After watching the demonstration shown by the teacher, I noticed a cloudy solution forming within 10 seconds after mixing the 50cmi?? of (0. 4 moldm-i?? ) Na2S2O3 was mixed with 5. 0 cmi?? HCL (2. 0 moldm-i?? ) diluted with 20cmi?? 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:  After a while I noticed a smell of SO2 as it diffused across the room.

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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 5cmi?? of Na2S2O3 and 1cmi?? 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 20cmi?? 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: HCl (cmi?? ) Na2S2O3 (cmi?? ) H2O (cmi?? ) Time (seconds)  In this reaction the concentration of Na2S2O3 will remain constant at 10 cmi?? , 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: Na2S2O3 (cmi?? ) HCl (cmi?? ) H2O (cmi?? ) Time (seconds) 2  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-i?? ) = moles Volume (dm-i?? ) Using the fact that 1cmi?? of 2. 0M HCl contains .

If 0. 002 moles in total are diluted with water with 9cmi?? of water to produce a total volume of 10cmi?? then the concentration (molarity ) = 0. 002/ 0. 01(dmi?? ) 0. 2M The concentrations for HCl can be calculated using the equation above: HCl (cmi?? ) H2O (cmi?? ) Concentration (moldm-i?? )  Similarly using the equation I can calculate the concentration for Na2S2O3, but this time 1cmi?? of 0. 4M Na2S2O3 contains 1/1000 moles in total are diluted with water with 9cmi?? of water to produce a total volume of 10cmi??then the concentration (molarity ) = 0. 04M Na2S2O3(cmi?? ) H2O (cmi?? )

Concentration (moldm-i?? ) Using these calculations I can plot a graph, rate of reaction against concentration of the reactants: Rate of Reaction (1/ time) Concentration of Reactant (H+ or S2O3i?? -) I already know that the rate equation will be as shown below: Rate = [H+ (aq)]? [S2O3i?? -(aq)]? As you can see in the rate equation, the order of each reactant is unknown. This will be worked out by the graph produced using the inverse of time recorded against the concentration of the reactant.

After plotting the graph for each reactant, if I get a graph as shown below this will indicate a zero order with respect to the reactant: Rate of Reaction (1/ time) Concentration of Reactant (H+ or S2O3i?? -) If the rate equation for a reactant is first order then I will obtain a graph as shown below: Rate of Reaction (1/ time) Concentration of Reactant (H+ or S2O3i?? -) If the rate equation for a reactant is second order then I will obtain a graph as shown below: Rate of Reaction (1/ time) Concentration of Reactant (H+ or S2O3i?? -) To confirm the reactant is second order, if the [Reactant]i?? then you should produce a graph like this:

Rate of Reaction (1/ time) Concentration of [Reactant]i?? (H+ or S2O3i?? -) From the type of graph achieved, I will be able to complete the rate equation for this reaction. Apparatus Stop clock Burettes 2 Measuring cylinder (10cmi?? ) Measuring cylinder (100cmi?? ) Beakers 100cmi?? Test tubes Test tube rack 0. 4 moldm-i?? of Na2S2O3 (90cmi?? ) 2. 0 moldm-i?? of HCL (90 cmi?? ) Water Ball point pen Sticky labels Funnel Clamp Stand 2 Diagram Safety Hydrochloric acid Contact with the eyes or skin can cause serious permanent damage therefore avoid contact by wearing gloves and avoid spillages.

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