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 N2O5 (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 M 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. 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. Your Answer: 0.693 = k t1/2 0.693 = k (5720) k = 1.21 x 10-4 ln [A] - ln [A]0 = - k t ln 19.8 - ln 100 = - 1.21 x 10-4 t t = 13, 384 years Question 4 Using the potential energy diagram below, state whether the reaction described by the diagram is endothermic or exothermic and spontaneous or nonspontaneous, being sure to explain your answer. Your Answer: The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous because it has relatively large Eact. Question 5 Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of H2O and corresponding amounts of CO, H2, and CH4. CO (g) + 3 H2 (g) CH4 (g) + H2O (g) Your Answer: 0.309 mole of H2O formed = 0.309 mole of CH4 formed 0.309 mole of H2O formed = 0.800 - 0.309 = 0.491 mole CO 0.309 mole of H2O formed = 3 x 0.309 mole H2 reacted = 2.40 - (3 x 0.309) = 1.473 mole H2 [CO] = 0.491 mole / 8.00 L = 6.1375 x 10-2 M [H2] = 1.473 mole / 8.00 L = 18.4125 x 10-2 M [CH4] = 0.309 mole / 8.00 L = 3.8625 x 10-2 M [H2O] = 0.309 mole / 8.00 L = 3.8625 x 10-2