Rates & Mechanisms of Chemical Reactions McMurray & Fay ch. 12

Rates of Reaction Different reactions occur at different speeds

How do we measure the rate of a reaction? Two ways: Measure disappearance of a reactant Measure appearance of a product

Rates are always measured as a function of time dollars per hour dollars per year miles per hour (miles/hr or miles hr-1) feet per second (feet/second or ft s-1)

Reaction rates measure changes in amounts of chemicals over time In solutions: molarity per time (M s-1, M hr-1, etc.) In other instances: moles per time grams per time

Relative rates Concentrations of reactants may change at different rates, for example: 2 H2 + O2  2 H2O For every O2 used, 2 H2 are used For every mole of O2, 2 moles of H2 are used For every 1 M the concentration of O2 changes, the concentration of H2 changes by 2 M

2 H2 + O2  2 H2O Water is appearing at twice the rate that oxygen is disappearing Hydrogen is disappearing at twice the rate so half of that rate is equal to that of oxygen Oxygen is disappearing at a certain rate

General rate law: aA + bB  cC + dD

2 H2O (l) + CaC2 (s)  C2H2 (g) + Ca(OH)2 (aq) If H2O is used up at 3.8 mol s-1... at what rate is CaC2 used up? at what rate is C2H2 produced?

But usually, reaction rate is expressed independently of concentration of any reactant or product C6H12O6  2 C2H5OH + 2 CO2 [C2H5OH] increases at 0.025 M hr-1 Reaction rate = 0.0125 M hr -1

Instantaneous reaction rates Reaction rates may change over time Rate dependent on concentrations Reactant concentrations decrease over time Can only measure rates at specific times

Instantaneous reaction rate

Initial rate of reaction Find concentration change at t=0

Definition: Rate Law Expression of reaction rate that includes concentrations of all reactants

The exponents in the rate law must be determined experimentally. Generic Rate Law For the reaction: A + B  products The reaction rate can be written as: rate = k [A]m[B]n The exponents in the rate law must be determined experimentally.

Example Rate Law Reaction: BrO3- + 5 Br- + 6 H+  3 Br2 + 3 H2O Rate = k [BrO3-] [Br -] [H+]2

Reaction Orders rate = k [A]m[B]n Each rate law includes the concentration of each reactant raised to a power: [A]1 = first order [A]2 = second order [A]0 = zeroth order

Reaction order affects how reaction rates change with concentration rate = k [A]m

First order rate = k [A]1 When [A] = 2x: rate = k (2x)1= 2{kx} First order rule: Increase conc by factor of n = Increase rate by factor of n When [A] = 2x: rate = k (2x)1= 2{kx} When [A] = 3x: rate = k (3x)1 = 3{kx}

Second order rate = k [A]2 When [A] = x: rate = k (x)2= {kx2} Second order rule: Increase conc by factor of n = Increase rate by factor of n2 When [A] = 2x: rate = k (2x)2[B]n= 4{kx2} When [A] = 3x: rate = k (3x)2= 9{kx2}

Zeroth order rate = k [A]0 When [A] = x: rate = k (x)0= {k} Zeroth order rule: Rate does not change if you change concentration When [A] = 2x: rate = k (2x)0= {k} When [A] = 3x: rate = k (3x)0= {k}

Overall Reaction Order The rate law rate = k [A]m[B]n Is said to be reaction order m + n It is mth order with respect to component A and nth order with respect to component B

So, for BrO3- + 5 Br- + 6 H+  3 Br2 + 3 H2O Rate = k [BrO3-] [Br-] [H+]2 This reaction is: First order with respect to [BrO3-] and [Br-] Second order with respect to [H+]

Experimental determination of reaction order Vary initial concentrations Measure initial rate of formation of products Compare how concentrations change with how rate changes

Determining Rate Laws from Experimental Data rate = k [A]m[B]n 1. Find two data sets where concentration of first reagent changes but other concentrations remain constant. Use the rules on the right to determine rate order for first reagent First order rule: Increase conc by factor of n = Increase rate by factor of n Second order rule: Increase conc by factor of n = Increase rate by factor of n2 2. Repeat step 1 for subsequent reagents as necessary Zeroth order rule: Rate does not change if you change concentration 3. Plug experimental data and rate order exponents determined in steps 1&2 into rate law. Solve for k.

2 NO (g) + O2 (g)  2NO2 (g) Experiment # [NO] (M) [O2] (M) Initial Rate (M s-1) 1 0.015 0.048 2 0.030 0.192 3 0.096 4 0.384 1. Find two data sets where concentration of first reagent changes but other concentrations remain constant. Use the rules on the right to determine rate order for first reagent First order rule: Increase conc by factor of n = Increase rate by factor of n Second order rule: Increase conc by factor of n =Increase rate by factor of n2 Zeroth order rule: Rate does not change if you change concentration

2. Repeat step 1 for subsequent reagents as necessary 2 NO (g) + O2 (g)  2NO2 (g) Experiment # [NO] (M) [O2] (M) Initial Rate (M s-1) 1 0.015 0.048 2 0.030 0.192 3 0.096 4 0.384 2. Repeat step 1 for subsequent reagents as necessary First order rule: Increase conc by factor of n = Increase rate by factor of n Second order rule: Increase conc by factor of n =Increase rate by factor of n2 Zeroth order rule: Rate does not change if you change concentration

2 NO (g) + O2 (g)  2NO2 (g) Experiment # [NO] (M) [O2] (M) Initial Rate (M s-1) 1 0.015 0.048 2 0.030 0.192 3 0.096 4 0.384 3. Plug experimental data and rate order exponents determined in steps 1&2 into rate law. Solve for k.

H2O2 (aq) + 3 I- (aq)  I3- (aq) + 2H2O (l) Experiment # [H2O2] (M) [I-] (M) Initial Rate (M s-1) 1 0.100 1.15 x 10-4 2 0.200 2.30 x 10-4 3 4 4.60 x 10-4 1. Find two data sets where concentration of first reagent changes but other concentrations remain constant. Use the rules on the right to determine rate order for first reagent First order rule: Increase conc by factor of n = Increase rate by factor of n Second order rule: Increase conc by factor of n =Increase rate by factor of n2 Zeroth order rule: Rate does not change if you change concentration

H2O2 (aq) + 3 I- (aq)  I3- (aq) + 2H2O (l) Experiment # [H2O2] (M) [I-] (M) Initial Rate (M s-1) 1 0.100 1.15 x 10-4 2 0.200 2.30 x 10-4 3 4 4.60 x 10-4 First order rule: Increase conc by factor of n = Increase rate by factor of n 2. Repeat step 1 for subsequent reagents as necessary Second order rule: Increase conc by factor of n =Increase rate by factor of n2 Zeroth order rule: Rate does not change if you change concentration

H2O2 (aq) + 3 I- (aq)  I3- (aq) + 2H2O (l) Experiment # [H2O2] (M) [I-] (M) Initial Rate (M s-1) 1 0.100 1.15 x 10-4 2 0.200 2.30 x 10-4 3 4 4.60 x 10-4 3. Plug experimental data and rate order exponents determined in steps 1&2 into rate law. Solve for k.

NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l) Experiment # [NH4+] (M) [NO2-] (M) Initial Rate (M s-1) 1 0.24 0.10 7.2 x 10-6 2 0.12 3.6 x 10-6 3 0.15 5.4 x 10-6 1. Find two data sets where concentration of first reagent changes but other concentrations remain constant. Use the rules on the right to determine rate order for first reagent First order rule: Increase conc by factor of n = Increase rate by factor of n 2. Repeat step 1 for subsequent reagents as necessary Second order rule: Increase conc by factor of n =Increase rate by factor of n2 3. Plug experimental data and rate order exponents determined in steps 1&2 into rate law. Solve for k. Zeroth order rule: Rate does not change if you change concentration

Integrated Rate Laws

First-order reactions rate = k[A] - d[A] dt = k[A] d[A] [A] = - k dt

First-order reactions ln [A]t – ln [A]0 = -kt ln [A]t = -k t + ln [A]0 y = m x + b

First-order reactions graph ln [A] vs. t k = -slope = -.0034

Second order reaction -d[A] dt = k [A]2 d[A] [A]2 = - k

Second-order reactions 1 [A] 1 [A]0 - = kt 1 [A] 1 [A]0 = k t + y = m x + b

Second order reactions graph 1/[A] vs. T k = -slope = -.0033

Zeroth-order reactions Rate = k [A]0 Rate = k -d[A] dt = k d[A] = -k dt

Zeroth-order reactions [A] = -k t + [A]0 y = m x + b Graph [A] vs. time slope = -k

Finding reaction order experimentally Zeroth order: [A] vs. t

Finding reaction order experimentally Is this zeroth, first, or second order?

Finding reaction order experimentally

You can use integrated rate laws to solve for: Time at which a given concentration is present Concentration at a given time Rate constant

Using integrated rate laws to determine elapsed time or rate constant First order: ln [A]t – ln [A]0 = -kt ln = -kt [A]t [A]0

H2O2 in a lake decays with a rate constant of 0. 10 hr-1 H2O2 in a lake decays with a rate constant of 0.10 hr-1. How long will it take H2O2 to decay to 5% of its present value? ln = -kt [A]t [A]0 1. Start with the integrated rate law 2. Plug the ratio [A]t/[A]0 into the rate law. We want [A]t to be 5% of the starting value [A]0, so [A]t/[A0] =0.05 3. Plug this value into the rate law. 4. Solve for t.

The concentration of 32P in a cell decreases from 9. 53 at t0 to 7 The concentration of 32P in a cell decreases from 9.53 at t0 to 7.85 at t = 4 days. Find the rate constant for the disappearance of 32P. ln = -kt [A]t [A]0 1. Start with the integrated rate law 2. Plug the concentrations into the rate law. 3. Add the value for time elapsed. 4. Solve for k.

A reaction obeys second-order kinetics with rate constant 4. 5 M day-1 A reaction obeys second-order kinetics with rate constant 4.5 M day-1. How long does it take for the concentration of reactant A to drop from 2.1 M to 1.3 M? 1 [A]t - = kt [A]0 1. Start with the integrated rate law 2. Plug [A]t and [A]0 into the rate law. 3. Solve for t.

251Cf decays via first-order kinetics with a rate constant of 7 251Cf decays via first-order kinetics with a rate constant of 7.7 x 10-4 yr. A metal sample has a 251 Cf concentration of 1.23 %. What will the concentration be in 100 years? ln = -kt [A]t [A]0

An enzyme breaks down by zeroth order kinetics An enzyme breaks down by zeroth order kinetics. At first, the concentration of the enzyme is 0.0035 M. Ten minutes later, the concentration of the enzyme is 0.0024 M. What is the rate constant for the decay? [A] - [A]0 = -k t

Definition: Half-life The amount of time required for half of the reactant to disappear To find: use [A]t = ½ [A]0

Half life for first order reaction ln [A]t = -k t + ln [A]0 ln (1/2[A]0) = -k t1/2 + ln [A]0 ln [A]0- ln (1/2 [A]0) = k t1/2 [A]0 (1/2 [A]0) ln = k t1/2

Half-life for first order reaction ln 2 = k t1/2 ln 2 k t1/2 =

ln 2 k t1/2 =

Half-life for zeroth order reactions [A] = -k t + [A]0 1/2[A]0 = -k t1/2 + [A]0 t1/2 = (1/2k) [A]0 Half-life depends on initial concentration so is NOT CONSTANT

t1/2 = (1/2k) [A]0

Half-life for second order reactions 1 1/2[A]0 1 [A]0 - = kt1/2 Half-life depends on initial concentration so is NOT CONSTANT 1 [A]0 = kt1/2 1 k[A]0 t1/2 =

First-Order Reactions For practical reasons, we usually only talk about half-lives for First-Order Reactions

For first-order reactions: ln 2 k 0.693 k t1/2 = =

For first-order reactions you can also rearrange to get: 0.693 t1/2 ln 2 t1/2 k = =

A compound decays by first-order kinetics and has a half-life of 1 A compound decays by first-order kinetics and has a half-life of 1.6 hours. What is the rate constant, k, of the reaction?

All radioactive isotopes decay via first-order kinetics. Radioactive decay All radioactive isotopes decay via first-order kinetics.

Radioactive decay Not always easy to measure concentration of radioisotopes. Substitute one of following into the first-order rate law for concentration: activity (number of decays per second, N) isotope ratios

131I is used as a metabolic tracer in hospitals 131I is used as a metabolic tracer in hospitals. It has a half-life, t½ = 8.07 days. How long will it take before the activity falls to 1% of the initial value? ln = -kt [A]t [A]0

Vocabulary: CARBON DATING Method of using 14C/12C ratios to determine the age of an object

Carbon Dating: How it works Radioactive 14C (t1/2 = 5730 years) is constantly produced in the atmosphere 12C is not radioactive 14C:12C ratio in atmosphere constant at 1.2 x 10-12 Living organisms constantly exchange carbon through respiration so 14C:12C ratio similar to atmosphere After death, this exchange stops; amount of 14C decays by first-order kinetics

A fossil is found to have about 0. 36 of the normal 14C:12C ratio A fossil is found to have about 0.36 of the normal 14C:12C ratio. How old is the fossil?

Collision Theory: What affects the rate of a reaction?

Reaction: 2 HI  H2 + I2 H I H I Before new bonds can be formed, the old ones must be broken. Old bonds don’t just break for no reason—it takes an input of energy to break them

H I H I

I H H I

For a reaction to occur… Reactants must COLLIDE… …with SUFFICIENT ENERGY …at the right ANGLE ...to break the old bonds

Factors that affect the ability to react also affect the speed at which a reaction occurs. Chemical nature of the reactants Ability of the reactants to move around Concentrations of the reactants Temperature Catalysts

Concentrations affect collision rates The higher the concentration of the reactants, the higher the likelihood of a good collision

Temperature The higher the temperature, Reactants must collide with enough energy to break bonds Temperature is a measure of the average kinetic energy The higher the temperature, the higher the likelihood of a good collision

The more reactants can move, the more likely they are to collide For a reaction to occur, reactants must come in contact with each other (Most reactions occur in liquid or gas state) The more reactants can move, the more likely they are to collide

The stronger the bonds holding a molecule together, the more energy a collision must have to break those bonds The number of collisions with sufficient energy to cause a reaction depends on the chemical properties of the reactants

Vocabulary: CATALYST A substrate that allows reactants to interact more efficiently Frequently hold reactant molecules so collisions occur from the correct angle

(An effective collision is one that actually gives product molecules.) Collision Theory The rate of a reaction is proportional to the number of effective collisions per second amongst reactant molecules. (An effective collision is one that actually gives product molecules.)

Vocabulary: ACTIVATION ENERGY The minimum amount of kinetic energy needed in a collision to cause a reaction to happen

Vocabulary: TRANSITION STATE The moment in a reaction when the old bonds are partially broken and the new bonds are partially formed

Reaction: 2 HI  H2 + I2 H I H I

Reaction: 2 HI  H2 + I2 H I H I H I

Arrhenius equation Temperature (in Kelvin) Rate constant Gas constant (frequency factor or pre-exponential factor) Gas constant (8.314 J/mol K) Activation Energy

Arrhenius equation: conversion to logarithmic form

Can plot ln k vs. 1/T for experimental data ln k = ln A – Ea/RT y = b + m x Can plot ln k vs. 1/T for experimental data Slope of line = -Ea/R

Experimental Determination of Ea Run reaction at different temperatures Determine k at each temperature Plot ln k vs. 1/T Multiply slope of best-fit line by R

Determination of Ea (Actual Experimental Data)

Determination of Ea (Actual Experimental Data)

Arrhenius Plot Ea=42801 J/mol

Alternate application of Arrhenius law: Two-point equation

For the reaction shown previously (Ea = 42 For the reaction shown previously (Ea = 42.8 kJ/mol), the rate constant k is 0.0031 s-1 at 10.5 oC. What will the rate constant be at 35 oC?

Rate constants for the decomposition of gaseous dinitrogen pentoxide are 3.7 x 10-5 s-1 at 25oC and 1.7 x 10-3 s-1 at 55oC. What is the activation energy for this reaction in kJ/mol?

Reaction Mechanisms

O2 + 4 H+ + 4 e-  2 H2O O2 + e-  O2-• O2-• + H+  HO2• HO2• + e-  HO2- HO2- + H+  H2O2 H2O2 + e-  H2O2-• H2O2-• + H+  H2O + HO• HO• + e-  HO- HO- + H+  H2O

Reaction Mechanisms Balanced chemical equations represent a net process Reactions actually occur in steps called elementary processes which only involves one or two reactants A reaction mechanism shows each of these steps in order

Kinetics and reaction mechanisms Each step in a reaction mechanism has its own rate coefficient How can we use these to determine the overall reaction rate?

Which of the following is an elementary process Which of the following is an elementary process? (There is more than one correct answer) C3H8 + 5 O2  3CO2 + 4 H2O 2 NO + H2  N2O + H2O NO + O3  NO2 + O2 3 H2 + N2  2 NH3 N2O2 + H2 N2O + H2O

How fast can you drive on a highway?

rate-determining step The rate at which a reaction can occur is limited by the slowest step Slowest step (elementary process) is the rate-determining step

Rate Laws for Elementary Processes Elementary processes will be unimolecular or bimolecular Unimolecular processes are proportional to the concentration of that molecule Bimolecular processes are proportional to the concentration of each molecule

Example: O3  O2 + O rate = k[O3] This elementary process is unimolecular, so rate = k[O3]

Example: CH3Br + OH-  Br- + CH3OH This elementary process is bimolecular, so rate = k[CH3Br][OH-]

Reactions where the first step is slow NO2 (g) +NO2 (g) NO(g) + NO3 (g) NO3 (g) + CO (g) N2O (g) + CO2 (g) Slow Faster

Reactions where the first step is fast Reversible Slow Fast 2 NO (g) N2O2 (g) N2O2 (g) + H2 (g) N2O (g) + H2O (g) N2O (g) + H2 (g) N2 (g) + H2O (g)

Catalysts

Catalyst Definition: a substance that changes the rate of a chemical reactant without being used up Divided into two types: Homogeneous catalysts (same phase as the reactants) Heterogeneous catalysts (different phase from the reactants)

Homogeneous catalysis Catalysts are in the same phase as reactants, e.g. all aqueous or all gaseous 2 H2O2 (aq)  2 H2O (l) + O2 (g)

Enzymes as catalysts