Portage learning chem 108 module 5 exam newest | with complete solutions

Module 5 Exam 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. 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). rate= ∆ [O2] /∆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. Late instantaneous rate is smaller, the concentrations of reactants is lowest in the late stage of reaction. Answer Key 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