1. Chapter 12 AP Kinetics worksheet #2

2. #1 The reaction between NO and H2 is believed to occur in the following three-step process.   NO + NO <===> N2O2 (fast) N2O2 + H2 N2O + H2O (slow) N2O + H2  N2 + H2O (fast)   (a) Write a balanced equation for the overall reaction.
2NO + 2H2  2H2O + N2

3. (b) Identify the intermediates in the reaction. Explain your reasoning.
N2O2 and N2O are the intermediates. They are both products of one step of the reaction that are used up in a later step. Neither is a product or a reactant in the overall reaction.

4. (c) From the mechanism represented above, a student correctly deduces that the rate law for the reaction is rate =k[NO]2[H2]. The student then concludes that : (1) the reaction is third-order and (2) the mechanism involves the simultaneous collision of two NO molecules and an H2 molecule. Are conclusions (1) and (2) correct? Explain.
(1) Correct. The rxn is 3rd order because the sum of the orders of the individual reactants is 3. (2) Incorrect. The possibility of 3 particles colliding with the correct orientation at the same instant is extremely low. (or --No step in the mechanism involved 2 NO molecules and 1 H2 molecule.)

5. (d) Explain why an increase in temperature increases the rate constant, k, given the rate law in (c).  
Increasing temperature will increase the avg. kinetic energy of the particles. Higher KE will result in more collisions and more particles possessing the activation energy and will increase the rate of the rxn. Or---- If concentrations remain constant and rate of reaction increases, k must increase.

6. (a) Complete the potential-energy diagram for reaction II on the graph above.
#2 A2 + B2  2 AB X2 + Y2  2 XY  Two reactions are represented above. The potential-energy diagram for reaction I is shown below. The potential energy of the reactants in reaction II is also indicated on the diagram. Reaction II is endothermic, and the activation energy of reaction I is greater than that of reaction II.  
(a) Complete the potential-energy diagram for reaction II on the graph above.
Line must have lower activation energy and have products higher than reactants.

7. (b) For reaction I, predict how each of the following is affected as the temperature is increased by 20°C. Explain the basis for each prediction. (i) Rate of reaction (ii) Heat of reaction      
At temperature is increased the rate of rxn will increase greatly. There will be more collisions and more of the collisions will have the activation energy.
The heat of reaction does not change because the energy of products and energy of reactants does not change.

8. (c) For reaction II, the form of the rate law is rate = k[X2]m[Y2]n
(c) For reaction II, the form of the rate law is rate = k[X2]m[Y2]n. Briefly describe an experiment that can be conducted in order to determine the values of m and n in the rate law for the reaction.    
Multiple trials can be performed. 1st, vary the concentrations of X2 while holding the concentration of Y2 constant. Measure affects on the rate. 2nd, vary the concentrations of Y2 while holding the concentration of X2 constant. Determine the affect on the rate. Use this data to determine the orders of the reactants.

9. (d) From the information given, determine which reaction initially proceeds at the faster rate under the same conditions of concentration and temperature. Justify your answer.
Rxn II will occur at a faster rate because it has a lower activation energy than Rxn I. More particles will have enough energy to produce products.

10. #3 2 A + B  C + D  The following results were obtained when the reaction represented above was studied at 25 °C Experiment Initial[A] Initial[B] Initial Rate of Formation of C (mol L¯1 min¯1) x 10¯ x 10¯ x 10¯ ?? x 10¯3   a) Determine the order of the reaction with respect to A and B. Justify your answer.  
A and B are both 1st order. In Exp. 1 & 2, conc. of B was held constant while conc. of A was tripled. This caused the rate to triple. Since they both did the same thing, A is 1st order. In Exp. 2 & 3, conc of A and B each doubled, causing the rate to be 4 times greater. We previously determined that doubling A would double the rate, so doubling B must also cause the rate to double (resulting in a rate 4 times greater).

11. b) Write the rate law for the reaction
b) Write the rate law for the reaction. Calculate the value of the rate constant, specifying units.
Rate = k[A][B] 4.3 x 10¯4 = k (0.25)(0.75) k = L/(mol.min)

12. c) Determine the initial rate of change of [A] in Experiment 3.
Experiment Initial[A] Initial[B] Initial Rate of Formation of C (mol L¯1 min¯1) x 10¯ x 10¯ x 10¯ ?? x 10¯3  
Since the rate of formation of C is 5.3 x 10-3 mol/(L.min), the rate of change of A will be twice that (there are 2 A reacting for every C formed- see balanced equation). Rate of change of A = x 10-2 mol/(L.min) It is negative because reactant conc. decreases.

13. d) Determine the initial value of [B] in Experiment 4
d) Determine the initial value of [B] in Experiment 4. Experiment Initial[A] Initial[B] Initial Rate of Formation of C (mol L¯1 min¯1) x 10¯ x 10¯ x 10¯ ?? x 10¯3  
Rate = k[A][B] 8.0 x 10-3 = (1.75)[B] [B] = 2.00 mol/L

14. e) Identify which of the reaction mechanisms represented below is consistent with the rate law developed in part (b). Justify your choice A + B  C + M Fast M + A  D Slow B <===> M Fast equilibrium M + A  C + X Slow A + X D Fast   A + <===> M Fast equilibrium M + A C + X Slow X  D Fast
2 is the best mechanism. The steps add up to the overall equation. The slow step is equivalent to the rate law because M is equivalent to B (from the first step) and can be substituted in for B.

15. #4 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)  Experiments conducted to study the rate of the reaction represented by the equation above. Initial concentrations and rates of reaction are given in the table below. (a) Determine the order for each of the reactants, NO and H2, from the data given and show your reasoning.
NO is 2nd order and H2 is 1st order. In exp. 1 & 2, conc. of NO is held constant while conc. of H2 is doubled. This results in the doubling of the rate. Since they both did the same thing, the order of H2 is 1. In exp. 3 & 4, conc. of H2 is held constant while conc. of NO is doubled. This results in the rate being 4 times greater. Since 22 = 4, the order of NO is 2.

16. (ii) Write the overall rate law for the reaction.
Rate = k[NO]2[H2]

17. (b) Calculate the value of the rate constant, k, for the reaction
(b) Calculate the value of the rate constant, k, for the reaction. Include units.
Rate = k[NO]2[H2]
Using data from Exp. 1,
1.8 x 10-4 mol/(L.min) = k (0.006 mol/L)2(0.001 mol/L)
k = 5000 L2/(mol2.min)

18. (c) For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H2 has been consumed.  
2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g) Before Change After The concentration of NO is 0.005M

19. d) The following sequence of elementary steps is a proposed mechanism for the reaction.   I. NO + NO <===> N2O II. N2O2 + H2  H2O + N2O III. N2O + H2  N2 + H2O  Based on the data present, which of the above is the rate-determining step? Show that the mechanism is consistent with:   (i) the observed rate law for the reaction, and (ii) the overall stoichiometry of the reaction.  
Step II is the rate determining step because it is equivalent to the rate law. It includes one H2 and one N2O2 (which is shown to be equivalent to 2 NO in step I).

20. #5  H2(g) + I2(g)  2 HI(g) For the exothermic reaction represented above, carried out at 298 K, the rate law is as follows. Rate = k [H2] [I2] Predict the effect of each of the following changes on the initial rate of the reaction and explain your prediction. (a) Addition of hydrogen gas at constant temperature and volume.
Addition of hydrogen gas would increase the rate of the reaction because the reaction is first order in hydrogen. Rate = k [H2] [I2]

21. (b) Increase in volume of the reaction vessel at constant temperature.
Increasing the volume will decrease the concentration of both reactants in a gaseous system. This will decrease the reaction rate.

22. (c) Addition of a catalyst
(c) Addition of a catalyst. In your explanation, include a diagram of potential energy versus reaction coordinate.  
The rate of reaction will incr. because an alternate pathway with a lower activation energy is provided.

23. (d) Increase in temperature
(d) Increase in temperature. In your explanation, include a diagram showing the number of molecules as a function of energy
The rate of rxn will increase. High temp. results in higher KE, more collisions and more collisions having the activation energy.

24. #6 Consider the following general equation for a chemical reaction
#6 Consider the following general equation for a chemical reaction A(g) + B(g)  C(g) + D(g) (a) Describe the two factors that determine whether a collision between molecules of A and B results in a reaction.
Reactants must collide with sufficient energy. If they don’t have the EA, no rxn occurs.
Reactants must have the proper molecular orientation when they collide (they must fit together properly)

25. (b) How would a decrease in temperature affect the rate of the reaction shown above? Explain your answer.
A decrease in temp. would decrease the rate of the rxn. Fewer particles would collide with the activation energy and there would also be fewer collisions.

26. (c) Write the rate law expression that would result if the reaction proceeded by the mechanism shown below A + B <===> [AB] (fast) [AB] + B C + D (slow)
Rate = k[A][B]2 The first step shows that [AB] is equivalent to [A][B]. Substituting them in for [AB] results in the above rate law.

27. (d) Explain why a catalyst increases the rate of a reaction but does not change the value of the equilibrium constant for that reaction.  
A catalyst is not a product or a reactant. Only concentrations of products and reactants are used in the calculations of equilibrium constants. Catalysts work by providing a pathway with lower activation energy. (Don’t worry about this question, yet.)

28. 7. 2 ClO2(g) + F2(g)  2 ClO2F(g) The following results were obtained when the reaction represented above was studied at 25 °C.
Experiment
Initial [ClO2]
Initial [F2]
Initial Rate of Increase of ClO2F in mol/Ls 1
0.010
0.10
2.4 x 10-3 2
0.40
9.6 x 10-3 3
0.020
0.20
Write the rate law expression for the reaction above.
Rate = k[ClO2] [F2]

29. 2.4 x 10-3 = k[0.010] [0.10] K = 2.4 M-1s-1 or 2.4 L/(mol.s)
ClO2(g) + F2(g)  2 ClO2F(g) The following results were obtained when the reaction represented above was studied at 25 °C.
Experiment
Initial [ClO2]
Initial [F2]
Initial Rate of Increase of ClO2F in mol/Ls 1
0.010
0.10
2.4 x 10-3 2
0.40
9.6 x 10-3 3
0.020
0.20
(b) Calculate the numerical value of the rate constant and specify its units.
2.4 x 10-3 = k[0.010] [0.10]
K = 2.4 M-1s-1 or 2.4 L/(mol.s)

30. ½ the rate of increase of ClO2F Rate = 4.8 x 10-3 mol/Ls
ClO2(g) + F2(g)  2 ClO2F(g) The following results were obtained when the reaction represented above was studied at 25 °C.
Experiment
Initial [ClO2]
Initial [F2]
Initial Rate of Increase of ClO2F in mol/Ls 1
0.010
0.10
2.4 x 10-3 2
0.40
9.6 x 10-3 3
0.020
0.20
(c) In experiment 2, what is the initial rate of decrease of [F2]?
Rate of decrease of F2 =
½ the rate of increase of ClO2F
Rate = 4.8 x 10-3 mol/Ls

31. ClO2(g) + F2(g)  2 ClO2F(g) The following results were obtained when the reaction represented above was studied at 25 °C.
Experiment
Initial [ClO2]
Initial [F2]
Initial Rate of Increase of ClO2F in mol/Ls 1
0.010
0.10
2.4 x 10-3 2
0.40
9.6 x 10-3 3
0.020
0.20
(d) Which of the following reaction mechanisms is consistent with the rate law developed in (a)? Justify your choice.
I. ClO2 + F2  ClO2F2 (fast)
ClO2F2  ClO2F + F (slow)
ClO2 + F  ClO2F (fast)
II. F2  2 F (slow)
2 (ClO2 + F  ClO2F) (fast)
Since the rate law is Rate = k[ClO2] [F2], the mechanism must have ClO2 and F2 (or something equivalent) in the slow step. This eliminates II. Choice I works because the steps add up to the overall equation and the slow step is equivalent to the rate law. In the first step, [ClO2F2] is equivalent to [ClO2][F2]