Reaction Kinetics

Reaction RatesThe rate of a reaction is usually expressed in terms of a change in concentration of one of the participants per unit time.

rate = [ ]            t

It is the speed at which reactants are converted into products

A rate must be specified with a specific time unit.

A + B ---> AB

The Collision Theory Molecules must collide in order to react If 2 or more molecules collide but are not

orientated correctly then no reaction will take place.

Most collisions are 2 particle collisions (1 particle and wall or 2 particles)

3 particles collisions are rare since they occur only when 3 particles are at precisely the same location @ precisely the same time.

No 4 body collisions are known

Not all collisions are effective. What determines if they are?

Energy of colliding particles (sufficient energy required to break bonds is needed) Orientation of colliding

particles

Factors That Affect Reaction Rates 1. Concentration

↑ molecules = ↑ chance of collisions = ↑ chance of reactions

Therefore, the rate of a chem. rxn. is increased as the [reactants] is increased. (visa versa)

2. Particle SizeThe smaller the particle size, the larger the surface area available to collide with, the greater the chance of reactions. (visa versa)This only applies to heterogenous rxn., which are rxn. w/ mixed phases.

eg. Zn(s) + 2 HCl(aq) ----> ZnCl2(aq) + H2(g)

3. TemperatureAn ↑ in the kinetic energy leads to an ↑ in molecular collisions and a ↑ chance of rxn.(visa versa) chem. rxn. occur only when collisions

have enough energy to cause a rearrangement of atoms high energy molecular collisions cause

“molecular damage” called chemical reactions

4. Catalysts chemicals that speed up the rate of a chem. rxn. w/o becoming a part of the product it provides a low-energy pathway from the reactants to products do NOT get used up in the rxn. inhibitors reduce rxn. rates by preventing the rxn. from occurring. The inhibitor may combine w/ a reactant or a catalyst to form a complex that is stable at a temp. so that the rxn. does not occur

5. Nature of Ion Being UsedThe nature of the reactants affects the rate of a rxn. Generally, a more complicated species will react more slowly than a simple ion.

Reaction Mechanisms 2 NO(g) + F2(g) → 2 NOF(g)

balanced eqn. indicates the substance present at the beginning and end of a chem. rxn.

it does not indicate the actual details of the rxn.

chem. rxn. do not necessarily occur in one step, they are usually the result of several steps made up of simple chem. rxn.

a series of steps that leads from reactants to products is called a reaction mechanism

A reaction mechanism… describes the order in which bonds

break & atoms rearrange throughout the course of a chem. rxn. (some are simple, some complex)

each step individually is called an “elementary step”

Eg. 2 NO(g) + F2(g) → 2 NOF(g)

Elementary Step 1:NO(g) + F2(g) → NOF2(g)

Elementary Step 2: NOF2(g) + NO(g) → 2 NOF(g)

Adding the elementary steps always gives the net balanced eqn.

2 NO(g) + F2(g) → 2 NOF(g)

** Notice that NOF2(g) does not appear as either a reactant or product in the net eqn. This is b/c it is produced in step 1 and consumed in step 2.

NOF2(g) “Intermediate product”

Multi-step rxns always involve 1 or more intermediate products.

Each elementary step occurs @ its own rate. One step might be fast, another slow. The rate of the overall rxn. is limited by the rate of the slowest step AKA “the rate-determining step”.

Energy in Chem. Rxn.Recall 2 major types of energy:

Kinetic Energy (KE)- energy of motioni.e. athlete running, soccer ball flying

Potential Energy (PE)- stored energyi.e. gas in car, food you eat,

chandelier hanging from ceiling

The total amount of E in a system is conserved (i.e. no E is lost, rather one form can be converted to another.)

E required to break bond comes from KE

During a collision KE is converted to PE as the reactant particles are damaged & bonds are rearranged(KE depends on mass & velocity)

Svante Arrhenius (1888) suggested that particles must possess a certain minimum amt. of KE in order to react

Activation Energy (Ea)This is the energy that must be reached by 2 colliding molecules before a reaction can take place.

Why is there an Activation Energy Barrier?

During the course of a reaction considerable redistribution of electrons may occur.

Example: The reaction of CH3Br with Cl- in water at 298K.

As the bimolecular reaction occurs… (i) - there is angle bending

- initially pyramidal CH3 grouping become planar(ii) - there is bond-making and breaking

- a partial CBr bond is weakened - energy released by the formation of the partial CCl bond won’t fully compensate for the other 2 (endothermic) changes and yet there is no lower energy pathway from reactants to products - the reactants can get to the point of highest potential energy (the "activated complex" or "transition state" - in curly braces above) only if they initially have sufficient kinetic energy to turn into the potential energy of the activated complex - the activated complex can not be isolated; it is that arrangement of reactants which can proceed to products without further input of energy.

Typical Potential Energy Diagramsreaction path

Ereactants

Eproducts

(E released)

When particles collide w/ sufficient E (@ least equal to Ea) existing bonds are broken & new bonds form

*

*

Activated Complex A very short lived molecular complex in

which bonds are in the process of being broken and formed (neither reactant or product)

Occurs @ peak of Ea in the “transition state” This complex can go forward products

or return reactants Ea = energy required to form activated

complex

*activated complex

*activated complex

Each elementary step of a rxn. mechanism requires its own Ea in order to proceed. It will also produce its own activated complex, & may produce intermediate products.

NOTE: activated complex intermediate product

Putting Rxn. Rate & Rxn. Mechanisms Together… reactant particles must not only collide w/ proper

orientation, but also w/ enough E to overcome the Ea barrier to become products

only a small fraction of reactant particles have enough Ea @ any given time, thus they can’t all get over the barrier @ once

as rxn. proceeds more & more reactant particles gain Ea as they move & collide

Rxn. rate = rate @ which reactant particles cross energy barrier

Ea = # of collisions w/ enough E = rxn. rate

Example:

4HBr + O2 → 2H2O + 2Br2

Effect of Catalysts on the Activation Energy

Catalysts provide… a new rxn. pathway from reactants

to products w/ a lower Ea an alternative rxn. mechanism NOTE: ∆H is the same for

catalyzed rxn. and uncatalyzed rxn.

Catalyst lowers Ea, thus less E is required from the molecules colliding to get over the barrier. Thus, rxn. rate ↑.

Reactants

Products

∆H =

There are many examples of catalysts in our everyday life. Here are a few examples:

1. A catalytic converter is a catalyst - exhaust contains CO & NO, NO2, etc.

catalyst

exhaust w/ 2CO + O2 → 2CO2

rhodium

platinum

2. Enzymes- protein catalysts that control many

human physiological rxn.E.g. Lactase is a digestive enzyme that speed up the breakdown of lactose (milk sugar)

3. The Breakdown of Ozone by Freon

Kinetic Energy Diagramlower temp. higher temp.

More particles @ lower KE (↓ temp.)

Increasing temp. shifts graph (rxn. pathway) to the right

↑ temp results in…

- ↑ KE of particles (thus molecules move faster)-↑ # of particles w/ sufficient E

- ↑ # of collisions w/ sufficient E to be effective (i.e. cross Ea barrier)

What would a catalyst do to the KE diagram?

before

w/ catalyst

Catalysts shifts Ea line to the left (i.e. Ea barrier requires less KE to surmount, thus more molecules get through)

Reaction Rate LawDefinition: The rate, r, will always be

proportional to the product of the initial concentrations of the reactants.

For a general reaction … aX + bY ------->

… the rate law expression is: r = k[X]m[Y]n

[X] & [Y] = molar concentrations of X & Y m & n = powers to which the concentrations must

be raised (can only be determined empirically)

k = constant of proportionality known as the rate constant (determined empirically, only valid for a specific reaction @ at specific temp.)

data show that k is not affected by [] changes

k does vary with temp. changes values of m & n are NOT the

stoichiometric numbers obtained from the balanced equation (unless it is a one-step rxn.)

m and n may be zero, fractions or integers

sum of the exponents is called the reaction order

Example:H2(g) + I2(g) -----> 2 HI(g)

The experimentally determined rate law eqn. is:

r = k[H2][I2] This is a second order reaction b/c the sum

of the exponents is 2. the values of m & n happens to be the

same as the stoichiometric numbers in the balanced eqn.

this must be a one-step rxn. (i.e. there’s only one rxn. step needed to convert reactants into products)

Experimental Determination of Reaction Order

Example: Find the rate law and the rxn. order for the following rxn.

NO + H2 -----> HNO2 (could balance eqn., but the balanced eqn.

won’t necessarily give the exponential values).

before determining the value of k, we need to find the exponential value for each participant

to find the relationship of one reactant it is necessary to keep the other reactant(s) constant

Rates of reaction between NO and H2 at 800°C Initial Rate of

Exp. [NO] [H2 ] Rxn. Number mol/L mol/L mol/Lsec 1 0.001 0.004 0.002 2 0.002 0.004 0.008 3 0.003 0.004 0.018 4 0.004 0.001 0.008 5 0.004 0.002 0.016 6 0.004 0.003 0.024

From the equation we can write a partial rate law as rate = k[NO]m[H2]n

Using exp. 1 & 2 … [NO] jumps from 0.001 to 0.002 mol/L (i.e. doubled) the rate jumps from 0.002 to 0.008 mol/Lsec (i.e.

quadrupled) m for the [NO] is the mathematical relationship b/t

these two values i.e. 2m = 4, thus m = 2, b/c 22 = 4

Confirm this with exp. 1 & 3

The rate law expression can now be updated to: rate = k[NO]2[H2]n

Using exp. 4 & 5 … [H2] doubles and the rate doubles n for the [H2] is the mathematical

relationship b/t these two valuesi.e. 2n = 2, thus n = 1, b/c 21 = 2

Confirm this with exp. 4 & 6

The rate law expression can be rewritten as:

rate = k[NO]2[H2]1 Sum of the exponents indicates that this

is a third order reaction

Finding k Using the rate law, fill in the values from the data

table. 0.002 mol/L sec = k(0.001 mol/L)2 • (0.004 mol/L) 0.002 mol/L sec = k•(0.000001 mol2 /L2) • (0.004 mol/L) 0.002 mol/L sec = k • 0.000 000 009 mol3/L3

k = 0.002 mol/L sec . 0.000 000 004 mol3/L3 = 500,000 sec/mol2L2 or sec mol-2 L-2

The rate law expression can be rewritten as: rate = 5 x 105 sec mol-2 L-2 [NO mol/L]2[H2 mol/L]

Graphs can be used to determine the order of reaction with respect to a particular reactant

Zeroth-Order Reaction

First-Order Reaction

Second-Order Reaction