Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chemical Kinetics Chapter 13

What is Kinetics? What do you think of when you hear the word kinetics? ◦ Thermodynamics: does a reaction take place? ◦ Kinetics: how fast does a reaction proceed? What makes a reaction go faster?

Factors affecting rate of reaction: Temperature ◦ Why? Concentration of reactants ◦ Why? Pressure (of a gas) ◦ Why? Surface area ◦ Why? Catalysts ◦ Why?

Collision Theory: When do molecules react? When they collide!!! Collision = Reaction (More collisions =Faster Reaction) Molecules must also have sufficient energy for a reaction to occur.

Reaction rate: change in the concentration of a reactant or a product with time (M/s). A B rate = -  [A] tt rate =  [B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative. Chemical Kinetics

A B rate = -  [A] tt rate = [B][B] tt time As the concentration of A decreases, the concentration of B increases

How do we measure the rate? Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time Change in color with time

393 nm light Detector  [Br 2 ]   Absorption How to more accurately measure the change in concentration: Measures the change in absorption over time Change in light absorption with time

Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = - overall from start to finish  [Br 2 ] tt = - [Br 2 ] final – [Br 2 ] initial t final - t initial instantaneous rate = rate for specific instance in time Instantaneous Average Rate

2A B Two moles of A disappear for each mole of B that is formed. rate =  [B] tt rate = -  [A] tt 1 2 aA + bB cC + dD rate = -  [A] tt 1 a = -  [B] tt 1 b =  [C] tt 1 c =  [D] tt 1 d Reaction Rates and Stoichiometry

Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = -  [CH 4 ] tt = -  [O 2 ] tt 1 2 =  [H 2 O] tt 1 2 =  [CO 2 ] tt Ex: Combustion of Methane

RATE LAW

aA + bB cC + dD Rate = k [A] x [B] y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall The Rate Law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers (no more Δ t)

rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] = rate constant = 3.50 x s -1 Example Rate Law

F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y When: Double [F 2 ] with [ClO 2 ] constant  Rate Doubles x = 1 When: Quadruple [ClO 2 ] with [F 2 ] constant  Rate Quadruples y = 1 rate = k [F 2 ][ClO 2 ]

F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] 1 Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactants (not products) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.

Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq)  2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) x x x rate = k [S 2 O 8 2- ] p [I - ] q rate = k [S 2 O 8 2- ][I - ]

Find the rate constant k = rate [S 2 O 8 2- ][I - ] = 2.2 x M/s (0.08 M)(0.034 M) = 0.08/M s Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) x x x S 2 O 8 2- (aq) + 3I - (aq)  2SO 4 2- (aq) + I 3 - (aq) Unit depends on reactant order

Reaction Order Zero order: the reactant has no impact on the rate of the reaction First order: the change in concentration directly affects the change in rate ◦ The concentration doubles, the rate doubles Second order: the change in concentration has a power affect on the rate ◦ The concentration doubles, the rate quadruples

HALF LIFE

½ Life half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. Basic example: ◦ How much of a 600g sample remains after 4 half lives?

OrderRate Law Concentration-Time Equation Half-Life rate = k k unit: M/s rate = k [A] k unit: 1/s rate = k [A] 2 k unit: 1/M·s ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Graph

½ Life First Order For first order reactions, half-life is independent of initial concentration of the reactant ◦ It is not in the ½ life equation

What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x s -1 ? t½t½ ln2 k = x s -1 = = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s -1 ) ½ Life Example

REACTION MECHANISMS

2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 + Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism.

Intermediates species that appear in a reaction mechanism but not in the overall balanced equation. formed in an early elementary step and consumed in a later elementary step.

Reaction Mechanism Example Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 +

Rate Laws and Elementary Steps Writing plausible reaction mechanisms: o The sum of the elementary steps must give the overall balanced equation for the reaction. o The rate-determining step should predict the same rate law that is determined experimentally. rate-determining step: the slowest step in the sequence of steps leading to product formation.

The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate = k[NO 2 ] 2. The reaction is believed to occur via two steps: Step 1:NO 2 + NO 2 NO + NO 3 Step 2:NO 3 + CO NO 2 + CO 2 1. What is the equation for the overall reaction? NO 2 + CO NO + CO 2 2. What is the intermediate? NO 3 3. What are the relative rates of steps 1 and 2? rate = k[NO 2 ] 2 is the rate law for step 1 so step 1 must be slower than step 2

What if reactions are in equilibrium with each other? Proposed Mechanism: 2A  B B + C  E + F C + F  G + E If the rate law is: Rate = k[A] 2 [C] What must be true about substances in equilibrium? What is the rate determining step in the mechanism? Explain.

REACTION ENERGY

A + B C + D Exothermic Reaction activation energy (E a ): minimum amount of energy required to initiate a chemical reaction. Reaction Energy Energy Released

Reaction Energy Endothermic Reaction A + B C + D Energy Gained

substance that increases the rate of a chemical reaction without itself being consumed EaEa k uncatalyzedcatalyzed rate catalyzed > rate uncatalyzed E a < E a ‘Catalysts

uncatalyzed enzyme catalyzed Catalysis effect on Reaction Progress intermediate