Chemical Kinetics Chemistry: The Molecular Science Moore, Stanitski and Jurs The study of speeds of reactions and the nanoscale pathways or rearrangements by which atoms and molecules are transformed to products Chapter 3: Chemical Kinetics: Rates of Reactions Chemical kinetics is also called reaction kinetics or kinetics. 2008 Brooks/Cole 2008 Brooks/Cole 2 Reaction Rate Factors affecting the speed of a reaction: Properties of reactants and products especially their structure and bonding. Concentrations of reactants (and products). Temperature Catalysts and if present, their concentration. Reaction Rate Combustion of Fe(s) powder: Reactions are either: omogeneous - reactants & products in one phase. eterogeneous - species in multiple phases. 2008 Brooks/Cole 3 2008 Brooks/Cole 4 Reaction Rate Change in [reactant] or [product] per unit time. Cresol violet (Cv + ; a dye) decomposes in NaO(aq): Cv + (aq) + O - (aq) CvO(aq) change in concentration of Cv rate = + = elapsed time [Cv + ] Reaction Rate Average rate of the Cv + reaction can be calculated: Time, t [Cv + ] Average rate (s) (mol / L) (mol L - s - ) 0.0 5.000 x 0-5 3.2 x 0-7 0.0 3.680 x 0-5 9.70 x 0-7 20.0 2.70 x 0-5 7.20 x 0-7 30.0.990 x 0-5 5.30 x 0-7 40.0.460 x 0-5 3.82 x 0-7 50.0.078 x 0-5 60.0 0.793 x 0-5 2.85 x 0-7 80.0 0.429 x 0-5.82 x 0-7 00.0 0.232 x 0-5 0.99 x 0-7 Sample calculation Avg. rate = [Cv + ] = (5.00x0-5 3.68x0-5 ) mol/l (0.0 0.0) s = -.32 x 0-6 mol L - min - Table shows positive rates explained soon 2008 Brooks/Cole 5 2008 Brooks/Cole 6
Reaction Rates and Stoichiometry Cv + (aq) + O - (aq) CvO(aq) Stoichiometry: Loss of Cv + Gain of CvO Rate of Cv + loss = Rate of CvO gain Another example: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) Reaction Rates and Stoichiometry For any general reaction: a A + b B c C + d D The overall rate of reaction is: Rate = [A] = [B] = + [C] = + a b c d [D] Loss of 2 N 2 O 5 Gain of O 2 Rate of N 2 O 5 loss = 2 x (rate of O 2 gain) Reactants decrease with time. Negative sign. Products increase with time. Positive sign Negative rate Positive rate Rate of loss of N 2 O 5 divided by -2, equals rate of gain of O 2 2008 Brooks/Cole 7 2008 Brooks/Cole 8 Reaction Rates and Stoichiometry For: 2 (g) + I 2 (g) 2 I (g) the rate of loss of I 2 is 0.0040 mol L - s -. What is the rate of formation of I? Rate = [ 2 ] = [I 2 ] = + 2 [ Rate = 2 ] = (-0.0040) = + 2 [I] So = +0.0080 mol L - s - [I] [I] Average Rate and Instantaneous Rate Graphical view of Cv + reaction: [Cv + ] (mol/l) 5.0E-5 4.0E-5 3.0E-5 2.0E-5.0E-5 0 0 20 40 60 80 00 t (s) Average rate (from 0 to 80 s) = slope of the blue triangle but the avg. rate depends on interval chosen. 2008 Brooks/Cole 9 2008 Brooks/Cole 0 Average Rate and Instantaneous Rate Effect of Concentration on Reaction Rate Rate may change when [reactant] changes. Cv + example shows this. For Cv + the rate is proportional to concentration. t [Cv + ] Rate of Cv + Rate/[Cv + ] (s) (M) loss (M / s) (s - ) 0 5.00 x 0-5.54 x 0-6 0.0308 80 4.29 x 0-6.32 x 0-7 0.0308 Instantaneous rate = slope of a line tangent to the curve. t = 0 s and t = 80 s have different instantaneous rates rate = k [Cv + ] 2008 Brooks/Cole 2008 Brooks/Cole 2 2
Rate Law and Order of Reaction A general reaction will usually have a rate law: rate = k [A] m [B] n... where k rate constant m, n order for A & B, respectively m + n + overall order of the reaction The orders are usually integers (-2, -, 0,, 2 ), but may also be fractions (, ) Determining Rate Laws from Initial Rates Rate laws must be measured. They cannot be predicted from reaction stoichiometry. Initial Rate Method To find the order for a reactant: Run the experiment with known [reactant] 0. Measure the initial rate of reaction (slope at t = 0). Change [reactant] 0 of reactant; keep all others constant. Remeasure the initial rate. The ratio of the two rates gives the order for the chosen reactant. 2008 Brooks/Cole 3 2008 Brooks/Cole 4 Determining Rate Laws from Initial Rates Data for the reaction of methyl acetate with base: C 3 COOC 3 + O - C 3 COO - + C 3 O Initial concentration (M) Expt. [C 3 COOC 3 ] [O - ] Initial rate (M/s) 0.040 0.040 2.2 x 0-4 2 0.040 0.080 4.5 x 0-4 3 0.080 0.080 9.0 x 0-4 Rate law: rate = k [C 3 COOC 3 ] m [O - ] n Determining Rate Laws from Initial Rates Initial concentration (M) Expt. [C 3 COOC 3 ] [O - ] Initial rate (M/s) 0.040 0.040 2.2 x 0-4 2 0.040 0.080 4.5 x 0-4 3 0.080 0.080 9.0 x 0-4 Dividing the first two data sets: 4.5 x 0-4 M/s = k (0.040 M) m (0.080 M) n Thus: 2.2 x 0-4 M/s = k (0.040 M) m (0.040 M) n 2.05 = () m (2.00) n 2.05 = (2.00) n and n = It is st order with respect to O -. raised to any power = 2008 Brooks/Cole 5 2008 Brooks/Cole 6 Determining Rate Laws from Initial Rates Use experiments 2 & 3 to find m: So: 9.0 x 0-4 M/s = k (0.080 M) m (0.080 M) n 4.5 x 0-4 M/s = k (0.040 M) m (0.080 M) n 2.00 = (2.00) m () n 2.00 = (2.00) m and m = Also st order with respect to C 3 COOC 3. Determining Rate Laws from Initial Rates The rate law is: rate = k [C 3 COOC 3 ][O - ] Overall order for the reaction is: m + n = + = 2 The reaction is: 2 nd order overall. st order in O - st order in C 3 COOC 3 2008 Brooks/Cole 7 2008 Brooks/Cole 8 3
Determining Rate Laws from Initial Rates If a rate law is known, k can be determined: k = Using run : rate [C 3 COOC 3 ][O - ] k = 2.2 x 0-4 M/s (0.040 M)(0.040 M) k = 0.375 M - s - = 0.375 L mol - s - Could repeat for each run, take an average But a graphical method is better. The Integrated Rate Law Calculus is used to integrate a rate law. Consider a st -order reaction: A products [A] rate = = k [A] d [A] = = k [A] dt (as a differential equation) Integrates to: ln [A] t = k t + ln [A] 0 y = m x + b (straight line) If a reaction is st -order, a plot of ln [A] vs. t will be linear. 2008 Brooks/Cole 9 2008 Brooks/Cole 20 The Integrated Rate Law Zeroth-order reaction The Integrated Rate Law The reaction: A products doesn t have to be st order. Some common integrated rate laws: Order Rate law Integrated rate law Slope [A] ln[a] slope = -k time t First-order reaction slope = -k time t Rate data for the decomposition of cyclopentene C 5 8 (g) C 5 6 (g) + 2 (g) were measured at 850 C. Determine the order of the reaction from the following plots of those data: 0 rate = k [A] t = -kt + [A] 0 -k rate = k[a] ln[a] t = -kt + ln[a] 0 -k 2 rate = k[a] 2 [A] t [A] 0 +k /[A] Second-order reaction slope = k y The most accurate k is obtained from the slope of a plot. x time t The reaction is first order (the only linear plot) k = - x (slope) of this plot. 2008 Brooks/Cole 2 2008 Brooks/Cole 22 alf-life t /2 = Time for [reactant] 0 to fall to [reactant] 0. alf-lives are only useful for st -order reactions. Why? t /2 is independent of the starting concentration. only true for st order reactions (not 0 th, 2 nd ) t /2 is constant for a given st -order reaction. alf-life For a st -order reaction: ln[a] t = -kt + ln[a] 0 When t = t /2 [A] t = [A] 0 Then: ln( [A] 0 ) = -kt /2 + ln[a] 0 ln( [A] 0 /[A] 0 ) = -kt /2 {note: ln x ln y = ln(x/y)} ln( ) = -ln(2) = -kt /2 {note: ln(/y) = ln y } t ln 2 0.693 /2 = = k k 2008 Brooks/Cole 23 2008 Brooks/Cole 24 4
[cisplatin] (mol/l) alf Life t /2 of a st -order reaction can be used to find k. For cisplatin (a chemotherapy agent): the cisplatin lost after 475 min. (0.000 M 0.0050 M) 0.00 [cisplatin] halves every 475 min 0.008 0.006 0.004 0.002 k = ln 2 = 0.693 t /2 475 min =.46 x 0-3 min - Calculating [ ] or t from a Rate Law Use an integrated rate equation. Example In a st -order reaction, [reactant] 0 = 0.500 mol/l and t /2 = 400.s. Calculate: a) [reactant],600.s after initiation. b) t for [reactant] to drop to /6 th of its initial value. c) t for [reactant] to drop to 0.0500 mol/l. 0 0 400 800 200 600 2000 t (min) 2008 Brooks/Cole 25 2008 Brooks/Cole 26 Calculating [ ] or t from a Rate Law In a st -order reaction, [reactant] 0 = 0.500 mol/l and t /2 = 400.s (a) Calculate [reactant],600.s after initiation. Calculating [ ] or t from a Rate Law In a st -order reaction, [reactant] 0 = 0.500 mol/l and t /2 = 400.s (b) Calculate t for [reactant] to drop to /6 th of its initial value. st order: k = ln 2/ t = 0.693/(400. s) =.733x0-3 s - and ln [A] t = -kt + ln [A] 0 so ln[a] t = -(0.00733 s - )(600 s) +ln(0.500) ln[a] t = -2.773 + -0.693 = -3.466 [A] t = e -3.466 = 0.032 mol/l [reactant] 0 [reactant] 0 2 t /2 [reactant] 0 [reactant] 2 0 4 t /2 4 [reactant] 0 [reactant] 8 0 t /2 [reactant] 0 [reactant] 8 6 0 t /2 4 t /2 = 4 (400 s) = 600 s Note: part (a) could be solved in a similar way. 600 s = 4 t /2 so 0.500 0.250 0.25 0.0625 0.033 M. 2008 Brooks/Cole 27 2008 Brooks/Cole 28 Calculating [ ] or t from a Rate Law In a st -order reaction, [reactant] 0 = 0.500 mol/l and t /2 = 400.s (c) Calculate t for [reactant] to drop to 0.0500 mol/l? From part (a): k =.733 x 0-3 s - Nanoscale View: Elementary Reactions Individual molecules undergo: unimolecular reactions a single particle (atom, ion, molecule) rearranges into or 2 different particles. bimolecular reactions two particles collide and rearrange. then ln [A] t = -kt + ln [A] 0 ln (0.0500) = -(0.00733 s - ) t + ln(0.500) -2.996 = -(0.00733 s - ) t 0.693 t = -2.303-0.00733 s - t =.33 x 0 3 s Both are elementary reactions what actually occurs at the nanoscale. Observed reactions (macroscale) may be: elementary directly occur by one of these two processes, or complex occur as a series of elementary steps. 2008 Brooks/Cole 29 2008 Brooks/Cole 30 5
Elementary reactions Example Label these elementary reactions as unimolecular or bimolecular: unimolecular bimolecular Unimolecular Reactions 2-butene isomerization is unimolecular: 3 C C 3 C=C (g) 3 C C=C (g) C 3 cis-2-butene trans-2-butene unimolecular unimolecular 2008 Brooks/Cole 3 2008 Brooks/Cole 32 Unimolecular Reactions Potential energy (0-2 J ) 500 400 300 200 00 E a = 435 x 0-2 J 0 E = -7 x 0 Initial state -2 J Final state -30 0 30 60 90 20 50 80 20 Reaction Progress (angle of twist) Exothermic overall cis-trans conversion twists the C=C bond. This requires a lot of energy (E a = 4.35x0-9 J/molecule = 262 kj/mol) Even more (4.42x0-9 J/molecule) to convert back. transition state or activated complex E a is the activation energy, the minimum E to go over the barrier. Transition State During isomerization, 2-butene passes through a transition state or activated complex. Occurs at the top of the activation barrier Exists for very short time (few fs, fs = 0-5 s). Falls apart to form products or reactants. If an exothermic reaction products are at lower E than the reactants the reverse reaction will be slower. 2008 Brooks/Cole 33 2008 Brooks/Cole 34 Bimolecular Reactions e.g. Iodide ions reacting with methyl bromide: Bimolecular Reactions I - (aq) + C 3 Br(aq) IC 3 (aq) + Br - (aq) transition state I - must collide in the right location to cause the inversion. I - must collide with enough E and in the right location to cause the inversion. unless the I - hits as shown it cannot drive out the Br - only a small fraction of the collisions have this orientation. the fractional factor is called the steric factor 2008 Brooks/Cole 35 2008 Brooks/Cole 36 6
Bimolecular Reactions Potential energy (0-2 J ) 50 20 90 60 30 0 transition state Also has an activation barrier (E a ). Forward and back E a are different. ere the forward reaction is endothermic. E a = 26 x 0-2 J Products (final state) E = 63 x 0-2 J Reactants (initial state) Reaction Progress (changing bond lengths and angles) Temperature and Reaction Rate Increasing T will speed up most reactions. igher T = higher average E k for the reactants. = larger fraction of the molecules can overcome the activation barrier. number of molecules 25 C 75 C kinetic energy E a Many more molecules have enough E to react at 75 C, so the reaction goes much faster. 2008 Brooks/Cole 37 2008 Brooks/Cole 38 Temperature and Reaction Rate Reaction rates are strongly T-dependent. Data for the I - + C 3 Br reaction: Temperature and Reaction Rate The Arrhenius equation shows how k varies with T k (L mol - K - ) 0.00 0.0 0.20 0.30 250 300 350 400 T (K) T (K) k (L mol - K - ) 273 4.8 x 0-5 290 2.00 x 0-4 30 2.3 x 0-3 330.39 x 0-2 350 6.80 x 0-2 370 2.8 x 0 - -E a / RT k = A e Quantity Name Interpretation and/or comments A Frequency factor ow often a collision occurs with the correct orientation. E a Activation energy Barrier height. e -Ea/RT Fraction of the molecules with enough E to cross the barrier. T Temperature Must be in kelvins. R Gas law constant 8.34 J K - mol -. 2008 Brooks/Cole 39 2008 Brooks/Cole 40 Determining Activation Energy Take the natural logarithms of both sides: Determining Activation Energy The iodide-methyl bromide reaction data: ln ab = ln a + ln b -E a / RT ln k = ln A e -E a / RT ln k = ln A + ln e ln k = ln A + ln k = E a R E a RT T ln e + ln A ln e = A plot of ln k vs. /T is linear (slope = E a /R). ln k 28 intercept = 23.85 8 8-2 slope = -9.29 x 0 3 K -2 0 0.00 0.002 0.003 0.004 /T (K - ) E a = -(slope) x R = -(-9.29 x0 3 K) 8.34 J K mol = 77.2 x 0 3 J/mol = 77.2 kj/mol A = e intercept = e 23.85 A = 2.28 x 0 0 L mol - s - (A has the same units as k) 2008 Brooks/Cole 4 2008 Brooks/Cole 42 7
Rate Laws for Elementary Reactions Elementary reactions Occur as written and their rate laws are predictable. Unimolecular reactions are always st -order. Bimolecular reactions are always 2 nd -order. Complex reactions Do not occur as written. They are carried out in a series of elementary steps. Reaction Mechanisms A complex reaction example: 2 I - (aq) + 2 O 2 (aq) + 2 3 O + (aq) I 2 (aq) + 4 2 O(l) When [ 3 O + ] is between 0-3 M and 0-5 M, rate = k [I - ][ 2 O 2 ] It must be complex Exponents in the rate law do not match the stoichiometry. Five reactants do not collide, form a transition state, and break into I 2 and 4 2 O. 2008 Brooks/Cole 43 2008 Brooks/Cole 44 Reaction Mechanisms Reaction Mechanisms A reaction mechanism lists the series of elementary steps that occur. shows how reactants change into products. 2 I - (aq) + 2 O 2 (aq) + 2 3 O + (aq) I 2 (aq) + 4 2 O(l) Mechanism OO + I - slow O - + OI Shows the bonding in 2 O 2 OI + I - fast O - + I 2 2{ O - + 3 O + fast 2 2 O } overall 2 I - + 2 O 2 + 2 3 O + I 2 + 4 2 O 2008 Brooks/Cole 45 2008 Brooks/Cole 46 Reaction Mechanisms Rate-limiting step The slowest step in the sequence Overall reaction rate is limited by, and equal to, the rate of the slowest step. A good analogy is supermarket shopping: You run in for item (~ min = fast step), but The checkout line is long (~0 min = slow step). Time spent is dominated by the checkout-line wait. In a reaction, a slow step may be thousands or even millions of times slower than a fast step. Reaction Mechanisms Step one OO + I - O - + OI Slow. It determines the overall rate. Steps 2 & 3 OI + I - O - + I 2 2{ O - + 3 O + 2 2 O } Fast. Will not affect the rate. The overall rate is expected to be rate = k [ 2 O 2 ][ I - ] as observed! 2008 Brooks/Cole 47 2008 Brooks/Cole 48 8
Mechanisms with a Fast Initial Step Consider: 2 NO (g) + Br 2 (g) 2 NOBr (g) The generally accepted reaction mechanism is: Step one NO + Br 2 NOBr 2 fast Step two NOBr 2 + NO 2 NOBr slow 2 NO + Br 2 2 NOBr Mechanisms with a Fast Initial Step Step 2 is rate limiting: rate = k 2 [NOBr 2 ] [NO] But NOBr 2 is an intermediate. It is difficult (sometimes impossible) to measure its concentration. NOBr 2 can form and fall apart many times (step ) before it converts to products. Step is reversible (double arrows). NO + Br 2 NOBr 2 fast, reversible 2008 Brooks/Cole 49 2008 Brooks/Cole 50 Mechanisms with a Fast Initial Step Equilibrium is established. k NO + Br 2 NOBr 2 reversible, equilibrium k - At equilibrium: rate forward = rate back k [NO][Br 2 ] = k - [NOBr 2 ] [NOBr 2 ] = k [NO][Br 2 ] k - Mechanisms with a Fast Initial Step The earlier rate law: becomes: rate = k 2 [NOBr 2 ] [NO] k [NO][Br 2 ] rate = k 2 [NO] k - k k rate = 2 [Br k 2 ][NO] 2 - rate = k' [Br 2 ][NO] 2 Now only contains starting materials - can be checked against experiment. 2008 Brooks/Cole 5 2008 Brooks/Cole 52 Summary Elementary reactions: the rate law can be written down from the stoichiometry. unimolecular rate = k[a] bimolecular rate = k[a] 2 or rate = k[a][b] You can t work backwards: Rate law matches stoichiometry: the reaction may be elementary or complex. The reaction must be complex if: rate law doesn t match the stoichiometry a species has a non-integer order. the overall order is greater than 2. Catalysts and Reaction Rate A Catalyst is a substance that: increases reaction rate without being consumed (reactants are consumed). changes the mechanism for the reaction. provides a lower E a in the rate limiting step. A catalyst does not change the products or their relative proportions. 2008 Brooks/Cole 53 2008 Brooks/Cole 54 9
Catalysts and Reaction Rate 2-butene isomerization is catalyzed by a trace of I 2. 3 C C=C C 3 (g) 3 C No catalyst: rate = k [cis-2-butene] C=C A trace of I 2 (g) speeds up the reaction, and: rate = k [ I 2 ] [cis-2-butene] k uncatalyzed k C 3 (g) Catalysts and Reaction Rate I 2 is not in the overall equation, and is not used up. The mechanism changes! Mechanism: step : {I 2 2 I } 3 C C 3 step 2: I + C=C I 2 splits into 2 atoms. Each has an unpaired e -. (shown by the dot) I attaches and breaks one C-C bond 3 C C 3 I C C 2008 Brooks/Cole 55 2008 Brooks/Cole 56 Catalysts and Reaction Rate step 3: step 4: Rotation around C-C 3 C C 3 I C C 3 C I C C C 3 step 5: {2 I I 2 } 3 C 3 C I C C C 3 C=C C 3 Loss of I and formation of C=C I 2 is regenerated + I Catalysts and Reaction Rate I 2 dissociates to I + I Reactants (initial state) Rotation around C-C I adds to cis-2-butene, (double single bond) E a = 262 kj/mol Transition state for the uncatalyzed reaction E a = 5 kj/mol E = -4 kj/mol Reaction Progress I leaves; double bond reforms I + I regenerates I 2 Products (final state) 2008 Brooks/Cole 57 2008 Brooks/Cole 58 Catalysts and Reaction Rate Key points: I 2 dissociates and reforms. Not consumed, not in the overall reaction equation. The activation energy is much lower. 5 kj/mol vs. 262 kj/mol The catalyzed reaction is 0 5 times faster at 500 K. 5 step mechanism = 5 humps in energy diagram. Catalyst and reactant are both in the same phase. omogeneous catalysis The initial and final energies are identical is the same whether catalyzed or not! Enzymes: Biological Catalysts Enzymes are: Usually very large proteins (often globular proteins) Very efficient catalysts for or more chemical reactions can increase rates by factors of 0 9-0 9 can catalyze millions of reactions per minute. ighly specific react with or a small number of substrates (the molecule undergoing the reaction). May require a cofactor to be present before they work Small organic or inorganic molecule or ion. Many use NAD + (nicotinamide adenine dinucleotide ion) 2008 Brooks/Cole 59 2008 Brooks/Cole 60 0
Enzyme Activity and Specificity Enzyme Activity and Specificity Enzymes are often large globular proteins, but only a small part (the active site) interacts with the substrate. The induced fit stretches and bends the substrate (and active site). 2008 Brooks/Cole 6 2008 Brooks/Cole 62 Enzyme Activity and Specificity Enzymes are effective catalysts because they: Bring and hold substrates together while a reaction occurs. old substrates in the shape that is most effective for reaction. Can donate or accept + from the substrate (act as acid or base) Stretch and bend substrate bonds in the induced fit so the reaction starts partway up the activation-energy hill. Enzyme kinetics Potential energy, E Formation of the enzyme-substrate complex Reactants (initial state) E a E' a E Transition state for the uncatalyzed reaction Products (final state) Reaction Progress Transformation of the substrate to products Activation energy E' a is much smaller than E a and so the enzyme makes the reaction much faster 2008 Brooks/Cole 63 2008 Brooks/Cole 64 Enzyme Activity and Specificity Enzyme catalyzed reactions: have unusual T dependence. speed up as T increases but will denature (change gross shape) and stop working. have a maximum reaction rate. [enzyme] limiting. can be blocked by an inhibitor binds, but doesn t react and release. Catalysis in Industry Catalysts are used extensively in industry. Many are heterogeneous catalysts usually a solid catalyst with gas or liquid reactants & products. acetic acid is prepared using solid rhodium(iii) iodide: C 3 O(l) + CO(g) RhI 3 C 3 COO(l) auto exhausts are cleaned by catalytic converters: 2 CO(g) + O 2 (g) Pt-NiO 2 CO 2 2 C 8 8 (g) + 25 O 2 Pt-NiO 6 CO 2 (g) + 8 2 O(g) 2 NO(g) catalyst N 2 (g) + O 2 (g) 2008 Brooks/Cole 65 2008 Brooks/Cole 66
Controlling Automobile Emissions forms a bond with the Pt surface Converting Methane to Liquid Fuel Methane is hard to transport. It can be converted to methanol: NO approaches the Pt surface dissociates into N and O atoms (each bonded to Pt) they form N 2 and O 2 and leave the surface. N and O migrate on the surface until they get close to like atoms C 4 (g) + O 2 (g) CO(g) + 2 2 (g) CO(g) + 2 2 (g) C 3 O(l) A Pt-coated ceramic catalyst allows the st reaction to occur at low T. 2008 Brooks/Cole 67 2008 Brooks/Cole 68 2