1 Meztizahn

Iodination Of Propanone Coursework Definition

 

Chemistry Individual Investigation Yann Perusset 13F

2

To increase the number of particles with enough energy for a successful collision the temperature canbe increased. This concept is again shown in the distribution curve shown above, withT1being thecolder temperature andT2being the warmer. The area under the curve past the E

A

shows thenumber of particles with enough energy to form a successful collision. Basically, reactions go faster athigher temperatures because a larger proportion of the colliding molecules have the minimumactivation enthalpy needed to react.Many factors can affect the overall rate of a reaction, the main factors being concentration,temperature, particle size and the presence of a catalyst. When I come to working out the rateequation and I increase the concentration of one of the reactants it is important to keep all otherfactors the same to determine what effect that particular solution has on the rate without any outsideinfluences. An increase in the concentration of a solution increases the amount of particles within it,the more particles there is, the more likely a successful collision.

Order of reaction, rate constants and rate equations

2+3

Rate Equations and rate constant

With a rate equation, it is possible to see how the concentration of each of the reactant affects therate of a reaction. An example of a simple rate equation is shown below.

Rate=k[A]

x

[B]

y

The total order of a reaction is the sum of the orders of the individual reactants. So for this reaction itwould be the sum of x and y.From the rate equation it is also possible to find out the rate-determining step. From looking at theorder of each of the substances it will tell you the relative number of moles of each substanceinvolved in the rate determining step. For example, assume substance A in the equation above is firstorder, this means that there will be one substance in the rate-determining step. From this, it wouldthen be possible to create a mechanism for the reaction.The rate constant can be calculated by rearranging the rate equation. Below is an example of the rateequation above but rearranged in terms of the rate constant, k:

=







To obtain a reliable value for the rate the initial rates method would be used (shown below)Once the rate equation is worked out, it can be easily linked to the reaction mechanism by looking atthe rate determining step.

Colourimetery

4

The rate of a reaction is the change in concentration of a reactant or product divided by the timetaken for the change to occur. Therefore to calculate the rate, you have to be able to measure theconcentration of a solution. To do so I shall use the process of Colourimetery. A colorimeter can beused to measure the change in colour of a reaction. It works on the principle that coloured solutions(like iodine that I will be using) absorb certain wavelengths of light. The amount of light that isabsorbed by the solution is known as the absorbance of the solution.The absorbance is proportional to the concentration of the solution used (Abs Conc)Known concentrations of the coloured solution are used to produce a calibration curve which canthen be used to find the concentration of any absorbance value that has that coloured solution withinit.

Some sample reactions

The catalytic decomposition of hydrogen peroxide

This is a simple example of measuring the initial rate of a reaction producing a gas.

A simple set-up to do this might be:

The reason for the weighing bottle containing the catalyst is to prevent introducing errors at the beginning of the experiment. Since this is the part of the reaction you are most interested in, introducing errors here would be stupid!

You have to find a way of adding the catalyst to the hydrogen peroxide solution without changing the volume of gas collected. If you added it to the flask using a spatula, and then quickly put the bung in, you might lose some gas before you got the bung in. Alternatively, as you pushed the bung in, you might force some air into the measuring cylinder. Either way, it makes your results meaningless.

To start the reaction, you just need to shake the flask so that the weighing bottle falls over, and then continue shaking to make sure the catalyst mixes evenly with the solution.

You could also use a special flask with a divided bottom, with the catalyst in one side, and the hydrogen peroxide solution in the other. They are easy to mix by tipping the flask.

If you use a 10 cm3 measuring cylinder, initially full of water, you can reasonably accurately record the time taken to collect a small fixed volume of gas.

You could, of course, use a small gas syringe instead.

If you were looking at the effect of the concentration of hydrogen peroxide on the rate, then you would have to change its concentration, but keep everything else constant.

The temperature would have to be kept constant, so would the total volume of the solution and the mass of manganese(IV) oxide. You would also have to be sure that the manganese(IV) oxide used always came from the same bottle so that its state of division was always the same.

You could, of course, use much the same apparatus to find out what happened if you varied the temperature, or the mass of the catalyst, or the state of division of the catalyst.

The thiosulphate-acid reaction

If you add dilute hydrochloric acid to sodium thiosulphate solution, you get the slow formation of a pale yellow precipitate of sulphur.

There is a very simple, but very effective, way of measuring the time taken for a small fixed amount of precipitate to form. Stand the flask on a piece of paper with a cross drawn on it, and then look down through the solution until the cross disappears.

So . . . you put a known volume of sodium thiosulphate solution in a flask. Then you add a small known volume of dilute hydrochloric acid, start timing, swirl the flask to mix everything up, and stand it on the paper with the cross on. Time how long it takes for the cross to disappear.

Then repeat using a smaller volume of sodium thiosulphate, but topped up to the same original volume with water. Everything else should be exactly as before.

If you started with, say, 50 cm3 of sodium thiosulphate solution, you would repeat the experiment with perhaps, 40, 30, 20, 15 and 10 cm3 - each time made up to a total of 50 cm3 with water.

The actual concentration of the sodium thiosulphate doesn't have to be known. In each case, you could record its relative concentration. The solution with 40 cm3 of sodium thiosulphate solution plus 10 cm3 of water has a concentration which is 80% of the original one, for example. The one with 10 cm3 of sodium thiosulphate solution plus 40 cm3 of water has a concentration which is 20% of the original one.

When you came to plotting a rate against concentration graph, as we looked at further up the page, you would plot 1/t as a measure of the rate, and volume of sodium thiosulphate solution as a measure of concentration. Alternatively, you could plot relative concentrations - from, say, 20% to 100%. It doesn't actually matter - the shape of the graph will be identical.

You could also look at the effect of temperature on this reaction, by warming the sodium thiosulphate solution before you added the acid. Take the temperature after adding the acid, though, because the cold acid will cool the solution slightly.

This time you would change the temperature between experiments, but keep everything else constant. To get reasonable times, you would have to use a diluted version of your sodium thiosulphate solution. Using the full strength solution hot will produce enough precipitate to hide the cross almost instantly.

Iodine clock reactions

There are several reactions which go under the name "iodine clock". They are all reactions which give iodine as one of the products. This is the simplest of them, but only because it involves the most familiar reagents.

The reaction we are looking at is the oxidation of iodide ions by hydrogen peroxide under acidic conditions.

The iodine is formed first as a pale yellow solution darkening to orange and then dark red, before dark grey solid iodine is precipitated.

There is a very clever way of picking out a when a particular very small amount of iodine has been formed.

Iodine reacts with starch solution to give a very deep blue solution. If you added some starch solution to the reaction above, as soon as the first trace of iodine was formed, the solution would turn blue. That doesn't actually help!

However, iodine also reacts with sodium thiosulphate solution.

If you add a very small amount of sodium thiosulphate solution to your reaction mixture (including the starch solution), it will react with the iodine that is initially produced, and so the iodine won't affect the starch, and you won't get any blue colour.

However, when that small amount of sodium thiosulphate has been used up, there is nothing to stop the next lot of iodine produced from reacting with the starch. The mixture suddenly goes blue.

Leave a Comment

(0 Comments)

Your email address will not be published. Required fields are marked *