Chemical Kinetics

Chapter 14AP 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– 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:1. Disappearance of the reactant2. 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…

a) Double conc of A (everything else the same)?

b) Double conc of B (everything else the same)?

c) Triple conc of A and double conc of B?

d) Order of A? B? Overall?

Rate = k[A]2[B]• What happens to the rate if we…

a) Double conc of A (everything else the same)?

b) Double conc of B (everything else the same)?

c) Triple conc of A and double conc of B?

d) Order of A? B? Overall?

Rate = k[A]0[B]3

• What happens to the rate if we…

a) Double conc of A (everything else the same)?

b) Double conc of B (everything else the same)?

c) Triple conc of A and double conc of B?

d) Order of A? B? Overall?

Rate = k[A]m[B]n

• Two ways to determine the rate law1. Initial rate method– Can have many reactants

2. 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

• Determinethe rate law:

2 NO(g) + 2 H2(g) N3(g) + 2H2O(g) Experiment [NO] [H2] Initial Rate (M/s)

1 0.10 0.10 1.23 x 10-3

2 0.10 0.20 2.46 x 10-3

1 0.20 0.10 4.92 x 10-3

• Determinethe 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 0.30 4.00 x 10-4

1 0.80 0.60 1.60 x 10-3

• Determinethe rate law:

S2O82-

(aq) + 3 I1-(aq) 2 SO4

2-(aq) + I3

1-(aq)

Experiment [NO] [H2] Initial Rate (M/s) 1 0.080 0.034 2.2 x 10-4

2 0.080 0.017 1.1 x 10-4

1 0.160 0.017 2.2 x 10-4

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:

• Expressed how rate depends on concentration.

• Integrated Rate Law

• Integrated form of the differential. Has specific variables.

𝑅𝑎𝑡𝑒= −∆ሾ𝐴ሿ∆𝑡 = 𝑘[𝐴]

ln[𝐴]𝑡 = −𝑘𝑡+ ln[𝐴]0

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

• 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

– 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

SummaryOrder Rate Law Integrated Rate Law Half Life

Zero Rate = k [A]t = -kt + [A]0 t1/2 = [𝐴]02𝑘

First Rate = k[A] ln[A]t = -kt + ln[A]0 t1/2 = 0.693𝑘

Second Rate = k[A]2 1[𝐴]𝑡 = 𝑘𝑡+ 1[𝐴]0 t1/2 = 1𝑘[𝐴]0

** 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. 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. 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 BA A

B BA B + A B

‡

Arrhenius Equation• 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

• 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 P2A PA + B PA + 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]