Chemical kinetics lecture

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Course Title: General and Inorganic

Chemistry

TOPIC 3: CHEMICAL KINETICS

Lecturer and contacts

Mr. Vincent Madadi

Department of Chemistry, University of Nairobi

P. O. Box 30197-00100,

Nairobi, Kenya

Chemistry Dept. Rm 114

Tel: 4446138 ext 2185

Email: [email protected], [email protected]

Website: http://www.uonbi.ac.ke/staff/vmadadi

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Introduction

• Kinetics is the study of rates of chemical reactions and

the mechanisms by which they occur.

• The reaction rate is the increase in concentration of a

product per unit time or decrease in concentration of a

reactant per unit time.

• A reaction mechanism is the series of molecular steps by which a reaction occurs.

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Thermodynamic vs kinetics of reaction

• Thermodynamics determines if a reaction can occur.

Kinetics determines how quickly a reaction occurs

• Some reactions that are thermodynamically feasible

are kinetically so slow as to be imperceptible

• The Rate of a Reaction

Cdiamond + O2(g)→ CO2(g) ΔG°= -396kJ

Very Slow

H+(aq) + OH-(aq) → H2O(l) ΔG°= -79kJ

Very Fast1/12/2011 3mov

Rate of chemical reaction

• 2 N2O5 → 4 NO2 + O2

• 2 moles of N2O5 disappear for every 4 moles of NO2

and 1 mole of O2 formed.

• Reaction rates are the rates at which reactants

disappear or products appear.

• This movie is an illustration of a reaction rate.

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Reaction rate

• Reaction rate is the change of concentration of a

reactant or product per unit time

aA + bB → cC + Dd

Reaction rate = ΔConcentration

Δtime

• Rate is expressed either as rate of appearance of product or rate of disappearance of reactant

• E.g.aA + bB→ cC + dD

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Reaction rate

2 NO2(g) → 2 NO(g) + O2(g)

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Reaction rate

• Mathematically, the rate of a reaction can be written as:

• Square brackets [ ] are often used to express molarity(i.e.[HCl] means Molarity of HCl)

• The relative rates of consumption of reactants and formation of products depend on the reaction stoichiometry

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Reaction rate

• The relative rates of consumption of reactants and formation

of products depend on the reaction stoichiometry

• For the reaction

2HBr (g) → H2 (g) + Br2 (g)

• two moles of HBr are consumed for every one mole of H2

which is formed

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Reaction rate

• Experimental Rate Law: the rate of a reaction is

proportional to the product of the concentrations of

the reactants raised to some power.

• For a reaction

aA + bB → products,

the rate law is the equation

rate = k[A]x[B]y

• Relationships Between Rate and Concentration

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Reaction rate

• x and y are the orders of the reaction in [A] and [B]

respectively

• The overall order of the reaction is x + y

• x and y are usually small integers, but may be zero,

negative, or fractions

• k is the specific rate constant

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Rate constant k

• Units depend on overall reaction order

• Value does not change with concentration

• Value does not change with time

• Valid for a specific temperature

• Dependent on presence or absence of a catalyst

• Value must be determined experimentally

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Factors that affect rate of reaction

• They are six key factors that affect rate of

reaction:

1) Nature of reactants and products

2) Concentration

3) Temperature

4) Catalyst

5) Surface area

6) Light radiation

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Factors that affect rate of reaction

1. Nature of reactants and products

• Chemical reactions involve rearrangement of bonds: Bonds in

reactants are broken and new bonds are formed in products

• In organic or molecular reactions, large number of bonds are

broken in reactants and formed in products-hence reactions are

slow

• E.g. Hydrolysis of cane sugar:

C12H22O11 + H2O C6H12O6 + C6H12O6

Cane sugar glucose fructose

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Factors that affect rate of reaction...

• Inorganic reactions involve ions, hence no bonds to be broken in

reactants . The reactions are faster.

• E.g.

Ag+(aq) + NO3

-(aq) + Na+Cl-(aq) AgCl(aq) + NaNO3(aq)

• Concentration

• Based on the rate law of mass action, the rate of reaction is

directly proportional to the product of the concentration of the

reactants at a particular temperature

• For reaction aA + bB cC + dC

Rate = K[A]a[B]b1/12/2011 14mov

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Factors that affect rate of reaction ...

3) Temperature

• In most reactions, the rate of reaction doubles with 10 K increase

in temperature

• But with 10 k increase in temperature, collision frequency (Z)

increases by a factor of 1.016

I.e. Z α T½

Z2/Z1 = (T2/T1)1/2 = (310/300)1/2 = 1.016

• However, the rate increases by almost 100%

• This is because increasing temperature by 10 K increases the

number of active molecules (molecules with E > Ea) which

increases the rate drastically1/12/2011 15mov

Factors that affect rate of reaction ...

Effect of increase in temperature by 10 K

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• 4) Catalyst

• A catalyst is a substance that alters the rate of reaction without

itself getting consumed

• There are two types of catalysts:

• Positive catalyst: It increases the rate of reaction e.g.

MnO2

2KClO3 2KCl + 3 O2

300 ⁰C

• MnO2 acts as a positive catalyst since uncatalysed reaction

takes place at 700 ⁰C and is slower

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• Negative catalyst: This is the catalyst which retards

the rate of reaction e.g. Oxidation of chloroform is

retarded by ethanol

1% ethanol

4CHCl3 + 3 O2 4COCl2 + 2Cl2 + H2O

Mechanism of catalysis

• A catalyst alters the rate of reaction by providing a

path with lower or higher activation energy

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5) Surface area

• The rate of homogeneous reaction is influenced by

the surface area of the reactants

• Particle size decreases, surface area increases for the

same mass because of creation of new surfaces.

Hence the rate of reaction increases

• E.g. Powdered zinc reacts faster with dilute HCl than a

block of zinc1/12/2011 19mov


6) Light radiation

• The rate of photochemical reaction is affected by light

radiation

• Photons (E = hv) supply the necessary energy of

activation to the reactant molecules to form products

λ = 400 nm

• E.g. H2 + Cl2 2HCl

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Order of reaction

• This is the sum of the powers of the concentration

term in an experimentally established rate law

• Example

aA + bB → cC + dD

• The theoretical rate law is:

Rate = K[A]a[B]b

• But experimentally determined rate expression is:

Rate = K[A]α[B]β

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Order of reaction cont.

• Where α and β are the actual moles of A and B and

may not be necessarily be equal to “a” and “b”

• Overall order of reaction is given by the sum of the

individual orders i.e.

• Overall order = α + β

• Order of reaction can be zero, +ve, -ve or fraction.

But higher orders of reaction are rare 1/12/2011 22mov

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Order of reaction cont.

• In complex reactions the order of reaction is

determined by the slowest step of the reaction which

is also called the rate determining step

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Molecularity of reaction

• This is the total number of molecules in the step

leading to chemical reaction

• For any reaction, the least number of molecules is one

• Thus molecularity cannot be zero or fraction

• Types of molecularity:

1) Unimolecular reactions:

• Reactions involving one molecule of the reactants1/12/2011 24mov

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Molecularity of reaction

• Example:

PCl5 PCl3 + Cl2

2) Bimolecular reactions

CH3COOC2H5 + H2O CH3COOH + C2H5OH

3) Thermolecular reactions

2NO + O2 2NO2

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Pseudo uni-molecular reactions

• The reaction in which the order is one but molecularity

is 2

• E.g.

CH3COOC2H5 + H2O CH3COOH + C2H5OH

• Order is one because water is in excess, hence

concentration does not change during the reaction

• Thus, the rate is independent of the conc. Of water but

only dependent on conc. of ester1/12/2011 26mov

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Rate law

• Exercise

1) The rate of reaction 2NO + O2 2NO2

follows the rate law Rate = K[NO]2[O2]

If K = 2x 10-6 mol-1L2, What is the rate of the reaction

when [NO] = 0.04 molL-1 and [O2] = 0.2 molL-1

[ Ans = 6.4 x 10-10 molL-1s-1]

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Rate law

2. The rate of the reaction

2NO + O2 2NO2

is doubled when the concentration of O2 is doubled,

but increases by factor of 8 when the concentration

of both reactants is doubled.

Determine the:

a) Order of reaction wrt NO and O2

b) Overall order of reaction

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Activation energy• This is the additional amount of energy that reactant

molecules must acquire in order to react and form products

• It is defined as the amount of energy that the reactants must

absorb to pass over the activation energy barrier to form

products

• Activation energy diagram

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Reactant

Product

Ea

ERE Th

Activation Energy cont.

• E = Eth

–ER

= Threshold Energy –Energy possessed by molecules

• The activation energy is related to the rate constant K and

temperature T according to Arrhenius equation

K = Ae-Ea/RT

Where A = Frequency factor

K = rate constant

T = temperature in Kelvin

R = Gas constant

E = Activation energy

e = base of natural logarithm1/12/2011 30mov

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Derivation of Arrhenius equation

• Arrhenius equation is derived from the Vant’

Hoffs reacton isochore

dlnKc/dT = ΔE/RT2

and the reaction dynamic equilibrium

A + B = C + D

• Kc for the reaction,

Kc = Kf/K

b= [C][D]/[A][B]

Kf[A][B] = K

b[C][D]

• Hence, Kc

= Kf/K

b1/12/2011 31mov

Derivation of Arrhenius equation cont.

• If ΔE is written as Ef –Eb, then from equation 1 and 2

then,

• dlnKf/dT – dlnKb/dT = Ef/RT2 – Eb/RT2

• or dlnk/dT = E/RT2

� dlnk = EdT/RT2

• Integrating the equation gives,

lnk = -E/RT + C

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Derivation of Arrhenius equation cont.

• Provided E is a constant, the equation can be written

as:

k = Ae-E/RT

• This is the Arrhenius equation

• Application of activation energy

1) To determine activation energy

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Determination of activation Energy

• There are two methods for determining of activation

energy Ea

1) Graphical method

2) Rate constant method

1) Graphical method

• From Arrhenius equation, K = Ae-E/RT

• Applying lo to both sides of the equation,1/12/2011 34mov

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Determination of activation Energy cont.

• lnK = lnA – Ea/RT

• >2.303logK = 2.303logA – Ea/RT

• >logK = logA – Ea/2.303RT

• >logK = -Ea/2.303RT + logA

Ξ y = mx + C

• Plotting logK against 1/T give a straight line and Ea can

be calculated

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Graphical Determination of activation

energy

• Graph

Log K

Slope = -Ea/2.303RT

Log A

Ea = slope x -2.303 R

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Graphical Determination of activation

energy...

• Slope = -Ea/2.303xR

• Ea = -2.303 x R x Slope

• Log A = Intercept

A = Antilog (Intercept)

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2. Determination of activation energy

from rate constant method

• From K = Ae-Ea/RT

• lnK = lnA – Ea/RT

• Let at Temperature T1 and rate constant K1; and at T2

and rate constant K2

• For small change in temperature, the change Ea and

A do not significantly change

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Determination of activation energy from rate

constant method ...

• Thus,

lnK1 = lnA – Ea/RT1 3

lnK2 = lnA – Ea/RT2 4

• Subtracting 3 from 4,

• LnK2 – lnk1 = Ea/RT1 – Ea/RT2 5

• logK2/K1 = Ea [1/T1 – 1/T2]

2.303

• logK2/K1 = Ea [(T2-T1)/T1T2]

2.303R1/12/2011 39mov

Determination of activation energy from rate

constant method...

• Log K2/K1 = Ea [(T2-T1)/T1T2] Eq. 6

2.303R

Since K1, K2, T1 and T2 are known, Ea can be calculated

from the equation 6

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Rate Laws• A rate law shows the relationship between the reaction rate and

the concentrations of reactants. Exponents tell the order of the reaction with respect to each reactant.

• This reaction is

First-order in [NH4+]

First-order in [NO2−]

• The overall reaction order can be found by adding the exponents on the reactants in the rate law.

• This reaction is second-order overall.

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• For gas-phase reactants use PA instead of [A]

Integrated Rate Laws

• These are expressions which relate the concentration of

the reactants with time

• Application:

1) Used to predict amount of reactant or product at a

particular time

2) Predict how long the reaction will take

3) To predict when a toxic chemical can be disposed

• Can be classified into Zero, first, second and third

order reactions1/12/2011 mov 42

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Integrated Rate Laws cont.

1) Zero order reactions

• This is a reaction where the rate of reaction is

independent of the concentration of the reactants

• Derivation:

• Let A Product

• Initial conc. (mol/l) t = 0, a 0

• At time t = t, a-x x

• Where x is the concentration of the reactant (A)

undergoing decomposition1/12/2011 mov 43

Zero order reactions cont.

• Rate expression for zero order reaction:

• dx/dt α (a-x)0 ........................................................Eq.1

• Or dx/dt = Ko(a-x)o = ko .......................................Eq. 2

• K0 is the rate constant for zero order reaction

• Integrating the equation,

• ∫dx = ∫Kodt = Ko ∫dt

x = Kot + C, .............................................................Eq. 3

• Where C is the integration constant

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• But when t = 0 and x = 0

• Thus,

0 = Ko x 0 + C .............................................Eq. 4

• Hence C = 0

• Substituting the value of C into equation 3 gives,

x = Kot

• Therefore, Ko = x/t ...........................................Eq. 5

• Eq.5 is the rate constant equation for zero order

reaction

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• Units for rate constant:

• Ko = x/t = conc/time = molL-1/s = molL-1s-1

• Half-life –This is the time duration in which half the

concentration of reactants is transformed into products

• Thus at t = t½, x = a/2,

• Substituting these values into equation 5

K0 = a/2t½ t ½ = a/2k0 t½ α a

• This shows that half-life of a zero order reaction is

directly proportional to the initial concentration1/12/2011 mov 46

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• Graph for zero order reaction

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Rate

Concentration of reactants

The rate of reaction is independent of the concentration of reactants

First Order Reactions

• First Order Reactions

• These are reactions where the sum of the powers of

concentration of the exponential term in an

experimentally established rate law is one

• Thus the rate of reaction is dependent on the single

power of the concentration term of the reactants

• Example:

• For the reaction, A Product

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First Order Reactions cont.

• Initial conc (molL-1) at t=0, a o ......Eq. 1

• Conc (molL-1) at t=t a-x x .......Eq. 2

• Where x moles of A have decomposed into products in

time “t”

• Deferential, -d(a-x)/dt or dx/dt α(a-x) ...............EQ. 3

• Or dx/dt = k1(a-x) ...............................................Eq. 4

• Where, K1 is the rate constant for first order reaction.

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• Rearrangement,

• dx/(a-x) = k1dt ....................................................Eq. 5

• Integrating, [identity ∫dx/x = ln x]

∫dx/(a-x) = k1 ∫dt

-ln(a-x) = k1t + C ..............................................Eq. 6

• Where C = constant of integration

• But at t=0, x = o

• Substituting the values into equation 6,

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• -lna = C .............................................................Eq. 7

• Substitute the value of C into equation c,

• -ln(a-x)= k1t –lna ................................................Eq. 8

• Rearranging

• k1t = lna – ln(a-x)

• K1 = ln[a/(a-x)]x 1/t

• Change ln to log10 [lnx = 2.303logx]

• K1 = (2.303/t)log(a/a-x) .......................................Eq.8

• This is the expression for rate constant for first order

reaction1/12/2011 mov 51


• Unit of the rate constant

• From Eq. 8

• K1 = (2.303/t)log(a/a-x) = conc/(time xconc)

k1 = 1/time = s-1 ...................................................................Eq. 9

Graph for first order reaction

• From equation 8

• -ln(a-x)= k1t –lna

• Multiply though by -1

• ln(a-x)= -k1t + lna1/12/2011 mov 52

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• Introduce log10

• 2.303log(a-x) = -k1t + 2.303 x log(a)

• Or log(a-x) = -k1 x t + log(a) ...........................Eq. 10

2.303

Ξ y =mx +c

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Log(a-x)

t

Slope = -k1/2.303

Log(a)

Second order reactions cont.• From the graph,

• K1 = -2.303 x slope

• Half-life of first order reaction

• From Eq. 8, K1 = 2.303 Log (a/a-x)

t

• At t = t ½ , x = a/2

• Substituting the values into the equation

• K1 = 2.303 Log (a/a - a/2) = 2.303 log 2

t ½ t ½

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• Making t ½ the subject,

• t½ = 2.303 log 2 = 2.303 x 0.301 = 0.693

k1 k1 k1

• Thus, t½ = 0.693/k1

• This means that for 1st order reactions, half-life is

independent of the initial concentration of reactants

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3. Second order reaction

• These are reactions in which the sum of the powers of

the of the concentration term in an experimentally

established rate law is 2

• There are two cases of second order reaction

• Case 1: 2A Product

• Thus, Rate law = dx/dt = K[A]2 Order = 2

• Case 2: A + B Product

• Rate law = dx/dt = k[A][B] order = 2

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Second order reaction cont.

• Case 1: Occurs when the concentration of both reactants

is the same

• Case 2: Occurs when the concentration of both reactants

is NOT the same

• 1) Case 1: Concentration of both reactants is the same

• Let the reaction, A + A Product

• Initial conc. t=0 a a 0 ..Eq. 1

• Conc. At t = t a-x a-x x ...Eq. 21/12/2011 mov 57


• Conc. is in mol L-1

• Thus, x molL-1 of the reactant A decomposes in time t

• Deferential,

• dx/dt α (a-x)(a-x) or dx/dt α (a-x)2 ..................Eq. 3

• Introduce rate constant k2

• dx/dt = k2 (a-x)2 ......................................Eq. 4

• Where k2 is the rateconstant for the second order

reaction1/12/2011 mov 58

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• Rearranging,

• dx/ (a-x)2 = k2 dt .....................................................Eq. 5

• Integrating,

• ∫dx/(a-x)2 = k2∫dt note [∫xndx = xn+1/n+1]

• Hence,

• 1/(a-x) = k2t + C ...................................................Eq. 6

• Where C = constant of integration

• But when t= 0, x= 0

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• Substituting the values into the equation,

• 1/a = C

• Substitute the value of C into equation 6

• 1/(a-x) = k2t + 1/a .............................Eq. 7

• Thus, K2t = 1/(a-x) - 1/a

• Hence, K2 = (1/at) x [x/(a-x)] ...................Eq. 8

• Eq.8 is the expression for rate constant for second order reaction

where initial conc. of the reactants is the same1/12/2011 mov 60

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• Case 2: When concentration of both reactants is

different

• Let the reaction, A + B Product

• Initial conc. t=0 a b 0 ..Eq. 1

• Conc. At t = t a-x b-x x ...Eq. 2

• Where x mol/L of A and B decomposed in time t to

form product

1/12/2011 mov 61


• Differential

• dx/dt α (a-x)(b-x)

• dx/dt = K2(a-x)(b-x)

• Where K2 is the rate constant for second order

reaction

• Separating the variables,

• dx/[(a-x)(b-x)] = K2 dt ..........................................Eq. 3

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• Integration by parts

• ∫ dx = ∫ A dx + ∫B dx = ∫k2dt ...Eq.4

(a-x)(b-x) (a-x) (b-x)

• 1 = A + B ......................Eq.5

(a-x)(b-x) (a-x) (b-x)

• Multiply through by (a-x)(b-x)

• 1 = A(b-x) + B(a-x) ..............................................Eq. 6

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• When x = a, 1 = A(b-a)

Thus, A = 1/(b-a)

• When x = b, 1 = B(a-b)

Thus, B = 1/(a-b)

Or

• ∫ dx = 1 ∫dx + 1 ∫dx = k2 ∫ dt ..Eq. 7

(a-x)(b-x) (b-a) (a-x) (a-b) (b-x)

• Factorise the middle term by 1/(a-b)

• = ∫ dx = 1 [ ∫dx - ∫dx ] = k2 ∫ dt ...Eq. 8

(a-x)(b-x) (a-b) (b-x) (a-x)1/12/2011 mov 64

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• ∫ dx = 1 [ -ln(b-x) + ln(a-x)] = k2 ∫ dt ..Eq. 9

(a-x)(b-x) (a-b)

Therefore,

• = 1 [ ln (a-x) ] = k2t + C ......................Eq. 10

(a-b) (b-x)

When t = 0, x = 0,

Substituting into the equation 14,

1/12/2011 mov 65


• = 1 [ ln (a ] = C .....................................Eq. 11

(a-b) (b)

• Substitute the value of C in eq. 15 into eq. 14

• = 1 [ ln (a-x) ] = k2t + 1 [ ln (a ]..........Eq. 12

(a-b) (b-x) (a-b) (b)

Thus

• K2 = 1 [ ln b(a-x) ] ............................Eq. 13

(a-b)t a(b-x)

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• Applying log10

• K2 = 2.303 log b(a-x) .............................................Eq. 14

(a-b)t a(b-x)

Characteristics:

Unit : K2 = 1 x x

at (a-x)

= 1/(conc x time)

K2 = [conc. X time]-1 .............................................................Eq. 15

1/12/2011 mov 67


• K2 = [mol-1s]-1 = mol-1Ls-1

• When one of the reactants is in excess,

• i.e. Let a>> b

• a-b = a, a-x =a

• Thus b and x can be neglected in comparison to a

• Hence, K2 = 2.303 logb(a-x) ................................Eq. 16

(a-b)t a(b-x)

• k2 = = 2.303 log(b/b-x)) .....................................Eq. 17

at 1/12/2011 mov 68

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• k2a = = 2.303 log(b/b-x)) ................................Eq. 18

t

• k’ = = 2.303 log(b/b-x)) .................................Eq. 20

t

Where K2a = k’ = rate constant for the first order

reaction

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• Graph for second order reactions

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1 lnb(a-x)(a-b) a(b-x)

t

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• Half life

• From equation 8, case 1

• K2 = 1 x x

at (a-x)

• t = t 1/2 , x = a/2

• K2 = 1 x a/2

at ½ (a – a/2)

• t ½ = 1/k2.a .....................................................Eq. 211/12/2011 mov 71


• Thus t ½ is directly proportional to 1/a

1/12/2011 mov 72

Chemical kinetics lecture

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Transcript of Chemical kinetics lecture