Lab Report 1

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Name: Teiyuri Aoshima Quiz Section: BFID Number: 1124136 Lab Partner: Lina Do

Grading: 60 pts. (Lab notebook pages are worth 5 pts - total for report and notebook is 65 pts)

Chem 162 Experiment #2: Chemical KineticsPart I: A Clock ReactionPart II: Crystal Violet-Hydroxide Reaction Note: All sections of this report must be typed.

By signing below, you certify that you have not falsified data, that you have not plagiarized any part of this lab report, and that all calculations and responses other than the reporting of raw data are your own independent work. . Failure to sign this declaration will result in 5 points being deducted from your report score.Signature:

Purpose and MethodPart I: Clock Reaction (3 pts) (purpose, reactions/rate law, method)

The pupose part of one of lab is to use the method of initial rates in order to determine the orders for BrO3-, I-, and H+ in the respectful chemical reaction as well as the rate constant and activation energy of the reaction. This reaction thiosulfate added so that the time of reaction may be recorded. The orders are found by through experimentation. Three sets of four reactions are conducted in which each set has only one species that does not remain constant. By finding the rate of each of these twelve reactions and comparing the log of this rate to the log of the species whose concentration changes, we can obtain the order of the species in the reaction by setting it equal to the slope of log(rate) versus log(concentration). Once these orders are found, the rate constant can be found by plugging in the values from any run, but in this lab we use runs 2, 6, and 10 because they are identical runs with slightly different outcomes. The activation energy is then found through the Arrhenius equation that compares the change in k as a dependance on temperature. The activation energy for the reaction with a catalyst is found by using the same Arrhenius equation on the two runs that had a catalyst added to it.

Part II: Crystal Violet-Hydroxide Reaction (3 pts) (purpose, reactions/rate law, method)

In part two of this experiment, the goal is to find the complete rate law through the use of integrated rate laws. In part two of this lab, the rate law for the reaction between crystal violet and OH- will be determined through the use of Beers law and our knowledge of integrated rate laws. Because integrated rate laws work for systems with one reactant, this part of the lab requires the use of pseudo rate constants in which the concentration of OH- is assumed to be constant. With this in mind, the absorbance vs concentration of CV+ data will generate a straight line. The slope of this line is the molar absorptivity that is needed to calculate the concentration of CV+ at each time in the abosrbance data for experiments run one to four. Once the concentration for CV+ are determined one can calculate the natural log of the concentrations as well as one divided by the concentrations and then make threee graphs of time vs concentration, time vs ln[concentration] and time vs 1/ concentration. Of these three plots the one that gives the straight line will determine the order of CV+ for that particular run. The pseudo rate constant is then equal to either the positive or negative slope depending on which order CV+ is in the particular run. The order of OH- is then found by plotting ln(k') vs ln(OH-) and the slope of this line is the order of OH- while the antilog of the y intercept is the value of the rate constant.

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Stock Solutions

Soln. Conc. Units

0.0005 M

0.010 M

0.040 M

HCl 0.100 M

Experiment Test Tube #1 Test Tube #2 Temp. D

Run Celsius Time Rate

# mL mL mL mL mL C sec. M/s

1 0.75 0.50 0.25 0.50 0.50 26.7 150.06

2** 0.50 0.50 0.50 0.50 0.50 25.98 94.125

3 0.25 0.50 0.75 0.50 0.50 25.475 70.67

4 0.00 0.50 1.00 0.50 0.50 25.4 45

5 0.75 0.50 0.50 0.25 0.50 23.5 178.3

6** 0.50 0.50 0.50 0.50 0.50 24.57 106.5

7 0.25 0.50 0.50 0.75 0.50 24.9 57.25

8 0.00 0.50 0.50 1.00 0.50 24.85 43.7

9 0.75 0.50 0.50 0.50 0.25 24.67 298

10** 0.50 0.50 0.50 0.50 0.50 24.4 85.3

11 0.25 0.50 0.50 0.50 0.75 24.85 41.47

12 0.00 0.50 0.50 0.50 1.00 24.7 28

13 0.50 0.50 0.50 0.50 0.50 0 280

14* 0.75 0.50 0.50 0.50 0.25 27 14

15* 0.75 0.50 0.50 0.50 0.25 0 28

* Measurements made with a drop of 1.0 mM Ammonium Molybdate

** Repeated measurements for calculating k and Ea

A drop of 1% Starch indicator solution is included in each run.

S2O32-

I-

BrO3-

H2O S2O32- I- BrO3

- H+

1.1 x 10-7

1.8 x 10-7

2.4 x 10-7

3.7 x 10-7

9.3 x 10-8

1.6 x 10-7

2.9 x 10-7

3.8 x 10-7

5.6 x 10-8

2.0 x 10-7

4.0 x 10-7

6.0 x 10-7

6.0 x 10-8

1.2 x 10-6

6.0 x 10-7

Data, Calculations and Graphs

Type a sample calculation for the "Rate" here: (2 pts)For run 1:rate = ∆[S2O3

2-] 6∆t0.005M x 0.005L = 2.5 x 10-7 mol S2O3

2-

[S2O32-] = 2.5 x 10-7 mol

7.5 x 10-4 + 5.0 x 10-4 + 2.5 x 10-4 + 5.0 x 10-4 + 5.0 x 10-4

[S2O32-] = 1.0 x 10-4 M

rate = 1.0 x 10-4 M 6 x 150.06 s

rate = 1.1 x 10-7 M/s

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Experiment Temp Initial Concentrations (M) Time Rate

Run # Celsius sec M/sec

Run 1 26.7 0.020 150.06 1 pt

Run 2 25.98 0.020 94.125

Run 3 25.475 0.020 70.67

Run 4 25.4 0.020 45

1 pt

X-axis Y-axis

log(Rate)

-3.0 -7.0-2.7 -6.7-2.5 -6.6-2.4 -6.4

2 pts

1 pt

S2O32- I- H+ BrO3

-

1.0 x 10-4 1.0 x 10-3 8.0 x 10-3 1.1 x 10-7

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 1.8 x 10-7

1.0 x 10-4 3.0 x 10-3 8.0 x 10-3 2.4 x 10-7

1.0 x 10-4 4.0 x 10-3 8.0 x 10-3 3.7 x 10-7

log(I-)

Reaction Order Determination for I-

Type your calculation for the concentration of BrO3- here:

0.040M x 0.0005L = 2 x 10-5 mol 2 x 10-5 mol 0.0025 L (total volume of all species used)= 8 x 10-3 M

HINT BOXRecall that since BrO3- and H+ where held constant, the rate law takes on the following form: Rate = B[I-]i

Therefore, a plot of log(Rate) vs. log(I-) will yield a straight line with a slope equal to i, the order of the reaction with respect to I-.

Place your plot relating to the reaction order for I- here (cover this instruction box so that your graph is an appropriate size).

Refer to the box above for hints.

Properly label your graph (labels for axes, including units, and a title).

Include a trendline and its equation and R2 value on your graph. This is done by right-clicking on one of the data points on your graph and choosing "Add Trendline" from the drop down menu. The first tab asks what type of trendline you wish to use, and the Options tab allows you to include the trendline equation and R2 value.

Order with respect to I-: 1 (round to nearest whole number)

-3.1 -3.0 -2.9 -2.8 -2.7 -2.6 -2.5 -2.4 -2.3 -2.2

-7.1

-7.0

-6.9

-6.8

-6.7

-6.6

-6.5

-6.4

-6.3

-6.2

-6.1

f(x) = 0.928571428571428 x − 4.21428571428571R² = 0.965714285714287

log(I-) versus log(rate) for Iodide Ion

log (I-)

log

(rat

e)

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Run 5 23.5 0.020 178.3 2 pts

Run 6 24.57 0.020 106.5

Run 7 24.9 0.020 57.25

Run 8 24.85 0.020 43.7

X-axis Y-axis

log(Rate)

-2.4 -7.0

-2.1 -6.8

-1.9 -6.5

-1.8 -6.4

2 pts

1 pt

S2O32- I- H+ BrO3

-

1.0 x 10-4 2.0 x 10-3 4.0 x 10-3 9.3 x 10-8

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 1.6 x 10-7

1.0 x 10-4 2.0 x 10-3 1.2 x 10-2 2.9 x 10-7

1.0 x 10-4 2.0 x 10-3 1.6 x 10-2 3.8 x 10-7

log(BrO3-)

Reaction Order Determination for BrO3-

Place your plot relating to the reaction order for BrO3- here (cover this instruction box so that

your graph is an appropriate size).



Order with respect to BrO3-: 1 (round to the nearest whole number)

-2.5 -2.4 -2.3 -2.2 -2.1 -2.0 -1.9 -1.8 -1.7

-7.1

-7.0

-6.9

-6.8

-6.7

-6.6

-6.5

-6.4

-6.3

-6.2

-6.1

f(x) = 1.02380952380952 x − 4.57619047619048R² = 0.967556253270539

log(BrO3-) versus log(rate) for Bromate Ion

log(BrO3-)

log(

rate

)

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Run 9 24.67 0.010 298 2 pts

Run 10 24.4 0.020 85.3

Run 11 24.85 0.030 41.47

Run 12 24.7 0.040 28

X-axis Y-axis

log(Rate)

-2.0 -7.3

-1.7 -6.7

-1.5 -6.4

-1.4 -6.2

2 pts

1 pt

S2O32- I- H+ BrO3

-

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 5.6 x 10-8

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 2.0 x 10-7

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 4.0 x 10-7

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 6.0 x 10-7

log(H+)

Reaction Order Determination for H+

Place your plot relating to the reaction order for H+ here (cover this instruction box so that your graph is an appropriate size).



Order with respect to H+: 2 (round to the nearest whole number)

-2.1 -2.0 -1.9 -1.8 -1.7 -1.6 -1.5 -1.4 -1.3 -1.2

-7.4

-7.2

-7.0

-6.8

-6.6

-6.4

-6.2

-6.0

-5.8

-5.6

f(x) = 1.80952380952381 x − 3.66428571428572R² = 0.996549344375431

log(H+) versus log(rate) for Hydrogen Cation

log(H+)

log(

rate

)

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Experiment Initial Concentrations (M) Time Rate k

Run # sec M/sec

Run 2 0.020 94.125

Run 6 0.020 106.5

Run 10 0.020 85.3

. Average:

Std Deviation:

Units:

2 pts

Activation Energy: Uncatalyzed



Ave * 24.98 0.020 95.31 1 pt

Run 13 0 0.020 280

* Average value from runs # 2,6, & 10.

2 pts

29.8 kJ 2 pts

Activation Energy: Catalyzed with Ammonium Molybdate

Calculated from catalyzed runs at different temperatures



Run 14 27 1.0 x 10-4 0.010 14 1 pt

Run 15 0 1.0 x 10-4 0.010 28

17.50 kJ 2 pts

S2O32- I- H+ BrO3

-

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 1.8 x 10-7 28 M-3s-1

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 1.6 x 10-7 25 M-3s-1

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 2.0 x 10-7 31 M-3s-1

28 M-3s-1

3.0 M-3s-1

M-3s-1

S2O32- I- H+ BrO3

-

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 1.8 x 10-7

1.0 x 10-4 2.0 x 10-3 8.0 x 10-3 6.0 x 10-8

Ea of the uncatalyzed reaction: Ea =

S2O32- I- H+ BrO3

-

2.0 x 10-3 8.0 x 10-3 1.2 x 10-6

2.0 x 10-3 8.0 x 10-3 6.0 x 10-7

Ea of the catalyzed reaction: Ea =

Hint: Use the rounded values for the rxn orders from your conclusions and not the linear regression values when determining k.

Calculation of Rate Constant, k

Determination of Activation Energy, Ea

Type your calculation of "k" here:For run 2:rate = k[BrO3

-][I-][H3O+]2

1.8 x 10-7M/s = k[8.0 x 10-3M][2.0 x 10-3M][0.020M]2

k = 28. M-3s-1

Type your calculation of Ea here:

ln(k2/k1) = (-Ea/R)(1/T2-1/T1)

ln(6.0 x 10-8/1.8 x 10-7) = (Ea/-8.3145Jmol-1k-1)(1/273.15k - 1/298.13k)-1.099- = 0.000037 J-1mol (Ea)

Ea = 29778J/molEa = 29.8 kJ/mol

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Part II: Crystal Violet-Hydroxide Reaction

Concentration of stock solutions

1 x 10-4 M 565 nm

0.10 M

Molar Absorptivity Data

to make 25 mL of 1.0 x 10-5 M CV+ is Dilution FactorAbsorbance

2.5 mL 10.0 1.0E-06 5.5E-02

4.0 2.5E-06 1.2E-01

2.0 5.0E-06 2.4E-01

1.0 1.0E-05 5.0E-01 1 pt

slope 50142

y-int 2.900E-03 50142 2 pts

Run Number NaOH (mL) DI water (mL) mLTotal

1 2 0.5 0.5 3 5.00E-06 0.0667

2 1.5 1 0.5 3 5.00E-06 0.0500

3 1 1.5 0.5 3 5.00E-06 0.0333

4 0.5 2 0.5 3 5.00E-06 0.0167

*Final [CV+] and [NaOH]

CV+ l max, CV+

OH-

Volume of CV+ stock solution needed

[CV+] (M)

CV+

Slope and Y-intercept forAbsorbance vs. [CV+]

Molar Absorptivity, e M-1cm-1

CV+ (mL) [CV+]0, M [NaOH]0, M

HINT BOX:Before you proceed, make sure you understand what you are doing here. Please refer to the "Intro" document in the Prelab section of the Experiment 2 website for a detailed explanation of the approach we are taking for this part of the lab. You will first evaluate the data for the CV+ standards to obtain a value for Molar Absorptivity that you will then use to convert Absorbance data to [CV+] (Remember: A=ebc)

When working with linear relationships, i.e. straight line plots, one can utilize the "slope" and "intercept" functions in Excel. These are a part of a linear regression function, and calculate the slope of the "best fit line" for a given set of data and the point at which the best fit line for the data will cross the y-axis (y-intercept). To use these functions, click on the cell in which you want the result to appear, B337 for the slope and B338 for the y-intercept, and enter the following:

=slope(H305:H308,G305:G308)

=intercept(H305:H308,G305:G308)

H305:H308 represent the cells that contain your y-data, and G305:C:308 represent your x-data. This is a short cut to creating an actual plot of the data and generating the trendline equation of y = mx + b.

You will create three plots on page 9, using the data from run 1 below, to determine the order of the reaction with respect to CV+. If the plot of [CV+] vs t is the most linear, the order will be 0. If plotting ln[CV+] vs t is the most linear, then the reaction is first order with respect to CV+. The third plot that you will make is 1/[CV+] vs t. If this last plot is the most linear of the three plots, then the reaction is second order with respect to CV+.

Once you know the order of the reaction in terms of CV+, you will perform the necessary "slope" functions for the other three runs to determine k' for each run. Be sure to replace the "?" in the label at the top of the data column so that it's clear which calculation you are performing. When you get to the section for determining the order of the reaction with respect to OH-, you will also need to use the "intercept" function mentioned above.

0.0E+00 2.0E-06 4.0E-06 6.0E-06 8.0E-06 1.0E-05 1.2E-050.0E+00

1.0E-01

2.0E-01

3.0E-01

4.0E-01

5.0E-01

6.0E-01

f(x) = 50141.9009370817 x − 0.00290629183400268R² = 0.998029061962931

Concentration of Crystal Violet versus Abosrbance for Beer's Law

[CV-] (M)

Abs

orba

nce

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column "I" next to the appropriate plot and update any column/table titles that contain "?").

Run 1 Run 2 2 pts

Time (s) Absorbance Time (s) Absorbance

7 0.567 1.13E-05 -11.391 8.850E+04 8 0.230 4.59E-06 -12.292

17 0.565 1.13E-05 -11.391 8.850E+04 18 0.228 4.55E-06 -12.300

27 0.518 1.03E-05 -11.483 9.709E+04 28 0.206 4.11E-06 -12.402

37 0.437 8.72E-06 -11.650 1.147E+05 38 0.188 3.75E-06 -12.494

47 0.401 8.00E-06 -11.736 1.250E+05 48 0.172 3.43E-06 -12.583

57 0.353 7.04E-06 -11.864 1.420E+05 58 0.157 3.13E-06 -12.674

67 0.311 6.20E-06 -11.991 1.613E+05 68 0.143 2.85E-06 -12.768

77 0.275 5.48E-06 -12.114 1.825E+05 78 0.132 2.63E-06 -12.849

87 0.243 4.85E-06 -12.237 2.062E+05 88 0.121 2.41E-06 -12.936

97 0.223 4.45E-06 -12.323 2.247E+05 108 0.102 2.03E-06 -13.107

107 0.196 3.91E-06 -12.452 2.558E+05 128 0.086 1.72E-06 -13.273

117 0.173 3.45E-06 -12.577 2.899E+05 148 0.071 1.42E-06 -13.465

Psuedo-rate constant Psuedo-rate constantslope -0.0114 k', Run 1 1.14E-02 slope -0.0086 k', Run 2 8.60E-03 2 pts

For the data below place your plots to determine the order with respect to [CV+] on the next page and besure to identify which one you are using to determine the order of the reaction (enter the order in

[CV+]t Ln[CV+]t 1/[CV+]t [CV+]t ln[CV+]t

ln([CV+]) vs. time ln([CV+]) vs. time

Type an example calculation for determining the CV+ concentration from the absorbance data:A=Ebc where E is the molar absortivityfor run one, time 7s0.567 = (50142M-1cm-1)(1cm)(c)c= [CV+] = 1.13 x 10-5 M

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2 pts

Place your plot of 1/[CV+] vs time here (cover this instruction box so that your graph is an appropriate size).



Place your plot of ln[CV+] vs time here (cover this instruction box so that your graph is an appropriate size).



Place your plot of [CV+] vs time here (cover this instruction box so that your graph is an appropriate size).



0 20 40 60 80 100 120 1400.00E+00

2.00E-06

4.00E-06

6.00E-06

8.00E-06

1.00E-05

1.20E-05

f(x) = − 7.74685314685315E-08 x + 1.18863822843823E-05R² = 0.975526176625003

Plot of [CV+] vs Time for Run One

Time (s)

[CV+

]t (M

)

0 20 40 60 80 100 120 140

-13.000

-12.500

-12.000

-11.500

-11.000

-10.500

f(x) = − 0.0114234265734266 x − 11.2258308857809R² = 0.993432177882323

Plot of ln[CV+] vs Time for Run One

Time (s)

ln[C

V+] (

M)

0 20 40 60 80 100 120 1400.000E+00

5.000E+04

1.000E+05

1.500E+05

2.000E+05

2.500E+05

3.000E+05

3.500E+05

f(x) = 1840.18216783217 x + 50582.7889277389R² = 0.961114095913829

Plot of 1/[CV+] vs Time for Run One

Time (s)

1/[C

V+] (

M)

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Run 3 Run 4

Time (s) Absorbance Time (s) Absorbance

9 0.136 2.71E-06 -12.819 7 0.183 3.65E-06 -12.521 2 pts19 0.180 3.59E-06 -12.537 17 0.223 4.45E-06 -12.323

29 0.163 3.25E-06 -12.637 27 0.210 4.19E-06 -12.383

39 0.149 2.97E-06 -12.727 37 0.198 3.95E-06 -12.442

49 0.136 2.71E-06 -12.819 47 0.188 3.75E-06 -12.494

59 0.125 2.49E-06 -12.903 57 0.178 3.55E-06 -12.549

69 0.115 2.29E-06 -12.987 67 0.165 3.29E-06 -12.625

89 0.095 1.89E-06 -13.179 77 0.157 3.13E-06 -12.674

109 0.075 1.50E-06 -13.410 87 0.150 2.99E-06 -12.720

129 0.065 1.30E-06 -13.553 107 0.139 2.77E-06 -12.797

149 0.052 1.04E-06 -13.776 127 0.126 2.51E-06 -12.895

147 0.114 2.27E-06 -12.996

167 0.103 2.05E-06 -13.098

187 0.092 1.83E-06 -13.211

207 0.084 1.68E-06 -13.297

227 0.076 1.52E-06 -13.397

247 0.068 1.36E-06 -13.508

Psuedo-rate constant Psuedo-rate constantslope -0.0084 k', Run 3 8.40E-03 slope -0.0048 k', Run 4 4.80E-03 2 pts

Run # k' ln(k')1 0.0667 1.14E-02 -2.708 -4.474

2 0.0500 8.60E-03 -2.996 -4.756

3 0.0333 8.40E-03 -3.402 -4.780

4 0.0167 4.80E-03 -4.092 -5.339

2 pts

slope 0.5842 1

y-int -2.9096 Rate Constant, k 5.450E-02

Units on k

2 pts

[CV+]t ln[CV+]t [CV+]t ln[CV+]t

ln([CV+]) vs. time ln([CV+]) vs. time

[OH-] ln[OH-]

ln(k') vs. ln([OH-])Order w.r.t. [OH-], o

M-1s-1

Reaction Order Determination for OH-

HINT BOXRecall: rate = k[CV+]c[OH-]o = k'[CV+]c, where k'=k[OH-]o

therefore, ln(k') = ln(k) + ln([OH-]o)and since ln([OH-]o) = o*ln[OH-] the following equation will yield a straight line: ln(k') = o*ln([OH-]) + ln(k)Comparing to y = mx+b y = ln(k'), x = ln([OH-]), b = ln(k)The slope of the line gives the order of the rxn m = o

and the rate constant is found by taking the anti-log of the y-intercept.

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Results and Discussion Part I1. Based on your data, write out the complete rate law including value and units for the rate constant. (1 pt)

Rate = 28M-3s-1[BrO3-][I-][H3O+]2

2. The literature values for the reaction orders are 1 for BrO3- and I- and 2 for H+ Compare the accuracy

of your orders to these literature values. (Calculate the % error.) Discuss your most likely sources of error. (2 pts)When the experimental values of my orders are rounded to the nearest number then the percent error of my orders are zero. However when the non rounded values are used:I-: abs((0.9286-1)/1) x 100= 7.14%; BrO3-: abs((1.0238-1)/1) x 100= 2.38%H+: abs((1.8095-2)/2) x 100= 9.53%The percent errors for the threee species are not too large although I- and H+ are bigger than the ideal 5%. My largest source of error would be stopping the stop watch after the reaction has turned more blue than the color indicator. This would cause a larger time to be recorded and thus when calculating the rate, a slower rate would be calculated and thus would affect the determination of the order when graphing the natural log of the concentration and the natural log of the rate. Another source of error would be to stop the stop watch too early and thus this would have the opposite effect in that a fast time would recorded and thus this would lead to the calculation of a faster rate which would affect the determination of orders.

3. How do the activation energies for the catalyzed and uncatalyzed reaction compare (include a % difference in your discussion)? Is this in line with what is expected? (2 pts) %difference = value 1 - value 2/ (average)%difference = (29.8 kJ- 17.5 kJ) /(23.65kJ)x 100 = 52.0%There is a 52% difference between the catalyzed an uncatalyzed reactions and this is in line with what is expected. A catalyst is a substance that will lower a reaction's activation energy by finding a pathway that requires less energy. For this reason it is expected that the catalyzed reaction will have a much lower activation energy than the uncatalyzed reaction and in this case the difference of 52% is valid.

Part II1. Based on your data, write out the complete rate law including value and units for the rate constant. The literature values of the orders with respect to CV+ are and OH- are 1 and 1 respectively. How do your values compare? (2 pts)Rate = 5.42 x 10-2 M-1s-1[CV+][OH-]My order for CV+ is one for all four of the runs made during part two of the lab and thus the order is one for CV+ and this is exactly what the literature value is. However, my order for OH- is one only when I round it up from 0.5842. In this case this percent error= abs((0.5842-1)/1) x 100 which equals 41.58% which is a very larger percent error.

2. Discuss your largest source of error in Part II. (2 pts)The largest source of error I could have made in part two of this lab occured during the part where every ten seconds, the absorption values for CV+ is recorded. Most likely, my timing between recording these values and recording the time at which the values are recorded were off and thus this would affect the slope of the graphs of the fours runs which would determind the value of K'. If the value of K' is affected by this then when graphing the ln(k') versus the ln[OH-] one will end up with a graph in which the value of the slope is affected thus giving is large error in the order of OH-. Another source of error would be an error in creating the diluted CV+ solutions which would cause a discrepency in the absorption data.

Laboratory Waste Evaluation (1 pt)Laboratory waste is considered anything generated during an experiment that is disposed of down the sewer drain, thrown in the garbage, collected in a container for disposal by the UW Environmental Health & Safety department, or released into the environment. Based on the written lab procedure and your actions during the lab, list the identity and approximate amount (mass or volume) of waste that you generated while performing this experiment. .

9mL of 0.0005M Na2S2O3; 9mL of 0.010 M KI

9mL of 0.040 M KBrO3; 9mL of 0.100 M HCl3mL of color comparison solution; about 14 drops of starch indicatorabout two drops of molybdate; about 250mL beaker worth of ice6mL of 1.0 x 10-4M CV+ solution; 10mL of 0.1M NaOHabout 200mL of DI water

Lab Report 1

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