Hi, I need help in Chemistry assignment

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timer Asked: Jan 24th, 2019
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Question Description

Hi, I need help in Chemistry assignment, You will need to use Excel

Please see the file

Experiment #1: Online Kinetic Simulation Purpose: To familiarize the student with the Integrated Rate Law equations and with Microsoft Excel. Assignment: Use the simulated rate data located at Your unknown will be ..\..\..\..\..\..\Desktop\data.txt Use the equations for first, second and third order rate laws (found in your textbook) to determine the order and kobs for your unknown data set. These can be found in the kinetics chapter of any general chemistry textbook. Recognize that these equations should lead you to prepare three different graphs. A graph that is consistent with the order of reaction should appear as a straight line. So, even though you prepare three graphs, only one should be consistent with the order of reaction, and you can determine kobs from the slope of the graph. Your Report: The report should be include: 1. The report should fit on the front of one sheet of paper. 2. your name; 3. the title of the assignment; 4. the unknown number 5. the order of reaction for that data set; 6. the value of kobs for the data set; and 7. the three graphs you prepared to determine the order and kobs (cut-and-paste the graphs from Excel into a Word document). Include a sentence that identifies which graph represents the correct order of reaction (which graph was used to determine the order). Cut-and-paste graphs from Excel into Word. Include a title for each graph. Scale the graph appropriately so that your report fits on one sheet of paper.

Tutor Answer

dnjones1
School: New York University

Attached.

Zero-Order Kinetics
The rate law expression for a zero-order reaction is
Rate = k[A]0 = k
and its integrated rate law is given by
[A] = -kt + [A]0
When compared to the equation for a straight line, y = mx + b, it is evident that
a plot of [A] vs Time will produce a straight line with a slope equal to the
negative rate constant, -k. This means that if the concentration of the reactant A
is halved, the reaction rate remains the same as shown by the
Rate vs Concentration graph. As shown in the Rate vs Time graph, the reaction
rate is independent of time.

0.98
0.98
0.98
0.98
0.98
0.98
0.98
0.98
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99

-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06

0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
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0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06

k=

0.06

3

Rate = k[A]m = k[A]0 = k
Units of k = M•s-1

Rate vs Time

Rate vs Concentration

1.00
0.80
0.60
0.40
0.20
0.00

Rate

Rate
(M•s-1)

0

0.2

0.4

1.00
0.80
0.60
0.40
0.20
0.00
0.98

Time (s)

Concentration vs Time
0.99
0.99
0.99
0.98
0.98

y = -0.06x + 1

Concentration vs Time

Concentration (M)

Slope
(-M•s-1)

Rate

0.3
0.2943
0.2883
0.2819
0.2763
0.2715
0.2668
0.2613
0.257
0.2519
0.2478
0.2435
0.2394
0.2355
0.2319
0.2283
0.2246
0.2209
0.2175
0.2149
0.2115
0.2086
0.2057
0.2026
0.1992
0.1965
0.1943
0.1914
0.1889
0.1867
0.1839

[A]
(M)

Concentration (M)

Time
(s)

0.99
0.99
0.99
0.98
0.98

-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06
-0.06

0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06
0.06

0

0.2

Time (s)

0.4

Concentration (M)

0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
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0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99

Concentration (M)

0.182
0.1793
0.1768
0.1753
0.1733
0.1709
0.1692
0.1669
0.1649
0.1631
0.1613
0.1591
0.1578
0.1562
0.1542
0.1522
0.1511
0.1493
0.1476
0.146
0.1444
0.1429
0.1423
0.1405
0.139
0.138
0.1361
0.1356
0.1344
0.1329
0.132
0.1303
0.1296
0.1275
0.1271
0.1257
0.1249
0.1239
0.123
0.1212
0.1208
0.119
0.1187
0.1178
0.1163
0.116
0.1148
0.1141
0.1125
0.1118
0.1114
0.1101

0.98
0.98
0.98
0.98
0.98

0.98
0.98
0.98
0.98
0.98

0.109
0.109
0.1082
0.1073
0.1064
0.1058
0.1049
0.1042
0.1034
0.1028
0.1019
0.1007
0.1006
0.0999
0.0985
0.0978
0.0976
0.0966
0.0962
0.0958
0.0953
0.0946
0.0933
0.0928
0.0921
0.0914
0.0912
0.0904
0.09
0.0893
0.0885
0.0883
0.0881
0.0877
0.0872
0.0859
0.0858
0.0857
0.0847
0.0844
0.0842
0.083
0.083
0.082
0.0818
0.0817
0.0808
0.0802
0.0801
0.08
0.0792
0.0782

0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0.99
0...

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Review

Anonymous
Excellent job

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