Reaction Order and Rate Law Essay Example for Free

Reaction Order and Rate Law establishData, Calculations, and QuestionsA. Calculate the initial and final concentrations 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 Final Final Reaction quantify (sec) Reaction Well HCl Water Na2S2O3 HCl Na2S2O3 HCl Na2S2O3 Trial 1 Trial 2 Avg Rate (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 reply magazine for each reply by adding the times for the two trials and dividing by 2. C. Calculate the reaction commit by taking the inverse of the average reaction time, i.e., 1 divided by the average reaction time.1. Use table 1 to nail down 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 evaluate when you ikon the concentration of one reactant eyepatch the second reactant remains unchanged. In Part 1, we varied 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 areexperimental information and results will be different from some of the nice, even numbers you saw on textbook problems. For example, in this experiment you may double the concentration of a reactant but the reaction rate may accession 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 is one (1).D. Write the rate virtue for the reaction.E. Using the rate law, the rate, and the appropriate concentration(s) from one (or more) of your experiments calculate k.F. What are the potential errors in this experiment?Laura TitusDone 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 enough if my numbers are correct or not. Rounding correctly, documenting at right time.