1. Things to know……

2.  Rate depends on temperature  Temp is the avg. KE  Order depends on rxn mechanism  Rate is determined by the slow step  Temp affects k  Increase temp 10oC, rate doubles  Rxns occur when collisions have sufficient Ea and correct orientation  Factors that can affect rate:  Nature of reactants, surface area, concentration, temp, catalyst, pressure

3. Reaction Rates How we measure rates. Rate Laws How the rate depends on amounts of reactants. Integrated Rate Laws How to calc amount left or time to reach a given amount. Half-life How long it takes to react 50% of reactants. Arrhenius Equation How rate constant changes with T. Mechanisms Link between rate and molecular scale processes.

4.  A plot of concentration vs. time for this reaction yields a curve like this.  The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

5.  The reaction slows down with time because the concentration of the reactants decreases. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

6.  What if the ratio is not 1:1? H 2 (g) + I 2 (g)  2 HI (g) Only 1/2 HI is made for each H 2 used.

7.  3. 2 A(g) + B(g) 2 C(g)  When the concentration of substance B in the reaction above is doubled, all other factors being held constant, it is found that the rate of the reaction remains unchanged. The most probable explanation for this observation is that  (A) the order of the reaction with respect to substance B is 1 (B) substance B is not involved in any of the steps in the mechanism of the reaction (C) substance B is not involved in the rate-determining step of the mechanism, but is involved in subsequent steps (D) substance B is probably a catalyst, and as such, its effect on the rate of the reaction does not depend on its concentration (E) the reactant with the smallest coefficient in the balanced equation generally has little or no effect on the rate of the reaction

8.  Step 1) N 2 H 2 O 2 N 2 HO 2 ¯ + H + fast equilibrium  Step 2) N 2 HO 2 ¯ ---> N 2 O + OH¯ (slow)  Step 3) H + + OH¯ ---> H 2 O (fast)  4. Nitramide, N 2 H 2 O 2, decomposes slowly in aqueous solution. This decomposition is believed to occur according to the reaction mechanism above. The rate law for the decomposition of nitramide that is consistent with this mechanism is given by which of the following?  (A) Rate = k [N 2 H 2 O 2 ] (B) Rate = k [N 2 H 2 O 2 ] [H + ] (C) Rate = (k [N 2 H 2 O 2 ]) / [H + ] (D) Rate = (k [N 2 H 2 O 2 ]) / [N 2 HO 2 ¯] (E) Rate = k [N 2 H 2 O 2 ] [OH¯]

9.  A rate law shows the relationship between the reaction rate and the concentrations of reactants.  For gas-phase reactants use P A instead of [A].  k is a constant that has a specific value for each reaction.  The value of k is determined experimentally. “Constant” is relative here- k is unique for each rxn k changes with T (section 14.5)

10.  When ln P is plotted as a function of time, a straight line results.  The process is first-order.  k is the negative slope: 5.1  10 -5 s -1.

11. Graphing ln [NO 2 ] vs. t yields: Time (s)[NO 2 ], Mln [NO 2 ] 0.00.01000-4.610 50.00.00787-4.845 100.00.00649-5.038 200.00.00481-5.337 300.00.00380-5.573 The plot is not a straight line, so the process is not first-order in [A]. Does not fit:

12. A graph of 1/[NO 2 ] vs. t gives this plot. Time (s)[NO 2 ], M1/[NO 2 ] 0.00.01000100 50.00.00787127 100.00.00649154 200.00.00481208 300.00.00380263 This is a straight line. Therefore, the process is second-order in [NO 2 ].

13. A graph of 1/[NO 2 ] vs. t gives this plot. Time (s)[NO 2 ], M1/[NO 2 ] 0.00.01000100 50.00.00787127 100.00.00649154 200.00.00481208 300.00.00380263 This is a straight line. Therefore, the process is second-order in [NO 2 ].

14. 10. The graph above shows the results of a study of the reaction of X with a large excess of Y to yield Z. The concentrations of X and Y were measured over a period of time. According to the results, which of the following can be concluded about the rate of law for the reaction under the conditions studied? A) It is zero order in [X]. B) It is first order in [X]. C) It is second order in [X]. D) It is the first order in [Y]. E) The overall order of the reaction is 2.

15. Experiment Initial [NO] (mol L¯ 1 Initial [O 2 ] (mol L¯ 1 Initial Rate of Formation of NO 2 (mol L¯ 1 s¯ 1 ) 10.10 2.5 x 10¯ 4 20.200.105.0 x 10¯ 4 30.200.408.0 x 10¯ 3 The initial-rate data in the table above were obtained for the reaction represented below. What is the experimental rate la for the reaction? (A) rate = k[NO] [O 2 ] (B) rate = k[NO] [O 2 ] 2 (C) rate = k[NO] 2 [O 2 ] (D) rate = k[NO] 2 [O 2 ] 2 (E) rate = k[NO] / [O 2 ]

16.  Half-life is defined as the time required for one-half of a reactant to react.  Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0.

17. For a first-order process, set [A] t =0.5 [A] 0 in integrated rate equation: NOTE: For a first-order process, the half-life does not depend on [A] 0.

18. For a second-order process, set [A] t =0.5 [A] 0 in 2nd order equation.

19. First orderSecond order Rate Laws Integrated Rate Laws complicated Half-life complicated

23.  KE converts to PE during collisions to break bonds.  The transition state (aka activated complex) is the unstable intermediate that forms at the peak of the PE diagram.  Increase Ae, k decreases, and therefore rate decreases.  When temp doubles, all particles speed up (way more than double); therefore, the relationship is not linear, but rather exponential.

24.  When to use it? When given k and time or asked to solve for Ea.  Plot ln k vs 1/T = linear graph  Slope of the line = -Ea/R\  Therefore, Ea = -R x slope  R = 8.31 J/K mol