same time. NOx oxidises CO to CO2, and is itself reduced to harmless N2 gas. 2NO(g) + 2CO(g) → N2(g) + 2CO2(g) CO and CxHy are oxidised by air: 2CO(g) + O2(g) → 2CO2(g) C7H16(g) + 11O2(g) → 7CO2(g) + 8H2O(g)

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For all three of these reaction to happen, you need a „three-way converter‟. An oxygen monitor is fitted to the engine. This checks the quantity of oxygen going into the engine to make sure there is enough to carry out the oxidation reactions. The overall result of passing exhaust gases through this kind of catalyst system is to convert NOx, CO and CxHy to relatively harmless N2, CO2 and H2O. The catalytic reactions do not start working until the catalyst has reached a temperature of about 200⁰C, so they are not effective until the engine has warmed up. Catalyst systems of this type cost several hundred dollars, mainly because of the high cost of the precious metals they contain. The catalyst is „poisoned‟ by lead, so unleaded fuel must always be used. Unfortunately, the major environmental impact of motor vehicles cannot be solved by catalytic converters – Climate Change. Worked Example 1) Natural gas (methane) and air react exothermically in a Bunsen flame. Explain why mixtures of these gases do not ignite spontaneously at room temperature. The activation energy barrier for this combustion reaction is too high at room temperature. Colliding molecules have insufficient energy to react. Once ignited, the Bunsen flame heats arriving gases before they react so that enough molecules have energy equal to or greater than the minimum required for reaction. Effect of Temperature on Rates When the temperature increases, the rate of a reaction increases too because the rate constant increases. The rate constant k is only a constant for a particular temperature. Changing the temperature changes the value of k because the proportion of molecules that have the required energy (greater than the activation energy) is increased and the colliding particles have a greater average energy. The Arrhenius equation shows the relationship between the rate constant k and the temperature T (in kelvin). The logarithmic (ln) form of this equation is: lnk =

−𝐸𝐴 𝑅𝑇

+ a constant Page 16 of 20

• EA is the activation energy for the reaction • R is the gas constant and has the value 8.31 J K-1 mol-1 • T is the kelvin temperature (absolute temperature) − • Gradient =

Examiner‟s Tip

Remember – units for activation energy are kJ mol-1

Working out the value of an activation energy. This shows that a plot of lnk against 1/T gave a straight line whose −𝐸 gradient is 𝑅 𝐴

Specification Reference: 

Investigate the activation energy of a reaction, e.g. oxidation of iodide ions by iodate (V)

Investigating the Activation Energy of a Reaction If we know the rate equation for a reaction, it is easy to calculate the rate constant k using the rate of reaction for a known concentration of reactants. Calculating the activation energy requires the results from experiments at a range of different temperatures, to give values of the rate constant k at each temperature. A suitable reaction to study is the oxidation of iodide ions by iodate (V) ions in acidic solution. The equation is: IO3-(aq) + 5I-(aq) + 6H+(aq) → 3I2(aq) + 3H2O(l) A small known amount of sodium thiosulfate is added at the start of the reaction together with starch as the indicator. The iodine released by the oxidation of the iodide ions first reacts with the thiosulfate, so the mixture remains colourless initially but then suddenly turns starch blue-black when all the thiosulfate has reacted. This is an example of a clock reaction with a built-in time delay that depends only on the concentration of the iodide and iodate ions. The relative initial rate of production of iodine can be found from 1/t, where t is the time delay. The value for the activation energy can be found graphically as described above.

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Specification Reference: 

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Apply knowledge of the rate equations for the hydrolysis of halogenoalkanes to deduce the mechanisms for primary and tertiary halogenoalkane hydrolysis and to deduce the mechanism for the reaction between propanone and iodine. Demonstrate that the mechanisms proposed for the hydrolysis of halogenoalkanes are consistent with the experimentally determined orders of reactions, and that a proposed mechanism for the reaction between propanone and iodine is consistent with the data from the experiment in 4.3e. Use kinetic data as evidence for SN1 or SN2 mechanisms in the nucleophilic substitution reactions of the halogenoalkanes.

Hydrolysis of Halogenoalkanes There are three different types of halogenoalkane. They can all be hydrolysed (split) by heating them with sodium hydroxide – but they react using different mechanisms.

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Primary Halogenoalkanes – halogen is joined to a carbon with one alkyl group attached.

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Secondary Halogenoalkanes – halogen is joined to a carbon with two alkyl groups attached.

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Tertiary Halogenoalkanes – halogen is joined to a carbon with three alkyl groups attached.

Two substitution mechanisms are possible when an iodoalkane reacts with aqueous alkali: R-I + OH- → R-OH + IOnly the experimental rate data can show which mechanism actually takes place.

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The single step illustrated for the substitution of 1-iodopropane, a primary halogenoalkane, involves two different species – both the hydroxide ion and the primary halogenoalkane.  The reaction will be second order – it depends on the concentration of both the hydroxide ion and the primary halogenoalkane.  The hydroxide ion joins onto the central carbon atom at the same time as the halogen atom was leaving.  Part of the energy required to break the C-Halogen bond was supplied by the energy released on producing the C-OH bond.  Calculations show that the approach of the hydroxide ion along the line of centres of the carbon an bromine atoms is that of lowest energy requirement. So, rate = k[RI][OH-] We call this an SN2 mechanism, meaning substitution/nucleophilic/second order. Tertiary halogenoalkanes hydrolyse by the alternative SN1 mechanism, meaning substitution/nucleophilic/first order.

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The C-halogen bond heterolytically breaks first (slow step) forming a tertiary carbocation – which are relatively stable. Followed by attack by the hydroxide ion (fast step). The slow step involves only one species and does not depend on the concentration of hydroxide ions. Hence the reaction is first order overall.

The mechanism must be consistent with the evidence:  If the reaction is second order overall it must involve two different species  If the reaction is first order overall (only one species in the rate equation) then this is the rate-determining step and it must be a two-step reaction. Nucleophile – Electrophile – Substitution Hydrolysis -

a species that is attracted to an electron-deficient centre, of an atom, molecule or ion that donates a pair of electrons to form a new covalent bond. an atom, molecule or ion that accepts a pair of electrons to form a new covalent bond. An electrophile is therefore a species that bonds to an electron-rich site in a molecule. one atom or group of atoms is replaced by another atom or group of atoms. when a molecule is broken down by the addition of water. This may involve water, acid or alkali as the inorganic reagent.

Summary Page 19 of 20

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Primary Halogenoalkanes only react by SN2 mechanism. Secondary Halogenoalkanes can react by both the SN1 and SN2 mechanisms – it depends what mood they‟re in. Tertiary Halogenoalkanes only react by the SN1 mechanism. It is an SN2 mechanism if the rate is dependent on the concentration of both the reactants and the order with respect to each is 1. It is an SN1 mechanism if the rate is only dependent on the concentration of the halogenoalkane.

Other Experimental Methods to Determine Rate of Reaction Titration If the concentration of a reactant or product can be estimated by a titration, the reaction can be followed using this technique:  Measure out samples of the reactants with known concentration.  Mix them together, start a clock and stir the mixture thoroughly.  At regular time intervals, withdraw samples using a pipette and quench (stop) the reaction. Quenching can usually be achieved either by adding the solution from the pipette to ice-cold water or to a solution that reacts with one of the reactants, to prevent further reaction from taking place. The time at which half the contents of the pipette have been added to the quenching solution is noted.  The quenched solution is then titrated against a suitable standard solution. The titre is proportional to the concentration of the reactant or product being titrated. This method can be used when an acid, alkali or iodine is a reactant or product. Acids can be titrated with a standard alkali, alkalis with a standard acid and iodine with a standard solution of sodium thiosulfate. Infrared Spectroscopy Infrared spectroscopy can be used in a similar way to colorimetry. The spectrometer is set at a particular frequency and the amount of infrared radiation absorbed at that frequency is measured at regular time intervals. The oxidation of propan-2-ol to propanone by acidified potassium dichromate (VI) can be followed by setting the spectrometer at 1700cm-1 (the absorption frequency due to the stretching of the C=O bond) and measuring the increase in absorption as the CHOH group is oxidised to the C=O group. Conductimetric Analysis If the number of ions changes, so will the electrical conductivity.

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