Advanced Chemistry Practice Problems

Transcription

1 Kinetics: Rate of Chemical Reactions The diagram below depicts the progress of a reaction. Each shape and color represents a different substance. The three boxes represent the concentrations of each substance as the indicated time elapses. Refer to the diagram to answer questions 1 4. V Time = 0 seconds Time = 15 seconds Time = 30 seconds 1. Select all images that represent reactants. There may be more than one reactant. a. b. c. 2. Which statement is true? a. The rate of change of substance is twice the magnitude as the rate of change of substance. b. The rate of change of substance is equal to the rate of change of substance. c. The rate of change of substance is twice the magnitude as the rate of change of substance. d. The rate of change of substance is equal in magnitude but opposite in sign to the rate of change of substance. 3. If each colored image represents 0.10 M of the substance, determine the rate (in M/s) of change of substance over the first 15 seconds.

2 Kinetics: Comparing Rate of Change for Reactants and Products 1. Consider the following reaction: 2N2O5(g) 2N2O4(g) + O2(g) If, at some point during the reaction, the rate of disappearance of N2O5 is 0.15 M/s, what is the rate of appearance of O2? 2. Consider the following reaction 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) At some point during the reaction, the rate of appearance of NO is M/s. What is the rate of disappearance of O 2 at this same point in the reaction?

3 1. The rate law of the reaction Kinetics: The Rate Law 2H2(g) + 2NO(g) N2(g) + 2H2O(g) is rate = k[h2][no] 2. Which of the following statements is/are false? a. The reaction is 3 rd order overall. b. The reaction is 2 nd order in H2. c. The reaction is 2 nd order in NO. d. The reaction is 1 st order in H2O. 2. The rate law of the reaction 2H2(g) + 2NO(g) N2(g) + 2H2O(g) is rate = k[h2][no] 2. What will be the effect on the rate of the reaction if the concentrations of both H2 and NO are doubled?

4 Kinetics: Rate of Chemical Reactions The diagram below depicts the progress of a reaction. Each shape and color represents a different substance. The three boxes represent the concentrations of each substance as the indicated time elapses. Refer to the diagram to answer questions 1 4. V Time = 0 seconds Time = 15 seconds Time = 30 seconds 1. Question: Select all images that represent reactants. There may be more than one reactant. a. b. c. Answer: The amount of reactant will decrease as the reaction progresses. Therefore answer b is the reactant.

5 2. Question: Which statement is true? a. The rate of change of substance is twice the magnitude as the rate of change of substance. b. The rate of change of substance is equal to the rate of change of substance. c. The rate of change of substance is twice the magnitude as the rate of change of substance. d. The rate of change of substance is equal in magnitude but opposite in sign to the rate of change of substance. Answer: The rate of change is determined by the change in concentration of the substance divided by the change in time. Answer a. is true. In the first 15 second time interval twice as much of is formed than. Therefore [ ]/15 s will be twice as big as [ ]/15 s. 3. Question: If each colored image represents 0.10 M of the substance, determine the rate (in M/s) of change of substance over the first 15 seconds. Answer: The rate of change for substance is determined by ( at 15 seconds - at 0 seconds)/15 seconds. Remember, each is equal to 0.10 M. Therefore the rate of change of = (0.4 M 0.8 M)/15 s = M/s

6 Kinetics: Comparing Rate of Change for Reactants and Products 1. Question: Consider the following reaction: 2N2O5(g) 2N2O4(g) + O2(g) If, at some point during the reaction, the rate of disappearance of N2O5 is 0.15 M/s, what is the rate of appearance of O2? Answer: For the reaction given in the problem: Rate N O N O O2 2t 2t The term rate of disappearance of N2O5 is represented by: rate of formation of O2 is represented by: expression that is required for this problem: O 2 t N O 2 5 O 2t t 2 t N t 2 O 5. The. Use only the portion of the Put in the value given for the rate of disappearance of N2O5. And solve for M O M/s 2 s t O 2 t

7 2. Question: Consider the following reaction 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) At some point during the reaction, the rate of appearance of NO is M/s. What is the rate of disappearance of O 2 at this same point in the reaction? Answer: For the reaction given in the problem: Rate 4t NH O NO H O 3 2 5t 4t The term rate of appearance for NO is represented by disappearance of O2 is represented by O 2. t 2 6t NO. The rate of t Put in the given information for the rate of appearance of NO and solve for O 2. t O 2 t O 2 5t O 2 5t M 4 s M 4 s M M/s 4 s

8 Kinetics: The Rate Law 1. Question: The rate law of the reaction 2H2(g) + 2NO(g) N2(g) + 2H2O(g) is rate = k[h2][no] 2. Which of the following statements is/are false? a. The reaction is 3 rd order overall. b. The reaction is 2 nd order in H2. c. The reaction is 2 nd order in NO. d. The reaction is 1 st order in H2O. Answer: The power to which the concentration is raised in the rate law determines the order. Therefore, the reaction is first-order in H2 and 2 nd order in NO. This means that b is false and c is true. The overall order is determined by adding the two powers together. Since, = 3, the reaction is third-order, overall. Also, d is false because products are not included in the rate law. 2. Question: The rate law of the reaction 2H2(g) + 2NO(g) N2(g) + 2H2O(g) is rate = k[h2][no] 2. What will be the effect on the rate of the reaction if the concentrations of both H2 and NO are doubled? Answer: One way to determine the effect of concentration changes on the rate is to do two separate calculations. In the second calculations use twice the concentration of the two reactants as used in the first calculation. In the example below, 1 M concentrations were used in the first calculation and 2 M was used for the second calculation. Rate = k(1 M)(1 M) 2 = 1 M 3 k Rate = k(2 M) (2 M) 2 = 8 M 3 k Therefore as the concentrations of each substance are doubled, the rate is increased by a factor of eight (8).

9 Kinetics: Obtaining a Rate Law from Experimental Data 1. Question: Determine the rate law and rate constant for the following reaction at 450 K using the data in the table provided. X + 2Y Z Initial rate of the reaction at 450 K (M/s) [X], M [Y], M Answer: The generic form for the rate law is: Rate = k[x] x [Y] y. The powers, x and y, are determined from the experimental data and NOT from the coefficients of the equation. To determine x, find two experiments where [Y] is held constant. The 2 nd and 3 rd sets of data meet that criteria. Set up a ratio of rates and concentrations as follows: rate rate exp3 exp [X] [ X] exp3 exp Now ask the question, What power would I have to raise the concentration by, that is, 2; to get the observed effect on rate, that is, 4. 2 x = 4, x = 2. The reaction is second-order in X.

10 To determine y, find two experiments where [X] is held constant. The 1 st and 2 nd sets of data meet that criteria. Set up a ratio of rates and concentrations as follows: rate rate exp2 exp [X] [X] exp2 exp Now ask the question, What power would I have to raise the concentration by, that is 3, to get the observed effect on rate, that is 3. 3 y = 3, y = 1. The reaction is first-order in Y. This gives a rate law of Rate = k[x] 2 [Y] To determine the rate constant at this temperature, take a set of experimental data, plug in the values, and solve for k. In this solution, the first set of experimental data is used. Rate = k[x] 2 [Y] M/s = k[0.01 M][0.01 M] M/s = k( M 3 ) s -1 M -2 = k The best value for k would be obtained by performing the above calculation for each set of experimental data and averaging the values together.

11 2. Question: Determine the rate law for the following reaction at 250 o C using the data in the table provided. 2A + B C + D Initial rate of the reaction at 250 o C, M/s [A], M [B], M Answer: Compared to the previous problem, this problem has a complicating fact. There is no experimental results for holding [B] constant. The rate law will have the generic form Rate = k[a] x [B] y. Begin by determining the order of B (that is, determining the value of y). To do this, we will select experiments 1 and 2 because these two hold [A] constant. See that the effect the change in [B] has on the rate. rate rate exp2 exp [B] [B] exp2 exp What power would the concentration of 2 need to be raised to in order to produce a change in rate of 4? 2 y = 4, y = 2. The reaction is 2 nd order in B. To determine A, we would like to find an experiment where B is held constant. There is not one. Take the rate law, with the known order and solve for the substance for which you do not know the order: rate = k[a] x [B] 2

12 rate k B 2 A x Now, take two experiments (experiments 2 and 3 are demonstrated here) and see how a change in concentration of A affects the quantity on the left side of the equation. (Note: units were removed for ease in following the solution.) k k x 2 = 2 x Therefore x = 1 and it is first-order in A. The rate law is therefore: Rate = k[a][b] 2

13 Kinetics: First-Order Kinetics and the Integrated Rate Law 1. Question: Consider the data collected for the reaction A products [A], M Time, s Determine the order for the reaction. Answer: The most common order for a reactant is first order. When given concentration and time information for a reaction, begin by plotting ln[a] versus t. If this produces a linear graph, the reaction is first-order. If not, the reaction is not first-order. Then a plot of 1/[A] v. t can be produced to see if the reaction is second-order. Take the natural log of each of the concentration values natural log [A] versus time ln [A] Time (s) The reaction is first-order.

14 2. Question: The rate constant of a certain first-order reaction is s 1. The reaction starts with a concentration of reactant of M. What is the concentration of the reactant after 1.50 minutes? Answer: The first-order integrated rate law equation is: [A] t ln [A] 0 kt According to the wording of the problem, [A]0 = M. The rate constant, k, is in units of s -1 therefore, the time needs to be in seconds as well: t 60 s 1.50 min 90.0 s 1min Plug the values given into the rate law and solve for [A]t, the concentration of the reactant at time of 90.0 seconds. [A] t ln ( M [A] t ln M 3 1/s) (90.0 s) [A] t ( 0.472) e M [A] t M [A] t (0.624) (0.250 M) M

15 Kinetics: First-Order Kinetics and Half-Life 1. Question: The half-life for the first-order conversion of cyclopropane to propene at 773 K is 17 min. What is the rate constant (in units of s 1 ) for the reaction at the same temperature? Answer: For a first-order reaction the relationship between half-life and the rate constant is given by: t k Watch carefully the units. The half-life time is given in minutes and the units of s -1 are needed for k. Convert time to seconds: Solve the half-life equation for k: Plug in the value for half-life gives: k s 0 4 s t 60 s 17 min 1min k t s

16 2. The half-life for the first-order conversion of cyclopropane to propene at 773 K is 17 min. Determine the amount of time required (in min) to convert 30% of the cyclopropane to propene. Answer: Using the rate constant, k, obtained in the previous question, the amount of time required to decrease the reactant from 100% down to70% (30% reacted) can be determined using the first-order integrated rate law as follows: [A] ln [A] t 0 kt 70% ln 100% ( s 1 ) t ( s -1 ) t Now, convert the time to minutes: 4 s -1 t 525 s t 1 min 525 s 8.7 min 60 s Note: The problem could have been worked by determining the rate constant in min -1, instead of caring the rate constant forward (in s -1 ) from the previous problem.

17 3. A certain first-order reaction required 456 s to reduce the amount of reactant to 1/16 of its original concentration at 450 K. What is the half-life of the reaction at this temperature? Answer: The reduction to 1/16 of the original amount is 4 half-lives. (1/2, 1/4, 1/8, 1/16). The total time is 456 s. Dividing this time by 4 will give the time for 1 halflife. 456 s 114 s 4