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Unformatted text preview: CHEM 614 Unit 13 Discussion Questions 1) The first‐order rate constant for the photodissociation of chemical A is 5.74 h‐1. Calculate the time needed for the concentration of A to decrease to: a) 1/8([A]o) = / = ‐ (5.74 h‐1)t t = 0.362 h b) 5% of its original value = . = ‐ (5.74 h‐1)t t = 0.522 h c) 1/10([A]o) = / = ‐ (5.74 h‐1)t t = 0.401 h 2) The second‐order rate constant for the decomposition of NO2 into NO and O2 at 573 K is 0.54 L mol‐1 s‐1. Calculate the time for an initial NO2 concentration of 0.20 M to decrease to a) ½ of its initial concentration = t = . . . = 9.26 s b) 1/9 of its initial concentration = t = . . . = 74.1 s c) 1/16 of its initial concentration = t = . . . = 139 s 3) The following data were collected for the reaction H2(g) + I2(g) 2HI(g) at 780 K: Time, s [HI], x 10‐3 M 0 1.00 1 0.43 2 0.27 3 0.20 4 0.16 a) Determine the order of the reaction. Compare graphs of ln[HI] vs time and 1/[HI] vs time ln [HI}
0
0 1 2 3 4 5 ‐0.5
‐1 ln [HI}
Linear (ln [HI}) ‐1.5
‐2
‐2.5 y = ‐0.4431x ‐ 0.2329
R² = 0.9287 1/[HI]
7
y = 1.3174x + 1.021
R² = 0.9997 6
5
4 1/[HI] 3 Linear (1/[HI]) 2
1
0
0 1 2 3 4 5 It would seem that the best straight line is for the plot of the reciprocal versus time. This is a second order reaction. b) Determine the rate constant of the reaction. . = 1.313 M‐1 s‐1 which would also be the slope of the line in the second graph 4) For the reaction A products, the following data were obtained: Time, hours [A], M Time, hours [A], M 0 1.24 6 0.442 1 0.960 7 0.402 2 0.775 8 0.365 3 0.655 9 0.335 4 0.560 10 0.310 5 0.502 a) Make appropriate plots to test the data for fitting zero‐, first‐, and second‐order rate laws and determine the overall order of the reaction. [A], M Zero Order Plot
1.4
1.2
1
0.8
[A], M 0.6 Linear ([A], M) 0.4
y = ‐0.0819x + 1.0044
R² = 0.8595 0.2
0
0 2 4 6 8 10 12 ln [A], First Order Plot
0.4
0.2
0
‐0.2 0 2 4 6 8 10 12 ‐0.4 ln [A] ‐0.6 Linear (ln [A]) ‐0.8
‐1
y = ‐0.1329x + 0.0486
R² = 0.9645 ‐1.2
‐1.4 1/[A], Second Order Plot
3.5
y = 0.242x + 0.8032
R² = 0.9999 3
2.5
2 1/[A] 1.5 Linear (1/[A]) 1
0.5
0
0 2 4 6 8 10 12 Second order provides the best fit linear line – second order reaction b) Determine the rate constant for the reaction. . . = 0.242 M‐1 s‐1 which would also be the slope of the line in the second graph c) Using the rate constant you have determined calculate the half‐life of the reaction. = / . . . d) At what time will the concentration of A be 0.380 M? = t = . . = 7.54 s . 5) A reaction A + B C is first order in both A and B and second order overall with a rate constant of 0.0500 M‐1 s‐1. If the initial concentration of A is 1.00 M, the initial concentration of B is 0.500 M, and no product is initially present, what is the concentration of the product C after 2 minutes? .
. . . . . . . . .
. . .500 – x = ( .
.500 – x = .
.
.500 – .
.487 M )(1.00 – x) – . x After 10 minutes? .
. . . . . . . . . . . .500 – x = (1.53 x 10‐7)(1.00 – x) .500 – x = 1.53 x 10‐7 – 1.53 x 10‐7x .500 – 1.53 x 10‐7 = 0.500 M 6) If the half‐life for the reaction C2H5Cl C2H4 + HCl Is the same when the initial concentration of C2H5Cl is 0.0050 M and 0.0078 M, what is the rate law for this reaction? Rate = k[C2H5Cl] This must be a first order reaction because the initial concentration has no impact on the half‐life 7) Determine the time required for each of the following reactions to occur: a) A fourth order reaction (A B + C) has the initial concentration of A decrease from 0.10 M to 0.080 M given that k = 0.015 L3 mol‐3 s‐1. . . . . = (4.5 x 10‐2 L3 mol‐3 s‐1)t + 1.00 x 103 M‐3 t = 2.1 x 104 s b) A fifth order reaction (A 2B + C) has an initial concentration of A = 0.35 M. If during the reaction the concentration of B changes from 0 M to 0.50 M, and given k= 0.0035 L4 mol‐4 s‐1, how much time has elapsed? [A]t = 0.35 M – 0.25 M = 0.10 M because of stoichiometry . . . . = (1.4 x 10 L mol s )t + 66.6 M 5
t = 7.1 x 10 s 8) Ethane forms methyl radicals at 700 oC in a first order reaction, for which k = 1.98 h‐1. a) What is the ½ life for the reaction? ‐2 . / = . 4 ‐4 ‐1 ‐4 = 0.35 h . b) Calculate the time for the amount of ethane to fall from 1.15 x 10‐3 mol to 2.35 x 10‐4 mol in a 500 mL reaction vessel at 700 oC. [ethane]o = . = 2.3 x 10‐3 M . [ethane]t = . = 4.7 x 10‐4 M . = .
. = ‐ (1.98 h‐1)t t = 0.802 h c) How much of a 6.88 mg sample of ethane in a 500 mL reaction vessel at 700 oC will remain after 45 minutes? [ethane]o = . = 4.45 x 10‐4 M . . . . . [A]t = 1.01 x 10‐4 M 1.01 x 10‐4 M x 0.500 L x = 1.51 mg 9) The rate law for the following reaction is Rate = k[NO2]2: 2NO2(g) N2O4(g) Which of the following changes will change the value of k? a) The pressure of NO2 is doubled. This will not change the rate constant. b) The reaction is run in an organic solvent. If the gases reacted while dissolved in the organic solvent there would be a different rate constant because the environment of the reaction has changed. c) The volume of the container is doubled. The volume will change the concentrations, but not the rate constant. d) The temperature is decreased. A temperature change will change the rate constant. e) A catalyst is added to the container. A catalyst added will change the rate constant. 10) Assume for the reaction A products, the initial concentration is 0.075 M and k = 0.054 (units will vary). Calculate the final concentration after 10 minutes and ½ life if the order of the reaction is (use the same value of k for each part, but allow the units to change as described): a) Zero order reaction (units for k are M s‐1) [A]t = ‐kt + [A]o = ‐(0.054 M s‐1)(600 s) + 0.075 M = ‐32.3 M indicating that there is no A left. = / . = 0.694 s . b) First order reaction (units for k are s‐1) .
[A]t = 6.37 x 10‐16 M / . . . . . c) Second order reaction (units for k are M‐1 s‐1) = (0.54 M‐1 s‐1)(600 s) + . . / . . d) Third order reaction (units for k are M‐2 s‐1) .
. . = / . . = 4938 s e) Second order reaction (A + B products) the initial concentration of A = 0.075 M, the initial concentration of B = 0.095 M and k = 0.054 M‐1 s‐1 .
. . . . . . . . . . . . 0.075 – x = .
0.075 – x = .
. .
. )(0.095 – x) . . –. No ½ life equation f) Second order reaction (A + B products) the initial concentration of A = 0.075 M, the initial concentration of B = 5 M, and k = 0.054 M‐1 s‐1. Because one concentration is very large compared to the other we can use pseudo‐first order kinetics. In this case [B]o >>[A]o . [A]t = 3.31 x 10‐72 M No ½ life equation . ...
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 Spring '11
 MILLER
 Electrochemistry, pH, Kinetics

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