Last wee we covered the following material: Review Vapor Pressure with two volatile components Chapter 14, Chemical Kinetics (continued) Quizzes next wee will be on Chap 14 through section 14.5. 13.6 Colloids Chapter 14 Hydrophilic and Hydrophobic Colloids Removal of Colloidal Particles Chemical Kinetics 14.1 Reaction Rates Rates in Terms of Concentrations Reaction Rates and Stoichiometry 14.2 The Dependence of Rate on Concentrations Reaction Order Units of Rate Constants Using Initial Rates to Determine Rate Laws Wee Five Chemical Kinetics (cont) 14.3 The Change of Concentration with Time First-Order Reactions Half-Life Second-Order Reactions 14.4 Temperature and Rate The Collision Model The Orientation Factor The Arrhenius Equation 14.5 Reaction Mechanisms Elementary Steps Multistep Mechanisms Rate Laws of Elementary Steps Rate Laws of Multistep Mechanisms Mechanisms with and Initial Fast Step Results for a reaction of A B C 4 H 9 Cl(aq) + H 2 O (l) C 4 H 9 OH (aq) + HCl (aq) [ C4H 9Cl] [ C4H 9OH ] Rate = = + t t Note the signs!
Consider the reaction 2 HI(g) H 2 (g) + I 2 (g) It s convenient to define the rate as HI] [ H 2] [ I 2] rate = = + = + 2 t t t And, in general for aa + bb cc + dd In fact, the instantaneous rate corresponds to d[a]/dt A] B] C] D] Rate = = = = a t b t c t d t Sample exercise 14.2 The decomposition of N2O5 proceeds according to the equation 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) If the rate of decomposition of of N 2 O 5 at a particular instant in a vessel is 4.2 X 10-7 M/s, what is the rate of appearance of (a) NO 2 ; (b) O 2? N2O5 ] NO2 ] O2 ] Rate = = + = + 2 t 4 t 1 t i.e. the rate of the reaction is 2.1 x 10-7 M/s the rate of appearance of NO 2 is 8.4 x 10-7 M/s and the rate of appearance of O 2 is 2.1 x 10-7 M/s 2 N 2 O 5 = 4 NO 2 + O 2 (g) at T = 45 o C in carbon tetrachloride as a solvent Time t {N 2 O 5 ] [N2O5] - [N2O5]/ t min min mol/l mol/l mol/l-min 0 2.33 184 2.08 319 1.91 526 1.67 867 1.35 1198 1.11 1877 0.72 2 N 2 O 5 = 4 NO 2 + O 2 (g) at T = 45 o C in carbon tetrachloride as a solvent Time t {N 2 O 5 ] [N 2 O 5 ] - [N 2 O 5 ]/ t min min mol/l mol/l mol/l-min 0 2.33 184 0.25 1.36 x 10-3 184 2.08 135 0.17 1.26 x 10-3 319 1.91 207 0.24 1.16 x 10-3 526 1.67 341 0.32 0.94 x 10-3 867 1.35 331 0.24 0.72 x 10-3 1198 1.11 679 0.39 0.57 x 10-3 1877 0.72 The information on the previous slide is a bit of a nuisance, since the instantaneous rate eeps changing and you now how much we lie constant values or linear relationships! So let s try something rather arbitrary at this point. Let s divide the instantaneous, average rate by [N 2 O 5 ] and/or [N 2 O 5 ] 2
2 N 2 O 5 = 4 NO 2 + O 2 (g) at T = 45 o C in carbon tetrachloride as a solvent {N 2 O 5 ] [N 2 O 5 ] [N 2 O 5 ] - [N 2 O 5 ]/ t Avg rate Avg rate mol/l avg mol/l mol/l-min /[N2O5] av /[N2O5] 2 av 2.33 2.08 1.91 1.67 1.35 1.11 0.72 2.21 0.25 1.36 x 10-3 6.2 x 10-4 2.8 x 10-4 2.00 0.17 1.26 x 10-3 6.3 x 10-4 3.2 x 10-4 1.79 0.24 1.16 x 10-3 6.5 x 10-4 3.6 x 10-4 1.51 0.32 0.94 x 10-3 6.2 x 10-4 4.1 x 10-4 1.23 0.24 0.72 x 10-3 5.9 x 10-4 4.8 x 10-4 0.92 0.39 0.57 x 10-3 6.2 x 10-4 6.7 x 10-4 Notice the nice constant value!!! It s convenient to write this result in symbolic form: Rate = [N 2 O 5 ] where the value of is about 6.2 x 10-4 so that when [N 2 O 5 ] = 0.221, Rate = (6.2 x 10-4 )(0.221) = 1.37 x 10-4 which is the average rate we started with In fact, we really should tae into account the 2 in front of the N 2 O 5, in accordance with the rule we developed earlier. This leads us to the general concept of Reaction Order When Rate = [reactant 1] m [reactant 2] n we say the reaction is m-th order in reactant 1 n-th order in reactant 2 and (m + n)-th order overall. Other reactions and their observed reaction orders 2 N 2 O 5 = 4 NO 2 + O 2 (g) Rate = [N 2 O 5 ]!!! CHCl 3 (g) + Cl 2 (g) CCl4 (g) + HCl(g) Rate = [CHCl 3 ][Cl 2 ] 1/2 H 2 (g) + I 2 (g) 2 HI (g) Rate = [H 2 ][I 2 ] Be careful because these orders are NOT related necessarily to the stoichiometry of the reaction!!! The order must be determined experimentally!!! We ll see later that it depends on the Reaction Mechanism, rather than the overall stoichiometry. Let s explore the results for the result Rate = [N 2 O 5 ] This can be expressed as Rate = - ( [N2O5] / t = - d[n 2 O 5 ] / dt = [N 2 O 5 ] An example of the plots of concentration vs time for a First-Order Reaction or, in general for A products Rate = - [A] / t = d[a] / dt = [A] rearrangement and integration some time = t and t =0 gives the result ln[a] t - ln[a] o = -t or ln [A] t = -t + ln [A] o or ln ([A] t /[A] o = - t This is the expression of concentration vs time for a First-Order Reaction
The Change of Concentration with Time Half-Life Half-life is the time taen for the concentration of a reactant to drop to half its original value. That is, half life, t 1/2 is the time taen for [A] 0 to reach ½[A] 0. Mathematically, ln 1 0.693 t 2 1 = = 2 The Change of Concentration with Time For a First-Order Reaction The identical length of the first and second half-life is a SPECIFIC characteristic of First-Order reactions Consider now Second-Order Reactions Example of Second-Order Plots of conc vs time [ A] d[ A] Rate = = = [ A] t dt [ A] t d[ A] = [ A [ A] 2 0 ] 1 1 = t + [ A] t [ A t 0 ] 0 dt 2
Second-Order Reactions We can show that the half life t 1 = 2 1 [ A] 0 A reaction can have rate constant expression of the form rate = [A][B], i.e., is second order overall, but has first order dependence on A and B. t 1/2 = 0.693/0.4 = 1.73 sec t 1/2 = [(0.4)(0.5)] -1 = 5.0 sec t 1/2 = ([A] 0 ) -1 = [(0.4)(1.0)] -1 = 2.5 sec Wee Five Chemical Kinetics (cont) General Order of reaction First Order reactions Second Order reactions Integrated form of each Half lives of each 14.3 The Change of Concentration with Time First-Order Reactions Half-Life Second-Order Reactions 14.4 Temperature and Rate The Collision Model The Orientation Factor The Arrhenius Equation 14.5 Reaction Mechanisms Elementary Steps Multistep Mechanisms Rate Laws of Elementary Steps Rate Laws of Multistep Mechanisms Mechanisms with and Initial Fast Step MQ1 Results: If you would lie to be a part of one of OSU s greatest outreach programs, please join us at one of the Wonders of Our World volunteer meetings. Volunteers help elementary school students learn science through fun hands-on experiments. Only a few hours a month are needed and training, supplies and transportation are provided! Come to find out why so many people enjoy being a part of WOW and how you can too! February WOW Volunteer Meetings Sunday February 9th at 7pm in 2015 McPherson Tuesday February 11 th at 7pm in 2015 McPherson Hi 175/175 Lo 36/175 Mean: 125 (71.4 %) If you have questions or if you would lie more information, please email or call Louise Van Wey at lvanwey@chemistry.ohio-state.edu or 688-0501. Thans!
Wee Five Chemical Kinetics (cont) Chapter 14, Chemical Kinetics (continued) 14.3 The Change of Concentration with Time First-Order Reactions Half-Life Second-Order Reactions 14.4 Temperature and Rate The Collision Model The Orientation Factor The Arrhenius Equation 14.5 Reaction Mechanisms Elementary Steps Multistep Mechanisms Rate Laws of Elementary Steps Rate Laws of Multistep Mechanisms Mechanisms with and Initial Fast Step Note the DRAMATIC effect of temperature on Temperature and Rate Collision Frequency The Collision Model eg H 2 + I 2 The Collision Model The greater the frequency of collision the faster the rate for a given concentration. Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product. Add an Orientation Factor. The higher the temperature, the more energy available to the molecules and the faster the rate. In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products. Arrhenius: molecules must posses a minimum amount of energy to react. Why? In order to form products, bonds must be broen in the reactants. Bond breaage requires energy. Activation energy, E a, is the minimum energy required to initiate a chemical reaction.
Consider the rearrangement of acetonitrile: H 3 C N C N H 3 C H 3 C C N C In H 3 C-N C, the C-N C bond bends until the C-N bond breas and the N C portion is perpendicular to the H 3 C portion. This structure is called the activated complex or transition state. The energy required for the above twist and brea is the activation energy, E a. Once the C-N bond is broen, the N C portion can continue to rotate forming a C-C N bond. The change in energy for the reaction is the difference in energy between CH 3 NC and CH 3 CN. The activation energy is the difference in energy between reactants, CH 3 NC and transition state. The rate depends on E a. Notice that if a forward reaction is exothermic (CH 3 NC CH 3 CN), then the reverse reaction is endothermic (CH 3 CN CH 3 NC). Effective Collisions Consider the reaction between Cl and NOCl to produce Cl 2 and NO - maybe: If the Cl collides with the Cl of NOCl then the products are Cl 2 and NO. If the Cl collided with the O of NOCl then no products are formed. We need to quantify this effect. Effective & Ineffective Collisions
The Arrhenius Equation -- Finally Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation: Ea = Ae RT is the rate constant, E a is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K. A is called the frequency factor, and is a measure of the probability of a favorable collision (including the collision frequency). Both A and E a are specific to a given reaction. = Ae E a RT The Arrhenius Equation If we have a lot of data, we can determine E a and A graphically by rearranging the Arrhenius equation: Ea ln = + ln A RT If we do not have a lot of data, or if we don t now A, then we can use Sample Exercise 14.8 For CH 3 NC CH 3 CN Temp. / o C / (s -1 ) 189.7 2.52 x 10-5 198.9 5.25 x 10-5 230.3 6.30 x 10-5 251.2 3.16 x10-5 ln 1 E = R 1 T2 T1 2 a 1 Sample Exercise 14.8 For CH 3 NC CH 3 CN Sample Exercise 14.8 For CH 3 NC CH 3 CN Temp. ( o C) T (K) 1/T (K -1 ) (s -1 ) ln 189.7 2.52 x 10-5 198.9 5.25 x 10-5 230.3 6.30 x 10-5 251.2 3.16 x10-5 Temp. ( o C) T (K) 1/T (K -1 x 10 3 ) (s -1 ) ln 189.7 462.9 2.160 2.52 x 10-5 198.9 472.1 2.118 5.25 x 10-5 230.3 503.5 1.986 6.30 x 10-5 251.2 524.4 1.907 3.16 x10-5
Sample Exercise 14.8 For CH 3 NC CH 3 CN Temp. ( o C) T (K) 1/T (K -1 x 10 3 ) (s -1 ) ln 189.7 462.9 2.160 2.52 x 10-5 -10.589 198.9 472.1 2.118 5.25 x 10-5 -9.855 230.3 503.5 1.986 6.30 x 10-5 -7.370 251.2 524.4 1.907 3.16 x10-5 -5.757 Fig 14.18 For CH 3 NC CH 3 CN reaction From the graph we find the slope = -1.9 x 10 4 K But this is also equal to - E a /R Or E a = -(slope)(r) = -(-1.9x10 4 )(8.314 J mol -1 K -1 )(1 J / 1000 J) = 1.62 x 10 2 J/mol We can now use these results to calculate the rate constant at any temperature. ln 1 E = R 1 T2 T1 2 a 1 To calculate 1 for a temperature of 430.0 K, mae substitutions for all other parameters, after choosing a starting point: or 162 J/mol 2 = 2.52 x 10-5 s -1 T 2 = 462.9 K And T 1 = 430.0 K to obtain 1 = 1.0 x 10-6 s -1 = Ae E a RT Catalysts serve to reduce the activation energy, thus increasing.
Reaction Mechanisms Involve Elementary Steps or Reactions When we now these, we CAN write down directly the rate expressions!!! Furthermore, these elementary steps represent fundamental molecular processes and lead to the term molecularity of the elementary reaction. Elementary Steps Elementary steps must add to give the balanced chemical equation. Intermediate: a species which appears in an elementary step which is not a reactant or product. Rate Laws of Elementary Steps The rate law of an elementary step is determined by its molecularity: Unimolecular processes are first order, Bimolecular processes are second order, and Termolecular processes are third order. Rate Laws of Multistep Mechanisms Rate-determining step: is the slowest of the elementary steps. Therefore, the rate-determining step governs the overall rate law for the reaction. Mechanisms with an Initial Fast Step It is possible for an intermediate to be a reactant. Consider 2NO(g) + Br 2 (g) 2NOBr(g) Reaction Mechanisms Mechanisms with an Initial Fast Step 2NO(g) + Br 2 (g) 2NOBr(g) The experimentally determined rate law is Rate = [NO] 2 [Br 2 ] Consider the following mechanism Step 1: NO(g) + Br 2 (g) Step 2: NOBr 2 (g) + NO(g) NOBr 2 (g) for which the rate law is (based on Step 2): 1-1 2 2NOBr(g) (fast) (slow) Mechanisms with an Initial Fast Step The rate law is (based on Step 2): Rate = 2 [NOBr 2 ][NO] The rate law should not depend on the concentration of an intermediate because intermediates are usually unstable. Assume NOBr 2 is unstable, so we express the concentration of NOBr 2 in terms of NOBr and Br 2 assuming there is an equilibrium in step 1 we have Mechanisms with an Initial Fast Step By definition of equilibrium (see Chap 15): 1 [NO][Br 2 ] = -1 [NOBr 2 ] Therefore, the overall rate law becomes Rate = 2 1 [NO] 2 [Br2 ] 1 Note the final rate law is consistent with the experimentally observed rate law. [ ] 1 [NO][Br ] NOBr2 = 2 1