Exam 1 Kinetics Review

Exam #1 Kinetics Review 1. Consider the following reaction, 2 ClO2 (aq) + 2 OH- (aq) -> ClO3- (aq) + ClO2- (aq) + H2O (l) Write an expression that describes the relationship between the rates of disappearance of ClO 2 and OH- and the rates of appearance of ClO3-, ClO2- and H2O. Rate  2. 1 Δ[ClO2 ] 1 Δ[OH- ] 1 Δ[ClO3- ] 1 Δ[ClO2- ] 1 Δ[H2O]     2 Δt 2 Δt 1 Δt 1 Δt 1 Δt Consider the following reaction in aqueous solution, 5Br-(aq) + BrO3-(aq) + 6H+(aq) --> 3Br2(aq) + 3H2O(l) 3. If the rate of appearance of Br2 at a particular moment during the reaction is 0.025 M s-1, what is the rate of disappearance (in M s-1) of Br- at that moment? 1 Δ[Br  ] 1 Δ[Br2 ] Δ[Br ] 5 (0.025M) M Rate     0.042 o 5 Δt 3 Δt Consider the following reaction at 25 C, Δt 3 s s (CH3)3COH(l) + HCl(aq) --> (CH3)3CCl(l) + H2O(l) The experimentally determined rate law for this reaction indicates that the reaction is first-order in (CH3)3COH and that the reaction is first-order overall. Which of the following would produce an increase in the rate of this reaction? a. b. c. d. e. 4. increasing the concentration of (CH3)3COH increasing the concentration of HCl decreasing the concentration of HCl decreasing the concentration of (CH3)3CCl It is impossible to tell. Rate k[(CH3 )3 OH] A certain first-order reaction has a rate constant, k, equal to 2.1 x 10 -5 s-1 at 355 K. If the activation energy for this reaction is 135 kJ/mol, calculate the value of the rate constant (in s-1) at 550. K. k  E  1 1  l n 1   a     k2  R  T2 T1   2.1x10 5 s 1  135000J mol-1  1 1    l n   -1 1  k 8.314 J mol K 550.K 355K   2   ln(2.1x105 s1 )  ln(k2 )  16.22 ln(k2 )  5.45 k 2  230 s -1 eln(k2 )  e 5.45 5. Laughing gas, N2O, can be prepared (ha, ha!) from H2 and NO: H2(g) + 2 NO(g) --> N2O(g) + H2O(g) A study of initial concentration (ha, ha!) versus initial rate at a certain temperature yields the following data for this reaction (ha, ha!): Rate1 k  [A]1     Rate1 k  [A]2  x y  [B]1    ...  [B]2    [H2], M [NO], M initial rate, M s-1 0.1000 0.5000 2.560 x 10-6 0.2000 0.3000 1.843 x 10-6 0.1000 0.3000 9.216 x 10-7 0.2000 0.6000 7.373 x 10-6 Determine the rate law for the reaction. Determine the value of the rate constant, k. 7.373x10 6 M  s 1  0.2000M   0.6000M      1.843x10 6 M  s 1  0.2000M   0.3000M  x 1.843x10 6 M  s 1  0.2000M   0.3000M      9.216x0 7 M  s 1  0.1000M   0.3000M  x 6. y 4.001  (1)2.000y y 2 Rate k[H2 ][NO]2 y 2.000  (2.000)x (1) x 1 2.560x106  k[.1000][.5000]2 k  1.024x104 M2s1 The complex ion, [Cr(NH3)5Cl]2+, reacts with OH- ion in aqueous solution, [Cr(NH3)5Cl]2+(aq) + OH-(aq) --> [Cr(NH3)5(OH)]2+(aq) + Cl-(aq) The following data were obtained for this reaction at 25oC, time, min   Rate  [A]final  [A]initial t final  t initial [Cr(NH3)5Cl]2+, M 0 1.00 6 0.657 12 0.432 18 0.284 24 0.186 30 0.122 36 0.0805 Determine the average rate of the reaction from t0 to t36. Determine the average rate from t0 to t6. Rate   0.0805M  1.00M  0.026 M  min 1 36min  0min Rate   We use “-“ for reactants to keep rate positive 0.657M  1.00M  0.057 M  min 1 6min  0min 7. l n2 t 1/2  k Which of the following are the correct units for the rate constant, k, for a zero-order reaction? a. M s-1 b. M-1 s-1 Rate k for 0 order c. M-2 s-1 d. M-3 s-1 e. M 14.3 days  8.  [A]t     k t ln   [A]0  ln2 k k  0.0485 days-1 Radioactive phosphorus is used in the study of biochemical reaction mechanisms. The isotope phosphorus-33 decays by first-order kinetics with a half-life of 14.3 days. If a chemist initially has a 7.5 M solution of pure phosphorus-33, calculate the concentration (in M) of phosphorus-33 in the solution after 2.4 days. 6.7M  [A]t  1 ln    (0.0485days )(2.4days) 7.5M   [A]t  0.890 7.5M e  [A]t  ln    7.5M   e 0.116 [A]t  6.7M or This only works because 1st order half-lives are the same [A]t  [A]0 ( 2) 1 9. t t 1/2  2.4 days [A]t  (7.5M)(1 2) 14.2days  [A]t  6.7M Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H 2, and iodine, I2: 2 HI (g) -> H2 (g) + I2 (g) The value of the rate constant, k, for this reaction was measured at several different temperatures and the data are shown below: temperature, K k, M-1 s-1 555 6.23 x 10-7 575 2.42 x 10-6 645 1.44 x 10-4 700 2.01 x 10-3 Calculate the value of the activation energy (in kJ/mol) for this reaction. 180  1.44x10 4 M1s 1  Ea 1   1  l n   3 1 1  -1 1  2.01x10 M s 8.314 J  mol K 700.K 645K      2.64  Ea  1.22x104  8.314 J  mol-1K1 E  1.80x105 J  mol-1  1.80x102 kJ  mol1 10. Consider the following reaction: 2 NOBr(g) -> 2 NO(g) + Br2(g) 1 1  kt [A]t [A]0 The reaction is known to be second order with k = 0.80 L/mol s. If you start with a concentration of 0.086 mol/L of NOBr, what will be its concentration after 22 seconds? 1 1  0.80 M-1 s -1 (22 s ) [A]t 0.086M 1  3.4x10 -2 M 1 29.2M [A]t  1  29.2M1 [A]t [A]t  0.034M 11. The reaction 2 NO2 + O3 -> N2O5 + O2 obeys the rate law, Rate = kobserved[NO2][O3] Which of the following mechanisms is consistent with this experimental rate law? (a) NO2 + NO2 <=> N2O4 N2O4 + O3 -> N2O5 + O2 (fast equilibrium) (slow) (b) NO2 NO5 + + O3 <=> NO5 NO5 -> N2O5 5/2 O2 (fast equilibrium) (slow) (c) NO2 NO3 + + O3 -> NO3 + NO2 -> N2O5 (d) NO2 + NO2 -> N2O2 N2O2 + O3 -> N2O5 + O2 + (slow) (fast) O2 (slow) (fast) (e) None of these mechanisms are possible. l n2 t 1/2  k  [A]t     k t ln   [A]0  12. The radioactive element, thallium-201, is used medicinally as a radiotracer to study damage in heart tissue. If a patient is injected with a 0.950 g dose of pure thallium-201, calculate the amount of time that would be required for the amount of thallium-201 in the patient's body to reach 0.0500 g. Thallium-201 decays by a first-order process with a half-life, t1/2 = 73.0 hours 73.0 h   0.0500g    (0.00950hours)(t) ln   0.950g  l n2 k k  9.50x10 3 hours-1 -2.94  (0.00950hours)(t) t  309 hours or t [A]t  [A]0 (1 2) t1/2 t 0.0500g  (0.950g)(1 2) 73.0hours t  0.0500g  ln   ln[(1 2) 73.0hours ]  0.950g  t   - 2.94  (.693)  73.0 hours   t  310. hours t 1/2  l n2 k 13. Homes in certain parts of the country contain high levels of the radioactive isotope, radon-222. Radon-222 decays by first-order kinetics with a half-life of 3.82 days. Calculate how long it would take for 95.0% of a sample of radon-222 to decay.  [A]t     k t ln   [A]0  3.82 days  ln2 k  5.0%  1 ln    (0.181days )(t)  100.0%  -3.00  (0.181days1 )(t) k  0.181 days-1 t  17 days or [A]t  [A]0 ( 2) 1 t t1/2 100.0%  (5.0%)( 2) 1 t  5.0%  ln    ln[( 1 2 ) 3.82days ]  100.0%  t 73.0hours  t  - 3.00  (.693)  3.82 days   t  17 days 14. The reaction in which NO2 forms a dimer, 2NO2(g) <==> N2O4(g) has the following experimentally determined rate law: Rate = k[NO2]2 where k = 400 L mol-1 s-1 How long (in seconds) would it take for a sample of NO2 with an initial concentration of 0.50 M to decrease to a concentration of 0.010 M? 0.25 1 1  (400.M1 s -1 )(t)  0.010M 0.50M 1 1  0.010M 0.50M  t 400.M1 s -1 t  0.25s USE THE FOLLOWING INFORMATION TO ANSWER THE NEXT THREE (3) QUESTIONS. + Tl + 2Ce 4+ Mnn+ -------> Tl3+ + 2Ce3+ The experimentally determined rate law is: Rate = k[Ce4+][Mn2+] with the following proposed mechanism involving ions of manganese: Step 1: Ce4+ + Mn2+ --> Ce3+ + Mn3+ Step 2: Ce4+ + Mn3+ --> Ce3+ + Mn4+ Step 3: Tl+ + Mn4+ --> Tl3+ + Mn2+ 15. Which ion of manganese is the catalyst according to the information provided above? Mn2+ (it’s regenerated at the end) 16. Which step in the proposed mechanism is the rate-limiting step? Step 1….since the experimentally deteremined rate law agrees with it. 17. What is the overall order of the reaction? 2nd order overall