Reaction Order and Rate Law Essay

Data, Calculations, and QuestionsA. Calculate the initial and final soaking ups as needed to complete boards 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 response Time (sec) Reaction Well HCl Water Na2S2O3 HCl Na2S2O3 HCl Na2S2O3 Trial 1 Trial 2 Avg roam (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 h peerlessst reply time for each reply by adding the times for the two trials and dividing by 2.C. Calculate the reaction rate by taking the inverse of the h mavinst reaction time, i.e., 1 divided by the bonnie reaction time.1. Use table 1 to determine 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 double the concentration of one reactant while 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 aredata-based data and results will be different from some of the nice, even numbers you saw on standard problems. For example, in this experiment you whitethorn double the concentration of a reactant but the reaction rate may increase anywhere from 1.7 times to 2.4 times. This still means an approximate doubling of the reaction rate. On the some 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 law 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 Ti tusDone 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 full sure if my numbers are correct or not. Rounding correctly, documenting at right time.