CHEM 152 Lab Report 01 Kinetics II -Method of Initial Rates

Page 1 of 6


Name: Quiz Section:
ID Number: Lab Partner:

Chem 152 Experiment #1 Note: All sections of this
Kinetics II: Method of Initial Rates 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:


Grading: 60 pts. (Lab notebook pages are worth 10 pts - total for report and notebook is 60 pts)

Data, Calculations and Graphs


Stock Solutions
Soln. Conc. Units
S2O32- 0.0005 M
I- 0.010 M (4 pts)
BrO3- 0.040 M
HCl 0.100 M
Table of Recorded Reagent Volumes, Temperature, Time of Reaction, and Calculated Rate
Experiment Test Tube #1 Test Tube #2 Temp. D
Run H2O S2O32- I- BrO3- H+ Celsius Time Rate
# mL mL mL mL mL C sec. M/s
1 0.00 0.50 1.00 0.50 0.50 17.86666667 45 9.3*10^-6
2 0.25 0.50 0.75 0.50 0.50 18.13333333 66.33333333 1.3*10-6
3** 0.50 0.50 0.50 0.50 0.50 18.63333333 109.3333333 7.6*10^-7
4 0.75 0.50 0.25 0.50 0.50 18.85 212 3.9*10^-7
5 0.00 0.50 0.50 1.00 0.50 18.425 38.5 2.2*10^-6
6 0.25 0.50 0.50 0.75 0.50 18.125 64.75 1.3*10-6
7** 0.50 0.50 0.50 0.50 0.50 18.1 106.5 7.8*10^-7
8 0.75 0.50 0.50 0.25 0.50 18.53333333 156 5.3*10^-7
9 0.00 0.50 0.50 0.50 1.00 18.875 32.25 2.4*10^-6
10 0.25 0.50 0.50 0.50 0.75 18.575 37.5 2.2*10^-6
11** 0.50 0.50 0.50 0.50 0.50 18.725 77.25 1.1*10^-6
12 0.75 0.50 0.50 0.50 0.25 18.63333333 441 1.9*10^-7
13 0.50 0.50 0.50 0.50 0.50 0.955555556 254.9 3.3*10^-7
14* 0.75 0.50 0.50 0.50 0.25 17.85454545 56.45454545 1.5*10^-6
15* 0.75 0.50 0.50 0.50 0.25 0.727272727 90.09090909 9.2*10^-7
* 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.


Show your calculation for the "Rate" here using data for Run #1: (4 pts)

Rate = (1/6)([S2O ] /ΔT = (1/6)(5.0*10 /45) = 9.3*10^ M/s
2-
3 o
-4 -6




Reaction Order Determination for I-

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Reaction Order Determination for I-

Show your calculation for the concentration of BrO3- used in Runs 1-4: (3 pts)

Rate = -Δ[BrO3-]/ΔT
Δ[BrO3-] = - Rate*ΔT
Δ[BrO3-] = 9.3*10-6*45 = - 4.185*10-4 M
Δ[BrO3-] = - 1.3*10-3*66.3333 = - 0.0862 M
Δ[BrO3-] = - 7.6*10-7*109.3333 = - 8.309*10-5
Δ[BrO3-] = - 3.9*10-7*212 = - 8.268*10-5



Table of Recorded Temperature, Initial Concentrations, Time of Reaction, and Calculated Rate
Experiment Temp Initial Concentrations (M) Time Rate
Run # Celsius S2O32- I- H+ BrO3- sec M/sec
Run 1 17.86666667 0.50 1.00 0.50 0.50 45 9.3*10^-6
Run 2 18.13333333 0.50 0.75 0.50 0.50 66.33333333 1.3*10-6
Run 3 18.63333333 0.50 0.50 0.50 0.50 109.3333333 7.6*10^-7
Run 4 18.85 0.50 0.25 0.50 0.50 212 3.9*10^-7

READ THIS BEFORE PROCEEDING:
Recall that since BrO3- and H+ are held constant, the rate law takes on the X-axis Y-axis
following form: log(I-) log(Rate)
Rate = B[I-]i
0.0 -5.032
where "B" is the pseudo rate constant that includes k, [H+], and [BrO3-]. -0.1 -5.886
Therefore, a plot of log(Rate) vs. log([I-]) (y-axis vs x-axis) will yield a straight -0.3 -6.119
line with a slope equal to i, the order of the reaction with respect to I-.
-0.6 -6.409




Relationship Between I- and Rate of Reaction (5 pts)
-4.500
-0.7 -0.6 -0.5 -0.4 -0.3 -0.2 0.0 0.1


-5.000


f(x)==0.775134597711058
R² 2.00044052863436 x − 5.35138766519824 -5.500
log(Rate)




-6.000


-6.500


-7.000
log(I-)



What is the order with respect to I- as determined from your data? i=2 (2 pts)
Round to the nearest whole number.
If no trendline equation is provided on the plot, no credit will be given for your answer to this question.