overview of kinetics

7/26/2019 Overview of kinetics

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Rates of Reaction

OVERVIEW



Overview

• Reaction Rates – Definition of Reaction Rates

– Experimental Determination of Rate

– Dependence of Rate on Concentration

– Change of Concentration with Time

– Temperature and Rate; Collision and Transition-State

Theories.

– Arrhenius Equation

• Reaction Mechanisms

– Elementary Reactions – Rate Law and the Mechanism

– Catalysis



Reaction Rates

• Deal with the speed of a reaction and controlled by: – Proportional to concentrations of reactants

– Proportional to catalyst concentration; catalyst = a substance

that increases the rate of reaction without being consumed in

the reaction.

– Larger surface area of catalyst means higher reaction rate

(more sites for reaction to take place).

– Temperature: Higher temperature of reaction means faster.



Definition of Reaction Rate

• Reaction rate = increase in

concentration of product of areaction as a function of time ordecrease in concentration ofreaction as a function of time.

• Thus the rate of a reaction is:

12

12

changetime

Achangeconc

tt

] A[] A[

t

] A[Rate A

Concentration vs Reaction Time

A + 2B --> 3C

0.000

0.045

0.090

0 250 500

Time, s

C o n

c e n t r a t i o n ,

MInit

Rate

Ave.

Rate

Inst.

Rate

• Rates are expressed as positive numbers. For the reaction in

the graph we have:

t

] A[R A

t

]B[RB

t

]C[RC



Reaction Rates and Stoichiometry

• A + B C; RC = R A = RB.

• A + 2B 3C;

E.g.Calculate the rate of decomposition of HI in thereaction: 2HI(g) H2(g) + I2(g). Given: After a

reaction time of 100 secs. the concentration of HIdecreased by 0.500 M.

• For the general reaction: aA + bB cC + dD

E.g. For the reaction 2A + 3B 4C + 2D; determinethe rates of B, C and D if the rate of consumption of Ais 0.100 M/s.

CB A R3

1R2

1R

DCB A Rd

aR

c

aR

b

aR



Rate Laws and Reaction Order

• Rate Law – an equation that tells how the reaction rate depends on the

concentration of each reaction.• Reaction order – the value of the exponents of concentration terms in

the rate law.

• For the reaction: aA + bB cC + dD, the initial rate of reaction isrelated to the concentration of reactants.

• Varying the initial concentration of one reactant at a time produces

rates, which will lead to the order of each reactant.

• The rate law describes this dependence: R = k[A]m[B]n where k = rateconstant and m and n are the orders of A and B respectively.

– m = 1 (A varied, B held constant) gives R = k’[A]. Rate is directlyproportional to [A]. Doubling A doubles R

– m = 2 (A varied, B held constant) gives R = k’[A]2. The rate is proportional

to [A]2. Doubling A quadruples R.E.g. Determine order of each reactant:

HCOOH(aq) + Br 2(aq) 2H+(aq) + 2Br (aq) + CO2(g) R = k[Br 2]

E.g. The formation of HI gas has the following rate law: R = k[H2][I2].What is the order of each reactant?



Experimental Determination of a Rate Law: First

Order

• Varying initial concentration of reactants

changes the initial rate (usually all but

one held constant) like one with two

unknowns.

• Initial rate is the initial slope of the

graph shown.

• As the initial concentration of that

compound increases so does the rate.

– Initial rate vs. [A]o plotted.

– If straight line then reaction is first order

and slope is rate constant.

• Second order rate law determined in

like manner.

Concentration vs Reaction TimeA + 2B --> 3C

0.000

0.050

0.100

0 250 500

Time, s

[ A ] 0 ,

M

Initial Rate vs. [A]o

0.0000

0.0001

0.0002

0.0003

0.0004

0.0005

0.00 0.03 0.05 0.08 0.10

[A]o

R o



Rate Law for All Reactants

• Order for all components done same way.

E.g. Determine the reaction order for each reactant from the table.

(aq)+5Br (aq)+6H+(aq)3Br 2(aq)+3H2O(l)[

3BrO ]o [Br ]o [H+]o Ro

0.10 0.10 0.10 1.20.20 0.10 0.10 2.4

0.10 0.30 0.10 3.50.20 0.10 0.15 5.4

3BrO

Eg. 2: Determine the reaction orders for the reaction indicated fromthe data provided.

A + 2B + C Products.

[A]o [B]o [C]o Ro 2.06 3.05 4.00 3.70.87 3.05 4.00 0.660.50 0.50 0.50 0.0131.00 0.50 1.00 0.072



Integrated Rate Law: First –Order Reaction

• For a first order reaction, Rate = [A]/ t = k[A] or RA = d[A]/dt = k[A].

• Use of calculus leads to: or

• Allows one to calculate the [A] at any time after the start of the reaction.

E.g. Calculate the concentration of N2O remaining after its decompositionaccording to 2N2O(g) 2N2(g) + O2(g) if it’s rate is first order and [N2O]o = 0.20M,

k = 3.4 s1

and T = 780°C. Find its concentration after 100 ms.

• Linearized forms: or

• Plot ln[A] vs t.

• Slope of straight line leads to rate constant, k.

E.g. When cyclohexane(let's call it C) is heated to 500 oC, it changes intopropene. Using the following data from one experiment, determine the first orderrate constant.:

kt] A[

] A[ln

o

t303.2

k

] A[

] A[log

o

o] Aln[kt] Aln[ o] Alog[t303.2

k] Alog[

t,min 0.00 5.00 10.00 15.00[C],mM 1.50 1.24 1.00 0.83



Half-Life: First Order Reaction

• Half-l i fe of First o rder r eact ion ,

t1/2 = 0.693/k. the time required for

the concentration of the reactant to

change to ½ of its initial value.

i.e. at t1/2 , [A] = ½ [A]o

E.g. For the decomposition of N2O

5at 65 °C, the half-life was found to

be 130 s. Determine the rate

constant for this reaction.

k t

t k

t k A A

o

o

/693.0

2

1ln

][][2/1ln

2/1

2/1

2/1

• For n half-l ives t = n*t1/2 [A] = 2n [A]o

o

tn

nn

/

] A[

] A[

tkln

tnklnn

2

1

2

1

2

121



Second –Order Reactions: Integrated Rate

Law

• Rate law: R = k[A]

2

and the integrated rate equation is:

• Plot of vs. t gives a straight line with a slope of k.

• Half-life is:

E.g. At 330°C, the rate constant for the decomposition of

NO2 is 0.775 L/(mol*s). If the reaction is second-order,what is the concentration of NO2 after 2.5x102 s if the

starting of concentration was 0.050 M?

ot ] A[kt

] A[

11

t] A[

1

o2/1 ] A[k

1t



Reaction Mechanisms

• Give insight into sequence of reaction events leading to product

(reaction mechanism).• Each of the steps leading to product is called an elementary

reaction or elementary step.

• Consider the reaction of nitrogen dioxide with carbon dioxide whichis second order on NO2:

NO2(g) + CO(g) NO(g) + CO2(g) Rate = k[NO2]2.

• Rate law suggests at least two steps.• A proposed mechanism for this reaction involves two steps.

– NO3 is a reaction intermediate = a substance that is produced andconsumed in the reaction so that none is detected when the reaction is

finished.• The elementary reactions are often described in terms of their

molecularity. – Unimolecular One particle in elementary.

– Bimolecular = 2 particles and

– Termolecular = 3 particles

Step 1 2NO2(g) NO3(g) + NO(g)Step 2 NO3(g) +CO(g) NO2(g) + CO2(g)Overall NO2 + CO NO + CO2



Rate Laws and Reaction Mechanisms

• Overall reaction order is often determined by the rate determining step.

• Use rate law of limiting step; No intermediates!2NO2(g) NO3(g) + NO(g), R1 = k1[NO2]

2 Slow

NO3(g) +CO(g) NO2(g) + CO2(g) R2 = k2[NO3][CO] Fast

NO2 + CO NO + CO2 Robs = k[NO2]2

E.g. Determine the rate law for the following mechanism:

2*[N2O5(g) )g(NO)g(NO 32k 11k

] Fast

NO3(g) +NO2(g) )g(O)g(NO)g(NO 22k2

Slow

NO3 + NO )g(NO2 2k3

Fast

2N2O5(g) )g(O)g(NO4 22kobs

Use steady state approximation. at “equilibrium” rates of forward andreverse reactions are same. Use to eliminate intermediates from rate lawequations.

or 32152

11

1]NO][NO[k]ON[k

RR

]NO[

]ON[

k

k]NO[

2

52

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overview of kinetics

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