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Exam (elaborations) CHEM 108 Module 5 Exam Answers- Portage Learning Exam Page 1 In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below the following data table is obtained: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Data Table #1 Time (sec) [N2O$12.39
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Exam (elaborations) CHEM 108 Module 5 Exam Answers- Portage Learning Exam Page 1 In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below the following data table is obtained: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Data Table #1 Time (sec) [N2O
Exam (elaborations)
CHEM 108 Module 5 Exam Answers- Portage Learning
Exam Page 1 In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below the following data table is obtained: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Data Table #1 Time (sec) [N2O5] [O2] 0 0.200 M 0 300 0.182 M 0.009 M ...
exam elaborations chem 108 module 5 exam answers portage learning exam page 1 in the reaction of gaseous n2o5 to yield no2 gas and o2 gas as shown below the following data table is obtained 2 n2o5
Exam Page 1
In the reaction of gaseous N 2O5 to yield NO2 gas and O2 gas as shown below the following data table
is obtained:
2 N2O5 (g) → 4 NO2 (g) + O2 (g) Data Table #1 Time (sec)[N2O5] [O2]
0 0.200 M 0
300 0.182 M 0.009 M
600 0.166 M 0.017 M
900 0.152 M 0.024 M
1200 0.140 M 0.030 M
1800 0.122 M 0.039 M
2400 0.112 M 0.044 M
3000 0.108 M 0.046 M
Complete the following three problems:
Using the [O2] data from the table show the calculation of the average rate over the measured time interval from 0 to 3000 secs.
the average rate over the measured time interval from 0 to 3000 secs.
Rate= [O2]/t= (0.046-0)/3000-0 = 1.53 x 10-5 mole/L.s
Using the [O2] data from the table show the calculation of the instantaneous rate late in the reaction (2400 secs to 3000 secs).
the instantaneous rate late in the reaction (2400 secs to 3000 secs)
Rate = [O]/t = (0.046-0.044)/3000-2400 = 3.33 x 10-6 mol/L.s
Explain the relative values of the average rate and the late instantaneous rate.
The late instantaneous rate is larger since the concentrations of reactants is higher during the earlier
stages of the reaction
-2.0 points
Instructor Comments
The late instantaneous rate is smaller since the concentrations of reactants is higher during the earlier stages of the reactionEXAM PAGE 1 Answer Key
In the reaction of gaseous N 2O5 to yield NO2 gas and O2 gas as shown below the following data table
is obtained:
2 N2O5 (g) → 4 NO2 (g) + O2 (g) Data Table #1 Time (sec)[N2O5] [O2]
0 0.200 M 0
300 0.182 M 0.009 M
600 0.166 M 0.017 M
900 0.152 M 0.024 M
1200 0.140 M 0.030 M
1800 0.122 M 0.039 M
2400 0.112 M 0.044 M
3000 0.108 M 0.046 M
Complete the following three problems:
Using the [O2] data from the table show the calculation of the average rate over the measured time interval from 0 to 3000 secs.
The average rate over the measured time interval from 0 to 3000 secs is:
rate = ∆[O2] / ∆t = (0.046 - 0) / 3000 - 0 = 1.53 x 10-5 mol/L•s
Using the [O2] data from the table show the calculation of the instantaneous rate late in the reaction (2400 secs to 3000 secs).
The late instantaneous rate over the measured time interval from 2400 to 3000 secs is:
rate = ∆[O2] / ∆t = (0.046 - 0.044) / 3000 - 2400 = 3.33 x 10-6 mol/L•s
Explain the relative values of the average rate and the late instantaneous rate.
The late instantaneous rate is smaller since the concentrations of reactants is lowest during the late stages of the reaction.
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