Chemical Kinetics. Reaction Rate. Reaction Rate. Reaction Rate. Reaction Rate. Chemistry: The Molecular Science Moore, Stanitski and Jurs

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1 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 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 Brooks/Cole 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 - ) x x x x x x x x x x x x x x x x x 0-7 Sample calculation Avg. rate = [Cv + ] = (5.00x x0-5 ) mol/l ( ) s = -.32 x 0-6 mol L - min - Table shows positive rates explained soon 2008 Brooks/Cole Brooks/Cole 6

2 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 Brooks/Cole Brooks/Cole 8 Reaction Rates and Stoichiometry For: 2 (g) + I 2 (g) 2 I (g) the rate of loss of I 2 is mol L - s -. What is the rate of formation of I? Rate = [ 2 ] = [I 2 ] = + 2 [ Rate = 2 ] = ( ) = + 2 [I] So = 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 t (s) Average rate (from 0 to 80 s) = slope of the blue triangle but the avg. rate depends on interval chosen Brooks/Cole 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 - ) x x x x 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

3 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 Brooks/Cole 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) x x 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) x x 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 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 Brooks/Cole Brooks/Cole 8 3

4 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 = M - s - = 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 Brooks/Cole 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 Brooks/Cole 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 = = k k 2008 Brooks/Cole Brooks/Cole 24 4

5 [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 M) 0.00 [cisplatin] halves every 475 min k = ln 2 = 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 = 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 mol/l t (min) 2008 Brooks/Cole Brooks/Cole 26 Calculating [ ] or t from a Rate Law In a st -order reaction, [reactant] 0 = 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 = 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 = -( s - )(600 s) +ln(0.500) ln[a] t = = [A] t = e = mol/l [reactant] 0 [reactant] 0 2 t /2 [reactant] 0 [reactant] t /2 4 [reactant] 0 [reactant] 8 0 t /2 [reactant] 0 [reactant] 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 M Brooks/Cole Brooks/Cole 28 Calculating [ ] or t from a Rate Law In a st -order reaction, [reactant] 0 = mol/l and t /2 = 400.s (c) Calculate t for [reactant] to drop to 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) = -( s - ) t + ln(0.500) = -( s - ) t t = 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 Brooks/Cole Brooks/Cole 30 5

6 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 Brooks/Cole 32 Unimolecular Reactions Potential energy (0-2 J ) E a = 435 x 0-2 J 0 E = -7 x 0 Initial state -2 J Final state 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 Brooks/Cole 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 Brooks/Cole 36 6

7 Bimolecular Reactions Potential energy (0-2 J ) 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 Brooks/Cole 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 - ) T (K) T (K) k (L mol - K - ) x x x x x 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 Brooks/Cole 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 = slope = x 0 3 K /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 A = 2.28 x 0 0 L mol - s - (A has the same units as k) 2008 Brooks/Cole Brooks/Cole 42 7

8 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) O + (aq) I 2 (aq) 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 Brooks/Cole 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) O + (aq) I 2 (aq) O(l) Mechanism OO + I - slow O - + OI Shows the bonding in 2 O 2 OI + I - fast O - + I 2 2{ O O + fast 2 2 O } overall 2 I O O + I O 2008 Brooks/Cole 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 O 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 Brooks/Cole 48 8

9 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 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 Brooks/Cole 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 Brooks/Cole Brooks/Cole 54 9

10 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 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 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 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 Brooks/Cole 60 0

11 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) Brooks/Cole 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 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) O(g) 2 NO(g) catalyst N 2 (g) + O 2 (g) 2008 Brooks/Cole Brooks/Cole 66

12 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) (g) CO(g) (g) C 3 O(l) A Pt-coated ceramic catalyst allows the st reaction to occur at low T Brooks/Cole Brooks/Cole 68 2