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'.
PLANNING
AIM:
The purpose of this experiment is to 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'. I will be comparing the rates of hydrolysis of the primary substances 1-chlorobutane, 1-bromobutane, 1-iodobutane, and will investigate the rate of hydrolysis of the tertiary substance 2-bromo-2-methylpropane. From the results I will then go on to deduce a rate expression/equation and a possible mechanism for the reaction.
Calculations that are to be carried out include that of gradients and rates of the graph. This is needed because the gradient of the graph gives us the rate of reaction. A rate for each tangent taken needs to be worked out so that a rate graph can be constructed, which will give the order of the hydrolysis of the haloalkane.
Example:
Figure 1 (I)
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
The rate of reaction at the start is the initial rate, which is what we are interested in. We can find the initial rate by drawing a tangent to the curve at the point t = 0, and measuring the gradient of this tangent.
In this example,
Initial gradient = Y2
X2
= 5.1 (cm3)
10 (s)
= 0.51 cm3
Therefore the rate is 0.51 cm3 of O2/ s-1. This rate will then be drawn on a separate graph along with the other rate values calculated from the other concentrations of the haloalkane, and from that the order of the reaction can be concluded and thus a mechanism deduced.
The activation enthalpy of one of the haloalkanes at the end of the experiment will be investigated and then calculations for the activation enthalpy will be used.
The activation enthalpy of the haloalkane will be calculated by changing the temperature only, while keeping all other variables constant - the concentration of the haloalkane will remain constant during each experiment of different temperatures, and the concentration of the OH- will also remain constant.
The activation energy of a reaction can be calculated using the Arrhenius equation:
ln k = ln A - (Ea/r) x 1/T
Where:
k is the rate constant of a reaction at temperature T kelvin. The units of k will depend upon the order of the reaction,
A is a constant for a particular reaction with the same units as k,
Ea is the Arrhenius activation energy (in J mol-1) of the reaction. This is a constant and does not vary with temperature,
R is the gas constant (8.3145 J mol-1 K-1),
ln means log to the base e.
A plot of ln k against 1/T follows the form 'y = mx' and should be a straight line. The slope (gradient) of the line is equal to -Ea/R from which Ea may be calculated.
Example:
The rate constant for the hydrolysis of bromoethane by sodium hydroxide
C2H5Br (aq) + OH- (aq) C2H5OH (aq) + Br- (aq)
was measured at six temperatures. The results were:
T / K
k/mol-1 dm3 s-1
300
.123 x 10-4
310
3.574 x 10-4
320
.058 x 10-3
330
2.932 x 10-3
340
7.652 x 10-3
350
.891 x 10-2
Table 1
Calculate the activation energy of reaction.
Answer:
We start by calculating ln k and (1/T) as shown below:
T / K
k/mol-1 dm3 s-1
ln k
/ T
300
.123 x 10-4
-9.0940
3.333 x 10-3
310
3.574 x 10-4
-7.9366
3.226 x 10-3
320
.058 x 10-3
-6.8515
3.125 x 10-3
330
2.932 x 10-3
-5.8321
3.03 x 10-3
340
7.652 x 10-3
-4.8727
2.941 x 10-3
350
.891 x 10-2
-3.9682
2.857 x 10-3
Table 2
We then plot ln k against 1/T, and fit a 'best line' through the points:
Figure 2
The slope of the graph is the 'perpendicular distance' (shown as arrow A) divided by the horizontal distance (shown as arrow B). From the graph, these distances are -5.13 and 4.76 x 10-4 K-1 respectively.
Slope = -5.13/4.76 x 10-4 K-1 = -10780 K
The minus sign of the slope is a mathematical convention, meaning that the graph slopes from right to left. The units of the slop are kelvin, K.
According to the equation, the slope of the graph equals -E/R.
Therefore -10780 K = -E/R
Thus the Arrhenius activation energy for the hydrolysis of bromoethane is 89.6 kJ mol-1 (II).
BACKGROUND THEORY:
RATES OF REACTION
Studying reaction kinetics will help me to make predictions about the reactions that will take place in my experiment, and will help me understand and deduce the mechanisms of the chemical reactions.
There are many factors, which affect reaction rates. Different chemical reactions go at different rates. Some reactions, such as burning fuel in a cylinder of a car engine or precipitating silver chloride from solution, for instance, go very fast. On the other hand, others such as the souring of milk or the rusting of iron, are much slower.
The rate of a chemical reaction may be affected by the following:
- the concentration of the reactants. For example, the rate of reaction of chlorine atoms with ozone in the stratosphere increases as the concentration of chlorine atoms increases. In the case of solutions, concentration is measured in mol dm-3; in the case of gases, the concentration is proportional to the pressure.
- the temperature. Nearly all reactions go faster at higher temperatures.
- the intensity of radiation, if the reaction involves radiation. For example, ultraviolet radiation of certain frequency causes O2 molecules to dissociate into O atoms, and the reaction goes faster when the intensity of the ultraviolet radiation increases.
- the particle size of a solid. A solid such as magnesium reacts much faster when it is finely powdered than when it is in a large lump because there is a much larger surface area of solid exposed for the reaction to take place on.
- the presence of a catalyst.
The collision theory of reactions
The effect of these factors can be explained using a simple collision theory. The basic idea is that reactions occur when the particles of reactants collide, provided they collide with a certain minimum kinetic energy. For example, imagine two particles, say an ozone molecule and a chlorine atom, moving around in the stratosphere. For a reaction to occur, the two particles must first collide so that they come into contact with each other. This will happen more often if there are more particles in a given volume, so it is easy to say why increasing the concentration of the reactants speeds up the reaction (Figure 3 below).
Figure 3
Indeed, any factor, which increases the number of collisions, will increase the rate of reaction. But, for most reactions, simply colliding is not enough - not every collision causes a reaction. As the particles approach and collide, kinetic energy is converted into potential energy and the potential energy of the reactants rises, as shown below.
Figure 4
Existing bonds start to stretch and break and new bonds start to form. Only those pair with enough combined kinetic energy on collision to overcome the energy barrier or activation enthalpy for the reaction will go on to produce products.
The proportion of colliding pairs with sufficient kinetic energy to overcome the energy barrier depends very much on the temperature of the gas. At higher temperatures, a much larger proportion of colliding pairs has enough energy to react.
Temperature often has a dramatic effect on reaction rates; for example, methane and oxygen do not react at room temperature, but react explosively when heated.
Concentration
Normally this means the concentration of a solution, though the idea can be applied to gases. For a gas the concentration is normally measured in terms of its pressure.
It is useful to know how concentration affects rates, for many reasons. For instance, in this investigation, I may want to use changes of concentration of the primary or tertiary substance to alter the rate of a reaction. Knowing how concentration affects rate can also tell me a great deal about the way that reactions occur - their mechanisms.
The meaning of 'rate of reaction':
The rate of something means the rate at which some quantity changes. For instance, speed is rate of change of distance. Whenever you are measuring a rate you need to be clear about the units you are using. Speed is often measured in metres per second.
When we talk about the rate of reaction, we are talking about the rate at which reactants are converted into products. In the decomposition of hydrogen peroxide, for example, to form water and oxygen,
2H2O2(aq) 2H2O(l) + O2(g)
hydrogen peroxide
Figure 5
the rate of the reaction means the rate at which H2O(l) and O2(g) are formed, which is the same as the rate at which H2O(aq) is used up. The rate of this reaction could be measured in moles of product (water or oxygen) formed per second, or moles of hydrogen peroxide used up per second. Suppose it turns out that 0.0001mol of oxygen are being formed per second (as shown by the above diagram).
The rate of the reaction is
0.0001 mol (O2) s-1 or 0.0002 mol (H2O) s-1 or -0.0002 mol (H2O2) s-1
The rate in terms of moles of H2O is twice the rate in terms of moles of O2 - this is because two moles of H2O are formed for each mole of O2. The rate in terms of H2O2 has a negative sign - this is because H2O2 is getting used up instead of being produced.
In this case, the units for the rate of reaction are mol s-1, though they can also be mol min-1 or even mol h-1 (III).
Measuring rate of reaction:
When measuring rate of reaction, it is done so by measuring the change in amount of a reactant or product in a certain time. In practice, this means measuring a property that is related to the amount of substance. If a gas is given off, the gas could be collected and its volume measured at different times during the reaction, or the reaction could be carried out in a conical flask on a balance and the total mass of the reaction mixture and flask could be measured at different times. Alternatively, following the concentration of one reactant or product as the reaction proceeds may be done. Sometimes this can be done by measuring a property related to concentration, such as colour intensity (if one of the substances involved is coloured), or pH (if an acid or alkali is involved).
So the procedure for measuring reaction rate is:
Step 1 Decide on a property of a reactant or product, which can be measured, such as volume of gas produced or total mass of the reaction mixture.
Step 2 Measure the change in the property in a certain time.
Step 3 Find the rate in terms of change of property
time
Investigating how rate depends on concentration:
Looking more closely at the decomposition of hydrogen peroxide on the previous page, this reaction proceeds slowly under normal conditions, but it is greatly speeded up by catalysts. A particular effective catalyst is the enzyme catalase. The volume of oxygen is measured in the inverted burette.
Figure 6
Let's look at the kind of results, which may arise when investigating how the rate of oxygen formation depends on the concentration of hydrogen peroxide solution.
The total volume of oxygen given off at different times is measured from the start of the experiment and a graph of this against time is plotted. This allows the rate of the reaction in (cm3 of O2) s-1 to be calculated. This could be converted to mol (O2) s-1, as it is known that 1mol of oxygen occupies about 24,000 cm3 (or 24 dm-3) at room temperature. But what we are interested in is comparing rates, and for these purposes we can use (cm3 of O2) without bothering to convert to moles. The following graph shows the results that were obtained by starting with hydrogen peroxide of concentration 0.4mol dm-3:
Figure 7
Notice these points about the graph:
* The graph is steep at first. The gradient of the graph gives us the rate of the reaction - the steeper the gradient, the faster the reaction. The reaction is at its fastest at the start, when the concentration of hydrogen peroxide in solution is high, before any has been used up.
* The graph gradually flattens out. This is because, as the hydrogen peroxide is used up, its concentration falls. The lower the concentration, the slower the reaction. Eventually, the graph is horizontal: the gradient is zero, and the reaction has come to a stop.
The rate of the reaction at the start is called the initial rate. The initial rate can be found by drawing a tangent to the curve at the point t = 0, and measuring the gradient of this tangent. In the example in Figure 7 above, the gradient is 0.51, so the initial rate is 0.51 (cm3 of O2) s-1.
Figure 8 below shows some results that were obtained when the same experiment was done using hydrogen peroxide solutions of different concentrations. In each case, the concentration of the enzyme catalase was kept constant, as were all other conditions such as temperature. As expected, the graphs start off with differing gradients, depending on the initial concentration of hydrogen peroxide. Table 3 shows the initial rates of the experiments in Figure 8.
Figure 9
Concentration of hydrogen peroxide at start/mol dm-3
Initial rate/
(cm3 of O2(g)) s-1
0.40
0.51
0.32
0.41
0.24
0.32
0.16
0.21
0.08
0.10
Table 3
Now the question 'How does the rate of the reaction depend on the concentration of hydrogen peroxide?' may be answered. Figure 9 shows the initial rates plotted against concentration of hydrogen peroxide, and it can be seen that it is a straight line. This means that the rate is directly proportional to the concentration of hydrogen peroxide. This can be shown as,
rate ? [H2O2(aq)]
or
rate = constant x [H2O2(aq)]
Figure 10
Hydrogen peroxide is not the only substance whose concentration affects the rate of this reaction. It is also affected by the concentration of the enzyme catalase, but in the series of experiments shown in Figure 9, the concentration of catalase was kept constant. Another set of experiments could be carried out to find the effect of changing the concentration of catalase.
In other words,
rate = constant x [catalase]
If this is combined with the equation involving H2O2 then,
rate = constant x [H2O2(aq)] x [catalase]
or
rate = k [H2O2(aq)] [catalase]
This is called the rate equation for the reaction, and the constant k is called the rate constant. The value of k varies with temperature, so it must always be stated at what temperature the measurements were made when the rate, or the rate constant, is given of a reaction.
Order of reaction:
For any chemical reaction, a rate equation can be written - provided that an experiment can be done first to find out how the rate depends on the concentration of the reactants. For a general reaction in which A and B are the reactants,
A + B products
the general rate equation is,
rate = k [A]m [B]n
m and n are the powers to which the concentration needs to be raised: they usually have values of 0, 1 or 2. m and n are called the order of the reaction, with respect to A and B. For instance, in the hydrogen peroxide example earlier,
rate = k [H2O2] [catalase]
In this case, m and n are both equal to 1. The reaction is first order with respect to H2O2 and first order with respect to catalase. The overall order of the reaction is given by (m + n),so in this case the reaction is overall second order.
The following examples will help to explain the idea of reaction order:
Example1 - the reaction of Br radicals to form Br2 molecules
The equation for the reaction is,
2Br(g) Br2(g)
Experiments show that the rate equation for the formation of Br2 is,
rate = k [Br]2
So this reaction is second order with respect to Br. Since Br is the only reactant involved, the reaction is also second order overall.
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 ...
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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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