CHEM 104: MODULE 1 – MODULE 6 Exam 2026/2027 (Portage Learning)) (31pages)

Module 1: Question 1 In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following data table is obtained: → 4 NO2 (g) + O2 (g) 1. Using the [O2 ] data from the table, show the calculation of the instantaneous rate early in the reaction (0 secs to 300 sec). 2. Using the [O2 ] data from the table, show the calculation of the instantaneous rate late in the reaction (2400 secs to 3000 secs). 3. Explain the relative values of the early instantaneous rate and the late instantaneous rate. Your Answer: 1. rate = (0.014 - 0) / (300 - 0) = 4.67 x 10-5 mol/Ls 2. rate = (0.090 - 0.083) / (3000 - 2400) = 1.167 x 10-5 mol/Ls 3. The late instantaneous rate is smaller than the early instantaneous rate. 2 N2 O5 (g) Data Table #2 Time (sec) [N2O5 ] [O2 ] 0 0.300 M 0 300 0.272 M 0.014 M 600 0.224 M 0.038 M 900 0.204 M 0.048 M 1200 0.186 M 0.057 M 1800 0.156 M 0.072 M 2400 0.134 M 0.083 M 3000 0.120 M 0.090 Question 2 The following rate data was obtained for the hypothetical reaction: A + B → X + Y Experiment # [A] [B] rate 1 0.50 0.50 2.0 2 1.00 0.50 8.0 3 1.00 1.00 64.0 1. Determine the reaction order with respect to [A]. 2. Determine the reaction order with respect to [B]. 3. Write the rate law in the form rate = k [A]n [B]m (filling in the correct exponents). 4. Show the calculation of the rate constant, k. Your Answer: rate = k [A]x [B]y rate 1 / rate 2 = k [0.50]x [0.50]y / k [1.00]x [0.50]y 2.0 / 8.0 = [0.50]x / [1.00]x 0.25 = 0.5x x = 2 rate 2 / rate 3 = k [1.00]x [0.50]y / k [1.00]x [1.00]y 8.0 / 64.0 = [0.50]y / [1.00]y 0.125 = 0.5y y = 3 rate = k [A]2 [B]3 2.0 = k [0.50]2 [0.50]3 k = 64 Question 3 ln [A] - ln [A]0 = - k t 0.693 = k t 1/2 An ancient sample of paper was found to contain 19.8 Ccontent as compared to a present-day sample. The t1/2 for 14C is 5720 yrs. Show the calculation of the decay constant (k) and the age of the paper.