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The decomposition of NOBr is studied manometrically
Chapter 16, Problem 18P(choose chapter or problem)
Problem 18P
The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both NOBr and Br2 are reddish brown:
2NOBr(g) →2NO(g) + Br2(g) Use the data below to answer the following:
(a) Determine the average rate over the entire experiment.
(b) Determine the average rate between 2.00 and 4.00 s.
(c) Use graphical methods to estimate the initial reaction rate.
(d) Use graphical methods to estimate the rate at 7.00 s.
(e) At what time does the instantaneous rate equal the average rate over the entire experiment?
Time (s) |
[NOBr] (mol/L) |
0.00 |
0.0100 |
2.00 |
0.0071 |
4.00 |
0.0055 |
6.00 |
0.0045 |
8.00 |
0.0038 |
10.00 |
0.0033 |
Questions & Answers
QUESTION:
Problem 18P
The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both NOBr and Br2 are reddish brown:
2NOBr(g) →2NO(g) + Br2(g) Use the data below to answer the following:
(a) Determine the average rate over the entire experiment.
(b) Determine the average rate between 2.00 and 4.00 s.
(c) Use graphical methods to estimate the initial reaction rate.
(d) Use graphical methods to estimate the rate at 7.00 s.
(e) At what time does the instantaneous rate equal the average rate over the entire experiment?
Time (s) |
[NOBr] (mol/L) |
0.00 |
0.0100 |
2.00 |
0.0071 |
4.00 |
0.0055 |
6.00 |
0.0045 |
8.00 |
0.0038 |
10.00 |
0.0033 |
ANSWER:
Solution 18P
The decomposition of NOBr is given below,
2NOBr(g) →2NO(g) + Br2(g)
We have to use the data below to answer the following questions:
Time (s) |
[NOBr] (mol/L) |
0.00 |
0.0100 |
2.00 |
0.0071 |
4.00 |
0.0055 |
6.00 |
0.0045 |
8.00 |
0.0038 |
10.00 |
0.0033 |
Step 1
(a)We have to determine the average rate over the entire experiment.
Rate can be calculated as the ratio of change in concentration to the change in time.
Here the initial concentration at initial time t = 0.0 is 0.0100 mol/L
The final concentration at final time , t = 10.00 is 0.0033 mol/L
Rate = = 6.7
10-4 mol/L. S
Thus the rate of the entire reaction is found to be 6.7 10-4 mol/L. S.