Question

Student Name: Instructor Name: DATA (EXP #4) Part 1: Effect of Concentration Time (sec) mL NaHSO3/starch mL water mL KIO3 ai.lolo 50 mL 0 mL 50 mL 7.05 45.80 50 mL 15 mL 35 mL 50 mL 25 mL 25 mL Min 475e (07Seconds 50 mL 35 mL 15 mL Part 2: The Effect of Temperature Time (sec) Temperature Temperature KIO3 Instructor Name: Student Name: DATA ANALYSIS: Part 1: Effect of Concentration: 1. For each trial calculate: The concentration of KIOs in molarity. This initial concentration was 0.0024M and it was diluted by water and the NaHSO/starch solution in each trial The "rate", which is simply 1/time. Mind you, this is not the true rate, but should be proportional to rate and will suit our purposes in finding the order with respect to KIO Log(Rate) and Log(Concentration) 2. Use a spreadsheet program in the adjacent room or the Chemistry Resource Center to plot Log(Rate) as a function of Log(Concentration) 3. Fit your data with an appropriate trendline (right click on points) and find select "display equation on chart" Attach a copy of your graph with your report. 4. Volume KIO (mL) Time (sec) Rate (1/Time) Concentration KIOs (M) Log (KIO) Concentration) Log (Rate) (s1) Ql.lolosec 50 mL a7.26 sec 35 mL 45.80sec 25 mL 15 mL 107 sec mVEMava

Answer 1

The concentration of KIO3 can be calculated by multiplying the volume of KIO3 that is being added to the water by its initial concentration that is 0.0024M and then dividing by the total volume that is 50mL in each case.

For the calculation of the rate of reaction, we will just have to divide the change in the concentration of KIO3 by the total time taken for the reaction.

Vol.eonc (KIQS Rota (Rate) (1/Tine OOH6 1. 50m.0M - 2-62 -134 2. 35 O.O0168 -1-43 -2-77 O.O37 -2.92 3. 25 o.0012 O O22 -1.66 4 15m| o.0002 -3-14 -2 03