VCU Influence of New Media on Intercultural Communication Research Outline Post your outline here please. This is the skeleton of your paper/project (even

VCU Influence of New Media on Intercultural Communication Research Outline Post your outline here please. This is the skeleton of your paper/project (even with a video project there should be a script/plan) . The outline could be a bullet-point structure and should include a title, introduction: presenting main argument and some background information; Body: bullet points of what and how you will introduce and develop your topic; then a Conclusion: as well bullet point structure of what will go in this section. Please include in each section the main references that you will use for each main point. A reference page of the potential/preliminary references you will be using should be included. Chem 132
Quantitative Chemical Kinetics1,2,3
Spring 2020
INTRODUCTION
Reaction Rates and Rate Laws
We’ve looked at a wide range of chemical reactions and learned how to determine the
relative concentrations of reactants and products at equilibrium. Regardless of where the
final position of a chemical equilibrium lies, however, to ask how long it will take to
reach equilibrium is an entirely different question. The rate at which a chemical reaction
occurs depends on several factors: the chemical identities of the reactants, their
concentrations, the temperature, and the presence of catalysts. Each of these factors can
markedly influence the observed rate of reaction.
Some reactions at a given temperature are very slow—the oxidation of H2(g) or
gasoline is unquestionably spontaneous, but neither process proceeds appreciably at room
temperature, even in a hundred years. (Add a lit match and it’s a different story!) Other
reactions are essentially instantaneous. The precipitation of solid silver chloride when
aqueous silver ions and chloride ions are mixed, and the formation of water when acidic
and basic solutions are mixed are examples of extremely rapid reactions. In this week’s
experiment we’ll study a reaction that proceeds at a moderate, relatively easily measured
rate at room temperature so that we can quantitatively characterize the reaction kinetics.
For a given reaction, reaction rate typically increases with an increase in the
concentration of any reactant. For a general reaction,
aA + bB → cC + dD
(1)
the reaction rate can be expressed as the change in the concentration of the products or
reactants as a function of time:
(2)
Because the reaction rate is strictly positive, the negative sign for the reactant terms
indicates that the concentration of those species decreases over time (they are being
consumed), whereas the concentration of products increases (they are being produced).
The reaction rate can also be expressed in terms of the concentration of the reactants
by an equation we call the rate law for the reaction:
Rate = k [A]m [B]n
(3)
Here m and n are generally, but not always, integers (0, 1, 2, or possibly 3); [A] and [B]
are the concentrations of A and B; and k is a constant called the rate constant of the
reaction. Note that the values of m and n can only be determined experimentally and are
entirely unrelated to the stoichiometry of the reaction.
The numbers m and n are called the orders of the reaction with respect to A and to
B, respectively. For example, if m = 1 the reaction is said to be first order with respect
to the reactant A. If n = 2 the reaction is second order with respect to reactant B.
Winans, R; Brown, C. A. J. Chem. Educ. 1975, 52, 526-527.
Nicholson, L. J. Chem. Educ. 1989, 66, 725-726.
3
Kildahl, N.; Varco-Shea, T. Explorations in Chemistry (Wiley: NY, 1996); 191-199.
1
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Chem 132
Spring 2020
The rate law in Eq. 3 provides valuable insight into the underlying chemistry of the
reaction, because the rate law (that is, the values of k, m, and n, which we determine by
experiment) directly reflects the mechanism of the reaction. From the rate law, we can
often determine how individual reactants break bonds and collide to form new bonds
during the reaction, and the order in which that occurs. A rate law is thus a kind a
microscope: it can inform us what is happening during the reaction at the level of
individual atoms or molecules. That is, it can reveal the reaction mechanism.
The reaction we will study is the fading of color caused by the reaction between
hydroxide ion (OH–) and the colored form of an indicator, phenolphthalein, in aqueous
solution. The structures of the colored indicator and the colorless product are shown
below. Because the unreacted form of the indicator is very strongly colored and the
product is colorless, the change in concentration of unreacted indicator can be followed
directly by monitoring the absorbance of the solution by Beer’s Law: A = εbc, where A is
absorbance, ε is molar absorptivity, b is path length, and c is concentration.
PINK
COLORLESS
Figure 1. The reaction of phenolphthalein with hydroxide ion.
Experimental Determination of Rate Law by Graphical Method
To find the rate law for this reaction, we’ll use the graphical method to determine the
reaction order and rate constant. This is based on the fact that for reactions of different
order, a plot of the concentration of a reactant vs. time exhibits a different shape.
Zero Order Reactions
For a simple reaction, such as A → B, if it is zero order with respect to A then m = 0.
The rate can be expressed in terms of the concentration of the reactant A raised to the
zeroth power, which is equal to one:
(Note: “[A]0” is “[A] raised to the zero power”)
(4)
Rate = k [A]0 = k (1) = k
Combining Eq. 2 and Eq. 4 and using some calculus, we can integrate, which gives
[A]t = [A]o – k t
(Note: “[A]o” is “the concentration of [A] at time zero”)
(5)
where [A]t is the concentration of species A at time t, and [A]o is the concentration of A
initially (at t = 0). Eq. 5 can be rearranged to be written in the form, y = m x + b:
[A]t = – k t + [A]o
(6)
This means a plot of [A] versus time will be linear if the reaction is zero order with
respect to A; the slope will be equal to the negative of the rate constant (k = –slope) and
the y-intercept is the concentration of A initially, [A]o.
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Chem 132
Spring 2020
First Order Reactions
If instead the reaction is first order with respect to A, m = 1, and we have:
Rate = k [A]1 = k [A]
(7)
Combining with Eq. 7 with Eq. 2 and again using calculus to integrate, we get
ln[A]t = ln[A]o – k t
(8)
Rearranging Eq. 7 so that it is written in the form y = m x + b, we get:
ln[A]t = – k t + ln[A]o
(9)
Thus, a plot of ln[A] versus time will be linear if the reaction is first order with
respect to A. In this case, the slope of the line equals the negative of the rate constant
(k = –slope), although here the y-intercept corresponds to the natural logarithm of the
initial concentration of A (ln [A]o).
Second Order Reactions
Finally, if the reaction is second order with respect to A, m = 2, and the rate is:
Rate = k [A]2
Combining Eq. 2 and Eq. 10 and integrating, we get:
(10)
(11)
Rearranging Eq. 11 to write it as y = m x + b, we get:
(12)
Thus, a plot of the reciprocal of [A] versus time will be linear if the reaction is
second order in A. This time the rate constant is directly equal to the slope (k = slope),
and the y-intercept equal to the reciprocal of the initial concentration of A (1/[A]o).
A Few Practical Considerations
If we measure concentration of the indicator phenolphthalein, [Ind], versus time, we
can make three plots: [Ind] vs. t; ln[Ind] vs. t; and 1/[Ind] vs. t. Only one of these three
plots should be linear, and which one it is will determine the order of the reaction with
respect to the indicator (that is, the value of m). For this lab however, we won’t plot the
indicator concentration itself. By Beer’s Law, the absorbance of the solution is directly
proportional to the concentration of indicator. Therefore, we can plot Abs vs. t, ln(Abs)
vs. t, and 1/Abs vs. t, to find the order of the reaction with respect to the concentration of
indicator: simply identify which one of the three plots is linear!
However, be careful: any curve looks linear if you look at a small enough piece of it
(e.g., the Earth looks flat!), and so it’s important to observe the change in concentration
(or absorbance) over a large range. Practically speaking, this means you must observe
the absorbance change by at least a factor of four over the course of the experiment.
A complicating factor is that the reaction involves two reactants, phenolphthalein and
hydroxide ions, so the full rate law may depend on the concentrations of both of them, as
shown in Eq. 3. In deriving Equations 6, 9, and 12 we assumed that only one reactant
(the indicator) was involved, effectively ignoring the effect of OH–. To get the order of
the reaction with respect to the concentration of hydroxide, we’ll do two things:
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Chem 132
Spring 2020
1) We will use the “method of isolation” where one of the reactants (OH– in this
case) is present in much greater concentration (a thousand times or more) than the
other reactant (phenolphthalein, the indicator, Ind). So, while the indicator is
completely consumed by the reaction, the hydroxide concentration is essentially
constant throughout the experiment. Under these conditions, the true rate law,
Rate = k [Ind]m [OH–]n
(13)
is approximated as only depending on [Ind], m, and an apparent rate constant, k’:
Rate ≈ k’ [Ind]m
(14)
– n

where k’ ≈ k [OH ] . Under these conditions, by taking [OH ] to be
approximately constant during the reaction, the values of m and k’ can be
determined by the graphical method described above.
2) To determine the value of k, the true rate constant, and n, the order with respect to
OH–, we need to determine the apparent rate constant (k’) at two different
concentrations of OH–, and then apply some clever math. We’ll leave the math
for the analysis after lab. See below!
Finally, the reaction rates among ions depend on the strength of the interactions
among the ionic charges. These can be partially “screened” from each other by the
presence of other ions and thus the magnitude of the rate constant depends on the ionic
strength, which is a function of the total concentration of ions. To eliminate this effect
when we combine data from experiments with different hydroxide concentrations, we
need to add spectator ions to keep the total concentration of ions constant. To do this,
we’ll add the neutral salt, NaCl, as necessary. So that you are able compare data from
different people in the class, we’ll set a single total ion concentration to use class-wide.
EXPERIMENTAL PROCEDURE
You will set up two experiments in which you measure the absorbance of
phenolphthalein versus time as it reacts with hydroxide. The two experiments will be
otherwise identical except the [OH–]. As mentioned above, we will use a constant total
ion concentration. Therefore, [NaCl] + [NaOH] = 3.0 M for all experiments. To do this,
you will use 4.0 M solutions of both NaOH and NaCl. Thus,
 The volume of NaOH plus the volume of NaCl must always equal 7.50 mL.
 The volume of indicator plus the volume of water must always equal 2.50 mL
 The combined volume of all four components must always equal 10.00 mL.
1. Set up the LabQuest colorimeter as detailed on the separate instructions. Blank the
colorimeter with a 3.0 M NaCl solution (not with water).
2. Determine the volume of indicator stock solution to use that gives an absorbance of
0.8 to 1.0 in the final mixture. Do this by trial-and-error by combining:
 various volumes of phenolphthalein solution,
 7.50 mL NaCl stock solution,
 25 µL of NaOH stock solution,
 and then diluting with water to total volume of 10.00 mL.
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Chem 132
Spring 2020
Measure the solution’s absorbance promptly after preparing it. Any volume of
phenolphthalein that gets you in the desired absorbance range is fine. Use
micropipettes, good volumetric technique, and keep good records when preparing
your trial solutions. When you have an amount of dye that works, record that volume
in your notebook along with the corresponding absorbance reading. It doesn’t matter
exactly what the volume of the phenolphthalein solution is, but you should measure
and record its value as accurately as possible.
3. Determine an appropriate volume of NaOH by trial-and-error. Take the absorbance
you measured in Step 2 and divide it by four, giving you your target final absorbance
(that is, you’ll collect data until the absorbance drops to this value). Any volume of
NaOH that gets you to your target final absorbance value after 5–20 minutes is fine.
Remember: the volume of NaOH plus the volume of NaCl should always equal 7.50
mL as described above. Hint: Start your trials with a high [NaOH] then lower it as
needed; it’s quicker to determine the reaction is too fast than too slow!
4. Measure the absorbance vs. time for the reaction. For each kinetics run:
i. First, add the appropriate amounts of NaOH and NaCl to a 10.00-mL volumetric
flask. Then add about half of the amount of deionized water that you will need
to bring the final volume to 10.00 mL.
ii. Mix the solution thoroughly.
iii. Add the volume of indicator that you found in Step 2. Start LabQuest data
acquisition when you add the indicator (see colorimeter instructions).
Yes, the colorimeter will be empty for the first few seconds of the experiment!
iv. Add water to the calibration line of the volumetric flask, and thoroughly mix the
solution. Then, remove an aliquot and fill your cuvette about ⅔ full. Place the
cuvette into the colorimeter (which will have already begun collecting data!).
Collect data until the absorbance has dropped by at least a factor of four.
5. After collecting good kinetics data at one NaOH concentration, repeat the Step 4 at a
second NaOH concentration that is at least a factor of two different from the value
used in the first run. Adjust the volume of NaCl correspondingly.
CALCULATIONS
1. Using Excel or a similar spreadsheet program, plot your absorbance-vs.-time data (for
both concentrations of OH–). As described in the introduction, you should make
three different plots for each experiment: Abs vs. t; ln(Abs) vs. t; and 1/Abs vs. t.
Based on which plot is linear, determine the order of the reaction with respect to the
indicator, and report the value of m. Then, for the linear plot, use the linear
regression tool and report your slope and intercept to an appropriate number of
significant figures and with the correct units (Remember: you are using absorbance!).
2. Using the slope from your linear regression, report the value of the apparent rate
constants for each reaction run, k’expt1 and k’expt2 (fast and slow, measured at the two
different OH– concentrations).
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Chem 132
Spring 2020
3. For the two experiments, use the apparent rate constants, k’expt1 and k’expt2, and the
hydroxide concentrations, [OH–]expt1 and [OH–]expt2, to determine the reaction order
with respect to OH–, n, and the true rate constant, k (see instructions below).
How to Calculate the Order with Respect to OH– and the True Rate Constant
As described in the introduction, the true rate law is given by:
Rate = k [Ind]m [OH–]n
(15)
where Ind is the phenolphthalein indicator. In this experiment, we can treat [OH–]n as
being constant because [OH–] changes very little during the experiment. Therefore, we
group it with the rate constant to give an apparent rate constant, k’ where k’ ≈ k [OH–]n.
We do this because the indicator concentration, [Ind], changes a lot, but [OH–] does not
(because of the way we designed the experiment). The rate law is now simplified:
Rate ≈ k’ [Ind]m
(16)

You did two experiments where everything was the same except for the value of [OH ].
For both of these experiments, you should find the same reaction order with respect to the
indicator (m). From the graphs of the two experiments, you determined the observed rate
constants, k’expt1 and k’expt2, corresponding to [OH–]expt1 and [OH–]expt2, respectively.
Your task is now to find the true rate constant (k) and reaction order with respect to
hydroxide (n). To find n, we compare k’expt1 and k’expt2. Recall how they are defined:
(17)
(18)
You want to solve those equations for n, but you also don’t know the value of k yet. But,
you can make k cancel out of the equations by dividing one equation by the other:
(19)
Notice how the exponent n was be applied to the entire fraction. We now take the
logarithm of both sides, taking n out of the exponent. Remember: log (xn) = n log x.
(20)
Plug in your experimentally derived values into Eq. 20, rearrange and solve for n.
Round n to the nearest integer and this is the reaction order with respect to hydroxide.
Finally, to get the true rate constant rearrange equations 17 and 18 to solve for k using
the rounded value of n:
and
(21)
Note the values you get for k from the two forms of Eq. 21 will vary slightly due to
experimental error. There is only one correct value of the true rate constant, k, so report
the average of the two values from Eq. 21.
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Chem 132
Quantitative Kinetics: Report Sheet
Spring 2020
Name: _______________________
1. Fill in the table below for your two experiments. Report the concentration of the
reagent as it was in the 1-cm cuvette. Don’t forget to use correct sig figs!
Experiment #1
Experiment #2
Volume of phenolphthalein soln. (mL)
[Phenolphthalein] (M)
Volume of NaOH soln. (mL)
[NaOH] (M)
Volume of NaCl soln. (mL)
[NaCl] (M)
2. Make six plots in Excel of absorbance vs. time, ln(absorbance) vs. time, and
absorbance–1 vs. time for each of your two [OH–] values. Include only the two plots
that resulted in the “best linear” fit in your answer. These two plots/figures should be
professional quality, and should include informative captions. Note that multi-part
figures that are all described by a single caption are particularly effective and avoid
repetition! Include a linear regression and report the slope, intercept, and R2 value.
Note: you must analyze data in your own individual Excel file.
3. What is the reaction order with respect to phenolphthalein? Provide experimental
evidence. Briefly explain your reasoning.
4. What are the values of the apparent rate constant, k’, for each of the two
experiments? Include units.
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Chem 132
Spring 2020
5. What is the reaction order with respect to hydroxide ion? Show your work.
6. What is the value of the true rate constant for the reaction of phenolphthalein with
hydroxide? Include units. Show your work.
7. Write the full numerical rate law for the reaction. Include units.
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Chem 132
Spring 2020
BONUS #1 (+2 pts): Based on the rate law, propose a reasonable reaction mechanism
in terms of elementary reactions. Consider multiple steps, a rate-determining step, and/or
any initial equilibria that might occur. Briefly justify your choice on a separate page.
BONUS #2 (+2 pts): In this analysis, we have actually used absorbance instead of the
concentration of phenolphthalein (in units of M). Solve for the true rate constant, using
the molarity of the indicator, not its absorbance. What do you notice? Attach your work.
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Stock phenolphthalein concentration (M)
Stock NaOH concentration (M)
Stock NaCl concentration (M)
Total Volume Used (mL)
Volume of phenolphthalein used (mL)
Volume of NaOH used (mL)
Volume of NaCl soln. (mL)
7,00E-04
4,00
4,00
10,00
Experiment #1
1,500
1,000
6,500
Experiment #2
1,500
0,500
7,000
4,2
4,3
4,4
4,5
4,6
4,7
4,8
4,9
5
5,1
5,2
5,3
5,4
5,5
5,6
5,7
5,8
5,9
6
6,1
6,2
6,3
6,4
6,5
6,6
6,7
6,8
6,9
7
7,1
7,2
7,3
7,4
7,5
7,6
7,7
7,8
7,9
8
Time (min) Absorbance #1 (fast)
0
0,848
0,1
0,8222
0,2
0,8005
0,3
0,7804
0,4
0,7615
0,5
0,7435
0,6
0,7262
0,7
0,7097
0,8
0,6937
0,9
0,6783
1
0,6634
1,1
0,649
1,2
0,635
1,3
0,6214
1,4
0,6083
1,5
0,5954
1,6
0,583
1,7
0,5709
1,8
0,5591
1,9
0,5476
2
0,5364
2,1
0,5255
2,2
0,5148
2,3
0,5044
2,4
0,4943
2,5
0,4844
2,6
0,4748
2,7
0,4653
2,8
0,4561
2,9
0,4472
3
0,4384
3,1
0,4298
3,2
0,4214
3,3
0,4132
3,4
0,4052
3,5
0,3974
3,6
0,3897
3,7
0,3822
3,8
0,3749
3,9
0,3677
4
0,3607
4,1
0,3539
Time (min) Absorbance #2 (slow)
0
0,851
0,1
0,8376
0,2
0,8262
0,3
0,8156
0,4
0,8054
0,5
0,7956
0,6
0,7861
0,7
0,7769
0,8
0,7679
0,9
0,7591
1
0,7505
1,1
0,7421
1,2
0,7339
1,3
0,7258
1,4
0,7179
1,5
0,7101
1,6
0,7025
1,7
0,6949
1,8
0,6875
1,9
0,6803
2
0,6731
2,1
0,6661
2,2
0,6591
2,3
0,6523
2,4
0,6455
2,5
0,6389
2,6
0,6324
2,7
0,6259
2,8
0,6195
2,9
0,6133
3
0,6071
3,1
0,601
3,2
0,5949
3,3
0,589
3,4
0,5831
3,5
0,5773
3,6
0,5716
3,7
0,5659
3,8
0,5603
3,9
0,5548
4
0,5494
4,1
0,544
4,2
4,3
4,4
4…
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