Reaction Order and Rate Law Essay

Data, Calculations, and QuestionsA. Calculate the initial and final submergings as needed to complete Tables 1 and 2.Data Table 1 varying the Concentration of 1.0 M HCl Concentrations Drops Drops Drops Initial Drops Drops Drops Initial Initial closing Final Reaction Time (sec) Reaction Well HCl Water Na2S2O3 HCl Na2S2O3 HCl Na2S2O3 Trial 1 Trial 2 Avg regularize (sec-1) 1 8 0 12 1 M 0.3 M 0.4 0.18 18.4 16.3 17.35 0.0576 2 8 6 6 1 M 0.15 0.4 0.0045 37.1 37.9 37.5 0.0267 3 8 8 4 1 M 0.1 0.4 0.02 107.2 106.6 106.9 0.0093 B. Calculate the average chemical reply time for each reception by adding the times for the two trials and dividing by 2.C. Calculate the reaction number by taking the inverse of the average reaction time, i.e., 1 sepa judge by the average reaction time.1. Use table 1 to ensure the reaction order for HCl.2. Use table 2 to determine the reaction order for Na2S2O3.Remember, you want to see what happens to the reaction rate when you mental image t he concentration of one reactant while the second reactant remains unchanged. In Part 1, we alter the concentration of HCl while we kept the concentration of Na2S2O3 the same. In Part 2 we varied the concentration of Na2S2O3 while keeping the concentration of HCl the same. These are experimental data and results will be different from some of the nice, even rime you saw on textbook problems. For example, in this experiment you may double the concentration of a reactant but the reaction rate may profit anywhere from 1.7 times to 2.4 times. This still means an approximate doubling of the reaction rate. On the other hand, if you double a reactant concentration and the reaction rate increases by 0.7 to 1.3 times that probably means that the reaction rate multiplier factor is one (1).D. Write the rate law for the reaction.E. Using the rate law, the rate, and the give up concentration(s) from one (or more) of your experiments calculate k.F. What are the potential errors in this exper iment?Laura Titus do in the tableTime average=time trial 1+time trial 2/2HCl reaction is 1.36Na2S2O3 reaction is 0.84Rate law = kHCl1.36Na2S2O30.84Rate law=k0.0241.360.05760.84Rate law= k.03264.048384K=1/.00158K= 632.9?Me not fully sure if my numbers are ready or not. Rounding correctly, documenting at right time.