Exam 1 (Ch. 12)

Exam 1 (Ch. 12) 1. (8 points) T F Half-life is the amount of reactant left after the reaction is half-way completed. T F Catalysts increase the rate of reactions by lowering the activation energy. 2. (6 points) Determine the overall order for reactions with the following rate constants (k). min-1 __1st ____ 1 nd M yr __2 ____ 1 th M  s __6 ____ 5 3. (4 points) Consider the following reaction: (CH3)3COH(l) + HCl(aq) --> (CH3)3CCl(l) + H2O(l) The experimentally determined rate law for this reaction indicates that the reaction is firstorder 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. increasing the concentration of (CH3)3COH increasing the concentration of HCl decreasing the concentration of HCl decreasing the concentration of (CH3)3CCl not enough information to determine 4. (4 points) Assume there are three reactions: One that is 0th order overall, one that is 1st order overall, one that is 2nd order overall. Assuming all three reactions have the same first halflife, which reaction would have the longest second half-life? a. b. c. d. e. the 0th order reaction the 1st order reaction the 2nd order reaction both the 0th order and 1st order reaction all will have the same 2nd half-life 5. (5 points) Collision theory is used to explain how chemical species undergo a reaction. Using this theory and the kinetic molecular model, which of the following influences the rate of a chemical reaction? _x__Temperature. _x__Orientation of the collision. _x__The energy of the reactants at the point of collision. ___The concentration of products _x__The concentrations of the reactants. 6. (4 points) A particular chemical reaction involves a single reactant. What is the order of the reaction if the rate increases by a factor of eight when the concentration of the reactant is doubled? 2x = 8 So….x = 3…. 3rd order 7. (4 points) Hydrogen sulfide (H2S) reacts with oxygen gas according to the following overall equation: 2 H2S(g) + O2(g) --> 2 S(s) + 2 H2O(l) Which of the following statements is TRUE? a. b. c. d. e. The reaction is 3rd order overall. Remember, the overall rate law can only The rate law is given by rate = k[H2S]2[O2]. be determined by experimental data. The reaction is 2nd order overall. The rate law is given by rate = k[H2S][O2]. The rate law cannot be determined from the information given 8. (4 points) A 1st order reaction started with 100.g of a reactant. How much of the reactant was left after 5 half-lives? (100.g)(1/2)5 = 3.13g 9. (2.5 points) Would the amount of reactant be the same as you calculated above if it were a 2nd order reaction instead (circle one)? YES NO The general half-life formula is valid for any order, since there is no time given 10. (5.5 points) If a 1st order reactant has a k of 0.111 s-1, how many seconds would it take for only 5.5% of the reactant to remain?  [A]  l n t   kt  [A]0   5.5%  1 ln   (0.111s )t  100.0%  11. Consider the following reaction in aqueous solution. t=26s 1  A  1  B  1  C  Rateof reaction        1  t  2  t  1  t  5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) a. (4 points) Which chemical species has the fastest rate of appearance/disappearance from t=0 to t=25s? H+ b. (8 points) If the appearance of Br2 was at the rate of 0.0051M/s…  …what would be the rate of disappearance of H+?   1 0.0051 M/s   1 - x  3 6 x = 0.010 M/s …what would be the rate of disappearance of Br-?  1 0.0051 M/s   1 - x  x = 0.0085 M/s 3 5 12. (5 points) Give two explanations why an increase in temperature increases the rate of a reaction.  Higher temperature speeds up molecules/atoms. So…more collision per time  Higher temperature adds more energy to molecules/atoms. So…more particles reach the activation energy. 13. (2.5 points) On the graph to the left, draw a dotted line to represent the average rate from t=0s to t=4s. (4 points) Calculate the average rate of the disappearance of A during this time. Rate  Δ [reactant(s)]  (1M  16M)   4 M/s Δt 4s  0s (2.5 points) The rate of disappearance of A from t=2s to t=4s should be faster / slower (circle one) than the rate you calculated above. (4 points) What is the 1st half-life for the reaction in the graph? Since the reaction goes from 16M to 8M in one sec….one second. 14. Determine the order for each reactant using the experimental data below. [A] M 0.222 0.222 0.111 0.222 [B] M 0.222 0.999 0.999 0.999 [C] M 0.222 0.222 0.222 0.777 Rate M/s 7.3 148 148 516  148 M/s   0.222M   k   0.222M   7.3 M/s  x  148 M/s   0.222M   k  148 M/s  0.111M    x  0.999M     0.222M  y  0.999M     0.999M  y x z  0.222M     0.222M  z y 20.3 = 4.5y  516 M/s   0.222M   0.999M   0.777M   k     0.222M   0.999M   0.222M   148 M/s  ▪ (9 points) Order for A, B, C: Rate = k [A]0[B]2[C] ▪ (2 points) Rate constant, (k): 7.3 M/s = k (0.222M)0(0.222M)2(0.222M) k = 670 M-2s-1 15. Use the following elementary steps to answer the next items. O3 → O2 + O (slow) O + O3 → 2O2 (fast) (3 points) Give the overall chemical reaction: (3 points) Identity of the intermediate: (3 points) Give the overall rate law:  0.222M     0.222M  O3 → O2 + O O + O3 → 2O2 2O3 → 3O2 O Rate = k[O3] 1 = 2x y=2 x=0 z 3.5 = 3.5z z=1 (3 points) Another name for the slow step is… The rate-determining step