How does the activation enthalpy and the rate of the iodine-clock reaction vary with the concentration of reactants, catalysts, and the presence of different catalysts?

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Salter’s Chemistry A-Level Individual Investigation Report

How does the activation enthalpy and the rate of the iodine-clock reaction vary with the concentration of reactants, catalysts, and the presence of different catalysts?

Investigation

How does the activation enthalpy and the rate of reaction vary with the concentration of reactants, catalysts, and the presence of different catalysts?

Aim

In this investigation I have set out to run a few standard tests and attempt to explain the mechanism by which the transition metal catalysts work in this particular instance.

Theory

The iodine-clock reaction. This reaction has been known for quite a long time, and it involves the oxidation of iodide ions to iodine molecules, which are soluble in water and will show its presence by a pale brown colour. The addition of starch to the reaction mixture makes the colour change far more visible by forming a dark-blue complex with iodine. The reaction can be represented by the following half-equation:

2I-(aq)     --->      I2 (aq)    +   2e-                

E standard /V = +0.54

2 I-(aq)            >      I2 (aq) + 2 e-           E O  /V = +0.54     ........................ Equation 1

There are a range of oxidation agents available to carry out this reaction. In fact, almost all transition metal ions could be used to oxidise iodide ions. A quick scan through a table of electrode potentials will yield the necessary list of possible oxidation agents. However, in order to investigate the catalytic properties of the transition metal ions, another type of oxidation agent had to be chosen. The standard iodine-clock experimentA1 uses the reduction of the peroxy-disulphate ions (S2O82-), for various reasons which will be described later. The half-equation of this reaction is:

2e-     +     S2O82-(aq)     --->      2SO42-(aq)                 

E standard /V = +2.01

2 e- + S2O82-(aq)             >      2 SO42-(aq)            E O  /V = +2.01     ........................ Equation 2

Therefore, the overall equation of the reaction is:

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

2 I-(aq) + S2O82-(aq)             >      I2 (aq) + 2 SO42-(aq)            ........................ Equation 3

As we are using Potassium Salts of both anions, the equation with the spectator ions (K+) would be:

2KI (aq)   +   K2S2O8 (aq)        --->        I2 (aq)   +   2 K2SO4 (aq)

2 KI(aq) + K2S2O8(aq)             >      I2 (aq) + 2 K2SO4(aq)            ........................ Equation 4

This reaction involves two anions having sufficient energy to overcome the repulsive forces between them and colliding in the right manner to happen. It occurs uncatalysed but can be accelerated by a great extent with the introduction of a transition metal catalyst, the mechanism of which will be dealt with later in the text.

Introduction to the Rates of Reaction. We all know that the rate of reaction is affected by temperature. As a general rule of thumb, an increase of temperature by 10oC centigrade usually results in a doubling of the reaction rate. The reaction rate depend on a cocktail of factors, among them the rate of molecular collisions (which in turn depend on the energy that the molecule possesses - temperature, and the concentration of molecules.) This 'rate of molecular collisions', is represented by in the equation. The Arrhenius Equation, shown below, which governs all rates of reactions.

k   =   A exp( -EA / RT )            ........................ Equation 5

This equation applies where k is the 'rate constant', which gives the rate of reaction at a given temperature, EA is the activation enthalpy of the reaction, and R is the ideal gas constant. A is known as the 'pre-exponential factor', and it is a quantitative representation of the orientation of the particles when they collide. It is no good if the particles do not collide at the right position and in the right orientation, especially with larger molecules - no reaction will take place. This 'pre-exponential factor' on the whole greatly reduce the rate of reaction, and with smaller particles it will still reduce the rate but not by such a large amount.

The order of reaction with respective to chemicals. The rate of a particular reaction may or may not be dependent on the relative abundance of its chemical feedstock. To cite an everyday example, the rate of production of teddy bears is not dictated by the amount of wool available or by the capacity of the factory. It is directly proportional to the rate at which consumers consume teddy bears. Similarly some rates of reaction is proportional to the abundance of a chemical, or the concentration of it if a solution is used. Reactions in this category includes most uncatalysed inorganic reactions which have their rates dictated by the rate of molecular collisions. Others, notably enzyme-based biological processes, would start off appearing to be first order with respect to the concentration of the enzyme and the input, but will soon tail off and becomes zeroth order with respective to the input. This is due to the occupation of all the 'active sites' on the enzyme, not due to a lack of collisions between the reactants. In the real-life example, it would be a case of world-wide teddy shortage as the demand exceeds the supply and the factory (or the active-site) is working as fast as it can.

The Reaction Mechanism. The rate of reaction calculated using the above reaction refers to the rate of one particular step of a reaction. If the reaction only involves one step, the of course that is equal to the actual rate of reaction. For instance, the initial rate of reaction between dilute hydrochloric acid and magnesium can be calculated using the above method quite satisfactorily. However, with more complex reactions, notably in organic reactions, the reaction involves several steps, and this is where the reaction mechanism comes into play.

Back to our teddy bear analogy again, consider a production line of teddy bears. Supposing that the eyes can be fitted to the body at a rate of 5 pairs/minute, but legs can only be connected at a rate of 1 pair/minute, the teddy bears will only emerge from the end of the line at a maximum rate of 1 pair/minute. So if a chemical reaction involves a series of steps, some of which are faster than the other, the one which dictates the overall rate of reaction would be the slowest step. This step is known as the 'rate determining step'.

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Chemical reactions take this analogy even further. Supposing the rate of step one in a particular reaction is of first order with respect to the concentration of chemical A, but the rate of step two is of zeroth order with respect to chemical A, the overall rate will still be of zeroth order, because the intermediate product formed (input to step two) can only be processed at a certain fixed rate (hence it is of zeroth order with respect to chemical A). Any excess intermediate product will remain in the intermediate stage or if it is unstable will revert back ...

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