Chemical Kinetics Chapter 14 AP Chemistry

Chemical Kinetics Kinetics – the area of chemistry concerned with the rate (or speed) of a reaction. Kinetics vs. Thermodynamics Applications: Medicine, Chemical Engineering

Reaction Rate Factors Physical state of reactants Concentration Surface area Concentration Rate increases with concentration increase Temperature Rate doubles every 10oC increase Catalyst Increase the reaction rate w/o being used up

Reaction Rates Speed is the change in a particular quantity with respect to a change in time. In chemistry, we define the reaction rate The change in concentration of the reactants or products over time Units are usually M/sec Rate =

C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) A graph of concentration vs. time is often plotted. Slope of the tangent line at any point along the curve is the instantaneous rate. Rate decreases with time. Reactants decrease with time

Average Rate Because the rate of reaction changes with time, it is useful to consider an average rate. Rate = The average rate for a reaction is usually take as the early stages of a reaction.

Measuring Rates To determine the progress of a reaction, we can measure two quantities: Disappearance of the reactant Formation of products Reaction rate is a positive value. Reaction rate is the same, no matter the method of measurement.

aA + bB  cC + dD The rate of reaction is given by the following equalities: A: Rate = B: Rate = C: Rate = D: Rate = Rate =

H2O2(g)  H2(g) + O2 (g) Write the rate in terms of each species.

SO2(g) + O2(g)  SO3(g) Write the rate in terms of each species.

H2(g) + O2(g)  H2O(g) Hydrogen is burning at the rate of 0.85 M/sec. Rate of oxygen consumption? Rate of water vapor formation?

How? How is it possible to measure the concentration of reactants or products? There are a variety of methods. One of the more common methods is spectroscopy. Measures the ability to absorb/transmit light and converts the data to a concentration.

Spectrophotometer (Spec-20)

Beer-Lambert Law There is a linear relationship between the concentration of a sample and its absorbance. A = -logT Beer’s Law: A = εbC Standards to find slope Convert T to A to C

NH3(g)  N2(g) + H2(g) Nitrogen is forming at the rate of 0.264 M/sec. Rate of ammonia consumption? Rate of hydrogen formation?

Rate Law General Equation: aA + bB  cC + dD Rate = m is the order of A n is the order of B (m+n) is the overall reaction order k is the rate constant specific for a rxn, Temperature dependent

Rate = k[A]m[B]n The rate law must be experimentally determined. m and n are NOT the stoichiometric coefficient Unit of rate constant k M-p s-1 or 1/(Mp s ) p = (m+n) – 1 Rate depends on reactant conc…k does not depend on reactant conc.

Rate = k[A][B] What happens to the rate if we… Double conc of A (everything else the same)? Double conc of B (everything else the same)? Triple conc of A and double conc of B? Order of A? B? Overall?

Rate = k[A]2[B] What happens to the rate if we… Double conc of A (everything else the same)? Double conc of B (everything else the same)? Triple conc of A and double conc of B? Order of A? B? Overall?

Rate = k[A]0[B]3 What happens to the rate if we… Double conc of A (everything else the same)? Double conc of B (everything else the same)? Triple conc of A and double conc of B? Order of A? B? Overall?

Rate = k[A]m[B]n Two ways to determine the rate law Initial rate method Can have many reactants Graphical Method Can only have one reactant Solving a rate Law: Need to determine the orders of the reactants Need to determine the rate constant k

2 NO(g) + 2 H2(g)  N3(g) + 2H2O(g) Determine the rate law: 2 NO(g) + 2 H2(g)  N3(g) + 2H2O(g) Experiment [NO] [H2] Initial Rate (M/s) 1 0.10 1.23 x 10-3 2 0.20 2.46 x 10-3 4.92 x 10-3

a A(g) + b Bg)  c C(g) + d D(g) Determine the rate law: a A(g) + b Bg)  c C(g) + d D(g) Experiment [A] [B] Initial Rate (M/s) 1 0.40 0.30 1.00 x 10-4 2 0.80 4.00 x 10-4 0.60 1.60 x 10-3

S2O82-(aq) + 3 I1-(aq)  2 SO42-(aq) + I31-(aq) Determine the rate law: S2O82-(aq) + 3 I1-(aq)  2 SO42-(aq) + I31-(aq) Experiment [NO] [H2] Initial Rate (M/s) 1 0.080 0.034 2.2 x 10-4 2 0.017 1.1 x 10-4 0.160

Change in Conc. with time So far, we have considered rate based on the change in concentration and rate constants. Using calculus, we can convert these same equations to more useful forms. This is the graphing method to determining the order of a reaction. Specific for only one reactant: [A]

Rate Laws Differential Rate Law: Integrated Rate Law Expressed how rate depends on concentration. Integrated Rate Law Integrated form of the differential. Has specific variables.

Zero Order Reaction Rate only depends on the rate constant…not on the concentration of A Differential: Integrated:

First Order Reaction Rate only depends on the rate constant and on the concentration of A Differential: Integrated:

First Order Plot of [A] vs. t Plot of ln[A] vs. t

Second Order Reaction Rate only depends on the rate constant and on the square concentration of A Differential: Integrated:

Second Order Plot of ln[A] vs. t Plot of 1/[A] vs. t

Usefulness of the Integrated Rate Laws [A]t = -kt + [A]0 We can know the concentration at any time point for a given reaction. We can determine the order of a reaction. We can determine the half life of a reaction.

Determining the Order [A]t = -kt + [A]0 y = mx + b This tells us that a plot of concentration of A vs time will yield a straight line. Because this is the zero order rate equation, a plot of [A] vs. t will yield a straight line. y = mx + b

Determining the Order First Order: ln[A]t = -kt + ln[A]0 Second Order: A plot of ln[A] vs. t will give a straight line. Second Order: A plot of 1/[A] vs. t will give a striaght line.

Half life, t1/2 The time required for the concentration of a reactant to reach one-half its value: [A]t1/2 = ½[A]0 This is a convenient way to describe the rate of a reaction. A fast reaction will have a short half life.

Derivation: t1/2 of First Order ln[A]t = -kt + ln[A]0

Summary ** Note: The half life of a first order says that it does NOT depend the concentration of the reactant A. So, the concentration decreases by ½ each regular time interval, t1/2.

A first order reaction has k = 6.7 x 10-4 s-1. How long will it take for the conc to go from 0.25M to 0.15M? If the initial conc is 0.25M, what is the conc after 8.8 min?

A first order reaction has [A] = 2. 00M initially A first order reaction has [A] = 2.00M initially. After 126 min, [A] = 0.0250M. What is the rate constant k? What is the half life?

A second order reaction has k = 7.0 x 10-9 M-1s-1. If the initial conc is 0.086M, what is the conc after 2.0 min?

A first order reaction has t1/2 = 35.0 sec. What is the rate constant k? How long would it take for 95% decomposition of the reactant?

A first order reaction has a half life of 19. 8 min A first order reaction has a half life of 19.8 min. What is the reaction rate when [A] = 0.750M?

Temperature

Collision Model Based on Kinetic Molecular Theory Molecules must collide to react Greater the collisions, greater the rate As concentration increases, rate increases As temperature increases, rate increases

Orientation Most collisions do not lead to reactions Molecular orientation of collision is important

Still not enough Usually, a collision in the correct orientation is still not enough to cause a reaction. Kinetic energy of a collision must cause bonds to break. For a reaction to occur, there must be enough kinetic energy to be greater than some energy. Activation Energy, Ea, is the minimum energy required to initiate a reaction.

Activation Energy

Transition State Transition state is also called the activated complex High energy intermediate state The activation energy represents the higher energy state of the transition state A A + B B A A B B A B + A B ‡

Arrhenius Equation k = A e−Ea/RT This equation combines all factors contributing to the reaction rate. k = A e−Ea/RT k = rate constant Ea = activation energy R = Gas constant T = Temperature in Kelvin A = frequency factor “constant” probability of correctly oriented collisions

Other versions y = m * x + b Original Equation: k = A e−Ea/RT Linear Equation: If we know the Ea and k2 at T2, we can calculate k1 at T1: y = m * x + b

Catalysis Catalyst – speed up a reaction without being consumed Beneficial or harmful, ex. Body Homogenous catalyst – same phase as reactant Phase transfer catalyst Heterogeneous catalyst – different phase than reactant Saturation of alkenes and alkynes

Catalyst Lowers the activation energy Potential energy diagram is 3-D

Enzymes Biological Catalyst Reactants called “substrate” S + E  SE  P + E Active site – binding location SE = enzyme-substrate complex Lock and Key Model vs. Induced Fit Model

Reaction Mechanism Balanced rxn tells us the species before a rxn starts and after the rxn ends. Does NOT show how the reaction occurs. Reaction Mechanism – the steps a reaction progresses. The steps can be of varying speeds

Elementary Reactions Reactions occur because of collisions Must be correctly oriented Must have enough energy to reach TS A + B  C + D This is a single collision reaction Elementary reactions – rxn occurring in a single step

Molecularity Number of molecules that participate in a single reaction. Unimolecular, bimolecular, termolecular Probability decreases with molecularity A  P 2A  P A + B  P A + 2B  P

Multiple Steps Some reactions occur in multiple steps Multistep Mechanism – a reaction consisting of a series of elementary reactions Must add to give the overall reaction. Items that are produced within a mechanism and consumed within a mechanism are intermediates. Intermediates are not R or P.

NO2(g) + CO(g)  NO(g) + CO2(g)

Rate Laws Earlier we stated that rate laws can not be determined from a chemical equation. Why? There is a possibility for multistep mechanism We must consider the speed of each step However, we can derive the rate law from a the mechanism of a reaction.

It’s Elementary, my dear chemists! If a rxn is an elementary rxn, the rate law is based off the molecularity (coefficients) A  P Rate = 2 A  P Rate = A + B  P Rate = 3 A  P Rate = A + 2 B  P Rate = A + B + C  P Rate =

Rate-Determining Step (RDS) Let’s go shopping. Mechanisms can have a slow step. RDS = slowest step – governs rate law If first step is slow, intermediates short lived If later step is slow, there is a buildup if intermediates. Each step has its own rate constant

Slow Initial Step NO2(g) + CO(g)  NO(g) + CO2(g) Experimentally Determined Rate Law: Rate = k[NO2]2[CO]0

Fast Initial Step The slow step rate law still governs the rate law as before. However, because the slow step is the second step, there are intermediates in the rate law. Intermediates are short lived and can not be measured easily – not good in rate law

Fast Initial Step Make assumptions: there is a dynamic equilibrium established in the fast step The intermediate is more likely to decompose (reverse of fast step is also fast = k-1) than be consumed in step 2 (k2) The forward rxn rate in step one (k1) equals the rate of the reverse reaction: Rate forward = Rate Reverse

2 NO(g) + Br2(g0  2 NOBr(g) Experimental Rate Law: Rate = k[NO]2[Br2]