Find out how the rate of hydrolysis of an organic halogen compound depends on the identity of the halogen atom, and the nature of the carbon-hydrogen 'skeleton'.

Example 2 - the reaction of iodide ions, I-, with peroxodisulphate (VI) ions, S2O82-

The equation for the reaction is,

S2O82-(aq) + 2I-(aq) 2SO42-(aq) + I2(aq)

Experiments show that the rate equation for the reaction is,

rate = k [S2O82-(aq)] [I-(aq)]

So this reaction is first order with respect to S2O82-, first order with respect to I- and second order overall.

Notice that in Example 2, the order of the reaction with respect to I- is one, even though there are two I- ions in the balanced equation for the reaction. This raises an important point - the rate equation cannot be predicted for a reaction from its balanced equation. This may work (as in Example 1 above), but it often won't. The only way to find the rate equation for a reaction is by doing experiments to find the effect of varying the concentrations of reactants. This point becomes very clear in Examples 3 and 4.

Example 3 - the reaction between bromide ions, Br -, and bromate (V) ions, BrO3-

This reaction takes place in acidic solution:

BrO3-(aq) + 5Br-(aq) + 6H+(aq) 3Br2(aq) + 5H2O(l)

Experiments show that the rate equation for the reaction is,

rate = k [BrO3-] [Br-] [H+]2

So the reaction is first order with respect to both BrO3- and Br-, and second order with respect to H+. Notice that the orders do not correspond to the numbers in the balanced equation.

Example 4 - the reaction of propanone with iodine

This reaction is catalysed by acid,

CH3COCH3(aq) + I2(aq) acid catalyst CH3COCH2I(aq) + H+(aq) + I-(aq)

The rate equation found by experiments is,

rate = k [CH3COCH3] [H+]

(Notice that the acid catalyst, H+, appears in the rate equation even though it is not used up). So the reaction is first order with respect to both CH3COCH3 and H+. Notice that [I2] does not appear in the rate equation even though iodine is one of the reactants. The reaction is zero order with respect to I2 (IV).

Finding the order of reaction:

In order to find the order of reaction, experiments must be carried out. Most reactions involve more than one reactant, and in this case several experiments must be carried out, to find the order with respect to each reactant separately. The variables have to be controlled so that the concentration of only one substance is changing at a time, and all measurements must be taken to the same temperature.

Looking again at the decomposition of hydrogen peroxide example (in the presence of the enzyme catalase).

2H2O2(aq) catalase 2H2O(l) + O2(g)

If we are looking at the effect of changing the concentration of hydrogen peroxide, there is no need to worry about the catalase - it is an enzyme, so it doesn't get used up, and its concentration does not change. But if looking at the effect of varying the concentration of catalase, the concentration of hydrogen peroxide must be controlled so it is kept constant - otherwise there will be two variables changing at the same time and the results will therefore be two-tailed. The approach is to do several experiments to measure the initial rate of the reaction, keeping the concentration of hydrogen peroxide constant each time, while varying the concentration of catalase. Another way of controlling the concentration of a reactant is to have a large excess of it, so that over the course of the experiment, the concentration does not change significantly.

Once the set data has been collected (for the effect of changing the concentration of a particular reactant), there are several methods, which can be used to find the order with respect to that reactant.

- THE PROGRESS CURVE METHOD

A progress curve shows how the concentration of a reactant (or product) changes as the reaction proceeds. A progress curve for the hydrogen peroxide decomposition is shown below.

Figure 10

Figure 11 shows how a progress curve can be used to find the rate of the reaction for different concentrations. The gradient of the tangent gives the rate of reaction for a particular concentration of hydrogen peroxide. The order can then be found with respect to hydrogen peroxide as in the initial rates method (below).

Figure 11 (V)

2 - THE INITIAL RATE METHOD

a) By drawing tangents at the origin of different progress curves

This is the method used in the hydrogen peroxide investigation as shown previously in Figure 7. Several experimental 'runs' are carried out at different concentrations. For each run, the initial rate can be found graphically, as was shown in Figure 7.

Once the initial rate for different concentrations is known, the order can be found. If a graph of initial rate against concentration is a straight line (as in Figure 9), the reaction is first order. If a graph of initial rate against (concentration)2 is a straight line, the reaction is second order. If the rate does not depend on concentration at all, it is zero order.

b) Using the reciprocal of the reaction time as a measure of the rate

One way of measuring the initial rate of a reaction is to measure how long the reaction takes to produce a small, fixed amount of one of the products. The time taken is called the reaction time. If the rate is high (the reaction proceeds quickly), the reaction time will be small. If the rate is low (the reaction proceeds slowly), the reaction time will be large.

A good example of this method is the reaction between sodium thiosulphate solution and hydrochloric acid.

Na2S2O3(aq) + 2HCl(aq) 2NaCl(aq) + SO2(g) + S(s) + H2O(l)

sodium sulphur

thiosuphate

As the reaction proceeds, solid sulphur forms as a colloidal suspension of fine particles and the mixture becomes cloudy. The reaction flask is placed over a cross on a piece of white paper and viewed from above as shown in Figure 12. The cross is no longer visible when a certain amount of sulphur has formed. The reaction time to reach this point can be measured using different starting concentrations of sodium thiosulphate solution. The volume of each solution and the concentration of the hydrochloric acid must be kept constant in each experiment.

Figure 12

For each starting concentration,

average rate for this stage of the reaction = amount of sulphur needed to obscure the cross

reaction time for sulphur to form

The amount of sulphur needed to obscure the cross will be the same for each experiment, so the rate of reaction is proportional to the reciprocal of the reaction time for the cross to be obscured, 1/t.

average rate ? 1 / t

The shorter the time taken, the faster the reaction; the longer the time taken, the slower the reaction. If a graph of 1/t is plotted against the concentrations of the thiosulphate solutions used, a straight line plot is obtained, showing that the rate of the reaction is proportional to the concentration of sodium thiosulphate.

Rate equations and reaction mechanisms:

Once the rate equation for a reaction is known, it can be linked to the reaction mechanism. Most reaction mechanisms involve several individual steps. The rate equation gives information about the slowest step in the mechanism - the rate-determining step.

It has already been seen that the rate equation for a reaction cannot be predicted from the balanced chemical equation. For instance, considering the reaction of the substance 2-bromo-2-methylpropane with hydroxide ions,

CH3 CH3

? ?

CH3 ? C ? Br + OH- CH3 ? C ? OH + Br -

? ?

CH3 CH3

This reaction is found to be first order with respect to (CH3)3CBr and zero order with respect to OH-. In other words,

rate = k [(CH3)3CBr]

[OH-] is not involved in the rate equation. This suggests that OH- ions are not involved in the slow rate-determining step. The reaction cannot take place by direct reaction of 2-bromo-2-methylpropane itself with OH- ions. Chemists have studied this reaction in detail and they have found that it takes place in two steps.

First, the C?Br bond breaks heterolytically.

CH3 CH3

? ?

CH3 ? C ? Br CH3 ? C+ + Br - Step 1

? ?

CH3 CH3

Because this step only involves (CH3)3CBr, its rate depends only on [(CH3)3CBr], not on [OH-].

The second step involves reaction of the carbocation, (CH3)3C+, with OH-.

CH3 CH3

? ?

CH3 ? C+ + OH- CH3 ? C ? OH Step 2

? ?

CH3 CH3

Like most ionic reactions, this process is very fast, certainly faster than step 1. So the rate of step 1 controls the rate of the whole reaction. That is why the rate of the reaction depends only on the concentration of (CH3)3CBr; the reaction is first order with respect to (CH3)3CBr, but zero order with respect to OH-. Step 1 is the rate-determining step and its rate equation becomes the rate equation for the whole reaction. (VI)

The mechanism in this example involves two steps. Some simple reactions occur in a single step; other complex ones may involve more than two steps. But in every case, once the reaction has been broken down into steps, the rate equation can then be written for each step from its chemical equation. But the overall rate equation for the reaction can only be found by experiment.

The mechanism of enzyme-catalysed reactions:

When the substrate concentration is low, the rate equation for the reaction,

S enzyme P

substrate product

is,

rate = k [E] [S] where [E] is the concentration of the enzyme.

From this, it can be deduced that the rate-determining step must involve one enzyme molecule and one substrate molecule.

E + S rate-determining step ES

enzyme-substrate

complex

The steps that follow this are faster:

ES fast EP fast E + P

enzyme-product

complex

However, if the substrate concentration is high, the rate equation becomes,

rate = k [E]

This is because, with more than enough substrate around, the first step is no longer the rate-determining one. All the enzyme active sites are occupied, and [ES] is constant. What matters now is how fast ES or EP can break down to form the products. The rate of breakdown of ES depends on its concentration, which is effectively [E] since almost all the enzyme is present as ES (VII).

Rate-determining steps:

Some steps in a reaction are slow while the other steps are fast. One reason is that the steps have different energy barriers (activation enthalpies). When the energy barrier is big, few pairs of colliding molecules have enough energy to pass over it and the rate of conversion of reactants into products is slow. Figure 13 compares the enthalpy profiles for reactions with large and small activation enthalpies.

Figure 13

The enthalpy profile will be different for a reaction involving two steps. There will be two activation enthalpies - one for each step.

In this case of 2-bromo-2-methylpropane with hydroxide ions, step 1 (the rate-determining step) has a larger activation enthalpy (Figure 14). Usually, the chemicals in the middle of a reaction mechanism - the intermediates - are at a higher energy than the reactants or products, because they have unusual structures or bonding - like carbocation (CH3)3C+.

In any reaction with several steps, the rate-determining step will be the one with the largest activation enthalpy.

Figure 14

Temperature

Temperature has an important effect on the rate of chemical reactions and we make use of it constantly in our everyday lives, for instance the chemical industry depends heavily on it and there would be no Haber process for making ammonia without it.

When measuring the rate of reaction at different temperatures, for many reactions, the rate is roughly doubled by a temperature rise of just 10 ?C.

Extending the collision theory of reactions:

Thinking of the reaction between nitrogen and hydrogen, for example, to make ammonia in the Haber process:

N2(g) + 3H2(g) 2NH3(g)

The collision theory says that reaction can only occur when N2 and H2 collide. The more frequent the collisions, the faster the reaction. If the pressure is increased, for instance, the N2 and H2 molecules become closer together, so they collide and react more often (which is why the Haber process uses high pressures - to make ammonia more quickly).

What about the effect of temperature? When the temperature is raised, molecules move faster, and therefore collide more frequently. How much more frequently they collide can be calculated - the average speed of molecules is proportional to the square root of the absolute temperature. So for example, if the temperature is increased from 300 K to 310 K, the average speed of the molecules would be expected to increase by a factor of (310/300)1/2, which is approximately 1.016 i.e. an increase of 1.6%. Yet it is known that the rate actually increases by much more than this, sometimes by 200%-300%.

There is more to the effect of temperature than simply making the particles collide more frequently. What matters is not just how frequently they collide, but also with how much energy. Unless the molecules collide with a certain minimum kinetic energy, they just bounce off one another and remain unreacted. In fact, this is what happens to most of the molecules most of the time. In the reaction of N2 and H2 at 300 K, only 1 in 1011 collisions results in a reaction.

Therefore, the collision theory says that reactions occur when molecules collide with a certain minimum kinetic energy. The more frequent these collisions, the faster the reaction.

This energy is needed to overcome the energy barrier, the activation enthalpy, for the reaction. It is the energy that is needed to start breaking the bonds in the colliding molecules so that a collision can lead to a reaction (VIII).

The distribution of energies:

At any temperature, the speeds - and therefore the kinetic energies - of the molecules in a substance are spread over a wide range. Some have high kinetic energies, many have medium energies and some have low energies.

This distribution of kinetic energies in a gas (say mixture of N2 and H2 molecules) at a given temperature is shown in Figure 15 below, and is called the Maxwell-Boltzmann distribution.

Figure 15

As the temperature increases, more molecules move at higher speeds and have higher kinetic energies. Figure 16 shows how the distribution of energies changes when the temperature is increased by 10 ?C, from 300 K to 310 K. It can be seen that there is still a spread of energies, but now a greater proportion of molecules have higher energies.

Figure 16

The following example will show the significance of this for reaction rates.

I will take as my example a reaction whose activation enthalpy, Ea, is +50kJ mol-1, a value typical of many reactions.

I need to think about how many collisions have energy greater than 50kJ mol-1, because these are the collisions that can lead to a reaction. Indeed, each colliding pair of molecules will have energy far less than 50kJ: this is the energy possessed by 6 x 1023 (1 mole) of them.

Figure 17 shows the number of collisions with energy greater than 50kJ mol-1, for the reaction at 300 K. It is given by the shaded area underneath the curve. Only those collisions with energies in the shaded area can lead to a reaction. (The shape of the energy distribution curve for collisions is the same as for individual molecules.)

Figure 17

Looking at the graph in Figure 18, which shows the curves for both 300 K and 310 K, it can be seen that at the higher temperature, a significantly higher proportion of molecules (about twice as many) have energies above 50kJ mol-1. This means that twice as many molecules have enough energy to react - so the reaction goes twice as fast.

Figure 18 (IX)

Catalyst

What is a catalyst?

A catalyst is a substance, which alters the rate of reaction and you can always get them back to the end. They are not changed chemically, though sometimes they may be changed physically. For instance, the surface of a solid catalyst may crumble or become roughened. This suggests that the catalyst is taking some part in the reaction, but is being regenerated.

Only small amounts of a catalyst are usually needed. The catalyst does not affect the amount of product formed, but only the rate at which it is formed. Another interesting feature of catalysts is that they are often specific for one particular reaction. This is particularly true of biological catalysts, called enzymes. These are complex protein molecules, which catalyse the chemical reaction that take place in the living cells of plants and animals. They affect one particular biochemical reaction strongly, but leave a similar reaction almost unaffected.

Types of catalyst:

If the reactants and catalysts are in the same physical state (for example, both are in aqueous solution), the reaction is said to involve homogeneous catalysis. Enzyme-catalysed reactions in cells take place in aqueous solution and are examples of this type of catalysis.

In contrast, many important industrial processes involve heterogeneous catalysis, where the reactants and catalysts are in different physical states. This usually involves a mixture of gases or liquids reacting in the presence of a solid.

Heterogeneous catalysts:

When a solid catalyst is used to increase the rate of a reaction between gases or liquids, the reaction occurs on the surface of the solid. Figure 19 shows an example of heterogeneous catalysis. The reactants form bonds with atoms on the surface of the catalyst - they are adsorbed onto the surface. As a result, bonds in the reactant molecules are weakened and break. New bonds form between the reactants, held close together on the surface, to form the products. This, in turn, weakens the bonds to the catalyst surface, and the product molecules are released.

It is important that the catalyst has a large surface area for contact with reactants. For this reason, solid catalysts are used in a finely divided form or as a fine wire mesh. Sometimes the catalyst is supported on a porous material to increase its surface area and prevent it from crumbling (X).

Figure 19

Catalyst poisoning:

Catalysts can be poisoned so that they no longer function properly. In fact, many substances, which are poisonous to humans, operated by blocking an enzyme-catalysed reaction.

In heterogeneous catalysis, the 'poison' molecules are adsorbed more strongly to the catalyst surface than the reactant molecules and the catalyst becomes inactive. This is the reason why leaded petrol cannot be used in cars fitted with a catalytic converter; lead is strongly adsorbed onto the surface of the catalyst.

How catalysts work:

The collision theory and enthalpy profiles help us to understand how catalysts work.

In a chemical reaction, existing bonds in the reactants must first stretch and break. Then new bonds can form as the reactants are converted to products.

Bond breaking is an endothermic process. A pair of reacting molecules must have enough energy between them to pass over the activation enthalpy barrier before a reaction can occur. If the energy barrier is very high, relatively few pairs of molecules will have enough energy to overcome it and react to form the products - so the reaction is slow.

Catalysts speed up reactions by providing an alternative reaction pathway for the breaking and remaking of bonds that has a lower activation enthalpy.

In the following enthalpy profile, the energy barrier is slower and more pairs of molecules can pass over to form products. This means that the reaction proceeds more quickly. The enthalpy change, ?H, is the same for the catalysed and uncatalysed reactions. Figure 20 shows the enthalpy profile for a catalysed and uncatalysed reaction.

Figure 20

Catalysts and equilibrium:

Catalysts do not affect the position of equilibrium in a reversible reaction nor alter the composition of the equilibrium mixture, but only alter the rate at which the equilibrium is attained.

Homogeneous catalysts:

Homogeneous catalysts normally work by forming n intermediate compound with the reactants (which is why the enthalpy profile for the catalysed reaction above has two humps - one for each step). The intermediate then breaks down to give the product and reform the catalyst.

ORGANIC CHEMISTRY - HALOGENOALKANES

Organic halogen compounds:

Organic halogen compounds have one or more halogen atoms (F, Cl, Br or I) attached to a hydrocarbon chain. They do not occur in nature, but they are useful for all sorts of human purposes.

As with all functional groups, the halogen atom modifies the properties of the relatively unreactive hydrocarbon chain. The simplest examples are the halogenoalkanes (sometimes referred to as haloalkanes), with the halogen atom attached to an alkane chain.

Physical properties:

Figure 21

Looking at Figure 21 above, the carbon-hydrogen bond is polar, but not polar enough to make a big difference to the physical properties of the compound. For example, all haloalkanes are immiscible with water. Their boiling points depend on the size and number of halogen atoms present: the bigger the halogen atom and the more halogen atoms there are, the higher the boiling point, as can be seen from Table 4 below.

Table 4

The influence of halogen atoms on the boiling is important when it comes to designing halogen atoms for particular purposes. If you want a compound with a high boiling point, you have to include a larger halogen atom such as Cl or Br rather than a smaller one such as F: but it is these larger atoms that can cause the greater environmental damage.

Chemical reactions of haloalkanes:

Reactions of haloalkanes involve breaking the C ? Hal bond (Hal stands for any halogen atom). The bond can break homolytically or heterolytically.

Homolytic fission:

Homolytic fission forms radicals. One way this can occur is when radiation of the right frequency (visible or ultraviolet) is absorbed by the haloalkane. For example, with chloromethane:

H H

? ?

H ? C ? Cl + hv H ? C* + Cl*

? ?

H H

chloromethane methyl chlorine

radical radical

this is shortened to:

CH3 ? Cl + hv CH3* + Cl* (XI)

This kind of reaction occurs when haloalkanes reach the stratosphere, where they are exposed to intense ultraviolet radiation. This is how the chlorine radicals that cause so much trouble for ozone in the stratosphere have formed.

Heterolytic fission:

Heterolytic fission is more common under normal laboratory conditions when reactions of haloalkanes tend to be carried out in a polar solvent such as ethanol, or ethanol and water. The C ? Hal bond is already polar, and in the right situation it can break, forming a negative halide ion and a positive carbocation. For example, with 2-chloro-2-methylpropane:

CH3 CH3

? ?

CH3 ? C ? Cl CH3 ? C+ + Cl-

? ?

CH3 CH3

2-chloro-2-methylpropane carbocation chloride ion

Sometimes, ions are not formed by simple bond fission in this way. Instead, heterolytic bond fission is brought about by another, negatively charged, substance reacting with the positively polarised carbon atom, causing a substitution reaction.

Importance of reaction conditions:

For many molecules, the conditions under which the reaction is carried out can determine how a bond breaks. For example, when bromomethane is dissolved in a polar substance, such as a mixture of ethanol and water, the C ? Br bond breaks heterolytically to form ions. However, when it reacts in a non-polar solvent, such as hexane, or in the gas phase at high temperature or in the presence of light, the C ? Br bond breaks homolytically.

Different halogens, different reactivity:

Whether homolytic or heterolytic fission occurs, all reactions of haloalkanes involve breaking the C ? Hal bond. The stronger the bond is, the more difficult it is to break. Figure 22 gives the bond enthalpies of four different types of C ? Hal bond. Note, the higher the bond enthalpy, the stronger the bond.

Figure 22

The great strength of the C ? F bond makes it very difficult to break so fluoro compounds are very unreactive. As you go down the halogen group, the C ? Hal bond gets weaker and weaker, so the compounds get more and more reactive. Chloro compounds are fairly unreactive, and once they have been released into the troposphere they stay there quite a long time. This means that compounds such as CFCs can stay around long enough to get into the stratosphere and begin to damage the ozone layer.

Bromo and iodo compounds are fairly reactive.

Substitution reactions of haloalkanes:

Substitution reactions of haloalkanes are typical. For instance, a substitution reaction takes place between a haloalkane and hydroxide ions, in which the haloalkane is hydrolysed to form an alcohol. With bromobutane, for example:

CH3 ? CH2 ? CH2 ? CH2 ? Br + OH- CH3 ? CH2 ? CH2 ? CH2 ? OH + Br-

The C ? Br bond is polar.

?

? C?+ ? Br?-

?

The oxygen atom of the hydroxide ion is negatively charged.

The partial positive charge on the carbon atom attracts the negatively charged oxygen of the hydroxide ion. A lone pair of electrons on the O atom forms a bond with the C atoms as the C ? Br bond breaks (XII).

This reaction involves heterolytic fission - ions, rather than radicals, are formed. A free carbocation is not formed in this case, because the OH- ion attacks at the same time as the C ? Br bond breaks - it is a smooth, continuous process.

Curly arrows have been used to show the movement of electrons during the reaction. In this case a full-headed arrow is used to show the movement of a pair of electrons (the half-arrow shows the movement of single electrons in radical reactions).

Haloalkanes give substitution reactions with many different reagents, as well as hydroxide ions. What is needed is a group carrying a pair of electrons to start forming a bond to the carbon atom.

Attacking groups like these, which can donate a pair of electrons to a positively charged carbon atom to

form a new covalent bond, are called nucleophiles. They can be either negative ions like OH- or molecules with a lone pair of electrons available for bonding, for example, H2O and NH3. Table 5 shows some common nucleophiles.

Table 5

The carbon atom attacked by the nucleophile may be part of a carbocation and carry a full positive charge, or it may be part of a neutral molecule (as with the bromobutane reaction above) and carry a partial positive charge as a result of bond polarisation.

If we write X- as a general symbol for any nucleophile, the nucleophilic substitution process can be described by:

In general, when any nucleophile X- reacts with a general haloalkane R ? Hal, the reaction that occurs is

R ? Hal + X- R ? X + Hal-

(Where R represents the alkyl group, and X represents the halogen atom).

The structures of the substances, which were investigating are:

H H H H

? ? ? ?

Cl ? C ? C ? C ? C ? H C4H9Cl 1-chlorobutane (primary substance)

? ? ? ?

H H H H

H H H H

? ? ? ?

Br ? C ? C ? C ? C ? H C4H9Br 1-bromobutane (primary substance)

? ? ? ?

H H H H

H H H H

? ? ? ?

I ? C ? C ? C ? C ? H C4H9I 1-iodobutane (primary substance)

? ? ? ?

H H H H

CH3

?

H ? C ? Br C4H11Br 2-bromo-2-methylpropane (tertiary substance)

?

CH3

The structures of the other substances, which were used in the investigation are:

H H

? ?

H ? C ? C ? O ? H C2H5OH Ethanol

? ?

H H

Silver nitrate solution, AgNO3, will also be used.

Two different experiments were carried out - both of different methods. The first involved the primary substances, and the second involved the tertiary substance.

During the first experiment, the rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane were compared. The equations for the hydrolysis reactions that occurred are as follows:

-chlorobutane -

CH3 ? CH2 ? CH2 ? CH2 ? Cl + OH- CH3 ? CH2 ? CH2 ? CH2 ? OH + Cl-

The rate of reaction was followed by carrying it out in the presence of silver ions, so that any halide ions produced formed a silver halide precipitate:

Ag+(aq) + Cl-(aq) AgCl(s)

-bromobutane -

CH3 ? CH2 ? CH2 ? CH2 ? Br + OH- CH3 ? CH2 ? CH2 ? CH2 ? OH + Br-

Ag+(aq) + Br-(aq) AgBr(s)

-iodobutane -

CH3 ? CH2 ? CH2 ? CH2 ? I + OH- CH3 ? CH2 ? CH2 ? CH2 ? OH + I-

Ag+(aq) + I-(aq) AgI(s)

During the second experiment, the rate of reaction was measured via a conductivity experiment. The equations for the reactions that occurred are:

2-bromo-2-methylpropane -

CH3 CH3

? ?

CH3 ? C ? Br + OH- CH3 ? C ? OH + Br -

? ?

CH3 CH3

The chemical techniques that were used are that of a substitution reaction (for the hydrolysis of the primary haloalkanes), and a conductivity method (for measuring the rate of reaction for the tertiary haloalkane).

The chemical theory behind substitution reactions of haloalkanes is explained in detail on page ****. A general equation for the hydrolysis of haloalkanes is:

R ? X + H2O R ? OH + H+ + X-

(Where R = alkyl group; X = halogen atom).

The rate of the reaction was carried out in the presence of silver ions, so that any halide ions produced formed a silver halide precipitate.

Ag+(aq) + X-(aq) AgX(s)

Since halogenoalkanes are insoluble in water, ethanol was added to act as a common solvent for the halogeno-compounds and silver ions.

The conductance method of liquid level measurement is based on the electrical conductance of the measured material, which is usually a liquid that can conduct a current with a low-voltage source (normally <20 V). Hence the method is also referred to as a conductivity system. Conductance is a relatively low-cost, simple method to detect and control level in a vessel (XIII).

The following is a leaflet guiding how to operate the conductance meter:

(XIV)

Quantities of chemical substances used:

Experiment 1 - comparing the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane

- 1-chlorobutane - 4 pipette drops

- 1-bromobutane - 4 pipette drops

- 1-iodobutane - 4 pipette drops

- Ethanol - 6 cm3 (2 cm3 per test tube)

- Silver Nitrate, 0.05 M - 3 cm3 (1 cm3 per test tube)

Experiment 2 - rate of reaction of 2-bromo-2-methylpropane

- 2-bromo-2-methylpropane - 4.6 cm3

- Ethanol - 310 cm3

- Distilled water - 160 cm3

Large quantities of the substances were not used, as time was an important issue and if there are greater quantities then the rate of hydrolysis/reaction will also increase, resulting in less time to carry out the experiment.

Apparatus:

Experiment 1 -

- Beaker - 250cm3

- Graduated pipette (with pipette filler) - 10 cm3

- Thermometer - 0-100 ?C

- Test tube (fitted with cork) x 4

- Test tube rack

- Waterproof marker

- Measuring cylinder - 10 cm3

- Ethanol, C2H5OH

- 1-chlorobutane, C4H9Cl

- 1-bromobutane, C4H9Br

- 1-iodobutane, C4H9I

- Silver nitrate solution, AgNO3

- Stop-clock (digital display)

- Water-bath (thermostatically controlled) - set at 60 ?C

- Protective wear - eye/body/hand protection in the form of laboratory goggles, coat and protective plastic gloves

The equipment chosen were such that accuracy, precision and reliability are ensured. For example, a pipette was used because of its high degree of accuracy in quantitative analysis. It has an error of 0.06 cm3 if used correctly (i.e. if it is allowed to drain and retain the last drop).

Experiment 2 -

- 'Primo' conductivity meter and cell

- Calibration solution, HI 70032P

- Pipette (with pipette filler) - 25 cm3

- Volumetric flask (with stopper) - 250 cm3

- Boiling tube (with stopper) x 2

- Stop-clock (digital display)

- Ethanol, C2H5OH

- Distilled water

- 2-bromo-2-methylpropane, C4H11Br

- Glass stirring rod

- Protective wear - eye/body/hand protection in the form of laboratory goggles, coat and protective plastic gloves

Once again, the equipment chosen in this experiment were such that accuracy, precision and reliability are ensured. For example, a volumetric flask was used because of its high degree of accuracy in quantitative analysis. It has an error of 0.2 cm3, as quoted by British Standards. It is a low figure, and can therefore be discarded

RISK ASSESSMENT

-CHLOROBUTANE/1-BROMOBUTANE/1-IODOBUTANE

These substances are highly flammable substances, as well as harmful substances. They are harmful by inhalation, indigestion, and contact with the skin. 1-chlorobutane is narcotic in high concentration and may be irritating to the eyes and skin.

They are dangerous with sodium - an explosive reaction may occur if they come into contact with this substance.

If any of the substances are swallowed, the mouth should be washed out thoroughly and a glass or two of water should be given. Medical attention should also be sought.

If any vapours of the above substances are inhaled the victim must be removed to fresh air to rest, and kept warm. If this is not the case and it is more than a sniff, then medical attention should be sought. If the vapour gets into the eyes, the affected eye should be flooded with water for 10 minutes, and medical attention should be considered (XV).

If the substance has been spilt on the skin, the affected area should be washed thoroughly with soap and cold water. If this spill is on the clothes then the contaminated clothing should be removed and put outside.

If the spill is not in any of these areas but in the laboratory then the area of the spill should be ventilated as soon as possible. The contamined area should then be covered with mineral absorbent, and this solution cleared up and dispersing agent or washing-up liquid spread over the mixture working to an emulsion. The waste should finally be washed with excess water and poured down the foul-water drain.

These substances should be stored with Flammable Liquids (FL).

2-BROMO-2-METHYLPROPANE

2-bromo-2-methylpropane is also a highly flammable substance. Its NFPA rating (scale 0-4) is, health = 2, fire = 3, reactivity = 0 (XVI).

If this substance comes into contact with the eyes, the eyelids must be held apart and flushed promptly with constant flowing water for at least 20 minutes. Medical attention must be sought immediately.

If the substance is contacted with the skin, all contaminated clothing must be removed. The skin should be washed thoroughly with mild soap and plenty of water for at least 15 minutes. Clothing should be washed before re-used. Medical attention should be sought if irritation occurs.

In case of inhalation, the patient should be removed to fresh air. He/she should be kept quiet and warm. Artificial respiration should be applied if necessary and medical attention should be sought immediately.

If the substance has been indigested, do not induce vomiting.

If swallowed, the mouth should be washed thoroughly with plenty of water and the patient should be given water to drink. Medical attention should be sought immediately.

NOTE: Never give an unconscious person anything to drink.

After handling, wash before eating, drinking or smoking.

If the substance is spilt, the area of spill should be evacuated and personnel kept upwind. All sources of ignition should be shut off. The method for cleaning up the spill includes absorbing the substance on sand or vermiculite and it should be placed in a closed container for disposal. Non-sparking tools should be used if necessary. The area should also be ventilated and the spill site should be washed after material pickup is complete.

The substance should be handled in accordance with good industrial hygiene and safety procedures.

Non-sparking tools and explosion-proof electrical equipment and lighting should be used.

Finally, the substance should be stored in a dry, cool, shaded and ventilated area. It should be kept away from heat, sparks and open flames (XVII).

IMPLEMENTATION

PROCEDURE:

Experiment 1 -

2 cm3 of ethanol was poured into three test tubes marked with the letters A to C. A cross was also marked, at the base of the three test tubes - all three crosses were of the same colour, shape and approximately the same amount of ink was used to draw the crosses. The 2 cm3 of ethanol was inserted into the test tubes using a 25 cm3 graduated pipette. A pipette was used as this is the most accurate instrument for transferring liquids of certain measurements, with only a 0.06 cm3 error.

4 drops of 1-chlorobutane were then pipetted into test tube A, followed by 1-bromobutane into test tube B and finally 1-iodobutane into test tube C. The pipette was washed thoroughly with distilled water and dried after before pipetting the next substance into the next test tube to prevent mixing of the substances, which can affect the precision, accuracy and reliability of the experiment.

5 cm3 of silver nitrate solution, AgNO3, was pipetted into the fourth test tube. All the test tubes were then left to stand at room temperature with their corks fitted for exactly ten minutes.

Meanwhile, a thermostatically controlled water-bath was set to 60 ?C, and a 250 cm3 beaker was filled with distilled water was inserted into the water-bath.

Once the ten minutes had passed, the beaker was removed from the water-bath, 1 cm3 of aqueous silver nitrate was pipetted into test tube A, 1-chlorobutane, and the test tube was inserted into the beaker with a loosely fitted cork. The tube was shaken once to mix the contents. The cork was loosely fitted on the tube as this helped to reduce evaporation.

The stop-clock was immediately started, and a thermometer of 0-100 ?C was inserted into the beaker. A thermometer was inserted into the beaker to observe the temperature of water - the water must stay at a constant temperature of 60 ?C - the thermometer helped to decide whether the water needed to be re-heated to 60 ?C again.

The tube was watched continuously from a plan/birds-eye view, for about ten minutes. The cross is no longer visible when a precipitate has formed, which covers the entire cross, and this is the time that was noted in a table similar to the one below. The thermometer was also observed closely.

HALOALKANE

TIME FOR PRECIPITATE TO APPEAR (SECONDS)

OBSERVATIONS

A - 1-chlorobutane

B - 1-bromobutane

C - 1-iodobutane

The reaction was then carried out with test tube B, 1-bromobutane, and then C, 1-iodobutane. After each of the three experiments, new distilled water was used and pipettes were washed and dried before re-use. This ensured precision.

The results for the hydrolysis of the primary haloalkanes are shown in the table below:

HALOALKANE

TIME FOR PRECIPITATE TO APPEAR (SECONDS)

STARTING COLOUR OF SOLUTION

ENDING COLOUR OF SOLUTION

OBSERVATIONS

A -

-chlorobutane

573

Colourless

White

A white precipitate of silver chloride (AgCl(s)) forms.

B -

-bromobutane

66

Colourless

Cream

Turns cloudy. A cream precipitate of AgBr(s) forms.

C -

-iodobutane

45

Colourless

Yellow

A yellow precipitate of AgI(s) forms.

Each of the test tubes were left and later observed again. Silver chloride showed to have turned black in light, silver bromide turned yellow in light, and there was little or no change in the colour of silver iodide under light.

The rate of the reaction is calculated via the following:

For each haloalkane,

rate = amount of AgNO3 needed to obscure the cross

reaction time for precipitate to form

The amount of AgNO3 needed to obscure the cross was the same for each experiment, so the rate of reaction is proportional to the reciprocal of the reaction time for the cross to be obscured, 1/t.

average rate ? 1 / t

Therefore, for 1-chlorobutane:

rate = 1 / 573

= 0.00175 (3 s.f.)

For 1-bromobutane:

rate = 1 / 66

= 0.0152 (3 s.f.)

For 1-iodobutane:

rate = 1 / 45

= 0.0222 (3 s.f.)

The experiment was then repeated with 1-iodobutane. During the experiment, the concentration of 1-iodobutane was changed, with a constant concentration of OH- - at 2cm3, and the time for the hydrolysis was observed. The concentrations were 6 drops, 8 drops, and 10 drops. The rate graph for this experiment is shown on the next page.

The experiment, using exactly the same procedures, was repeated but this time the concentration of 1-iodobutane was kept constant, at 4 drops, and the OH- concentration was changed. The concentrations used were 2 cm3, 4cm3, 6 cm3 and 8 cm3. A rate graph of the results is shown on page 25.

Experiment 2 - altering the concentration of 2-bromo-2-methylpropane

Before the experiment commenced, the conductivity meter was set up and calibrated, following the steps under 'Operation' and 'Calibration' in the conductivity meter leaflet on page 19. The calibration results were 1382 ppm.

A solvent composed of 40 cm3 of ethanol, and 10 cm3 of distilled water was made up, using a volumetric flask. The volumetric flask was inverted six times so the solution is fully mixed.

The 50 cm3 solvent was then poured into a stoppered boiling tube, and was left to stand for fifteen minutes at room temperature. Meanwhile, the conductivity meter was switched on and mounted in a second, empty tube and was also left to stand at room temperature. The solvent was left for fifteen minutes, as it needs to reach thermal equilibrium. The conductivity was left to stand so the electronics in the meter could warm up.

0.2 cm3 of 2-bromo-2-methylpropane was then pipetted into the solvent, and the mixture was stirred vigorously with a glass stirring rod. This was done to ensure that the solution was fully mixed and that precision was secured.

The conductivity probe was then removed from its tube and was immersed in the mixture. The conductivity of the solution, in ppm (parts per million), was then measured at exactly thirty-second intervals for eight minutes. The results were recorded in a table similar to the one below.

The exact same procedure was then repeated for the remaining concentrations of 2-bromo-2-methylpropane. These concentrations were 0.4, 0.6, 0.8 and 1.0 cm3.

The results obtained are shown in the following tables:

0.2 cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

40

04.30

23

01.00

62

05.00

35

01.30

77

05.30

47

02.00

89

06.00

64

02.30

93

06.30

69

03.00

00

07.00

72

03.30

05

07.30

76

04.00

12

08.00

80

0.4 cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

18

04.30

443

01.00

80

05.00

468

01.30

220

05.30

490

02.00

259

06.00

522

02.30

298

06.30

545

03.00

335

07.00

571

03.30

370

07.30

590

04.00

407

08.00

617

0.6 cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

90

04.30

630

01.00

290

05.00

668

01.30

361

05.30

695

02.00

417

06.00

725

02.30

462

06.30

754

03.00

515

07.00

775

03.30

550

07.30

804

04.00

591

08.00

825

0.8cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

254

04.30

826

01.00

400

05.00

865

01.30

470

05.30

907

02.00

554

06.00

948

02.30

610

06.30

984

03.00

668

07.00

015

03.30

725

07.30

50

04.00

774

08.00

083

.0 cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

460

04.30

049

01.00

565

05.00

088

01.30

661

05.30

132

02.00

747

06.00

170

02.30

825

06.30

205

03.00

890

07.00

235

03.30

948

07.30

260

04.00

000

08.00

288

The initial rate for each concentration is calculated as follows:

Rate = Gradient

Gradient = Y2 - Y1

X2 - X1

= [reactants at Y2] - [reactants atY1] (where [ ] refers to the concentration)

t2 - t1 (where t refers to the time)

Rate = ?[reactants] (where ? refers to the change)

?t

The initial rate will be taken, therefore the gradient line will start at zero, and thus Y1 and X1 will equal 0. From this it can be shown that,

Rate = Y2 - 0

X2 - 0

= Y2

X2

Therefore, for concentration 0.2 cm3:

rate = 80 / 60

= 1.33 ppm

For concentration 0.4 cm3:

rate = 250 / 60

= 4.16 ppm

For concentration 0.6 cm3:

rate = 245 / 30

= 8.16 ppm

For concentration 0.8 cm3:

rate = 310 / 30

= 10.30 ppm

For concentration 1.0 cm3:

rate = 550 / 30

= 18.30 ppm

As can be seen from the rate graph, the line is not straight - the circled crosses indicate anomalies. This may have been due to procedural error or error in measuring quantities of the substances, or it may have been an error in the reading of the conductance meter (the meter may not have been positioned properly) - this was my fist time in handling a conductance meter. This is explained further in the evaluation of evidence page on 42. Therefore the experiment was repeated to eliminate all anomalies. The following results show the repeated experiment.

0.2 cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

90

04.30

294

01.00

37

05.00

310

01.30

70

05.30

326

02.00

96

06.00

344

02.30

217

06.30

355

03.00

243

07.00

365

03.30

262

07.30

374

04.00

280

08.00

383

0.4 cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

40

04.30

505

01.00

205

05.00

536

01.30

257

05.30

560

02.00

310

06.00

584

02.30

356

06.30

607

03.00

400

07.00

629

03.30

438

07.30

652

04.00

474

08.00

671

0.6 cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

205

04.30

662

01.00

301

05.00

700

01.30

380

05.30

735

02.00

446

06.00

769

02.30

500

06.30

800

03.00

545

07.00

827

03.30

585

07.30

850

04.00

626

08.00

875

0.8cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

295

04.30

826

01.00

393

05.00

865

01.30

470

05.30

907

02.00

554

06.00

948

02.30

610

06.30

984

03.00

668

07.00

015

03.30

725

07.30

050

04.00

774

08.00

083

.0 cm3 concentration:

MINUTES (30 SEC INTERVALS)

PPM

MINUTES (30 SEC INTERVALS)

PPM

00.30

396

04.30

049

01.00

540

05.00

088

01.30

661

05.30

132

02.00

747

06.00

170

02.30

825

06.30

205

03.00

890

07.00

235

03.30

948

07.30

260

04.00

000

08.00

288

The initial rate for each concentration is calculated as follows:

Rate = Gradient

Gradient = Y2 - Y1

X2 - X1

= [reactants at Y2] - [reactants atY1] (where [ ] refers to the concentration)

t2 - t1 (where t refers to the time)

Rate = ?[reactants] (where ? refers to the change)

?t

The initial rate will be taken, therefore the gradient line will start at zero, and thus Y1 and X1 will equal 0. From this it can be shown that,

Rate = Y2 - 0

X2 - 0

= Y2

X2

Therefore, for concentration 0.2 cm3:

rate = 196 / 60

= 3.20 ppm

For concentration 0.4 cm3:

rate = 398 / 60

= 6.63 ppm

For concentration 0.6 cm3:

rate = 294 / 30

= 9.80 ppm

For concentration 0.8 cm3:

rate = 395 / 30

= 13.16 ppm

For concentration 1.0 cm3:

rate = 516 / 30

= 17.20 ppm

Experiment 2 - altering the concentration of the OH-

The exact same procedures as those on page 26 were used, but this time the concentration of 2-bromo-2-methylpropane was constant, and the concentration of the ethanol and distilled water solvent was changed. The concentration of 2-bromo-2-methylpropane remained at 0.4 cm3, while the concentration of the OH- used was - 35 cm3 ethanol and 15 cm3 water, 30 cm3 ethanol and 20 cm3 water, 25 cm3 ethanol and 25 cm3 water, 20 cm3 ethanol and 30 cm3 water.

The results are shown in the following tables:

35 cm3 ethanol and 15 cm3 water concentration:

MINUTES (30 SEC INTERVALS)

PPM

00.30

86

01.00

290

01.30

382

02.00

468

02.30

540

03.00

621

03.30

697

04.00

766

04.30

840

05.00

911

30 cm3 ethanol and 20 cm3 water concentration:

MINUTES (30 SEC INTERVALS)

PPM

00.30

235

01.00

336

01.30

427

02.00

517

02.30

610

03.00

691

03.30

766

04.00

846

04.30

915

05.00

974

25 cm3 ethanol and 25 cm3 water concentration:

MINUTES (30 SEC INTERVALS)

PPM

00.30

270

01.00

390

01.30

500

02.00

596

02.30

681

03.00

755

03.30

829

04.00

903

04.30

976

05.00

031

20 cm3 ethanol and 30 cm3 water concentration:

MINUTES (30 SEC INTERVALS)

PPM

00.30

295

01.00

420

01.30

526

02.00

630

02.30

717

03.00

811

03.30

892

04.00

967

04.30

028

05.00

095

35 cm3 ethanol and 15 cm3 water concentration:

rate = 258 / 30

= 8.60 ppm

30 cm3 ethanol and 20 cm3 water concentration:

rate = 261 / 30

= 8.70 ppm

25 cm3 ethanol and 25 cm3 water concentration:

rate = 263 / 30

= 8.76 ppm

20 cm3 ethanol and 30 cm3 water concentration:

rate = 265 / 30

= 8.83 ppm

ANALYSING EVIDENCE AND DRAWING CONCLUSIONS

Experiment 1 -

From the results obtained from the experiment of the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane, it can be concluded that the rate of hydrolysis is as follows:

Thermodynamic data:

Bond Bond enthalpy/kJ mol-1

C ? Cl 346

C ? Br 290

C ? I 228

I < Br < Cl. This is because, since the bond strength of C ? I is the weakest amongst C ? Cl, C ? Br and C ? I, the iodide ion is a better leaving group. Iodide undergoes hydrolysis at the fastest rate resulting in the fastest formation of silver iodide, and hence a fast change in the colour of the solution is detected.

The C ? Br bond is stronger than the C ? I bond, and therefore the bromide ion has a weaker leaving group than the iodide ion. It undergoes hydrolysis more slowly than 1-iodobutane, resulting in a slower rate of formation of silver bromide, as indicated by a slower formation of precipitate/ colour change.

Lastly, the C ? Cl bond is the strongest amongst the three C ? X bonds. The chloride ion is the weakest leaving group and therefore chlorobutane does not undergo hydrolysis easily.

The rate graph on page 24 shows that the reaction of 1-iodobutane, with varying concentrations of 4,6,8 and 10 pipette drops on a constant concentration of OH-, to have a first order reaction as seen by the straight line on the graph.

A rate equation for this can now be deduced from the experimental evidence obtained. The general rate equation is,

rate = k [A]m [B]n

Thus, the rate equation for this reaction is,

rate = k [C4H9I]

The rate graph on page 25 shows that the reaction of 1-iodobutane with a constant concentration, with varying concentrations of OH- - at 2 cm3, 4 cm3, 6 cm3 and 8 cm3, to also have a first order reaction.

Therefore, the rate equation for this reaction is,

rate = k [OH-]

the reactions are found to be first order with respect to [C4H9I] and also first order with respect to [OH-]. Therefore, the overall order of the reaction is given by (m + n), and is overall second order, and the rate equation for this is,

rate = k [C4H9I] [OH-]

From this it is now possible to deduce a mechanism. The reaction is a one step reaction and requires both species to collide. The experiments show that the rate is affected by both reactants.

Experiment 2 -

The graph showing the conductance of different concentrations of 2-bromo-2-methylpropane against time in seconds, page 32, shows that the reaction of 2-bromo-2-methylpropane, at concentrations 0.2, 0.4, 0.6, 0.8 and 1.0 cm3 with a constant concentration of OH-, to be that of a first order reaction. This was clearly shown by its rate graph, page 34, which exhibited a straight line (positive correlation best-fit line) - the rate is directly proportional to the concentration.

A rate equation for this can now be deduced:

rate = k [(CH3)3CBr]

The graph showing the conductance of a constant concentration of 2-bromo-2-methylpropane (0.4 cm3) against different concentrations of OH-, page 36, shows the reaction of 2-bromo-2-methylpropane with different concentrations of OH- to be that of a zero order reaction. This was clearly shown by its rate graph on page 38, which showed that the rate is unaffected by how much of a substance is present.

The reactions are found to be first order with respect to [(CH3)3CBr] and zero order with respect to OH-. Thus the overall order of the reaction is second order.

This suggests that OH- ions are not involved in the slow rate-determining step. The reaction cannot take place by direct reaction of 2-bromo-2-methylpropane itself with OH- ions. The reaction takes place in two steps.

During the first step, the C ? Br bond breaks heterolytically, as shown below.

CH3 CH3

? ?

CH3 ? C ? Br CH3 ? C+ + Br - Step 1

? ?

CH3 CH3

Because this step only involves (CH3)3CBr, its rate depends only on [(CH3)3CBr], not on [OH-].

The second step involves reaction of the carbocation, (CH3)3C+, with OH-.

CH3 CH3

? ?

CH3 ? C+ + OH- CH3 ? C ? OH Step 2

? ?

CH3 CH3

This is an ionic reaction, and like most ionic reactions, this process is very fast - definitely faster than that of step 1. So step 1 is the slow step, and hence controls the rate-determining step, which is why the rate of the reaction depends only on the concentration of (CH3)3CBr - the reaction is first order with respect to (CH3)3CBr, but zero order with respect to OH-. Step 1 is the rate-determining step and its rate equation becomes the rate equation for the whole reaction.

EVALUATING EVIDENCE AND PROCEDURES

Experiment 1 -

Imperfections in experimental procedure (procedural error):

There were no anomalous results that occurred during the investigation of this section.

The main source of procedural error, which occurred, was that of determination of the colour change via eye judgement. Once the precipitate of the haloalkane had occurred I had to judge when the colour had changed, and then quickly stop the clock. So there was also an error in stopping the stop clock - there was a maximum of 0.5 seconds delay between the judgement of the colour change, and stopping the clock to note down the time.

Precision error:

Precision error involves imperfections in the measuring device.

Percentage error = error in measurement x 100

Actual measurement/ result

Apparatus Uncertainty value

Pipette - 25 cm3 0.02 cm3

Measuring cylinder - 25 cm3 0.5 cm3

Thermometer - 0-100 ?C 0.05 ?C

Therefore, for the pipette,

Percentage error = 0.02 cm3 x 100

2 cm3

= 1 %

For the measuring cylinder,

Percentage error = 0.5 cm3 x 100

20 cm3

= 2.5 %

For the thermometer,

Percentage error = 0.05 cm3 x 100

60 ?C

= 0.08 %

Therefore, the total percentage error for experiment 1 was 3.58 %. This is below 5 % and can therefore be discarded.

Atmospheric conditions may also have had an effect on the hydrolysis of the haloalkane. During the day at which the experiments were carried out, the temperature of the room had reached around 30 ?C. This could affect the rate of reaction, as it is known that the temperature has an affect on the rate, as was described in detail on page 11. It can increase the temperature of the solution, which can give particles more energy to collide, and therefore a faster hydrolysis of the haloalkane would occur - colour change appears quicker than usual.

Ways of improving the experiment includes using a colourimeter. This can give more accurate results as the colour change is detected immediately through frequency change - much quicker and more accurate than the human eye. This will therefore remove the need of eye judgement.

There were many steps that were taken during the experiments to ensure that accuracy, precision and reliability were ensured. For instance, the temperature of the heated water was constantly being watched while the experiment proceeded - this was done to create a fair test so all the experiments proceed under the same conditions. Any temperature change in the water was immediately rectified; this was done to ensure that accurate results are produced and that they are reliable.

The equipment used ensured precision. A pipette is the most accurate piece of apparatus that is available that is used to measure out liquids. It has an error of 0.02 cm3, which is very low.

Experiment 2 -

Imperfections in experimental procedure (procedural error):

There were anomalous results that occurred in the investigation. This occurred during the investigation of 2-bromo-2-methylpropane while altering its concentration. I strongly believe that this is due to inaccurate use of the conductance meter, as has been stated previously it is the first time, which I have worked with this type of apparatus. While the meter was in the solution being measured, it had not been constantly stirred, and therefore the mixture may not have been fully mixed throughout. It is important that the mixture is constantly being stirred as I am measuring in parts per million, and the reading for this could be inaccurate if the mixture is not fully and properly mixed - because there could be different concentrations in different areas of the solution.

This error was rectified during the investigation and was not repeated - this ensured that accurate results are obtained and that any error in the experiments, if any, is not due to procedural error. This particular experiment was repeated and the new method, stirring the conductance meter in the solution, was taken into account. The graphs obtained showed to be the same as those predicted and those of what previous experimenters found.

Precision error:

Percentage error can be calculated as follows:

Apparatus Uncertainty value

Volumetric flask - 250 cm3 0.2 cm3

Pipette - 25 cm3 0.02 cm3

Therefore, for the volumetric flask,

Percentage error = 0.20 cm3 x 100

50 cm3

= 0.4 %

For the pipette,

Percentage error = 0.02 cm3 x 100

50.00 cm3

= 0.04 %

Therefore, the total percentage error that occurred in experiment 2 was 0.44%. Since this is largely below 5%, and can therefore be discarded.

Once again, atmospheric conditions may also have had an effect on the rate of reaction. During the day at which the experiments were carried out, the temperature of the room had reached around 30 ?C. This could affect the rate of reaction, as it is known that the temperature has an affect on the rate, as was described in detail on page 11.

Ways of improving the experiment includes the use of controlled atmospheric conditions, as changes in the conditions of the atmosphere could affect the collisions of particles and therefore the rate of reaction. This was taken into account at the start of the investigation, but this kind of facility was not available to me.

There were many parts of my experiment, which were vital in ensuring that the evidence collected was accurate and reproducible, such as the chosen apparatus. For example, a 250 cm3 volumetric flask was used other than a 100 cm3 volumetric flask, as the larger the size of the apparatus, the more accurate it is. A 250 cm3 volumetric flask has a less percentage error than a 100 cm3.

BIBLIOGRAPHY:

(I) - Diagram taken from Chemical Ideas, Salters Advanced Chemistry, Second edition, Heinemann production, section 10.3 "The effect of concentration on rate", page 228.

(II) - Example taken from www.chem.Index.org/ReactionKin/Ea

(III) - Example taken from Chemical Ideas, Salters Advanced Chemistry, Second edition, Heinemann production, section 10.3 "The effect of concentration on rate", page 227.

(IV) - Examples taken from Chemical Ideas, Salters Advanced Chemistry, Second edition, Heinemann production, section 10.3 "The effect of concentration on rate", page 230-31.

(V) - Figures 8 and 9 taken from Chemical Ideas, Salters Advanced Chemistry, Second edition, Heinemann production, section 10.3 "The effect of concentration on rate", page 228.

(VI) - Theory taken from www.neon.chem.ox.ac.uk/vrchemistry

(VII) - Theory obtained from Chemical Storylines, Salters Advanced Chemistry, Second edition, Heinemann production, section EP6 "Enzymes", page 160-61.

(VIII) - Information obtained from worksheet MM L5-6 Atmosphere Topic, Module II, issued by Diane Graham, Esher Chemistry Department.

(IX) - Figures 15, 16, 17, 18 obtained from Chemical Ideas, Salters Advanced Chemistry, Second edition, Heinemann production, section 10.1 "The effect of temperature on rate", page 220.

(X) - Theory obtained from Advanced Level Chemistry, Second edition, Heinemann production, "Catalysts and their effect", page 167 - 68, John Atkinson and Carol Hibbert.

(XI) - Equations and homolytic fission theory obtained from Organic Chemistry, Palgrave Macmillan production, "Bond fission", page 689, K.Peter C. Vollhardt and Neil E. Schore.

(XII) - Example and diagram obtained from Chemical Ideas, Salters Advanced Chemistry, Second edition, Heinemann production, section 10.1 "Halogenoalkanes", page 300.

(XIII) - Conductance theory obtained from www.cs.unc.edu/jeffay/papers/Cond-99.pdf

(XIV) - Conductance meter leaflet obtained from Esher College Chemistry Department, chemistry technician, Cathy.

(XV) - Information obtained from 'Hazcards', Esher College, Chemistry Department, page 138, Diane Graham.

(XVI) - Figures obtained from www.chemada.com/cat1/items/TBUB.pdf

(XVII) - Information obtained from 'Hazards in the Chemical Laboratory', Second edition, Alligator production, page 298, G.D Muir.

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