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 If you ran the reaction using the amounts shown below, at room temperature, what would be the rate of the reaction, according to the results you calculated? How long do you predict it would take for the yellow color to disappear in the above trial? If you ran the above trial at 37oC, what would be the rate of the reaction? Experiment: Water (mL) 2.0 M HCl (mL) 1.0 M Acetone (mL) 0.0050 M I2 (mL) 5 10 5 5 2

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 If you ran the reaction using the amounts shown below, at room temperature, what would be the rate of the reaction, according to the results you calculated? How long do you predict it would take for the yellow color to disappear in the above trial? If you ran the above trial at 37oC, what would be the rate of the reaction? Experiment: Water (mL) 2.0 M HCl (mL) 1.0 M Acetone (mL) 0.0050 M I2 (mL) 5 10 5 5 2

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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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Question

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 25^oC)

Experiment 5 - 24 sec at 38^oC

Experiment 6 - 485 sec at 15^oC

If you ran the reaction using the amounts shown below, at room temperature, what would be the rate of the reaction, according to the results you calculated? How long do you predict it would take for the yellow color to disappear in the above trial? If you ran the above trial at 37^oC, what would be the rate of the reaction?

Experiment:	Water (mL)	2.0 M HCl (mL)	1.0 M Acetone (mL)	0.0050 M I₂ (mL)
5	10	5	5	2

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