-
Catalyst: catalyst can dramatically increase the rate of a chemical reaction. They are defined as substances, which will alter the rate of a chemical reaction while remaining chemically unchanged at the end of the reaction. Only a small amount of a catalyst is required, and it can affect the rate of reaction for a long time. Catalysts are specifying in their action; a catalyst that increases the rate of one reaction may have no effect on another. The mechanism of catalyst is not totally understood, but there appear to be two main types. Some catalyst increases the rate of reaction by the formation of intermediate compounds with the regeneration of the catalyst at the end of the reaction. Thus if A+B reacts slowly to give D, the presence of a catalyst C might increase the rate by allowing the reaction to take place in two fast stages:
A+C→AC fast
AC+B→D+C reactions
The catalyst C is continuously regenerated and so only a small amount is needed to speed up the reaction. The reactants in this type of reactions are usually all in the same phase and this is homogeneous catalysis. The speeding up of the thermal decomposition of potassium chlorate by the addition of manganese dioxide is thought to be an example of this.
In the other form of catalyst the reactants and the catalyst are not in the same phase; for example, finely divided solids. This is heterogeneous catalysis. The molecules of the reacting gases are absorbed on to the surface of the catalyst, bringing them into closer contact with one another and making them more reactive by weakening the bonds within the molecules. This effect is obtained in the reaction between nitrogen and hydrogen to make ammonia, for which the catalyst is finely divided iron.
N2 (g) + 3H2 (g) ⇔ 2NH3 (g)
Collision Theory
In a chemical reaction: -
-
Particles must collide
- The collision must have enough energy within it, for the particles to reach the transition state and form products.
- The particles must collide in the right orientation if products are to be formed.
The more the number of collisions, the more the chance that the collisions will have enough energy to reach the top of the activation energy to form products. In general most of the collisions either never have enough energy to form products or the particles do not collide in the right orientation, therefore only a few particles go on to form products (Maxwell Boltzmann Theory – A Levels!).
Rate equations
Guldberg and Waage first studied the effect of the concentration of the reactants in 1864, for simple one-step reactions. The formulated a law of mass action which stated that the rate of reaction, at constant temperature, was proportional to the product of the active masses of the reacting substances. Modern rate laws have been developed from their work. For simplicity, the active mass of a reactant A can be taken as the concentration of A in mole dm-3 and is written [A]. Thus for the simple reaction:
A+B→C
The rate of reaction, at constant temperature, is proportional to the molar concentration of A multiplied by the molar concentration of B:
Rate of reaction ∝ [A] × [B]
Rate of reaction = κ [A] [B]
This is the rate equation or rate expression, and κ is called the velocity constant or rate constant for the reaction.
Order of reaction
The way in which the rate of reaction depends on the concentration of the reactants is expressed in terms of the orders of the reaction. For the reaction:
A+B→C
the rate may be found to depend simply on the concentrations of A, and not on the concentration of B at all. In this case:
Rate of reaction = κ [A]
and we say that the reaction is first order with respect to A, zero order with respect of B, and first order overall.
In another case it may be found that the rate depends on the concentration of both A and B, so that:
Rate of reaction = κ [A] [B]
and the reaction is first order with respect to A, first order with respect to B and second order overall.
In amore complicated case the rate of reaction may be found to depend on the concentration of A and the square of the concentration of B. then the rate expression becomes:
Rate of reaction = κ [A] [B]2
and the reaction is first order with respect to A, second order with respect to B, and third order overall.
So the overall order of the reaction is the sum of the indices in the rate expression. All this has to be found experimentally because the orders of a equation do not necessary coincide with the figures used in a balanced equation. For example, potassium peroxodisulphate reacts with potassium iodide to form iodine. Ionically this equation is:
S2O82-+2I- →I2 + 2SO42-
The rate equation is found to be:
Rate of reaction = κ [S2O82-][I-]
So the reaction is first order with respect to both peroxodisulphate and iodide ions, although from the equation you might think that it was second order (involving two ions) for the iodine. Determination of the orders if a reaction can be used to find out how it takes place.
Finding the rate of reaction
The rate reaction at a given moment is usually expressed as the amount of reaction taking place in unit time, and to find it the change of the concentration of one component of the reaction mixture is measured. Such characteristics as change of colour, volume of gas evolved, or change of pressure may be used. The rate found should be quoted in terms of the change used. For example: “Rate of reaction with respect to concentration of iodine” or “Rate of reaction with respect to oxygen involved”.
Consider the decomposition of hydrogen peroxide, catalysed by manganese dioxide:
2H2O2 (aq) → 2H2O (l) + O2 (g)
The oxygen evolved can be collected in a syringe and its volume measured at set time intervals (say fifteen second). A graph of the results can be plotted.
When the concentration of hydrogen peroxide is high, at the star, the reaction is fast and the slope steep. As the concentration falls the reaction slows down, and the curve flattens out when all the hydrogen peroxide is decomposed. The rate, at a particular time (t), is given by the slope of the curve, and this can be found by drawing a tangent AB to the curve at t.
The units in this case are cm3s-1 of oxygen evolved. The volume of oxygen evolved is proportional to the amount of hydrogen peroxide decomposed, and the rate could have been determined by plotting the concentration of hydrogen peroxide remaining undecomposed.
Since the concentration of hydrogen peroxide is decreasing the curve will be of the same shape as that plotted for the evolution of oxygen, but inverted; the units are now mol dm-3s-1.
Activation energy
Kinetic studies of reactions in the gaseous phase show that there is an “energy barrier” to the overcome before reaction takes place. A given sample of gas at a fixed temperature will contain not only molecules with average kinetic energy but some with very low energy, and also a proportion with much higher than average energy. It can be shown that only those molecules having energy above a certain minimum value will reacts on collision; this minimum value is known as the activation energy. Thus the reaction paths for exothermic and endothermic reactions may be represented by the following diagrams:
For the reactions in the gaseous phase the number of molecules having energy higher than the activation energy E can be derived from Maxwell’s distribution of velocity law, expressed in a simplified form as:
n=n0e-E/RT
where E= the activation energy
R= the gas constant
T= temperature/ κ
n0= total number of molecules
n= number of molecules with energy greater than E
If reactions take place only between molecules having an energy greater than E then both the rate of reaction and the velocity constant will be proportional to n at a given concentration, hence
Rate of reaction ∝ κ ∝ Ae-E/RT
where A is a constant. n0 is fixed for a given concentration.
Then on taking logarithms a relationship between the velocity constant κ and the activation energy can be established:
Log κ= C- (E/2.3R) × (1/T)
when C is a constant and R=8.3 JK-1mol-1.
This can be used to find the activation energies. The rate of reaction κ is measured at various temperatures and log κ plotted against 1/T. The slope of the line will be given by E/2.3R. Thus if the angle of the slope is Δ then tanΔ= -E/ 2.3R and E= -2.3R tanΔ Jmol-1.
The effect of some catalyst may well be that, by forming intermediate compounds; they reduce the activation energy necessary for the overall reaction.
Less activation energy may be required for the two stages of the reaction involving the catalyst than is required if the reaction takes place without the catalyst.
Plan
Aim
In this investigation I will be looking at how increasing the volume of acid, the rate of reaction will increases. The acid makes a yellow cloudiness when it reacts with sodium thiosulphate making the water opaque. In this experiment we will put in a blank paper a cross with a mark pen, so if the if we put a volume of acid in the water, the cross will disappears within some second. That is depended of the volume of acid, which is proportional to the rate of reaction. So:
-
Volume of acid ∝ rate of reaction
-
Rate of reaction ∝ time to the cross for disappear
-
Volume of acid ∝ time to the cross for disappear
Na2S2O3 + HCL ➔ 2NaCl + SO2 + S + H2O
This Sulphur forms the yellow cloudiness in the reaction, which obscures the view of the cross.
Fair Test:
- Volume of Sodium thiosulphate
- Temperature (room temperature)
- The angle to which the cross is viewed
- Mixing technique
- Concentration of sodium thiosulphate should be the same
- Volume of hydrochloric acid used is the same
The main variable in this investigation will be the concentration of HCl
The time taken for the cross to disappear in each exp. will be different.
Maxwell Boltzmann Theory:
The Maxwell – Boltzmann curve shows the number of molecules in a closed vessel containing a certain amount of energy. The curve shows that only those molecules above the activation energy will go on to form products. This graph shows that only a very small fraction of molecules go on to form products if we either increase the temperature or increase the concentration the curve flattens hens allowing more molecules to have enough energy to form products (since there are more collisions).
Prediction:
As we increase the concentration of he acid we increase the rate of reaction. To make this statement more quantitative we can say that as we double the number of acid particles we double the rate of reaction. By doubling the concentration we double the number of acid particles per unit volume therefore double the number of collisions, therefore doubling the fact that more particles will reach the top of the activation energy to form products.
i.e. Rate is directly proportional to concentration
i.e. Rate ∝ concentration
Unit per time ∝ concentration
In this investigation the graphs I would be expected to see are: -
Procedure
We fill the measuring cylinder with sodium thiosulphate with an estimate quantity, and then we add an estimate volume of acid (HCl), and mixed then. So in a logical time the water will become cloudiness. Before of that we had put, in a blank paper a cross-written by a mark pen. After time, the water will become opaque, so we will not see the cross. What we need to count is the time taken for the cross to disappears, so the time will began counting when both substances, sodium thiosulphate and hydrochloric acid, are mixed.
Dilution / Range
In this investigation at least six different concentrations of HCl will be taken to make this a valid experiment.
Apparatus List
-
Two 100cm3 measuring cylinder
- Stop watch
- 1M HCl (1 litre)
- 0.1M sodium thiosulphate
- Pen & paper
-
250cm3 conical flask
-
50°C thermometer
Safety
- Goggles
- Lap coat
-
SO2 gas is dangerous it might cause asthma; therefore the experiment must be carried out in a well-ventilated room.
Obtaining Evidence
- Precise method.
- Tabular presentation
Analysis
- Two accurate graphs must be done by average observations
- One of them is a time taken (sec) – Concentration of HCl (M) graph
- It won’t be a perfect curve and not a line
- The other one is a Rate unit/time – concentration of HCl graph
- Comments on line… This graph shows that as the concentration increases the time taken for the cross to disappear decreases. (a simple explanation to make you gain two marks)
- For the other graph… This graph shows that as the concentration increases the rate of reaction increases.
- To get your A* you must make it sound better.
- Graphs must be accurate… units for each thing… title for each thing… Occupy as much space as possible in the graph paper.
- Next paragraph, (in my background information I said that something doubles… therefore this doubles as well). Say also that these graphs are showing exactly what I needed.
- On graph, pick four points, draw a line from the horizontal line until the graph’s line towards each point chosen, then see the correspondent on the vertical line, then prove that they double. So if the first point was 0.3, the second is the double :0.6. (link it to your prediction).