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The isolation method.

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The Isolation Method Simplifying nth Order Reactions to use Simple Integrated Rate Equations Introduction Many mechanisms of reactions are categorised by a rate-limiting step, thus the rate of the reaction is hindered by a slow step that the reactions must overcome. In this practical a catalyst (H2SO4-) is used to increase the rate of the reaction, without being consumed in the reaction. The reaction below is a reaction that uses a catalyst in order to react: CH3COCH3 + I3 � CH3COCH2I + 2 I- + H+ In theory, if the catalysed reaction is quicker than the uncatalysed one, then the equation can be written in the form below: Rate = k [CH3COCH3]m [H2SO4-]n [I3-]p Although the constants can be found by using initial rate equations, in this practical, iodine is isolated. Therefore, the disappearance of this reactant is monitored as the other reactant concentrations are kept constant. In order to isolate one reactant, the concentration is kept much smaller than those of the other reactants. ...read more.


be shown as: Run A 7.13 x 10-3 = k [CH3COCH3]m [H2SO4-]n Run B 6.7 x 10-3 = k [CH3COCH3]m [H2SO4-]n Run C 5 x 10-4 = k [CH3COCH3]m [H2SO4-]n However, in order to work out m and n, the concentrations for each run need to be worked out; the workings below are for Run A: Propanone has a liquid density of 0.79g cm-3, and it molecular mass can be worked out to be 58; therefore 1 cm3 can be worked out to be: Mol = Mass Molecular Mass Mol = 0.79 58 Mol = 0.0136 Therefore, in this run, the propanone has a concentration of 0.0272 mol dm-3. Therefore, the concentration of propanone in 20 cm3 can be worked out to be: Concentration = Mol Volume Concentration = 0.0272 0.020 Concentration = 1.36 mol dm-3 Also in Run A, the concentration of the sulphuric acid can be worked out to be: Concentration = 1 x 0.020 Concentration = 0.020 mol dm-3 For Run B, the propanone concentration can be worked out as 0.68 mol dm-3, the sulphuric acid remains the same and is 0.020 mol dm-3. ...read more.


[0.01]n 6.7 x 10-3 = k [0.020]n 0.0746 = 0.5 n = log 0.075 log 0.5 n = 3.73 n = 4 Therefore, using equation A, k can be worked out as: 7.13 x 10-3 = k [1.36]1 [0.020]4 k = 7.13 x 10-3 [1.36]1 [0.020]4 k = 7.13 x 10-3 1.6 x 10-7 k = 44562.5 mol dm-3 s-1 This being such a large number, would indicate that either the calculations have been worked out incorrectly or the practical results were incorrect. Questions 1. Given the form of the rate equation that you have worked out, what species do you think are involved in the rate determining step? The species that may be involved in the rate determining step, I feel, would be the propanone. This is due to the fact that when the concentration of the propanone increased the reaction quickened. 2. If you had doubled the I3- concentration, how would this have affected the measured k1? There would be no change in the k1 results due to the fact that the reaction would stay the same if the iodide concentration was doubled. ...read more.

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