Iodine in solution has a yellow-brown color, whereas the acetone and both products are colorless. This means that we can run the reaction using iodine as the limiting reagent, and when we see that the yellow-brown color of the iodine has disappeared, we know that the reaction has stopped and all of the iodine has been consumed.   Recall that to calculate a rate of reaction, we would like to know the change in concentration of a reactant or product during a fixed amount of time. We will choose the reactant iodine in this case, because its brown color allows us to determine when its concentration has fallen to zero.

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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I2 (aq) + CH3COCH3 (aq)  HI (aq) + ICH2COCH3 (aq)
Iodine in solution has a yellow-brown color, whereas the acetone and both products are colorless. This means that we can run the reaction using iodine as the limiting reagent, and when we see that the yellow-brown color of the iodine has disappeared, we know that the reaction has stopped and all of the iodine has been consumed.
 
Recall that to calculate a rate of reaction, we would like to
know the change in concentration of a reactant or product during a fixed amount of time. We
will choose the reactant iodine in this case, because its brown color allows us to determine
when its concentration has fallen to zero.
 
Rate = –Δ[I2] / Δt = – ([I2]final – [I2]initial) / ([tfinal – tinitial) = – (0 – [I2]initial) / (tfinal – 0) = [I2]initial / tfinal
 
Run several trials at different concentrations of reactants and use themethod of initial rates to obtain values for k, x, y, and z in the rate law.
 
Rate = k [I2]x [acetone]y [H+]z
 
Run the reaction at fixed concentrations but at different temperatures, so that the activation energy and frequency factor can be determined from the Arrhenius equation. This equation can be rearranged into the form below, which mirrors that of a straight
line:
ln k = (– Ea/R) (1/T) + ln A
y = m x + b
 
Thus once the value for k is known at several different temperatures, a graph of ln k vs (1/T)
should approximate a straight line, allowing the slope (which equals –Ea/R) and y-intercept
(which equals ln A) to be determined.
 
Experiment #: Water (mL) 1.0 M HCl (mL) 4.0 M Acetone (mL) 0.0050 M I2 (mL)
1 10 5 5 5
2 5 10 5 5
3 5 5 10 5
4 5 5 5 10
Data collected:

Experiment 1 - 58 sec

Experiment 2 - 34 sec

Experiment 3 - 31 sec

Experiment 4 - 130 sec

(Experiments 1-4 were at 25oC)

Experiment 5 - 24 sec at 38oC

Experiment 6 - 485 sec at 15oC

Calculation of the initial concentrations of reactants: [In this section, present the calculations that are pertinent for determining the initial concentrations of your reactants.  Include units! Explain the chemical principles that were involved in deciding to manipulate the numbers the way you did.  For instance, you would not simply say in words that you converted from mass to moles; you would explain why you needed moles instead of mass.]

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