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Chemical Reaction Kinetics Chemical reaction kinetics study the rate of chemical reactions

Definition of chemical reaction rate The number of moles of a reactant converted (consumed) in a reaction per unit time

for a reaction A + B C + D (mol/s)

When we say rate we always refer to ONE of the components in the reaction The minus sign refers to that the concentration of reactant decreases in reaction

Reaction rate equation (or kinetic equation) Many forms exist The most common one

where k reaction rate constant

A0 pre-exponential factor

Ea reaction activation energy

reaction orders with respect to A, B, C, D, respectively

R gas constant

T reaction temperature in Kelvin scale

Chemical Reactions

[A]

or [A][A]

12

12 dt

dr

ttr AA

RT

EAkk

dt

dr a

A

-expin which [D][C][B][A]

[A]0

Arrhenius equation

Kinetic parameters

2

Chemical Reaction Kinetics Meanings of kinetic parameters, k, A0, Ea,

Reaction rate constant, k It tells how fast a reaction can occur It is a constant dependent on temperature but independent of concentrations

Pre-exponential factor, A0

It refers to the frequency of collision between molecules, the higher frequency, the faster rxn

Reaction activation energy, Ea

It can be understood as the energybarrier for a reaction to overcome

The higher Ea value is, the more

difficult for a reaction to occur Reaction orders w.r.t. each component,

The magnitude of these values reflects the effectiveness of each component in the reaction

The values of can be positive or negative or zero

The values of can be integrals or fraction

All these kinetic parameters have to be determined experimentally

Chemical Reactions

RT

EAk a-

exp0

reaction process

Ea reactant

productener

gy

3

Chemical Reaction Kinetics Factors affecting the rate of a chemical reaction

Reaction temperature, T An increase T will lead to increasing k, thus reaction rate. The dependence of k on T is given by differentiating k expression,

the higher Ea value is, the more significant of the effect of increasing T on the reaction rate Concentration of reactants / products

The effect of increasing a concentration is positive if the respective order is positive The larger the value of order is the stronger the effect of increasing conc on the rate When order equals to zero, there is no effect of concentration on the rate.

The presence of a catalyst A catalyst can alter reaction rate (speeding up desired rxns or slowing down undesired rxns)

Note: we assume rate of mass transfer (to meet) is sufficient high comparing to rA.

Chemical Reactions

[D][C][B][A]-

exp0

RT

EAr a

A

200

ln

-lnln

-exp

RT

E

dt

kd

RT

EAk

RT

EAk aaa

4

Chemical Reaction Kinetics The reaction rate and mass transfer rate

When we discuss the reaction rate, it only makes sense if there are sufficient number reactant molecules can be delivered to the reaction site.

We say a reaction is kinetic control when the rate of mass transfer > the rate of rxn rA.

This means that molecules being transported

to the reaction site are ‘queuing’ for reaction

If the mass transfer rate is slower than the reaction rate, the overall rate we observed will be the rate of mass transfer, not the reaction rate - diffusion control

This means that molecules are waiting to be delivered

before reacting ‘queuing’ for reaction

The concept of rate determining step (r.d.s.)

The slowest step in a reaction process determine the overall rate of a reaction

Chemical Reactions

masstransfer reaction

masstransfer

reaction

5

Chemical Reaction Kinetics Catalysis and catalysts

Catalyst is a substance which can alter reaction rate without itself being destroyed or consumed (many other definitions and this is one of them)

95% of chemical industries apply one or more catalysts in their processes

e.g. polymerisation, air/water depolution, ammonia synthesis, cracking heavy oil to LPG, etc

A catalyst can be an acid, a base; can be a liquid or a solid. Most industrial catalysts are metals, metal oxides or a mixture of them formulated & made in special ways

Use of catalysts in industry Speeding up desired reactions thus increase the process output

Slowing down undesired reaction thus reduce the unwanted waste products

Altering reaction route by changing the relative speed of certain steps in a reaction network therefore realising certain products which would not be possible without catalysts.

Allowing some rxns to occur under mild conditions e.g. working with heat sensitive materials

Enzymes are catalysts that participate in bio-active processes

etc

Chemical Reactions

6

Chemical Reaction Kinetics Calculation of reaction rate

Example: a gas phase reaction 2N2O5=4NO2+O2 occurs at 300°C. The concentrations of N2O5 found in the reaction mixture at different time intervals are given below:

t h 0 1 2 3 5 7 9

[N2O5] mol/L 1.40 1.07 0.80 0.58 0.34 0.18 0.09

Calculate the rxn rates w.r.t. N2O5, NO2 & O2 1) betw. 0-1h; 2) betw. 3-5h; 3) average betw 0-9h.

N2O5 consumption rate NO2 formation rate O2 formation rate

Eqn’s to use:

1) 0-1 h

2) 3-5h

3) aver.

Chemical Reactions

12

1221

12

1224

12

12 [A][A] ;

[A][A] ;

[A][A]2252 tt

rtt

rtt

r ONOON

hmol/L 1650

01

401071 0.66;

01

401071 0.33;

01

401071 21

24

2252

...

r..

r..

r ONOON

hmol/L 060

01

580340 0.24;

01

580340 0.12;

35

580340 21

24

2252

...

r..

r..

r ONOON

hmol/L 0730

01

401090 0.291;

01

401090 0.146;

09

401090 21

24

2252

...

r..

r..

r ONOON

7

Chemical Reaction Kinetics More about reaction rate 2N2O5 = 4NO2 + O2

1) 0-1 h rN2O5=0.33 rNO2=0.66 rO2=0.165 mol/Lh

2) 3-5h rN2O5=0.12 rNO2=0.24 rO2=0.06 mol/Lh

3) aver. rN2O5=0.146 rNO2=0.291 rO2=0.073 mol/Lh

For the same reaction, the reaction rate expressed by different components varies with their stoichiometry coefficients

rN2O5 : rNO2 : rO2 = 2 : 4 : 1 or

Given reaction rate for one of the rxn components you should be able to calc others.

For the same reaction the reaction rate may vary with the time because of change of reactant conc’s with time & rate in general is proportional to [ ]’s. When rxn orders w.r.t. reactant > 0 (usually they are) the rxn rate rbeginning > rlater

As reaction rate is a function of temperature, the determination of reaction of reaction rate must be done at a constant temperature (you may need to determine the rate at a different T, or you may need to vary temperature to determine such as Ea

Don’t forget to put the correct unit to the reaction rate you determined

Chemical Reactions

1

4

22252 ONOON rrr

8

Chemical Reaction CalculationsQuestion 1. How many grams of water are produced in the oxidation of 1.0g of glucose,

C6H12O6? Reaction equation: C6H12O6 + 6O2 = 6CO2 + 6H2O

Step 1: Use molar mass of glucose to convert g to moles

1 mole C6H12O6=6x12(C)+12x1(H)+6x16(O)=180g/mol

number of moles C6H12O6=1.0g x (1mol/180g)=5.55x10-3 mol

Step 2: Use balanced equation to determine no. of moles of H2O produced

1 mole C6H12O6 produces 6 moles H2O

the no. of moles of H2O produced: 5.55x10-3 moles C6H12O6x6=0.033 moles H2O

Step 3: Convert moles of H2O to grams using molar mass

1 mole of H2O=2x1(H)+1x16(O)=18 g/mol

Grams of H2O produced: 0.033 mol of H2Ox(18g/mol)x = 0.6g H2O (answer)

Note: You cannot use the weight directly in the calculation. It has to be converted to moles.

Chemical Reactions

9

Chemical Reaction Calculations

Question 2. In a reactor one put 180g of glucose (C6H12O6) and 160g of O2. Can you

produce 108g of H2O. Why? What is the maximum amount of H2O which can be

produced? Reaction equation: C6H12O6 + 6O2 = 6CO2 + 6H2O

Step 1: Convert all components from grams to moles:

No.moles of C6H12O6=180g/190g/mol=1 mole of C6H12O6

No.moles of O2= 160/32g/mol=5 mole of O2

No.moles of H2O= 108/18g/mol=6 mole of O2

Step 2: Find out how much glucose AND O2 you need to produce 108g H2O.

To produce 108g which is 6 moles of H2O, you will need 1mole glucose AND 6 moles of O2. Do

we have enough glucose? - Yes. Do we have enough O2? - No.

Step 3: Every 6 molecules of O2 will burn 1 molecule of glucose, this will proceed UNTIL

one of the reactant consumed completely, in this case O2.

When all O2 is consumed the reaction will stop and the max. amount of H2O which can be

produced can be calculated from O2 available: 5moles of O2 gives 5moles (or 90g) of H2O

Note: When one of reactants is consumed completely the reaction will stop.

Chemical Reactions

10

Chemical Reaction CalculationsQuestion 3. As in Question 2, one puts 180g of glucose (C6H12O6) and 160g of O2. When

O2 is completely consumed, what is the glucose left & what is the percentage

conversion of glucose? Reaction equation: C6H12O6 + 6O2 = 6CO2 + 6H2O

Step 1: Convert all components from grams to moles:

No.moles of C6H12O6=180g/190g/mol=1 mole of C6H12O6

No.moles of O2= 160/32g/mol=5 mole of O2

No.moles of H2O= 108/18g/mol=6 mole of O2

Step 2: Find out how much glucose left after all O2 has consumed.

The molar ratio of glucose and O2 in the reaction=1:6. For a consumption of 5 moles of O2,

the amount glucose reacted will be 1x5/6=5/6 moles or 0.83 moles, or 0.83x180=150g.

The amount glucose left over=1-0.83=0.167moles or 180-150=30g

Step 3: The percentage conversion of glucose at complete conversion of O2

(try the weight base)

Note: The conversion (%) calculated based on moles is the same as that based on weight .

Chemical Reactions

%.%.

%conversion 3831001

83301100

[A]

[A][A](%)

in

outin

11

Chemical Reaction Calculations

Question 4. The brown gas NO2 can form colorless gas N2O4, 2NO2 N2O4. At 25°C the

concentrations of NO2 & N2O4 are 0.018 M & 0.055 M respectively when at

equilibrium. 1) Calculate the equilibrium constant Keq at 25°C. 2) If in

another equilibrium system of the same gases at the same temperature, the NO2

concentration is found to be 0.08 M, what is the concentration of N2O4?

Step 1: Determine the equilibrium constant

From equilibrium constant definition:

Step 2: When at equilibrium

Chemical Reactions

1700.018

0550

][NO

]O[N22

2

42 .

Keq

M 08811700.08]O[N 1700.08

]O[N

][NO

]O[N 2422

422

2

42 .Keq

12

Chemical Reaction CalculationsQuestion 5. The rate constants of a reaction are determined to be 3x10-5 mol/L.h at

200°C and 4x10-4 mol/L.h at 250°C. Estimate the reaction activation energy.

Arrhenius eqn relates the rate constant to activation energy

Mthd.1: lnk=lnA0+(-Ea/R)(1/T), A plot of lnk against 1/T will produce a

straight line, the slope of which is -Ea/R. So that Ea=slope x R

Ea=-12750 x 8.314=106,000 J/mol = 106 kJ/mol

Mthd 2:

Let A0,1=A0,2

T1=273+200 K, T2=273+250 K, k1=3x10-5 & k2=4x10-4 mol/L.h, R=8.314 JK/mol

Chemical Reactions

RT/EaeAk 0

2120

10

20

10

2

1

20

10

2

1

202101

lnlnln

&

2

1

2

1

21

RT

E

RT

E

A

A

eA

eA

k

k

eA

eA

k

k

eAkeAk

aa

,

,RT/E

,

RT/E,

RT/E,

RT/E,

RT/E,

RT/E,

a

a

a

a

aa

2

1

21

21

21

21

212

1 lnor ln k

k

TT

TRTE

TRT

TTE

RT

E

RT

E

k

ka

aaa

kJ/mol 106

104

103ln

250273200273

2502732002733148ln

4

5

2

1

21

21

.

k

k

TT

TRTEa

1/T

ln kslope= -12750

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Chemical Reaction CalculationsQuestion 6. Two catalysts A & B are compared for their catalytic activity for reaction RP.

When A is present it takes 10s for R to change from 2 to 0.5 moles and when B is present it takes 20s for R to decrease from 5 to 2 moles at the same

temperature and with the quantities of catalyst. Which catalyst is more active for the reaction concerned?

Answer: The activity of two catalysts can be compared based on the average

reaction rate when A & B presence separately.

The A catalyst is more active for the reaction concerned.

Chemical Reactions

mol/s 10020

5-3][-][

mol/s 15010

2-.50][-][

12

12

.t

RRr

.t

RRr

B

B,B,B,R

A

A,A,A,R

14

Chemical Reaction CalculationsQuestion 7. Verify that the rate constant of a reaction following second order rate law

rA=-k[A]2 can be determined from the slope of a line obtained by plotting 1/[A]t against reaction time t, where [A]t is the concentration of A measured at time

t.

Answer: Second order rate law: (1)

rearrange: (2)

Define boundary conditions: at t=0, [A]=[A]0 and at t=t, [A]=[A]t

integrate eqn (2), t from 0-t and [A] from [A]0 to [A]t

(3)

compare eqn (3) with linear eqn Y=aX+B, which is a straight line with slope a

Let

A plot of vs. t will give a straight line with slope=k.

Chemical Reactions

2[A][A]

k dt

drA

kdtd

k dt

d

22

[A]

[A][A]

[A]

0t0t

t

0

[A]

[A] 22 [A]

1

[A]

10

[A]

1

[A]

1[A]

[A]

1-

[A]

[A] t

0

kttkdtkdkdt

d

0t [A]

1 and

[A]

1 btX,ka,Y

t[A]

1

t

slope=k

1/[A]

15

Chemical Reaction Calculations

Question 8. Reaction RP follows the second order rate law rR=-k[R]2. Verify that the time required for the reactant R to fall to a half of its initial value is t1/2=1/(k[R]).

Answer: Second order rate law: (1)

After integration of eqn (1) with the boundary conditions:

at t=0, [R]=[R]0 & at t=t1/2, [R]=[R]t1/2=0.5[R]0

Chemical Reactions

2[R][R]

k dt

drR

021

000021

0021

021

00t

[R]

1[R]

1

[R]

21

[R]

1

0.5[R]

11

[R]

1

0.5[R]

1

[R]

1

0.5[R]

1

[R]

1

[R]

1

kt

kktkt

ktkt

/

//

/