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 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)/ = 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 reaction