week 10.1 chemical kinetics

Prepared by:

Mrs Faraziehan Senusi

PA-A11-7C

Collision Model

Catalysis

Chapter 5

Chemical Kinetics

Reaction Rates

Reference: Chemistry: the Molecular Nature of Matter and Change,

6th ed, 2011, Martin S. Silberberg, McGraw-Hill

Rate Laws

Reaction mechanism

Kinetics

• Chemical kinetics is the study of reaction rates,

the changes in concentrations of reactants (or

products) as a function of time.

• Chemical kinetics is the study of rates of

chemical reactions, the factors that affect

reaction rates, and the mechanisms (the series

of steps) by which reactions occur.

FACTORS THAT INFLUENCE

REACTION RATE

1. Physical State of the Reactants

– In order to react, molecules must come in

contact with each other.

– The more homogeneous the mixture of

reactants, the faster the molecules can

react.

2. Concentration of Reactants

– A reaction can occur only when the

reactant molecules collide.

– As the concentration of reactants increases,

the more molecules present in the

container, the more frequently they

collide, and the more often a reaction

between them occurs.

– Thus, reaction rate is proportional to the

concentration of reactants:

Rate α collision frequency α concentration

3. Temperature

– At higher temperatures, reactant

molecules have more kinetic energy,

move faster, and collide more often and

with greater energy.

• Presence of a Catalyst

– Catalysts speed up reactions by changing

the mechanism of the reaction.

– Catalysts are not consumed during the

course of the reaction.

Reaction Rates

• A rate is a change in some variable per unit of time.

• Rates of reactions can be determined by monitoring

the change in concentration of either reactants or

products as a function of time.

• Consider a general reaction, A B .

• We quickly measure the starting reactant

concentration (conc A1) at t1, allow the reaction to

proceed, and then quickly measure the reactant

concentration again (conc A2) at t2.

• The change in concentration divided by the change

in time gives the average rate:

• We use the minus sign simply to convert the

negative change in reactant concentration to a

positive value for the rate.

• If instead we measure the product to determine

the reaction rate, we find its concentration

increasing over time.

• That is, conc B2 is always higher than conc B1.

• Thus, the change in product concentration, Δ[B],

is positive, and the reaction rate for A B

expressed in terms of B is

In this reaction, the

concentration of butyl

chloride, C4H9Cl was

measured at various times.

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

The average rate of the

reaction over each

interval is the change in

concentration divided

by the change in time:

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

t

ClHC

94 rate average

• Note that the average

rate decreases as the

reaction proceeds.

• This is because as the

reaction goes forward,

there are fewer

collisions between

reactant molecules.

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

• A plot of concentration vs.

time for this reaction

yields a curve like this.

• The slope of a line tangent

to the curve at any point is

the instantaneous rate at

that time.

• All reactions slow down

over time.

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

Reaction Rates and Stoichiometry

• In this reaction, the ratio of C4H9Cl to C4H9OH is

1:1.

• Thus, the rate of disappearance of C4H9Cl is the

same as the rate of appearance of C4H9OH.

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

t

OHHC

t

ClHC

9494 rate

• What if the ratio is not 1:1?

2 HI(g) H2(g) + I2(g)

• Therefore,

t

I

t

HI

2

2

1 rate

• To generalize, then, for the reaction

aA + bB cC + dD

t

D

dt

C

ct

B

bt

A

a

1111 rate

Concentration and Rate

One can gain information about the rate

of a reaction by seeing how the rate

changes with changes in concentration.

Comparing Experiments 1 and 2,

when [NH4+] doubles, the initial rate doubles.

NH4+

(aq) + NO2−

(aq) N2(g) + 2 H2O(l)

Likewise, comparing Experiments 5 and 6,

when [NO2−] doubles, the initial rate doubles.

NH4+

(aq) + NO2−

(aq) N2(g) + 2 H2O(l)

• This means

Rate [NH4+]

Rate [NO2−]

Rate [NH+] [NO2−]

or

Rate = k [NH4+] [NO2

−]

• This equation is called the rate law, and k is the

rate constant.

Rate Laws

• A rate law shows the relationship between the

reaction rate and the concentrations of reactants.

• For a general reaction,

• the rate law has the form

• The exponents m and n, called the reaction orders,

define how the rate is affected by reactant

concentration.

• The overall reaction order can be found by adding

the exponents on the reactants in the rate law.

aA + bB + … cC + dD + …

Rate = k [NH4+] [NO2

−]

• This reaction is

First-order in [NH4+]

First-order in [NO2−]

• This reaction is second-order overall.

NH4+

(aq) + NO2−

(aq) N2(g) + 2 H2O(l)

Examples:

• If the rate doubles when [A] doubles, the rate depends on

[A] raised to the first power, [A]1, so m = 1.

• Similarly, if the rate quadruples when [B] doubles, the

rate depends on [B] raised to the second power, [B]2, so n

= 2.

• In another reaction, the rate may not change at all when

[A] doubles; in that case, the rate does not depend on [A],

or, to put it another way, the rate depends on [A] raised to

the zero power, [A]0, so m = 0.

• Keep in mind that the coefficients a and b in the general

balanced equation are not necessarily related in any way

to these reaction orders m and n.

A key point to remember is that the components of the rate law:

rate, reaction orders, and rate constant must be found by experiment;

Determining Reaction Orders

Experimentally

Example:

Consider the reaction between oxygen and nitrogen monoxide, a key

step in the formation of acid rain and in the industrial production of

nitric acid:

O2(g) + 2NO(g) 2NO2(g)

The rate law, expressed in general form, is Rate = k[O2]m[NO]n

• Table shows experiments that change one reactant concentration

while keeping the other constant.

• If we compare experiments 1 and 2, we see the effect of doubling

[O2] on the rate.

• First, take the ratio of their rate laws:

• Because k is a constant and [NO] does not change between these

two experiments, these quantities cancel.

• Substituting the values from Table, we obtain

• Dividing, we obtain and rounding the figure:

• The reaction is first order in O2 .

• To find the order with respect to NO, we compare experiments 3

and 1 , in which [O2] is held constant and [NO] is doubled:

• As before, k is constant, and in this pair of experiments [O2]

does not change, so these quantities cancel:

=

• we obtain and rounding the figure

• The reaction is second order in NO.

• Thus, the rate law is

Determining the Rate Constant

• The rate constant is specific for a particular reaction at a particular

temperature.

• The experiments with the reaction of O2 and NO were run at the

same temperature, so we can use data from any to solve for k.

From experiment 1,

The average value of k for the five experiments in Table is 1.72 x 103 L2/mo12·s

• Units for the rate constant :

Integrated Rate Laws

• The equation that relates concentration and time

is the integrated rate equation.

• We can also use it to calculate the half-life, t1/2,

of a reactant-the time it takes for half of that

reactant to be converted into product.

• The integrated rate equation and the half-life are

different for reactions of different order.

First-Order Processes

a represents the coefficient of reactant A

in the balanced overall equation.

• Using calculus to integrate the rate law for a first-order

process gives us

ln [A]t

[A]0 = −akt

Where

[A]0 is the initial concentration of A.

[A]t is the concentration of A at some time, t,

during the course of the reaction.

• Manipulating this equation produces…

ln [A]t − ln [A]0 = − akt

ln [A]t = − ak t + ln [A]0 …which is in the form

y = m x + b

Therefore, if a reaction is first-order, a plot of ln[A] vs. t

will yield a straight line, and the slope of the line will be –ak.

• Using ideal gas equation,

PV=nRT

0

0

00

lnln

ln

ln/

/ln

][

][ln

][

PktP

ktP

P

P

P

RTP

RTP

A

A

RT

PA

V

n

o

A

ln [A]t = -akt + ln [A]0

Consider the process in which methyl

isonitrile is converted to acetonitrile.

CH3NC CH3CN

Example:

This data was collected for this

reaction at 198.9°C.

• When ln P is plotted as a function of time,

a straight line results.

• Therefore,

The process is first-order.

k is the negative slope, k = 5.1 10 −5 s-1.

Second-Order Processes

For a simple second order reaction,

aktAA

dtaAd

adt

Ad

a

dt

Ad

t

Arate

a

0

2

2

2

][

1

][

1

k [A]

][

k[A] ][

k[A] rate

law, rate From

][][

productA

The decomposition of NO2 at 300°C is described

by the equation

NO2 (g) NO (g) + 1/2 O2 (g)

and yields data comparable to this:

Time (s) [NO2], M

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

Example:

• Graphing ln [NO2] vs. t yields:

Time (s) [NO2], M ln [NO2]

0.0 0.01000 −4.610

50.0 0.00787 −4.845

100.0 0.00649 −5.038

200.0 0.00481 −5.337

300.0 0.00380 −5.573

• The plot is not a straight line,

so the process is not first-

order in [A].

• Graphing 1/[NO2] vs. t,

gives this plot.

Time (s) [NO2], M 1/[NO2]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• Because this is a straight

line, the process is

second-order in [A].

Example

A reaction, A B + C is carried out at a particular temperature.

As the reaction proceeds, we measure the molarity of the reactant,

[A] at various time, as tabulated below:

a) Determine the order of the reaction.

b) State the rate-law expression for the

reaction.

c) Calculate the value of the rate

constant, k at this temperature.

Solution:

a) Determine the order of the reaction.

From graphical plots, the reaction is deduced to be 1st order

since plots of ln[A] vs. t gives straight line (linear).

b) State the rate-law expression for the reaction.

Rate law expression: r = k[A]

c) Calculate the value of the rate constant, k at this temperature.

For 1st order reaction, slope, m = -ak

rate constant, k = - (m/a) = 0.29 min-1

29.05.15.8

)2.0()83.1(

Zero -Order Processes

ta

adt

Ad

a

dt

Ad

t

Arate

a

k [A][A]

k ][

k[A] rate

law, rate From

][][

productA

0

0

• Half-life is defined as

the time required for

one-half of a reactant to

react.

• Because [A] at t1/2 is

one-half of the original

[A],

[A]t = 0.5 [A]0.

Half-Life

For a first-order process, this becomes

0.5 [A]0

[A]0 ln = −kt1/2

ln 0.5 = −kt1/2

−0.693 = −kt1/2

= t1/2 0.693

k NOTE: For a first-order

process, the half-life

does not depend on [A]0.

First-Order Processes

ln [A]t

[A]0 = −kt

For a second-order process,

1

0.5 [A]0 = kt1/2 +

1

[A]0

2

[A]0 = kt1/2 +

1

[A]0

2 − 1

[A]0 = kt1/2

1

[A]0 =

= t1/2 1

k[A]0

Second-Order Processes

ktAA

0][

1

][

1

2/10

2/10

2/100

0

2k

[A]

k0.5[A]

k [A]0.5[A]

k [A][A]

t

t

t

t

Zero -Order Processes

For a zero-order process,

Example

The decomposition of ethane (C2H6) to methyl radicals is

a first order reaction with a rate constant of 5.36 x 10-4 s-1

at 700°C.

C2H6 (g) 2CH3 (g)

Calculate the half-life of the reaction in minutes.

= t1/2

0.693

k

Answer:

t1/2 = 21.5 min

Example

The gas-phase decomposition of NOBr is second order in [NOBr],

with k = 0.810 M-1.s-1 at 10°C. We start with 4.00 x10-3 M NOBr in a

flask at 10°C. How many seconds does it take to use up 1.50 x 10-3 M

of this NOBr?