The determination of a rate equation

Authors Avatar

Abdul Basit Farhat                Chemistry

The determination of a rate equation

Introduction

After watching the demonstration shown by the teacher, I noticed a cloudy solution forming within 10 seconds after mixing the 50cm³ of (0.4 moldm-³) Na2S2O3 was mixed with 5.0 cm³ HCL (2.0 moldm-³) diluted with 20cm³ H2O. From this I can simply conclude that the reaction takes place quickly and further increasing concentration/ volumes would produce a faster reaction. This would make it difficult to measure the exact point when a cloudy solution is formed. If the time taken to produce the cloudy solution is faster than 10seconds this increases the chance of error and inaccuracy as the time taken for detection from the eye and time to respond by stopping the stop clock can lead to an error of judgement within +5 seconds, this would be too great if the reaction was carried out at a greater concentration/ volumes.

I know SO2 is produced from the equation below which is a gas:

2HCL         +        Na2S2O3                        2NaCl             +        SO2     +     S            +        H2O

After a while I noticed a smell of SO2 as it diffused across the room. This is important, if using large amounts of Na2S2O3 this will cause the production of more SO2 which can be dangerous and difficult to get rid of.

If I decide to use 5cm³ of Na2S2O3 and 1cm³ of HCL the reactions would be slow and varying the concentration would result in difficulty and a greater chance of error as measuring decimal places for reactants using equipment has greater percentage error

Therefore after consideration and thought of all these factors I have decided to keep a total volume of 20cm³ this will enable me to vary the concentration of HCL and Na2S2O3 and obtain suitable results via dilution using water. I will use 5 different concentration of one reactant and keep the other reactant constant. In using 5 concentrations this will give me enough pieces of data to plot a safe graph. If I use less than 5 concentrations it will be difficult to detect an anomalous, as it’s difficult to detect a trend using less data.

For example the concentration will be prepared like this for HCl:

In this reaction the concentration of Na2S2O3 will remain constant at 10 cm³, only varying the concentration of HCl by adding water. I think this will be safer as I will only be using small quantities of each reactant and the amount of sulphur dioxide produced will be in small quantities.

Similarly to obtain the rate of the reactant Na2S2O3, will be carried out in similar conditions keeping it a fair test. This time the variable HCl will be kept constant and I will use 5 concentrations of Na2S2O3 to determine the rate as shown below:

The rate of the reaction can be determined by calculating the amount of Sulphur produced in the time recorded. This is given by:

Rate = Amount of Sulphur

Time

This information can be used to determine the order of the reaction with respect to sodium thiosulphate.

Amount of Sulphur is assumed to be the same in each reaction so:

Rate of reaction =        1         .      

                                                    Time (seconds)

From this relationship I can calculate the rate for HCL and Na2S2O3 using the time taken for the production of a cloudy solution as an end point.

I can calculate the concentration of each reactant using the equation:

Concentration (moldm-³) = moles

                                 Volume (dm-³)

Using the fact that 1cm³ of 2.0M HCl contains 1/1000 * 2.0 = 0.002 moles. If 0.002 moles in total are diluted with water with 9cm³ of water to produce a total volume of 10cm³ then the concentration (molarity ) = 0.002/ 0.01(dm³)                 0.2M

Join now!

The concentrations for HCl can be calculated using the equation above:

Similarly using the equation I can calculate the concentration for Na2S2O3, but this time 1cm³ of 0.4M  Na2S2O3 contains 1/1000 * 0.4 =  4*10-4 moles. If 4*10-4 moles in total are diluted with water with 9cm³ of water to produce a total volume of 10cm³ then the concentration (molarity ) =4*10-4  / 0.01(dm³)                  0.04M

Using these calculations I can plot a graph, rate of reaction against concentration of the reactants:

Rate of

Reaction

...

This is a preview of the whole essay