Reaction Order and Rate Law Essay

Data, Calculations, and Questions A. 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 Time (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 reaction time for each reaction by adding the times for the two trials and dividing by 2. C. Calculate the reaction rate by taking the inverse of the average reaction time, i.e., 1 divided by the average 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 are  experimental data 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 increase 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 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 Titus Done in the table Time average=time trial 1+time trial 2/2 HCl reaction is 1.36 Na2S2O3 reaction is 0.84 Rate law = k[HCl]^1.36[Na2S2O3]^0.84 Rate law=k[0.024]^1.36[0.0576]^0.84 Rate law= k[.03264][.048384] K=1/.00158 K= 632.9? Me not fully sure if my numbers are correct or not. Rounding correctly, documenting at right time.