# Chapter 13: Kinetics Renee Y. Becker Valencia Community College.

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Chapter 13: Kinetics Renee Y. Becker Valencia Community College

Introduction 1.Chemical kinetics is the study of reaction rates 2. For a chemical reaction to be useful it must occur at a reasonable rate 3. It is important to be able to control the rate of reaction 4. Factors that influence rate a) Concentration of reactants (molarity) b) Nature of reaction, process by which the reaction takes place c) Temperature d) Reaction mechanism (rate determining step) e) Catalyst

Reaction rate Reaction rate- Positive quantity that expresses how the conc. of a reactant or product changes with time Gen eq. N 2 O 5(g)  2NO 2(g) + ½ O 2(g) 1. [ ] = concentration, molarity, mole/L 2. [N 2 O 5 ] decreases with time 3. [NO 2 ] increases with time 4. [O 2 ] increases with time 5. Because of coefficients the concentration of reactant and products does not change at the same rate 6. When 1 mole of N 2 O 5 is decomposed, 2 moles of NO 2 and ½ mole of O 2 is produced

N 2 O 5(g)  2NO 2(g) + ½ O 2(g) -  [N 2 O 5 ] =  [NO 2 ] =  [O 2 ] 2 ½ coef. From gen eq -  [N 2 O 5 ] because it’s concentration decreases, other positive because they increase Rate of reaction can be defined by dividing by the change in time,  t rate = -  [N 2 O 5 ] =  [NO 2 ] =  [O 2 ]  t 2  t ½  t

Reaction Rates Generic Formula: aA + bB  cC + dD rate = -  [A] = -  [B] =  [D] =  [C] a  t b  t d  t c  t

Reaction Rate and Concentration The higher the conc. of starting reactant the more rapidly a reaction takes place 1. Reactions occur as the result of collisions between reactant molecules 2. The higher the concentration of molecules, the greater the # of collisions in unit time and a faster reaction 3. As the reactants are consumed the concentration decreases, collisions decrease, reaction rate decreases 4. Reaction rate decreases with time and eventually = 0, all reactants consumed 5. Instantaneous rate, rate at a particular time 6. Initial rate at t = 0

Rate Expression and Rate Constant Rate expression / rate law: rate = k[A] where k = rate constant, varies w/ nature and temp. [A] = concentration of A

Order of rxn involving a single reactant General equationrate expression A  prod rate = k[A] m where m=order of reaction m=0 zero order m=1 first order m=2 second order m, can’t be deduced from the coef. of the balanced eq. Must be determined experimentally!

Order of rxn involving a single reactant Rate of decomposition of species A measured at 2 different conc., 1 & 2 rate 2 = k[A 2 ] m rate 1 = k[A 1 ] m By dividing we can solve for m, to find the order of the reaction Rate 2 = [A 2 ] m Rate 1 [A 1 ] m (Rate 2 /Rate 1 ) = ([A 2 ]/[A 1 ]) m

Example 1 CH 3 CHO (g)  CH 4(g) + CO (g) [CH 3 CHO].10 M.20 M.30 M.40 M Rate (mol/L s).085.34.761.4 Using the given data determine the reaction order

Example 1 Rate 2 = [A 2 ] m 4 = 2 m m = 2 Rate 1 [A 1 ] m second order rate = k[CH 3 CHO] 2 once the order of the rxn is known, rate constant can be calculated, let’s calculate the rate constant, k

Example 1 rate = k[CH 3 CHO] 2 rate =.085 mol/L s conc =.10 mol/L k = rate=.085 mol/L s = 8.5 L/mol s [CH 3 CHO] 2 (.10 mol/L) 2 now we can calc. the rate at any concentration, let’s try.55 M

Example 1 rate = k[CH 3 CHO] 2 rate = 8.5 L/mol s [.55] 2 = 2.6 mol/ L s Rate when [CH 3 CHO] =.55 M is 2.6 mol/L s

Order of rxn with more than 1 reactant aA + bB  prod rate exp: rate = k [A] m [B] n m = order of rxn with respect to A n = order of rxn with respect to B Overall order of the rxn is the sum, m + n

Key When more than 1 reactant is involved the order can be determined by holding the concentration of 1 reactant constant while varying the other reactant. From the measured rates you can calculate the order of the rxn with respect to the varying reactant

Example 2 (CH 3 ) 3 CBr + OH -  (CH 3 ) 3 COH + Br - Exp. 1 Exp. 2 Exp. 3 Exp. 4 Exp. 5 [(CH 3 ) 3 CBr].51.01.51.0 [OH - ].05.10.20 Rate (mol/L s).005.01.015.01 Find the order of the reaction with respect to both reactants, write the rate expression, and find the overall order of the reaction

Reactant concentration and time Rate expressionrate = k[A] Shows how the rate of decomposition of A changes with concentration More important to know the relation between concentration and time Using calculus: Integrated rate equations relating react conc. to time

For the following rate law, what is the overall order of the reaction? Rate = k [A] 2 [B] 1.1 2.2 3.3 4.4

OrderRate Expression Conc-Time Relation Half-lifeLinear Plot 0 Rate = k[A] 0 – [A] = kt [A] 0 2k [A] vs. t 1 Rate = k[A]ln [A] 0 = kt [A] 0.693 k ln [A] vs. t 2 Rate = k[A] 2 1 – 1 = kt [A] [A] 0 1 k[A] 0 1 vs. t [A]

First Order First Order:A  Products rate = k[A] ln [A] o /[A] = kt t½ =.693/k [A] o = original conc. of A [A] = Conc. of A at time, t k = first order rate constant ln = natural logarithm

Second Order Second order: A  Products rate = k[A] 2 1 – 1 = kt t½ = 1 [A] [A] 0 k[A] 0

Zero Order Zero order:A  Products rate = k [A] 0 – [A] = ktt½ = [A] 0 2k

Example 3 The following data was obtained for the gas-phase decomp. of HI Is this reaction zero, first, or second order in HI? Hint: Graph each Conc. Vs. time corresponding to correct [A], ln [A], or 1/[A] Time (h)0246 [HI]1.000.500.330.25

[HI] vs. time not linear so it is not zero order

ln [HI] vs. time not linear so it is not first order

1/[HI] vs. time is linear so it is second order

Activation Energy Activation Energy: E a (kJ) For every rxn there is a certain minimum energy that molecules must possess for collisions to be effective. 1. Positive quantity (E a >0) 2. Depends only upon the nature of reaction 3. Fast rxn = small E a 4. Is independent of temp and concentration

For the following reaction: A + B  C If I double the concentration of A and hold B constant, the rate doubles. What is the order of the reaction with respect to A? 1.0 2.1 3.2

Activation Energy

Reaction Rate and Temp 1.As temp increases rate increases, Kinetic Energy increases, and successful collisions increase 2. General rule for every 10°C inc. in temp, rate doubles

The Arrhenius Equation f = e -Ea/RT f = fraction of molecules having an En. equal to or greater than E a R = gas constantA = constantT = temp in K ln k = ln A –E a /RT plot of ln k Vs. 1/T linear slope = -E a /R Two-point equation relating k & T ln k 2 = E a [1/T 1 – 1/T 2 ] k 1 R

Example 4 For a certain rxn the rate constant doubles when the temp increases from 15 to 25°C. a) Calc. The activation energy, E a b) Calc. the rate constant at 100°C, taking k at 25°C to be 1.2 x 10 -2 L/mol s

If I increase the temperature of a reaction from 110 K to 120 K, what happens to the rate of the reaction? 1.Stay the same 2.Doubles 3.Triples

Reaction Mechanism Description of a path, or a sequence of steps, by which a reaction occurs at the molecular level. Simplest case- only a single step, collision between two reactant molecules

Reaction Mechanism “Mechanism” for the reaction of CO with NO 2 at high temp (above 600 K) CO (g) + NO 2(g)  NO (g) + CO 2(g) “Mechanism” for the reaction of CO with NO 2 at low temp NO 2(g) + NO 2(g)  NO 3(g) + NO (g) slow CO (g) + NO 3(g)  CO 2(g) + NO 2(g) fast CO (g) + NO 2(g)  NO (g) + CO 2(g) overall The overall reaction, obtained by summing the individual steps is identical but the rate expressions are different. High temp: rate = k[CO][NO 2 ] Low temp: rate= k[NO 2 ] 2

Reaction Mechanism Elementary Steps: Individual steps that constitute a reaction mechanism UnimolecularA  B + Crate = k[A] Bimolecular A + A  B + C rate = k[A][A] = [A] 2 TermolecularA + B + C  D + E rate = k[A][B][C] The rate of an elementary step is equal to a rate constant, k, multiplied by the concentration of each reactant molecule. You can treat all reactants as if they were first order. If a reactant is second order it will appear twice in the general equation.

Reaction Mechanism Slow Steps- A step that is much slower than any other in a reaction mechanism. Rate-determining step - The rate of the overall reaction can be taken to be that of the slow step Step 1:A  Bfast Step 2:B  Cslow Step 3:C  Dfast A  D The rate A  D (overall reaction) is approx. equal to the rate of B  C the slow step

Deducing a Rate Expression from a Proposed Mechanism 1.Find the slowest step and equate the rate of the overall reaction to the rate of that step. 2. Find the rate expression for the slowest step. NO 2(g) + NO 2(g)  NO 3(g) + NO (g) slow CO (g) + NO 2(g)  CO 2(g) + NO 2(g) fast CO (g) + NO 2(g)  CO 2(g) + NO (g) Rate of overall reaction = rate of 1 st step = k[NO 2 ] [NO 2 ] = k[NO 2 ] 2

Elimination of Intermediates Intermediate 1.A species produced in one step of the mechanism and consumed in a later step. 2. Concentration too small to determine experimentally 3. Must be eliminated from rate expression 4. The final rate expression must include only those species that appear in the balanced equation for the overall reaction

Example 5 Find the rate expression for the following reaction mechanism Step1:NO (g) + Cl 2(g)  NOCl 2(g) fast Step2:NOCl 2(g) + NO (g)  2NOCl (g) slow 2 NO (g) + Cl 2(g)  2NOCl (g)

Example 6 The decomposition of ozone, O 3, to diatomic oxygen, O 2, is believed to occur by a two-step mechanism: Step 1:O 3(g)  O 2(g) + O (g) fast Step 2:O 3(g) + O (g)  2 O 2(g) slow 2 O 3(g)  3 O 2(g) Find the rate expression for this reaction

Which is an intermediate for the following multi step mechanism? 2 A + 2 B  C + 2 D Step 1 2 B  E Step 2 E + A  D + F Step 3 F + A  C + D 1.E 2.F 3.E & F 4.A

Catalysts Catalysis A catalyst increases the rate of a reaction without being consumed by it. Changes the reaction mechanism to one with a lower activation energy. 1. Heterogeneous catalysis a) Catalyst is in a different phase from the reaction mixture. Most common, solid catalyst with gas or liquid mixture. b) Solid catalyst is easily poisoned, foreign material deposited on the surface during reaction reduce or destroy its effectiveness. 2. Homogeneous Catalysis a) Same phase as the reactants

Which is the catalyst for the following multi step mechanism? 2 A + 2 B  C + 2 D Step 1 2 B + G  E Step 2 E + A  D + F Step 3 F + A  C + D + G 1.E 2.G 3.E & F 4.A

Summary Problem and Homework Problems Chapter 12: Kinetics

Hydrogen peroxide decomposes to water and oxygen according to the following reaction H 2 O 2(aq)  H 2 O + ½ O 2(g) It’s rate of decomposition is measured by titrating samples of the solution with potassium permanganate (KMnO 4 ) at certain intervals. a) Initial rate determinations at 40  C for the decomposition give the following data: [H 2 O 2 ]Rate (mol/L min) 0.101.93 x 10 -4 0.203.86 x 10 -4 0.305.79 x 10 -4 1.Order of rxn? 2.Rate expression? 3.Calc. k @ 40°C 4.Calc. half-life @ 40°C

b) Hydrogen peroxide is sold commercially as a 30.0% solution. If the solution is kept at 40  C, how long will it take for the solution to become 10.0% H 2 O 2 ?

c) It has been determined that at 50  C, the rate constant for the reaction is 4.32 x 10 -3 /min. Calculate the activation energy for the decomposition of H 2 O 2

d) Manufacturers recommend that solutions of hydrogen peroxide be kept in a refrigerator at 4  C. How long will it take for a 30.0% solution to decompose to 10.0% if the solution is kept at 4  C?

e) The rate constant for the uncatalyzed reaction at 25  C is 5.21 x 10 -4 /min. The rate constant for the catalyzed reaction at 25  C is 2.95 x 10 8 /min. 1) What is the half-life of the uncatalyzed reaction at 25  C? 2) What is the half-life of the catalyzed reaction?

1) Express the rate of reaction 2HI (g)  H 2(g) + I 2(g) a) in terms of  [H 2 ] b) in terms of  [HI]

3) Dinitrogen pentaoxide decomposes according to the following equation: 2N 2 O 5(g)  4NO 2(g) + O 2(g) a) write an expression for reaction rate in terms of  [ N 2 O 5 ],  [NO 2 ], and  [O 2 ]

4) What is the order with respect to each reactant and the overall order of the reactions described by the following rate expressions? a) rate = k 1 [A] 3 b) rate = k 2 [A][B] c) rate = k 3 [A][B] 2 d)rate = k 4 [B] e)rate = k

5) Complete the following table for the reaction, which is first order in both reactants A (g) + B (g)  products [A][B]K (L/mol s)Rate (mol/L s).2.31.5.029.78.025.45.520.033

7) The decomposition of ammonia on tungsten at 1100  C is zero-order, with a rate constant of 2.5 x 10 -4 mol/L min a) write the rate expression b) calculate the rate when the concentration of ammonia is 0.080M c) At what concentration of ammonia is the rate equal to the rate constant?

8)In solution at constant H + concentration, I - reacts with H 2 O 2 to produce I 2 H + (aq) + I - (aq) + ½ H 2 O 2(aq)  ½ I 2(aq) + H 2 O The reaction rate can be followed by monitoring iodine production. The following data apply: [I - ][H 2 O 2 ]Rate (mol/L s).02 3.3 x 10 -5.04.026.6 x 10 -5.06.029.9 x 10 -5.04 1.3 x 10 -4 a)Order of I - b)Order of H 2 O 2 c)Calc. k d)Rate? When [I - ] =.01 M [H 2 O 2 ] =.03 M

9) In the first-order decomposition of acetone at 500  C it is found that the concentration is 0.0300 M after 200 min and 0.0200M after 400 min. H 3 C-CO-CH 3(g)  products Calculate a) The rate constant b) The half-life c) The initial concentration

10) The decomposition of hydrogen iodide is second-order. Its half-life is 85 seconds when the initial conc. is 0.15M HI (g)  ½ H 2(g) + ½ I 2(g) a) What is k for the reaction? b) How long will it take to go from 0.300M to 0.100M?

11) Write the rate expression for each of the following elementary steps: a) K + + HCl  KCl + H + b) NO 3 + CO  NO 2 + CO 2 c) 2NO 2  2NO + O 2

12) For the reaction 2H 2(g) + 2NO (g)  N 2(g) + 2H 2 O (g) the experimental rate expression is rate = k[NO] 2 [H 2 ] The following mechanism is proposed: 2NO  N 2 O 2 fast N 2 O 2 + H 2  H 2 O + N 2 Oslow N 2 O + H 2  N 2 + H 2 Ofast Show that the mechanism is consistent with the rate expression.