CHEM 104 – Week 1 Module 1 Problem Set | Portage Learning | Verified Questions & Correct Detailed Answers | 2025/2026 Edition

Portage learning CHEM 104 Week 1 Module 1: Problem Set
questions and answers 2025\2026 A+ Grade


Show your work and be sure to give answer with the correct units.

In the reaction of 0.200 M gaseous N2O5 to yield NO2 gas and O2 gas as shown below:



2 N2O5 (g) → 4 NO2 (g) + O2 (g)



the following data table is obtained:

Time (sec)

[N2O5]

[O2]

0

0.200 M

0

300

0.180 M

0.010 M

600

0.162 M

0.019 M

900

0.146 M

0.027 M

1200

0.132 M

, 0.034 M

1800

0.110 M

0.045 M

2400

0.096 M

0.052 M

3000

0.092 M

0.054 M



(a) Use the [O2] data from the table to calculate the average rate over the measured time interval from
0 to 3000 secs.

(b) Use the [O2] data from the table to calculate the instantaneous rate early in the reaction (0 secs to
300 sec).

(c) Use the [O2] data from the table to calculate the instantaneous rate late in the reaction (2400 secs to
3000 secs).

(d) Explain the relative values of the average rate, early instantaneous rate and late instantaneous rate.
- correct answer (a) The average rate over the measured time interval from 0 to 3000 secs is:

rate = ∆[O2] / ∆t = (0.054 - 0) / 3000 - 0 = 1.80 x 10-5 mol/L•s



(b) The instantaneous rate early in the reaction from 0 to 300 secs is:

rate = ∆[O2] / ∆t = (0.010 - 0) / 300 - 0 = 3.33 x 10-5 mol/L•s



(c) The instantaneous rate late in the reaction from 2400 to 3000 secs is:

rate = ∆[O2] / ∆t = (0.054 - 0.052) / 3000 - 2400 = 3.33 x 10-6 mol/L•s



(d) It can be seen that the early instantaneous rate is the largest since the concentrations of reactants is
highest during the earliest stages of the reaction and the late instantaneous rate is smallest since the
concentrations of reactants is lowest during the late stages of the reaction.