Dr. Charles J. Horn
Abstract: This two part experiment is designed to determine the rate law of the following reaction, 2I-(aq) + H2O2(aq) + 2H+I2(aq) + 2H2O(L), and to then determine if a change in temperature has an effect on that rate of this reaction. It was found that the reaction rate=k[I-]^1[H2O2+]^1, and the experimental activation energy is 60.62 KJ/mol.
The rate of a chemical reaction often depends on reactant concentrations, temperature, and if there’s presence of a catalyst. The rate of reaction for this experiment can be determined by analyzing the amount of iodine (I2) formed. Two chemical reactions are useful to determining …show more content…
Trial 1: (.005 L Na2SO4)(.02 moles Na2SO4/1.0L)(1 mol I2/2 mol Na2SO4)= .00005 mol I2
Use this method for all six trials 2) Find the reaction rate using moles of I2 produced, measured time in seconds, and Volume of total solution for all six trials
Trial 1: (.00005 mol I2/.0403L)=(.00124906 mol/L) /(692seconds)= .00000181mol/L(s)
Use this method for all six trials 3) Find the rate constant using the reaction rate, measured volumes used, stock concentrations, and the rate law of the main reaction.
Trial 1: K=(.00000181MOL/L(s))/((.01 L H2O2)(.1 M H2O2)/.0403L total))((.3MKI)(.006LKI)/.0403L total)=.00107
Use this method for all six trials 4) To graph, we must calculate Ln(k)