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.
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)
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
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
= 5.1 (cm3)
= 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
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.
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
1.123 x 10-4
3.574 x 10-4
1.058 x 10-3
2.932 x 10-3
7.652 x 10-3
1.891 x 10-2
Calculate the activation energy of reaction.
We start by calculating ln k and (1/T) as shown below:
T / K
k/mol-1 dm3 s-1
1 / T
1.123 x 10-4
3.333 x 10-3
3.574 x 10-4
3.226 x 10-3
1.058 x 10-3
3.125 x 10-3
2.932 x 10-3
3.03 x 10-3
7.652 x 10-3
2.941 x 10-3
1.891 x 10-2
2.857 x 10-3
We then plot ln k against 1/T, and fit a ‘best line’ through the points:
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).
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).
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.
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.
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)
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
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.
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:
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.
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