Why is it divided in half? I always thought we just plug it into the found rate. That's what I did with the others, and those were right. What's different?

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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Why is it divided in half? I always thought we just plug it into the found rate. That's what I did with the others, and those were right. What's different?

Data for the reaction
2 NO(g) + 02 (g) → 2 NO2(g)
are given (for a particular temperature) in the table.
Concentration (mol/L)
Initial rate
Experiment
[NO]
[02]
(mol NO/L· h)
3.6 х 10
5.2 x 10-3
4.6 × 10–8
1
3.6 x 10-4
1.04 x 10-2
9.2 x 10-8
1.8 x 10-4
1.04 x 10–2
2.3 x 10-8
3
1.8 x 10-4
5.2 x 10-3
4
a What is the rate law for this reaction?
Rate = k[NO][O2]
O Rate = k[NO]²[O2]
O Rate = k[NO]?[O2]?
Transcribed Image Text:Data for the reaction 2 NO(g) + 02 (g) → 2 NO2(g) are given (for a particular temperature) in the table. Concentration (mol/L) Initial rate Experiment [NO] [02] (mol NO/L· h) 3.6 х 10 5.2 x 10-3 4.6 × 10–8 1 3.6 x 10-4 1.04 x 10-2 9.2 x 10-8 1.8 x 10-4 1.04 x 10–2 2.3 x 10-8 3 1.8 x 10-4 5.2 x 10-3 4 a What is the rate law for this reaction? Rate = k[NO][O2] O Rate = k[NO]²[O2] O Rate = k[NO]?[O2]?
b Calculate the rate constant.
Rate constant = 65
L?/mol? · h
Calculate the rate constant from the first experiment. Note the initial rates are for the disappearan
ΔΝΟ
1
Rate =
2
1
(4.6 × 10-8 mol/L · h) = 2.3 × 10¬8 mol/L · h
At
Rate
2.3 x 10-8 mol/L · h
k
65 L² /mol? · h
%D
NO] [0,] (2.6 x 10-4 mol/L)* (5.2 x 10-8 mol/L)
Transcribed Image Text:b Calculate the rate constant. Rate constant = 65 L?/mol? · h Calculate the rate constant from the first experiment. Note the initial rates are for the disappearan ΔΝΟ 1 Rate = 2 1 (4.6 × 10-8 mol/L · h) = 2.3 × 10¬8 mol/L · h At Rate 2.3 x 10-8 mol/L · h k 65 L² /mol? · h %D NO] [0,] (2.6 x 10-4 mol/L)* (5.2 x 10-8 mol/L)
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