3. The decomposition of N2O5 in the gas phase was studied at constant temperature, 2N2O5 (g) → 4NO2 (g) + O2(g) The following results were collected: find nature log. (LN) N2O5] 0.1000 Ln[N2O5] -2.30 -2.65 -2.99 -3.69 -4.38 Time (s) 0.0707 0.0500 50 0.0250 100 0.0125 200 0.00625 300 400 .08 a. Complete the table. Using the data and graph paper, plot the [N2O5] versus time and Ln[N2O5] versus time. Determine the value of k. Which graph did graph *b. On your graph of [N2O5] versus time, highlight the times it takes for each halving of you use? the reactant concentration. What is the half-life? Using this half-life calculate k.

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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3. The decomposition of N2O5 in the gas phase was studied at constant temperature,
2N2O5 (g) → 4NO2 (g) + O2(g)
The following results were collected:
find
nature)
N2O5]
0.1000
Ln[N2O5]
-2.30
2.65
2.99
-3.69
-4.38
-5.08
a. Complete the table. Using the data and graph paper, plot the [N2O5] versus time and
Time (s)
10g.
(LN)
0.0707
0.0500
50
0.0250
100
0.0125
200
0.00625
300
400
Ln[N2O5] versus time. Determine the value of k. Which graph did
you use?
graph
*b. On your graph of [N2O5] versus time, highlight the times it takes for each halving of
the reactant concentration. What is the half-life? Using this half-life calculate k.
design experiments to find the activation energy,
Transcribed Image Text:3. The decomposition of N2O5 in the gas phase was studied at constant temperature, 2N2O5 (g) → 4NO2 (g) + O2(g) The following results were collected: find nature) N2O5] 0.1000 Ln[N2O5] -2.30 2.65 2.99 -3.69 -4.38 -5.08 a. Complete the table. Using the data and graph paper, plot the [N2O5] versus time and Time (s) 10g. (LN) 0.0707 0.0500 50 0.0250 100 0.0125 200 0.00625 300 400 Ln[N2O5] versus time. Determine the value of k. Which graph did you use? graph *b. On your graph of [N2O5] versus time, highlight the times it takes for each halving of the reactant concentration. What is the half-life? Using this half-life calculate k. design experiments to find the activation energy,
KInetics Problems Laboratory
1. Using the following chemical equation and data:
A=Second order reactant
B-first order reactant
Similar
to
3A + B+ 2C →D+2E
Experiment
notes
Exамple
Initial [A]
1.0 x 102
1.0 x 102
2.0 x 102
2.0 x 102
Initial [B]
4.0 x 103
8.0 x 103
8.0 x 10-3
8.0 x 10-3
Initial [C]
2.0 x 102
2.0 x 102
2.0 x 10,2
1.0 x 102
1
Initial Rate
2
5 mM/s
3
10 mM/s *
40 mM/s *
40 mM/s
4
Determine the rate equation.
b. Calculate the specific rate constant (k).
Determine the initial rate of reaction if the initial concentrations of A, B, C are
3 x 102 M, 5 x 10-3 M, 5 x 103M, respectively, for the reaction.
a.
с.
a.) Rate= [A]²CB]'
b.) K= rate
CAJ C6]' = 5 m/s
(10x (0-2) 2 (4.0x10-3)'=2M!5-1
%3D
C.)
2. The reaction between bromate ions and bromide ions in acidic aqueous solution is given
by the following equation:
BrO3 (aq) + 5Br (aq) +6H* (aq) → 3B12 (1) + 3H2O (1)
Transcribed Image Text:KInetics Problems Laboratory 1. Using the following chemical equation and data: A=Second order reactant B-first order reactant Similar to 3A + B+ 2C →D+2E Experiment notes Exамple Initial [A] 1.0 x 102 1.0 x 102 2.0 x 102 2.0 x 102 Initial [B] 4.0 x 103 8.0 x 103 8.0 x 10-3 8.0 x 10-3 Initial [C] 2.0 x 102 2.0 x 102 2.0 x 10,2 1.0 x 102 1 Initial Rate 2 5 mM/s 3 10 mM/s * 40 mM/s * 40 mM/s 4 Determine the rate equation. b. Calculate the specific rate constant (k). Determine the initial rate of reaction if the initial concentrations of A, B, C are 3 x 102 M, 5 x 10-3 M, 5 x 103M, respectively, for the reaction. a. с. a.) Rate= [A]²CB]' b.) K= rate CAJ C6]' = 5 m/s (10x (0-2) 2 (4.0x10-3)'=2M!5-1 %3D C.) 2. The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the following equation: BrO3 (aq) + 5Br (aq) +6H* (aq) → 3B12 (1) + 3H2O (1)
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