Kinetics. Consider an irreversible unimolecular reaction k. -d[a]/dt = k[a] Can also describe in terms of appearance of B.

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Tamsin Hudson
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1 Kinetic data gives insight into reaction mechanisms kinetic analysis will describe a relationship between concentrations of all chemical species before the rate determining step in a given reaction and the rate of the reaction Consider an irreversible unimolecular reaction k The rate equation will describe the change in concentration versus rate -d[]/dt = k[] Can also describe in terms of appearance of d[]/dt = k[] [] however is dependent on [] [] = [] o [] d[]/dt = k([] o []) Need to have variable of [], not in [] if solve d[]/dt [] o = concentration of at time 0 239

2 What about a reversible unimolecular reaction? k 1 k 2 [] = [] o, [] = 0 at time = 0 Disappearance of [] -d[]/dt = k 1 [] k 2 [] [] = [] o [] Cannot solve for d[]/dt having a variable of [] -d[]/dt = k 1 [] k 2 ([] o []) -d[]/dt = (k 1 + k 2 )[] k 2 [] o 240

3 Consider an irreversible bimolecular reaction k C -d[]/dt = k[][] [] o [] = [] o [] In order to make d[]/dt solvable, need to have only [] as a variable [] = [] [] o + [] o -d[]/dt = k[]([] [] o + [] o ) Consider a dimerization k -d[]/dt = k[] 2 Overall therefore a solvable rate equation should be relatively straightforward to write, just remember to understand the relationships present so the only variable is the term being integrated 241

4 The kinetic order of a rate equation is the sum of all exponents of the concentrations appearing in the term -d[]/dt = k[] Therefore first order -d[]/dt = k[][] Therefore second order Can also classify by the order of a given reactant The overall second order reaction above is thus second order overall, but first order with respect to [] and first order with respect to [] 242

5 n important experimental determination for a reaction is once a rate equation is written, how to determine the actual value of rate constant (k) Reconsider the irreversible first order reaction k -d[]/dt = k[] d[]/[] = -k dt Rearrange equation to place like terms on same side Need to integrate equation d[]/[] = -k dt ln[] ln[] o = -kt [] = [] o e -kt Therefore to determine the value of k need to measure the concentration of [] versus time 243

6 ln[] - ln[] o = -kt Kinetics Intercept = ln[] o ln [] Slope = -k time straightforward observation (if not always seen at first) is that if a graph of ln [] versus time yields a straight line, then the reaction must have been a first order reaction The slope of this graph will be the -k value and the intercept will be ln [] o (which can be used to double check accuracy since this value should be known) Can also use this data to determine the half-life (T ½ ) for a reaction t T ½, [] = ½[] o ln([] 0 /2) - ln[] 0 = -kt ½ T ½ = (1/k) ln ([] 0 /{[] o /2}) = (1/k) ln2 244

7 Consider how determination of rate constant changes if reaction is reversible k 1 k 2 -d[]/dt = (k 1 + k 2 )[] - k 2 [] o Need to integrate equation t equilibrium Leads to two equations ln - d[]/(k 1 + k 2 )[] - k 2 [] o = dt (k 1 + k 2 )[] - k 2 [] o k 1 [] o = -(k 1 + k 2 )t k 1 [] e = k 2 {[] o - [] e } 1) [] e = k 2 [] o k 1 + k 2 2) ln [] o - [] e [] - [] e = (k 1 + k 2 )t 245

8 1) [] e = k 2 [] o k 1 + k 2 2) ln [] o - [] e [] - [] e = (k 1 + k 2 )t In order to solve for the k 1 and k 2 values: (k 1 + k 2 ) can be determined from equation 2 by measuring [] versus time and knowing the [] o and [] e terms The individual k 1 and k 2 values are then determined from equation 1 This is for only a relatively simple reversible unimolecular reaction! * Can easily realize that this kinetic analysis will become very difficult as the complexity of the reactions increase * Need approximations to solve more complex reactions 246

9 s an example consider again a second order reaction k C Integrate equation -d[]/dt = k[]([] [] o + [] o ) 1 ln [] o ([] o - [] o + []) = kt [] o [] o [][] o What if we start reaction with one reagent in great excess? [] o >> [] o, which also means [] o >> [] If apply to integral 1 [] o ln [] o [] = kt 247

10 1 [] o ln [] o [] = kt ln [] o [] = [] o kt Let [] o k = k obs (often done in practice, all constants are combined into a k obs term which is the actually measured value) ln [] [] o = -k obs t * Identical to irreversible first order reaction (called pseudo first order kinetics) For second order reactions, if one reagent is used in great excess (need > 10 fold) then that reagent will not appear in rate equation 248

11 Excess reagent term, however, will affect the observed rate constant k C If [] is in excess -d[]/dt = k obs [] k obs = k 1 [] s [] increases, the k obs will also increase ln[] ln[] o = -k obs t ln [] Intercept = ln[] o Slope = -k obs With pseudo first order approximation, a second order reaction will follow first order behavior and the observed rate constant can be obtained from a graph of ln [] versus time time 249

12 nother common approximation to use in kinetic analysis is called a steady-state approximation Consider a reaction of the type shown rate equations k 1 k 3 k 2 C Step 1 -d[]/dt = k 1 [] - k 2 [] Step 2 d[c]/dt = k 3 [] Considering [] -d[]/dt = k 3 [] + k 2 [] - k 1 [] Now we assume d[]/dt = 0 Steady-state approximation Therefore [] does not change appreciably over time Can occur if either k 2 >> k 1 or k 3 >> k 1 250

13 Means that any reactive intermediate will not accumulate in concentration over time Change in concentration of is small compared to change in concentration of either or C over time -d[]/dt = 0 = k 3 [] + k 2 [] - k 1 [] Solve for [] Remember rate equations Can substitute for solvable integral k 1 [] [] = k 3 + k 2 -d[]/dt = k 1 [] - k 2 [] d[c]/dt = k 3 [] k 3 k 1 [] d[c]/dt = -d[]/dt = k 3 + k 2 Therefore -d[]/dt = k obs [] Obtain first order equation 251

14 Can also use steady-state approximation for nonunimolecular reactions Merely pay attention to properly write rate equations k 1 k 3 k 2 C -d[]/dt = k 1 [] - k 2 [] -d[]/dt = 0 = k 3 [][C] + k 2 [] - k 1 [] d[d]/dt = k 3 [][C] D Steady-state approximation for [] [] = k 1 [] k 2 + k 3 [C] d[c]/dt = -d[]/dt = k 3 [C]k 1 [] k 2 + k 3 [C] -d[]/dt = k obs [] The k obs term now includes [C] (which means you would need to run in excess), but otherwise is identical to steady-state approximation for unimolecular case 252

15 Examples of mechanistic questions answered by proper kinetic analysis 1) Does a reaction proceed through an intermediate? X X Or does starting material react directly to product? X X X X X X If the intermediate is reactive and cannot be detected spectroscopically, how can an experimentalist determine whether it occurs or not? 253

16 Presence of intermediate can be detected by the kinetic analysis Write the rate equations for the two possible mechanisms: With intermediate k 1 k 3 k 2 D P D represent alkene -d[]/dt = 0 = k 3 [][D] + k 2 [] - k 1 [] d[p]/dt = -d[]/dt = k 3 [D]k 1 [] k 2 + k 3 [D] -d[]/dt = k obs [] With excess of [D] Direct reaction k D P -d[]/dt = k[d][] -d[]/dt = k obs [] How to detect difference between direct reaction and pathway with intermediate? (both simplify to pseudo first order when approximations are applied) 254

17 Often an issue when trying to distinguish pathways, how to run experiment to differentiate between possible pathways? In this case, report experiments with a double reciprocal plot With intermediate Direct reaction k obs = k 3 [D]k 1 k 2 + k 3 [D] k obs = k[d] 1/k obs = k 2 k 1 k 3 [D] + 1/k 1 1/k obs = 1/(k[D]) 1/k obs Slope = k 2 /k 1 k 3 Intercept = 1/k 1 Slope = 1/k Whether the intercept passes through the origin (direct reaction) or has a finite value (with intermediate) in double reciprocal plot will distinguish possible mechanisms Intercept = 0 1/[D] 255

18 2) nother question that a proper kinetic analysis can answer is how to increase the rate of a reaction Consider the synthesis of benzhydrol in a sodium chloride solution Cl Cl H H HO H H 2 O k 1 k 2 k 3 C d[c]/dt = k 3 [H 2 O][] Use pseudo first order approximation with water d[c]/dt = k obs [] Since is a reactive intermediate, can apply steady-state approximation for the [] term 256

19 d[]/dt = 0 = k 1 [] - k 2 [][Cl-] - k 3 [H 2 O][] [] = k 1 [] k 3 [H 2 O] + k 2 [Cl-] d[c]/dt = -d[]/dt = k 3 [H 2 O][] -d[]/dt = k obs = k 1 k 3 [H 2 O][] k 3 [H 2 O] + k 2 [Cl-] k 1 k 3 [H 2 O] k 3 [H 2 O] + k 2 [Cl-] With excess of [H 2 O] Therefore rate becomes slower as concentration of chloride ion increases * Known as common ion rate depression 257

20 3) What does kinetics indicate if we change the type of trapping agent in question 1? ssume pathway with intermediate k 1 k 3 k 2 D k 3 [D]k 1 k obs = k 2 + k 3 [D] P 1/k obs = k 2 k 1 k 3 [D] + 1/k 1 Intercept = 1/k 1 1/k obs Slope = k 2 /k 1 k 3 If trapping agent is more reactive, then the k 3 rate will be faster Therefore slope in double reciprocal plot will be less, but the intercept will be the same 1/[D] 258

21 What occurs if no longer have steady-state conditions when an intermediate is formed? k 1 k 2 C The kinetics become much more complicated = [] o e -k 1 t Same as a normal unimolecular reaction = [] o k 1 k 2 -k 1 (e -k 1 t - e -k 2 t ) C = [] o {1 + 1/(k 1 -k 2 )[k 2 e -k 1 t - k 1 e -k 2 t ]} The concentration of, and C do not follow as simple of relationship as when a reactive intermediate is present 259

22 Concentration [C] [] Time [] If k 1 = 0.001/s and k 2 = /s, then [] max = 0.46 [] o and T []max = 15 min If we want to synthesize, but it can decompose to another product C, then we would need to determine the k 1 and k 2 rate constants and then find out when [] max occurs to stop reaction and then isolate 260

Lecture 12. Complications and how to solve them

Lecture 12. Complications and how to solve them Lecture 12 Complications and how to solve them 1. Pseudo Order An expression for second order reaction 2A Products Can be written as, -da/dt = k [A] 2 And the integration, 1/A 2 da = kdt 1/A t 1/A o =