Chm 152 – Kinetics

CHM 152 – Kinetics (Ch. 12) Reaction Rates 1. Write the rate expression for the reaction N2 + 3 H2  2 NH3 Rate =  [N 2 ] t   1 [H 2 ] 3 t  1 [NH 3 ] 2 t 2. According to the following unbalanced reaction, if the rate of appearance of oxygen gas is 4.00 x 10-2 M/s, what is the rate of disappearance of KClO3 (g)? 2 KClO3(g)  2 KCl(g) + 3 O2(g) (high Temp rxn) a. 2.67 x 10-2 M/s b. 6.00 x 10-2 M/s c. 2.00 x 10-2 M/s d. 3.00 x 10-2 M/s e. 5.33 x 10-2 M/s -[KClO3] / 2t = [O2] / 3t -[KClO3] / t = (2/3)(0.0400 M/s) = -0.0267 M/s, so choice A (disappearance implies the negative) Rate laws 3. What is the general rate law for this reaction? H3BO3 + 3 HCl  BCl3 + 3 H2O ____rate = k[H3BO3]x [HCl]y 4. What is the rate law for a zero order overall reaction? ____rate = k___ 5. Name one factor that can increase the success of reactant collisions. Explain how it increases the reaction rate.  Temp: increase temp, molecules move faster, more collisions so rate increases  Concentration: increase [ ], more molecules, more collisions so rate increases  180o angle of impact: more energy, greater chance of surpassing Ea, rate increases 6. The rate constant for the reaction HNO3 + NH3  NH4NO3 is 14.5 1/M.s If the concentration of nitric acid is 0.050 M and the concentration of ammonia is 0.10 M and both reactants are first order, what will the rate of this reaction be? Rate = k[HNO3][NH3] Rate = (14.5 L / mol.sec)(0.050 M)(0.10 M) Rate = 0.073 mol / L. sec 7. The rate law for the following reaction is rate = k[H2][F2]. If the rate is 3.15 x 10-4 M/s when [H2] = 0.084 M and [F2] = 0.25 M, calculate the rate when [H2] = 0.039 M and [F2] = 0.099 M. (Hint: find the rate constant k) 3.15 x 10-4 M/s = k (0.084M)( 0.25 M) so k = 0.015 M-1s-1 rate = 0.015 M-1s-1 (0.039M)(0.099M) = 5.79 x 10-5 M/s Initial Rates / Rate constant (k) 8. If the concentration of a reactant is doubled and the corresponding rate quadruples, what is the order with respect to that reactant? ____2nd _ 9. What are the units for the rate constant if the rate law is: rate = k[A][B]2[C]? b. 1/s b. 1/Ms c. 1/M2s d. M/s e. 1/M3s 10. When two compounds, A and B, are mixed together, they form compound C, by a reaction that’s not well understood. Fortunately, the following rate information was experimentally determined, as shown below: Experiment 1 2 3 [A] (mol/L) 0.050 0.10 0.050 [B] (mol/L) 0.050 0.050 0.10 Rate (mol/L.sec) 4.0 x 10-3 8.0 x 10-3 1.6 x 10-2 a) Determine the rate law for this reaction. When you double the concentration of A, the rate doubles, which suggests that the reaction is first order in A. When you double the concentration of B, the rate quadruples, which suggests that the reaction is second order in B. Using this to generate a rate equation, we get: Rate = k[A][B]2 b) Determine the rate constant for this reaction. Plugging the concentrations into the equation with the rate, we solve for k (incidentally, it doesn’t matter which trial we use – the numbers work out the same): 4.0 x 10-3 mol/L.sec = k(0.05 mol/L)(0.05 mol/L)2 k = 32 1/M2.s Note: Rate constants may have a variety of strange units using L, mol, and sec. This is so the rate will always work out in M/s. 11. The following data were collected for the reaction of BF3(g) + NH3(g)  F3BNH3(g): Experiment [BF3] (M) [NH3] (M) Rate (M/s) 1 0.250 0.250 0.2130 2 0.250 0.125 0.1065 3 0.200 0.100 0.0682 4 0.350 0.100 0.2384 5 0.175 0.100 0.0596 a) What is the overall rate law for this reaction? __rate = k[BF3]2[NH3]________ rate 1 / rate 2: constant [BF3], rate doubles and [NH3] doubles; [NH3] is 1st order rate 4 / rate 5: constant [NH3], rate quadruples and [BF3] doubles; [BF3] is 2nd order b) What is the overall order of the reaction? ___1+2 = 3rd order__________ c) What is the value of the rate constant (with correct units)? __k = 13.6 M-2 s-1_______________ experiment 1: 0.2130 M/s = k(0.250 M)2(0.250 M) Integrated Rate Laws 12. To graphically find the rate constant for a first order reaction you should plot __ln At___ versus time. For a second order reaction? __1/[A]t___ For a zero order reaction? __ [A]t__ Half life 13. What percentage of a sample would remain after 8 half lives? 0.39% (divide by 2 eight times) 14. What percentage of a sample remains after 5 half lives? __3.125%____ 15. What is the half-life for a first order reaction if the initial concentration of reactant is 1.25M and after 69.2 seconds the concentration has dropped to 0.955M? ln (.955 / 1.25) = -k (69.2s) k = 3.89 x 10-3 s-1 t1/2 = .693 / k = 178 s 16. What is the reactant concentration after 78.9 seconds for a second order reaction with a half-life of 3.l0 minutes if the initial concentration was 0.555M? 186 s = 1 / k (0.555M) k = 9.687 x 10-3 s-1M-1 1 / [A]t = k (78.9s) + 1/0.555 1 / [A]t = 2.566 [A]t = 0.390M 17. A researcher at GCC is running a new chemical reaction that obeys first order kinetics and discovers that after 24 hours only ½ of the reactants are turned into products. How long will it take in hours for 90% of the reactants to be reacted? (Hint: how much is remaining when 90% is reacted?) t1/2 = 24h = 0.693 / k so k = 0.028875 h-1 ln (10 / 100) = - 0.028875 h-1 (t) so t = 8.0 x 101 h Mechanisms 18. The slow step in a mechanism is also called the __rate determining__ step. 19. If an elementary step has two reactants, what is its molecularity? ___bimolecular____ 20. For this reaction: 2 NO(g) + Cl2(g)  2 NOCl(g) the real rate law is rate = k [NO] [Cl2] If the following steps are the mechanism, which one must be the rate determining step? ____1__ What is the intermediate in this mechanism? ___NOCl2__ What is the overall order? __2__ NO(g) + Cl2(g)  NOCl2(g) NO(g) + NOCl2(g)  2NOCl(g) step one step two 21. For the following reaction, the rate law is found to be Rate = k[Ce4+][Mn2+]. 2 Ce4+ + Tl+  2 Ce3+ + Tl3+ One mechanism for this reaction, containing the following elementary steps, is shown below: 1. Ce4+ + Mn2+ ---> Ce3+ + Mn3+ 2. Ce4+ + Mn3+ ---> Ce3+ + Mn4+ 3. Tl+ + Mn4+ ---> Tl3+ + Mn2+ a. There is no catalyst and no intermediate. b. There is no catalyst, but Mn2+ is the intermediate. c. Mn3+ is the catalyst and there is no intermediate. d. Mn2+ is the catalyst and Mn3+ and Mn4+ are the intermediates. e. Mn4+ is the catalyst and Mn2+ and Mn3+ are the intermediates. f. Ce3+ is the catalyst and Tl+ and Mn4+ are the intermediates. 22. Which of the following equations represents the rate law for the following elementary process: A + B  C + D? a. Rate = k[C][D] b. Rate = k[A] c. Rate = k[A][B]2 d. Rate = k[A][B] e. Rate = k[B] 23. Below is a possible mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. What is the overall reaction equation for this process? H+ + H2O2  H2O+-OH (rapid equilibrium) H2O+-OH + 2Br-  HOBr + H2O (slow) HOBr + H+  Br2 + H2O (fast) a. b. c. d. 2 H+ + H2O2 + Br- + HOBr  H2O+--OH + Br2 + H2 O H2O+--OH + H++ Br-  Br2 + H2O 2 H+ + H2O2 + 2 Br-  Br2 + 2 H2O 2 H2O+--OH + 2 Br-  H2O2 + Br2 + 2 H2O Arrhenius Equation 24. If a reaction has a rate constant of 3.7 x 10-3 s-1 at 25 oC, and an activation energy of 43.6 kJ/mol, what will be the rate constant for this reaction at 75 oC ? a. 3.0 x 10 -4 s-1 b. 4.6 x 10 -2 s-1 c. 3.7 x 10 -3 s-1 d. 4.8 x 10 -3 s-1 25. A second order reaction whose rate constant at 700 oC was found to be 4.0 x 10-3 L/mol.s has an activation energy of 35.0 kJ/mol. What is the rate constant for this reaction at 850 oC? a. 7.1 x 10-3 L/mol·s b. 5.25 x 10-6 L/mol·s c. 5.78 x 10-2 L/mol·s d. 1.8 x 10-3 L/mol·s e. 1.0 x 10-3 L/mol·s Catalysis 26. A catalyst increases the rate of a reaction by a. increasing the enthalpy of the reaction b. decreasing the enthalpy of the reaction c. lowering the activation energy of the reaction d. raising the activation energy of the reaction 27. A catalyst is effective because a. it supplies energy to the reactant molecules, allowing more of them to achieve energies in excess of the activation energy for the reaction b. it increases the temperature of the molecules in the reaction mixture c it increases the number of collisions between molecules d. it lowers the activation energy of the reaction by providing a lower energy mechanism or pathway 28. The reaction H2(g) + I2(g) ----> 2 HI(g) is exothermic and has the following rate law at 298K: rate = [H2][I2]. Addition of a catalyst would a. increase the rate of the forward reaction b. increase the rate of both the forward and reverse reactions c. increase the rate of the reverse reaction d. cause no increase or decrease in the rate of the reaction e. None of the above