Reaction Rates

--insert obligatory “slow children” joke

Definition

• A reaction rate is the rate at which a reaction takes place.

Definition

• A reaction rate is the rate at which a reaction takes place.

• Helpful, huh?

• Usually, a rate is expressed in moles/sec, or more commonly, M/s (mol/L/s, or mol L-1 s-1)

• Q: Moles of what?

• A: Anything convenient.

Rates decrease with time

0

0.1

0.2

0.3

0.4

0.5

0.6

1 3 5 7 9 11 13 15 17 19 21 23 25

But over short times, you can measure an initial rate.

0

0.5

1

1.5

2

2.5

1 3 5 7 9 11 13 15 17 19 21 23 25

Reactant 1 is used

Reactant 2 is used faster

Product is produced

• For the general reactionaA + bBcC + dD

--at a constant volume• Rate = -A] = - [B] = [C] = [D]

at bt ct dt

(notice that the reactants are disappearing, while products are being formed)

• For the general reactionaA + bBcC + dD

--at a constant volume• Rate = -A] = - [B] = [C] = [D]

at bt ct dt

(notice, also, that the rate is independent of which substance is used to measure it.)

N2 + 3H22NH3

• If, under some conditions, 2.40 moles of NH3 are produced each second, what is the rate of use of N2 and H2?

N2 + 3H22NH3

• If, under some conditions, 2.40 moles of NH3 are produced each second, what is the rate of use of N2 and H2?

• 2.40 mol NH3/s x -1 N2/2 NH3

=-1.20 mol N2/s

N2 + 3H22NH3

• If, under some conditions, 2.40 moles of NH3 are produced each second, what is the rate of use of N2 and H2?

• 2.40 mol NH3/s x -1 N2/2 NH3

=-1.20 mol N2/s

• 2.40 mol NH3/s x -3 H2/2 NH3

=-3.60 mol H2/s

N2 + 3H22NH3

• If, under some conditions, 2.40 moles of NH3 are produced each second, what is the rate of use of N2 and H2?

• You might say that the rate of the reaction is 1.20 mole/s as written, but we are usually interested in the rate of change of a substance

How can you speed up a reaction?

How can you speed up a reaction?

• --Heat it up.

• --Crush, grind or powder a solid reactant.

• --Increase pressure of a gaseous reactant

• --Increase concentrations of aqueous reactants

• --Add a catalyst (if known)

• (Stir or shake to bring reactants together.)

• Now, for convenience, measure the easiest rate: Changes in…– Color -- pressure

– Conductivity --Mass

– Gas volume --pH

Whatever changes can be measured

How would you speed up…

• Hydrochloric acid acts on tin metal to form hydrogen gas and aqueous tin (II) chloride

How would you speed up…

• Hydrochloric acid acts on tin metal to form hydrogen gas and aqueous tin (II) chloride

• Increase concentration of HCl

• Powder the tin

• Heat the reactants

How would you measure the rate of…

• Hydrochloric acid acts on tin metal to form hydrogen gas and aqueous tin (II) chloride

How would you measure the rate of…

• Hydrochloric acid acts on tin metal to form hydrogen gas and aqueous tin (II) chloride

• Rate of change in H2 volume

– loss of mass of tin, appearance of tin (II), increase in pH, and loss of conductivity are all harder

What happens to the rate after (mumble, mumble) seconds?

• Hydrochloric acid acts on tin metal to form hydrogen gas and aqueous tin (II) chloride

What happens to the rate after (mumble, mumble) seconds?

• Hydrochloric acid acts on tin metal to form hydrogen gas and aqueous tin (II) chloride

• It slows down.

• Why?

What happens to the rate after (mumble, mumble) seconds?

• Hydrochloric acid acts on tin metal to form hydrogen gas and aqueous tin (II) chloride

• It slows down.

• Why?

• --[H+] decreases as reaction proceeds

Rates decrease with time

0

0.1

0.2

0.3

0.4

0.5

0.6

1 3 5 7 9 11 13 15 17 19 21 23 25

• We measure the initial rates of reactions because

—there in the first instant—

that is the only time when we know all of the concentrations!

Reaction rates are affected by

• -Concentrations

• -Temperature

• -etc.

Production of product

0

0.5

1

1.5

2

2.5

3

3.5

1 3 5 7 9 11 13 15 17 19 21 23 25

Product appearance in a reaction

Product appearance in a reaction at a higher temperature

Product appearance in a reaction at a lower concentration

• So– we choose a temperature, control that temperature and

vary concentrations

to discover the effect that each concentration has on the rate.

Rate Laws

• Rate= k [A]x[B]y

• x and y are not related to the balancing of the reaction! (They are experimentally determined.)

You will be asked to…

• Write rate laws from empirical data.– Multiple trials show the exponents

• Calculate rate constants from the laws you write.– Watch your units! Units vary for different laws

• Calculate a rate from a known rate law, constant, and concentrations.– Use M/s or mol/s (mol L-1s-1 or mol s-1 )

The rate of decay of 3H (tritium)

• [3H] (M) Rate of Decay (nM/sec)

• .01 .018

• .02 .036

• .03 .054

• .04 .072

The rate is directly related to [3H]

0

0.02

0.04

0.06

0.08

0.1

0.12

0 0.02 0.04 0.06 0.08

• This is a first order reaction.

• The rate law is Rate=k[3H]

where k=1.78 x 10-9 /s

• A first order reaction has a rate of .024 M/s when [A] = .20 M

– What is the value of the rate constant, k?

• A first order reaction has a rate of .024 M/s when [A] = .20 M

– What is the value of the rate constant, k?– What are the units on the rate constant, k?

• A first order reaction has a rate of .024 M/s when [A] = .20 M

– What is the value of the rate constant, k?– What are the units on the rate constant, k?– What is the rate of the reaction when

[A]=.30M?

• A first order reaction has k=1.6 x 10-4 /s

– What is the rate when [A]= .050M?

• A first order reaction has k=1.6 x 10-4 /s

– What is the rate when [A]= .050M?– At what [A] will the rate be 5.4 x 10-6 M/s?

For example, the rate of IO3-

reduction by HSO3-

• [IO3-](mM) Rate of Reaction (M/s)

• .3 .015

• .6 .060

• .9 .135

• 1.2 .240

• When you double the concentration—the rate quadruples

• When you triple the original concentration, the rate increases by a factor of 9

4=22

9=32

This is a second order reaction

Rate = k [IO3-]2

where k=167/Ms

Concentrations of reactants fall

• In a first order reaction, the concentration of reactant decreases with time

• The decrease is faster with a larger k

0

0.2

0.4

0.6

0.8

1

1.2

1 3 5 7 9 11 13 15 17 19 21 23 25 27

Decrease of reactant

Decrease of reactant when k is larger

For a first order reaction…

• The amount of reactant left, [A], is

[A]= [Ao]e-kt —where

[Ao] is the original conc. (at time 0)

e is the base of the natural logs

k is the rate constant and

t is time

For a first order reaction…

• This equation is usually used as

ln[A]- ln[Ao]= -kt —where

[Ao] is the original conc. (at time 0)

e is the base of the natural logs

k is the rate constant and

t is time

• A first order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.

• A first order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.– What is k?

• A first order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.– What is k?– What is [A] after 120s?

• A first order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.– What is k?– What is [A] after 120s?– At what time will [A]=.020M?

• A first order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.– What is k?– What is [A] after 120s?– At what time will [A]=.020M?– What is the half life of this reaction?

For a second order reaction:

• The relationship between concentrations, k and t is

1 - 1 = kt

[A]t[A]o

• A second order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.

• A second order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.– What is k?

• A second order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.– What is k?– What is [A] after 120s?

• A second order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.– What is k?– What is [A] after 120s?– At what time will [A]=.020M?

• A second order reaction takes [A] from .080M to .055M in 85 s.– Write the rate law.– What is k?– What is [A] after 120s?– At what time will [A]=.020M?– What is the half life of this reaction

• For a first order reaction, the half life is constant.

• For a second order reaction, it increases

For a second order reaction:

• The equation

1 - 1 = kt

[A]t[A]o

Implies that the half life, t1/2, is the time where--

For a second order reaction:

• The equation

1 = kt1/2

[A]o

Or better yet,

t1/2 = 1

k [A]o

• For a second order reaction, if the initial half-life is 15 s when [A] = .90 M,– What is k?

• For a second order reaction, if the initial half-life is 15 s when [A] = .90 M,– What is k?– What are the units on k?

• For a second order reaction, if the initial half-life is 15 s when [A] = .90 M,– What is k?– What are the units on k?– Write the rate law.

• For a second order reaction, if the initial half-life is 15 s when [A] = .90 M,– What is k?– What are the units on k?– Write the rate law.– At what time will [A] = .30 M?

• For a second order reaction, if the initial half-life is 15 s when [A] = .90 M,– What is k?– What are the units on k?– Write the rate law.– At what time will [A] = .30 M?– What is the half-life at that time?

What is the rate law and constant?

• [A](M) [B](M) Rate (M/s)

• .10 .10 4.0

• .20 .10 16

• .10 .20 8.0

• .20 .20 ?

What is the rate law and constant?

• [A](M) [B](M) Rate (M/s)

• .10 .10 4.0 x 10-6

• .20 .20 3.2x 10-5

• .10 .30 1.2 x 10-5

• .30 .30 ?

What is the rate law and constant?

• [A](M) [B](M) [C](M) Rate (M/s)

• .10 .10 .10 3 x 10-5

• .20 .10 .10 6 x 10-5

• .10 .20 .10 6 x 10-5

• .10 .10 .20 1.2 x 10-4

• .30 .30 .30 ?

What is the rate law and constant?

• [A](M) [B](M) [C](M) Rate (M/s)

• .10 .10 .10 7.5 x 10-3

• .20 .20 .10 1.5 x 10-2

• .20 .10 .20 1.5 x 10-2

• .10 .30 .30 6.8 x 10-2

• .30 .30 .30 ?

…but wait! There’s More!

• The rate constant, k, is constant at a given temperature.

• What happens at a higher temperature?

• That’s right!

• k increases with temperature…

…but not all reactions are created equal

Think it through

-k indicates how fast a reaction occurs at given concentrations.

-it is related to the likelihood that any reactant will collide with another to react.

…and at a higher temperature…

…and at a higher temperature…

• 1) Particles move faster.

• 2) There are more collisions.

• 3) Those collisions have more energy.

To react, reactants must collide with enough energy, the activation

energy.

Cool, medium, warm

Notice:

1) The bell-shaped distributions

2) The average speed is higher at higher To

3) The speeds spread out more at higher To

What is “fast enough”?

More collisions will have enough energy to react at higher temperatures

What if this is “fast enough”?

What if this is “fast enough”?

When the activation energy is higher…

…the temperature has a far more dramatic effect on the rate constant.

The Arrhenius Equation

ln k = (-Ea/R) (1/T) + ln A

…where k is the rate constant

Eais the activation energy

R is the ideal gas constant

T is the absolute temp.

ln A is a constant that indicates the probability of an effective collision

The Arrhenius Equation

ln k

1/T

Slope is –Ea/R

Catalysis

• A catalyst speeds up a reaction

• This is done by lowering the energy barrier, Ea

• When the barrier is lower, more collisions are “fast enough”

• A catalyst does not…

• …increase the number of collisions

• …change the equilibrium positions—you just get there faster.

Reaction Mechanisms

• Most (?) reactions are not single step reactions.

• Consider: C2H5OH + 5O2 2CO2 + 3H2O

• Don’t try to tell me that 6 molecules collide in the right orientation all at the same point at the same time.

For example…

• 2N2O2N2 + O2

• Proceeds as:

N2ON2 + O

N2O +ON2 + O2

2N2O + O 2N2 + O2 +O

2N2O2N2 + O2

In other words...(or, “In words”)…

• One dinitrogen monoxide molecule decomposes into a nitrogen molecule and an oxygen atom.

• Then, this (highly reactive) oxygen atom reacts with another dinitrogen monoxide molecule to form another nitrogen molecule and one molecule of oxygen.

While…

2H2O2 2H2O + O2

• Proceeds as:

H2O2 +I- H2O + IO-

H2O2 + IO- H2O + O2 + I-

2H2O2 +I- + IO- 2H2O + O2 +I- + IO-

2H2O2 2H2O + O2

I-

Please notice:

(Reaction 1)

• The oxygen atom is created, then used.

This is an intermediate

(Reaction 2)

• The iodide ion is used, then re-created.

This is a catalyst

The intermediate

• 2N2O2N2 + O2

• Proceeds as:

N2ON2 + O

N2O +ON2 + O2

2N2O + O 2N2 + O2 +O

2N2O2N2 + O2

The catalyst

2H2O2 2H2O + O2

• Proceeds as:

H2O2 +I- H2O + IO-

H2O2 + IO- H2O + O2 + I-

2H2O2 +I- + IO- 2H2O + O2 +I- + IO-

2H2O2 2H2O + O2

I-

The other intermediate

2H2O2 2H2O + O2

• Proceeds as:

H2O2 +I- H2O + IO-

H2O2 + IO- H2O + O2 + I-

2H2O2 +I- + IO- 2H2O + O2 +I- + IO-

2H2O2 2H2O + O2

I-

In each reaction…

• There are two steps.

• The first step (coincidentally) is the slow step, the second is faster.

• The rate law reflects the stoichiometry of the slow step (and any previous steps)

• The slowest step is the one with the highest activation energy.

• It is called the rate-determining step (or rate-limiting step)

Rate- determining step

Energy

Reaction progress

Ea2Ea1

Reactants

Products

• Rate = k [N2O]

(not: Rate = k [N2O]2, as if the second step were slower)

• Rate = k [H2O2][I-]

(not: k [H2O2]2[I-], as if the second were slower)

Your turn

• What rate law would you expect for

NO2 + CO NO + CO2

(which proceeds as)

NO2 + NO2 NO + NO3

NO3 + CO NO2 + CO2

a) If the first step is slower.

b) If the second step is slower?

A catalyst…

• …can be saturated. When all of the catalyst is being used, concentrating other reactants has no effect.

• …can be poisoned. Many depend on surface effects. A contaminated surface won’t work

• …is necessary in biological reactions. An enzyme is a biological catalyst.

• …can save your profit margin (read as: “butt”) in industrial chemistry.

And finally…

• A catalyst affects the rate of the slowest step

• By… weakening bonds?• …making it more likely collisions will be in the

correct orientation? • …stabilizing intermediates? • …preventing side reactions?• ….?