Reaction Rate Lab

Objective 1. To determine how the concentration of a species can affect reaction rate in the determination of rate law and rate constant. 2. To determine how temperature affects reaction rate.
Introduction
Chemical kinetics deals with the speed, or rate, of a reaction and the mechanism by which the reaction occurs. We can think of the rate as the number of events per unit time. The rate at which you drive (your speed) is the number of miles you drive in an hour (mi/hr). For a chemical reaction the rate is the number of moles that react in a second. In practice, we usually monitor how much the concentration (the number of moles in a liter) changes in a second. Reaction rates are usually expressed in units of moles per liter per second,
Chemicals
0.20 M potassium iodide (KI solution), 0.0050 M sodium thiosulfate (Na2S2O3 containing 0.4% starch indicator), 0.10 M potassium peroxydisulfate (K2S2O8) .
Procedure
Part 1 : Finding the Rate Law and Rate Constant 1. The most accurate piece of glassware that allows for variable volumes to be dispensed for all volume additions was used. 2. The solution for each trial based on the volume as stated in table 3.1 was prepared. 3. Timing was beginning as soon as the peroxydisulfate is added and was immediately swirl the reaction to ensure good mixing. The timing was stopped when the solution changes to blue. 4. Each trial was performed at least two trial or until reproducible reaction times are obtained. The concentration of KI and K2S2O8 are such that a stoichiometric mixture for reaction was result from equal volumes. 5. To ensure uniform conditions for all runs, a constant total volume of 50.0 mL in a 250 mL conical flask was used and was maintained initial [S2O32-]0 = 0.0010 M for each run.

Volume used for the preparation each of
Find the value of x and y for the rate law equation 3 from these plot, rounding them to the nearest half-integer to reflect experience with “simple” reactions such as reaction equation 2.

Rate = K [I-]x [s2o8-2]y
Rate 1/ Rate 2 = K [0.08]x [0.04]y ÷ K [0.06]x [0.04]y
1.447 × 10-5/ 1.179 × 10-5= 1.33 x
= 1.33 x
Log 1.290 = X log 1.33
X = 0.892 ~ 1
Thus , the order of reaction is third order with respect to [I-].
Rate 1/ Rate 2 = K [0.04]x [0.04]y ÷ K [0.04]x [0.03]y
1.48× 10-5 / 1.09 × 10-5 = 1.33 Y
1.357 = 1.33 Y
Log 1.357 = Y log 1.33
Y = 1.070 ~ 1

Rate = K [I-]x [s2o8-2]y
Rate = K [I-]1[s2o8-2]1
X+y= 1+1=2

Thus , the order of reaction is the second order with respect to [s2o8-2]

So the overall order of reaction is second order
Rate = K [I-]x [s2o8-2]y
Trial 1: Rate = K [I-]x [s2o8-2]y
1.187 × 10-5 = K (0.08)(0.04) K = 3.709 × 10-3
Trial 2: Rate = K [I-]x