Kinetics Chemical kinetics is the study of the time dependence of chemical reactions.

Objectives Understand rates of reaction and the conditions affecting rates Derive the rate equation, rate constant and reaction order from experimental data Use integrated rate laws Understand the collision theory of reaction rates and the role of activation energy Relate reaction mechanisms and rate laws

Kinetics Reaction rate The change in concentration of a reactant or product per unit time

Rate of Reaction Rate of reaction may be described based on either the increase in concentration of a product or the decrease in concentration of a reactant per unit time Rates decrease with time Minus sign is required because the concentration of A decreases with time, and the rate is always expressed as a positive quantity. Instantaneous rate can be determined by finding the slope of a line tangent to a point representing a particular time.

Rate of Reaction Decomposition of N2O5 2N2O5  4 NO2 + O2 Rate can be expressed as

Rate of Reaction Rate could also be expressed in terms of the formation of NO2 or the rate of formation of O2. Rates expressed in these terms will have positive signs. Formation of NO2 is 2x rate of decomposition of N2O5 Formation of O2 is one half of the rate of decomposition of N2O5

Rate of Reaction Example Assume the disappearance of N2O5 between 40 min and 55 min is

Rate of Reaction Rate in terms of the appearance of NO2

Rate of Reaction Rate in terms of the rate at which O2 is formed

Rate of Reaction This procedure gives the average rate Graphing concentration vs. time does not give a straight line because the rate of the reaction changes during the course of the reaction. The instantaneous rate is given at a single point by drawing a line tangent to the concentration-time curve at a particular time and obtaining the slope of this line.

Rate of Reaction

Rate of Reaction For this reaction, if the rate is measured at a range of instantaneous concentration during the experiment, a Rate vs. Concentration graph can be plotted. The resulting graph is given below:

Rate of Reaction

Rate of Reaction This graph shows that the rate of reaction is directly proportional to the concentration of [N2O5]: Rate = constant x [N2O5] Rate = k[N2O5] This expression is called a rate equation, where k is called the rate constant

Rate of Reaction Summarizing the rate expressions To equate rates of disappearance or appearance, divide [reagent]/t by the stoichiometric coefficient in the balanced chemical equation

Rate of Reaction PROBLEM Give the relative rates for the disappearance of reactants and formation of products for the following reaction: 4 PH3(g)  P4(g) + 6 H2(g)

Rate of Reaction PROBLEM Data collected on the concentration of dye as a function of time are given in the graph below. What is the average rate of change of the dye concentration over the first 2 minutes? What is the average rate of change during the fifth minute (from t = 4 to t = 5)? Estimate the instantaneous rate at 4 minutes.

Rate of Reaction

Rate Laws Reversibility of Reactions For the reaction 2N2O5  4 NO2 + O2 the reverse reaction may also take place The reverse reaction effects the rate of change in concentration [N2O5] depends on the difference in the rates of the forward, kf, and the reverse, kb, reactions

Reaction Conditions and Rate Molecular collisions required for reaction to take place. Atoms and molecules are mobile (mixture of gases or using solutions of reactants) Under these conditions several factors affect the rate of reaction

Reaction Conditions and Rate Concentration Temperature Catalyst Surface area (if reactant is a solid)

Effect of Concentration Rate Equation Relationship between reactant concentration and reaction rate is expressed by the rate equation or rate law. For our example, N2O5 Rate of Reaction = k[N2O5]

Effect of Concentration k = proportionality constant = rate constant at a given temperature Rate equation indicates reaction rate is proportional to the concentration of the reactant [N2O5] doubles as the rate doubles

Effect of Concentration In general, aA + bB  xX Rate equation form Rate = k[A]m[B]n m and n are not necessarily the stoichiometric coeffcients (a and b). The exponents must be determined by experiment. m and n may be positive, negative, fractions or zero

Order of Reaction Order of a reaction is with respect to a particular reactant is the exponent of its concentration in the rate expression Total reaction order is the sum of the exponents

Order of Reaction Example Decomposition of H2O2 in the presence of iodide ion (clock reaction) Reaction rate = k[H2O2][I-] Reaction is first order with respect to H2O2 and I-; second order overall.

Order of Reaction So, what does it mean? Rate doubles if either H2O2 or I- is doubled and rate increases by a factor of 4 if both concentrations are doubled.

Order of Reaction Let’s consider the following rate law Rate = k[NO]2[Cl2] Reaction rate is second order in NO Reaction rate is first order in Cl2 Overall third order

Order of Reaction Let’s look at some data Exp [NO] M [Cl2] M Rate M∙s-1 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 11.4 x 10-6

Order of Reaction Exp 1 & 2 – [Cl2] held constant; [NO] is 2x; reaction rate increases by factor of 4 Exp 1, 3, 4 – Compare 1 & 3; [NO] held constant, [Cl2] 2x; rate doubles compare 1& 4; [NO] and [Cl2] 2x; rate 8x original value

Rate Constant k Reaction rates ([A]/t) have units of mol/L∙time when [ ] are given as moles/L Rate constants must have units consistent with the units for the other terms in the rate equation

Rate Constant k 1st order reactions: units of k are time-1 2nd order reactions: units of k are L/mol∙time zero order reactions: units of k are mol/L∙time

Determining a Rate Equation “Method of Initial Rates” Initial rate (instantaneous reaction rate at start of reaction (rate at t = 0) Mix reactants then determine [products]/t or -[reactants]/t after 1% or 2% of limiting reactant has been used up.

Determining a Rate Equation Example CH3CO2CH3(aq) + OH-(aq)  CH3CO2-(aq) + CH3OH(aq) Initial Concentrations Initial Rxn Rate Exp [CH3CO2CH3 ] M [OH-] M M∙s-1 1 0.050 0.00034  no change  x 2 2 0.10 0.00069 4 0.00137

Determining a Rate Equation When initial [reactant] is 2x while other [reactant] held constant, initial rxn rate 2x Reaction is directly proportional to the concentrations of both reactants Reaction is first order in each reactant Rate = k[CH3CO2CH3][OH-]

Determining a Rate Equation If rate equation is known, the value of k can be found by substituting values for the rate and concentration into rate equation

Determining a Rate Equation Using our example Substitute data from one of the experiments into the rate equation

Integrated Rate Laws What are they? Equations that describe concentration-time curves Why use them? Calculate a concentration at any given time Find length of time needed for a given amount of reactant to react

First Order Reactions If reaction ‘A  products’ is first order, the reaction rate is directly proportional to [A]1 A little calculus…

First Order Reactions Integrated rate equation Where [A]0 and [A]t are concentrations of reactants at time t = 0 and at a later time, t This ratio is the fraction of reactant that remains after a given time has elapsed.

First Order Reactions Negative sign (-) means ratio is less than 1 because [A]t is always less than [A]0. Logarithm of ratio is negative therefore other side of equation must be negative

First Order Reactions Equation is useful in three ways: If [A]t/[A]0 is measured after some amount of time has elapsed then k can be calculated 2. If [A]0 and k are known than [A]t­ can be calculated (amount remaining after a given amount of time 3. If k is known, then time elapsed until a specific fraction ([A]t/[A]0) remains can be calculated

First Order Reactions Note k for 1st order reactions is independent of concentration; k has units of time-1 Therefore any unit for [A]t and [A]0 can be chosen (M, mol, g, atoms, molecules, pressure)

First Order Reactions PROBLEM When heated cyclopropane rearranges to propene in a first order process Rate = k[cyclopropane] k = 5.4 x 10-2 h-1 If the initial concentration of cyclopropane is 0.050 mol/L, how much time (in hours) must elapse for its concentration to drop to 0.010 mol/L

First Order Reactions SOLUTION

Second Order Reactions If reaction ‘A  products’ is second order, the reaction rate equation is A little calculus again…

Second Order Reactions Integrated rate equation In this case k is the second-order rate constant (L/mol ∙ time)

Second Order Reactions PROBLEM The gas-phase decomposition of HI HI(g)  ½ H2(g) + ½ I2(g) has the rate equation Where k = 30. L/mol ∙ min at 443C. How much time does it take for the concentration of HI to drop from 0.010 mol/L to 0.0050 mol/L at 443C?

Second Order Reactions SOLUTION

Zero Order Reactions If reaction ‘A  products’ is zero order, the reaction rate equation is A little calculus again…

Zero Order Reactions Integrated rate equation Where k has the units mol/L ∙ s

Graphical Methods for Determining Reaction Order and the Rate Constant First Order [A]t = - kt + [A]0    y mx b

Graphical Methods for Determining Reaction Order and the Rate Constant Second Order 1/[A]t = + kt + 1/[A]0    y mx b

Half-Life and First Order Reactions Half-life t1/2 Time required for the concentration of a reaction to decrease to one-half its initial value; reactant R remaining is ½ Used primarily when dealing with 1st order reactions Indicates the rate at which a reactant is consumed – is reaction fast or slow? Longer the half-life – slower the reaction

Half-Life and First Order Reactions Where: [R]0 = initial concentration [R]t = concentration after the reaction is half complete

Half-Life and First Order Reactions Let’s evaluate t1/2 for 1st order reaction Substitute [R]t/[R]0 = ½ and t = t1/2 in integrated rate equation

Half-Life and First Order Reactions Rearrange equation (ln 2 = 0.693) Equations relates half-life to 1st order rate constant t1/2 is independent of concentration

Half-Life, Zero Order & Second Order Reactions Zero order reaction Second order reaction

Half-Life PROBLEM Sucrose, C12H22O11, decomposes to fructose and glucose in acid solution with the rate law Rate =k[sucrose] k = 0.208 h-1 at 25C What amount of time is required for 87.5% of the initial concentration of sucrose to decompose?

Half-Life SOLUTION After 87.5% of sucrose has decomposed, 12.5% remains (fraction remaining = 0.125). This will require 2 half-lives to reach this point. Time elapsed = 3 x 3.33 h = 9.99 h

NH2NO2(aq)  N2O(g) + H2O(l) Half-Life YOUR TURN The following reaction is first order with respect to [NH2NO2]. The rate constant, k, is 9.3 x 10-5 s-1. What is the half-life of this reaction? NH2NO2(aq)  N2O(g) + H2O(l) The following reaction at 400K is second order with respect to [CF3] and the value of the rate constant, k, is 2.51 x 1010 M-1s-1. If the initial [CF3] = 2.0M, what is the half-life of the reaction? 2 CF3(g)  C2F6(g)

Collision Theory of Reaction Rates Theory states that for any reaction to occur 3 conditions must be met The reacting molecules must collide with one another The reacting molecules must collide with sufficient energy to break bonds The molecules must collide in an orientation that can lead to rearrangement of the atoms

Collision Theory of Reaction Rates So, molecules must collide with one another. The rate of their reaction is primarily related to the number of collisions, which is in turn related to their concentration The number of collisions between the two reactant molecules is directly proportional to the concentration of each reactant, and the rate of the reaction shows a first order dependence on each reactant

Collision Theory of Reaction Rates

Temperature, Reaction Rate, & Activation Energy How and why does temperature affect reaction rates? In any sample of gas or liquid some molecules have very low energies, others have high energies, most have intermediate energies. As temperature increases average energy of the molecules in a sample increases as does the fraction having higher energies

Activation Energy Molecules require some minimum energy to react; ‘energy barrier’ Energy required to surmount the barrier is called the activation energy, Ea Barrier low, energy requirement (Ea) is low, a high proportion of molecules in a sample have sufficient energy to react – reaction is fast Barrier high, energy requirement (Ea) is high and only a few reactant molecules in a sample have sufficient energy to react – reaction is slow

Activation Energy Reaction Coordinate Diagram Consider HI  H2 + I2 Reaction can not occur without input of energy Energy reaches maximum at transition state; sufficient energy has been concentrated in bonds, bonds can break, reaction can move forward.

Activation Energy

Effects of Temperature Increase Raising the temperature always increases the reaction rate by increasing the fraction of molecules with enough energy to surmount the activation energy barrier

Effects of Molecular Orientation on Reaction Rate The lower the probability of achieving alignment, the smaller the value of k, the slower the reaction

The Arrhenius Equation Reaction rates depend on energy, frequency of collision between reactants, temperature, and correct geometry. Arrhenius equation summarizes these factors: R = gas constant 8.314 x 10-3 kJ/K∙mol A = frequency factor L/mol∙s (related to number of collision and fraction of collisions w/correct geometry. A is specific to each reaction and temperature dependent Fraction of molecules having the minimum energy required for reaction. Value is always <1

The Arrhenius Equation Use to calculate: Value of Ea Rate constant for a given temperature if Ea and A are known

The Arrhenius Equation A little rearrangement… y mx + b

The Arrhenius Equation PROBLEM Calculate Ea for the reaction 2N2O (a)  2N2 (g) + O2 (g) Exp T (K) K (L/mols) 1 1125 11.59 2 1053 1.67 3 1001 0.380 4 838 0.0011

The Arrhenius Equation Kinetics problems that deal with changing rates, or rate constants (k), and temperature changes require use of the Arrhenius equation

The Arrhenius Equation PROBLEM What is the activation energy of a reaction that has a rate constant of 2.50 x 102 kJ/mol at 325K and a rate constant of 5.00 x 102 kJ/mol at 375K?

Catalysis Catalyst increases the rate of a reaction without being consumed by it. Catalyst lowers the activation energy required thus speeding up the reaction

Heterogeneous & Homogeneous Catalysis Heterogeneous – catalyst is in a different phase from the reaction mixture; generally a solid that increases the rate of a gas phase or liquid phase reaction Au N2O(g)  N2(g) + ½ O2(g) N2O is chemically absorbed on the surface of the solid catalyst. A bond is formed between the N2O molecule and Au (covalent bonds) thus weakening the bond between nitrogen and oxygen making it easier for N2O to break apart.

Heterogeneous & Homogeneous Catalysis N  N – O(g) + Au(s)  N  N --- O --- Au(s)  N  N(g) + O2(g) + Au(s) Heterogeneous catalyst generally used in industrial processes: Preparation of ammonia sulfuric acid methanol Pt catalyst used to reduce automobile emissions

Heterogeneous & Homogeneous Catalysis Homogeneous – catalyst that is present in the same phase as the reactants. Speeds up reaction by forming a reactive intermediate that decomposes to give products thus providing an alternative path of lower activation energy for the reaction

Heterogeneous & Homogeneous Catalysis Clock reaction (decomposition of hydrogen peroxide) Uncatalyzed 2H2O2 (aq)  2H2O + O2 (g) Catalyzed Step 1:H2O2 (aq) + I- (aq)  H2O + IO- (aq) Step 2:H2O2 (aq) + IO- (aq)  H2O + O2 (g) + I- (aq) 2H2O2 (aq)  2H2O + O2(g)

Reaction Mechanisms Sequence of bond-making and bond-breaking steps during a reaction A ‘guess’ to help better understand the chemistry Nano level (atoms and molecules) Use rate equation to understand mechanisms

Reaction Mechanisms Most reactions occur in a sequence of steps Br2(g) + 2 NO(g)  2 BrNO(g) Written as a single step, this reaction would require the 3 reactants to collide simultaneously with the right velocity and orientation to react. Low probability of this happening

Reaction Mechanisms Using a sequence of steps involving only one or two molecules increases the probability of a collision (production of an intermediate) Step 1 Br2(g) + NO(g)  Br2NO(g) Step 2 Br2NO(g) + NO(g)  2 BrNO(g) Br2(g) + 2 NO(g)  2 BrNO(g)

Reaction Mechanisms Each step of mechanism is referred to as an elementary step Describes a single molecular event Each step has its own Ea and k Steps must add up to give overall balance chemical equation

Molecularity Classification of elementary steps based on # of reactant molecules colliding Unimolecular – one molecule in an elementary step O2(g)  O2(g) + O(g) Bimolecular – two molecules (same or different)  O2(g) + O(g)  2 O2(g)

Molecularity Termolecular – three molecules (same or different) O(g) + O2(g) + N2(g)  O3(g) + energetic N2(g)

Molecularity Elementary Step Molecularity Rate Equation A  product Unimolecular Rate = k[A] A + A  product Bimolecular Rate = k[A]2 A + B  product Rate = k[A][B] 2 A + B  product Termolecular Rate = k[A]2[B]

Molecularity The rate equation of an elementary step is defined by the reaction stoichiometry. The rate equation of an elementary step is given by the product of the rate constant and the concentrations of the reactants in that step.  The molecularity of an elementary step and its order are the same. This is not true for the overall reaction.

Rate Equations & Mechanisms Products of a reaction can never be produced at a rate faster than the rate of the slowest step Slowest step = Rate-determining step

Rate Equations & Mechanisms For the slowest step Rate = k1[A][B] Note: the rate law must be written with respect to the reactants only