Reaction Order and Rate Law Essay

Data, Calculations, and QuestionsA. Calculate the initial and last(a) stringencys as needed to complete Tables 1 and 2.Data Table 1 change the Concentration of 1.0 M HCl Concentrations Drops Drops Drops initial Drops Drops Drops Initial Initial Final Final reception Time (sec) Reaction Well HCl irrigate Na2S2O3 HCl Na2S2O3 HCl Na2S2O3 Trial 1 Trial 2 Avg assess (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 response snip for each reply by adding the clock for the two tribulations and dividing by 2.C. Calculate the reply rate by taking the inverse of the average response time, i.e., 1 carve up by the average reaction time.1. theatrical role submit 1 to determine the reaction order for HCl.2. Use table 2 to determine the reaction order for Na2S2O3.Remember, you essential to see what happens to the reaction rate when y ou three-fold the parsimoniousness of matchless reactant while the instant 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 provide be different from some of the nice, compensate metrical composition you saw on textual matter problems. For example, in this experiment you may reduplicate the concentration of a reactant but the reaction rate may increase anyplace from 1.7 times to 2.4 times. This still means an fierce doubling of the reaction rate. On the different hand, if you double a reactant concentration and the reaction rate increases by 0.7 to 1.3 times that credibly means that the reaction rate multiplier factor is one (1).D. Write the rate law of nature for the reaction.E. Using the rate law, the rate, and the appropriate concentration(s) from on e (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 0.84 prize 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 correct or not. move correctly, documenting at right time.