Chemical Kinetics Part - 2

8/18/2019 Chemical Kinetics Part - 2

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Chemistry / Chemical Kinetics

Collision Theory of Reaction Rates

The fundamental notion of the collision theory of reaction rates is that for reaction to occur, molecules, atoms,or ions must first collide. Increased concentrations of reacting species result in greater number of collisions per unit

time. However, not all collisions result in reaction; i.e., not all collisions are effective collisions. For a collision to

be effective, the reacting species must (1) possess at least a certain minimum energy necessary to rearrange outer

electrons in breaking bonds and forming new ones and (2) have the proper orientations toward each other at the

time of collision.

Collisions must occur in order for a chemical reaction to proceed, but they do not guarantee that a reaction will occur.

A collision between atoms, molecules, or ions is not like one between two hard billiard balls. Whether or not

chemical species “collide” depends on the distance at which they can interact with one another. For instance, the

gas phase ion-molecule reaction4 4 5 3CH CH CH CH

+ ++ → + can occur with a fairly long-range contact. Thisis because the interactions between ions and induced dipoles are effective over a relatively long distance. By

contrast, the reacting species in the gas reaction 3 3 2 6CH CH C H+ → are both neutral. They interact appreciablyonly through very short-range forces between induced dipoles, so they must approach one another very closely

before we could say that they “collide.”

Recall that the average kinetic energy of a collection of molecules proportional to the absolute temperature. At

higher temperatures, more of the molecules possess sufficient energy to react.

If colliding molecules have improper orientations, they do not react even though they may possess sufficient energy.

Figure 8 depicts collisions between molecules of NO and N2O. We assume that each possesses sufficient energyto react according to

2 2 2 NO N O NO N+ → +

(a)

O N

N

O

N

O N

N

O

N

O N

O

N

N

Product

Effectiveorientationof collision

(b)

O N

O N

O

N N

O N

O

N N

No reaction

Ineffectiveorientationof collision

N

O

N

Section - 3 REACTION MECHANISMS


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(c)

O N

O N

O

N N

O N

O

O

N

N N

No reaction

Ineffectiveorientationof collision

N

Figure 8: Some possible collisions between N2O and NO molecules in the gas phase. (a) A collision that could

be effective in producing the reaction. (b, c) Collision would be ineffective. The molecules must have the

proper orientations relative to each other and have sufficient energy to react.

Transition State Theory

Chemical reactions involve the making and breaking of chemical bonds. The energy associated with a chemical bond is a form of potential energy. Reactions are accompanied by changes in potential energy. Consider the

following hypothetical, one-step exothermic reaction at a certain temperature.

2A B AB B heat+ → + +Figure 9 shows a plot of potential energy versus reaction coordinate. In Figure 9(a) the ground state energy of the

reactants, A and B2, is higher than the ground state energy of the products, AB and B. The energy released in the

reaction is the difference between these two energies, E.∆ It is related to the change in enthalpy or heat contentthat you have studies in thermodynamics.

Quite often, for reaction to occur, some covalent bonds must be broken so that others can be formed. This can

occur only if the molecules collide with enough kinetic energy to overcome the potential energy stabilisation of the bonds. According to the transition state theory, the reactants pass through a short-lived, high-energy

intermediate state, called a transition state, before the product are formed.

22

transition statereactants productsABA +B AB + B

A B — B A B B A — B B+ → → +! !

A B B

E

forwardreaction

a

Ereversereaction

a

ProductsAB + B

!EreactionReactantsA + B2

Reaction coordinate

P o t e n t i a l

e n e r g y

(a)

A B B

E

forward

reaction

a

E

reversereaction

a

Products

AB + B

!EreactionReactantsA + B2

Reaction coordinate

P o t e n t i a l e n e r g y

(b)

Endothermic reactionExothermic reaction

The "reaction coordinate" representstheleading from reactants to products.This coordinate is sometimes labeled

"progress of reaction."

progress along the pathwayFigure 9: A potential energy diagram. (a) A reaction that releasesenergy (exothermic). An example of an exothermic gas-phase reacion is H + I HI + I

(b) A reaction that absorbs energy (endothermic). An example of anendothermic gas-phase reaction is I + H HI + I

2

2


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The activation energy, Ea, is the additional energy that must be absorbed by the reactants in their ground states

to allow them to reach the transition state. If A and B2 molecules do not possess the necessary amount of energy,

Ea, above their ground states when they collide, reaction cannot occur. If they do possess sufficient energy to

“climb the energy barrier” to reach the transition state, the reaction can proceed. When the atoms go from the

transition state arrangement to the product molecules, energy is released . If the reaction results in a net release of energy (Figure 9(a)), more energy than the activation energy is returned to the surroundings and the reaction is

exothermic. If the reaction results in a net absorption of energy (Figure 9(b)), an amount less than Ea is given off

when the transition state is converted to products and the reaction is endothermic. Thus, the activation energy must

be supplied to the system from its environment, but some of that energy is subsequently released to the surroundings.

The net release of energy is E∆ .

When the reverse reaction occurs, an increase in energy equal to the reverse activation energy, Ea reverse

, is

required to convert the AB product molecules to the transition state. As you can see from the potential energy

diagrams in Figure 9,

a forward a reverse reactionE – E E= ∆ As we shall see later in this unit, increasing the temperature changes the rate by altering the fraction of

molecules that can get over a given energy barrier. Introducing a catalyst changes the rate by lowering the

barrier.

As a specific example that illustrates the ideas of collision theory and transition state theory, consider the reaction

of iodine ions with methyl chloride.

– –

3 3I CH Cl CH I Cl+ → +

Many studies have established that this reaction proceeds as shown in Figure 10(a). The – I ion must approach the

CH3Cl from the “back side” of the C—Cl bond, through the any other angle would not lead to reaction. But acollision with the appropriate orientation could allow the new I—C bond to form at the same time that the C—Cl

bond is breaking. This collection of atoms, which we represent as

I C Cl

H

HH We can view this transitionstate as though carbon is only partially bonded to I and only partially bonded to Cl.

is what we call the transition state of this reaction (Figure 10(b)). From this state, either of two things could happen:

(1) the I—C bond could finish forming and the C—Cl bond could finish breaking with – Cl leaving, leading to products, or (2) the I—C bond could fall apart with I – leaving, and the C—Cl bond could re-form, leading back

to reactants.

(a) ClI – C

H

H

H

I C

HH

H

Cl

–

I C

HH

H

Cl –

Before collision Transition state After reaction


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(b)ClC

H

H

H

I –

ClC

H

H

H

I –

Figure 10: (a) A collision that could lead to reaction of – I + CH3Cl to give CH3I +

– Cl . The – I must approach

along the “back side” of the C—Cl bond. (b) Two collisions that are not in the “correct” orientation.

REACTION MECHANISMS AND THE RATE-LAW EXPRESSION

The step-by-step pathway by which a reaction occurs is called its mechanism. Some reactions take place in a

single step, but most reactions occur in a series of elementary steps.

The reaction orders are equal to thecoefficients for that step.

for any single elementary step

In many mechanisms, however, one step is much slower than the others.

A reaction can never occur faster than its slowest step.

This slow step is called the rate-determining step. The speed at which the slow step occurs limits the rate at

which the overall reaction occurs.

The balanced equation for the overall reaction is equal to the sum ofall the individual steps, including any steps that

might follow the rate-determining step. We emphasise again that the rate-law exponents do not necessarily match

the coefficients of the overall balanced equation.

For the general overall reaction

A B C D+ → +a b c d

the experimentally determined rate-law expression has the form

rate = k [A] x[B] y

The values of x and y are related to the coefficients of the reactants in the slowest step, influenced in some cases

by earlier steps.

Using a combination of experimental data and chemical intuition, we can postulate a mechanism by which a

reaction could occur. We can never prove absolutely that a proposed mechanism is correct. All we can do is

postulate a mechanism that is consistent with experimental data. We might later detect reaction-intermediate

species that are not part of the proposed mechanism. We must then modify the mechanism or discard it and propose a new one.


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As an example, the reaction of nitrogen dioxide and carbon monoxide has been found to be second order with

respect to NO2 ad zero order with respect to CO below 225° C.

2 2 NO (g) CO(g) NO(g) CO (g)+ → + rate = k[NO2]2

The balanced equation for the overall reaction shows the stoichiometry but does not necessarily mean that thereaction simply occurs by one molecule of NO

2 colliding with one molecule of CO. If the reaction really took place

in that one step, then the rate would be first order in NO2 and first order in CO, or rate = k[NO

2][CO]. The fact

that the experimentally determined orders do not match the coefficients in the overall balanced equation tells us that

the reaction does not take place in one step.

The following proposed two-step mechanism is consistent with the observed rate-law expression.

2 2 2 4

2 4 2 2

2 2

(slow)

(fast)

overall

(1) NO NO N O

2) N O CO NO CO NO(

NO CO NO CO

+ →+ → + +

+ → +

The rate-determining step of this mechanism involves a bimolecular collision between two NO2 molecules. This

is consistent with the rate expression involving [NO2]2. Because the CO is involved only after the slow step has

occurred, the reaction rate would not depend on [CO] (that is, the reaction would be zero order in CO) if this

were the actual mechanism. In this proposed mechanism, N2O

4 is formed in one step and is completely consumed

in a later step. Such a species is called a reaction intermediate.

However, in other studies of this reaction, nitrogen trioxide, NO3, has been detected as a transient (short-lived)

intermediate. The mechanism now thought to be correct is

2 2 3

23

2 2

(slow)

(fast)

overall

(1) NO NO NO NO2) NO CO NO CO(

NO CO NO CO

+ → ++ → +

+ → +

In this proposed mechanism two molecules of NO2 collide to produce one molecule each of NO

3 and NO. The

reaction intermediate NO3 then collides with one molecule of CO and reacts very rapidly to produce one molecule

each of NO2 and CO

2. Even though two NO

2 molecules are consumed in the first step, one is produced in the

second step. The net result is that only the NO2 molecule is consumed in the overall reaction.

Each of these proposed mechanisms meets both criteria for a plausible mechanism: (1) The steps add to give the

equation for the overall reaction, and (2) the mechanism is consistent with the experimentally determined rate-law

expression (in that two NO2 molecules and no CO molecules are reactants in the slow step). The NO

3 that has

been detected is evidence in favour of the second mechanism, but this does not unequivocally prove that mechanism;

it may be possible to think of other mechanisms that would involve NO3 as an intermediate and would also be

consistent with the observed rate law.

Important Points

1. In the reaction discussed above, the slowest (rate determining) step in the mechanism involves a collision

of two molecules. Such a step is said to be bimolecular. Likewise, depending upon thenumber of molecules,

ions or atoms taking part in the rate determining step, we classify the elementary reactions according to:

unimolecular : when only one species takes part

bimolecular : when two species take part

termolecular : when three species take part

The probability of a reaction where 4 molecules take part in a single step is almost nil.


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2. Molecularity (like order) cannot be predicted from the stoichiometric coefficients of a balanced equation.

It is determined theoretically after proposing the reaction mechanism (whereas order as we’ve seen is

determined experimentally).

Please Answer this Question before seeing its solution.

• You are a chemist of a research laboratory that is trying to increase the reaction rate for the balanced

chemical reaction: X 2Y Z.+ →

a. One of your researchers comes into your office and states that she has found a material that significantly

lowers the activation energy of the reaction. Explain the effect this will have on the rate of the reaction.

b. Another researcher states that after doing some experiments, he has determined that the rate law is

rate = k [ X ][Y ]. Is this possible?

c. Yet another person in the lab reports that the mechanism for the reaction is:

2Y I→ (slow)

X I Z+ → (fast)

Is the rate law from part b. consistent with this mechanism? If not, what should the rate law be?

Solution: a. Her finding should increase the rate since the activation energy, Ea, is inversely related to the rate

constant, k ; a decrease in Ea results in an increase in the value of k .

b. This is possible because the rate law does not have to reflect the overall stoichiometry of the

reaction.

c. No. Since the rate law is based on the slow step of the mechanism, it should be Rate = k[Y]2.

Pseudo first order reactions.

Consider the following reaction:

CH3 COO C

2H

5 + H

2O H

+

→ CH3 COOH + C2 H5 OH

t = 0 0.02 M 100 M 0 M 0 M

t 0 M 99.98 M 0.02 M 0.02 M

This reaction is a bimolecular elementary reaction. Therefore, its rate law can be written as:

Rate = K [CH3 COO C

2H

5] [H

2O].

But observing the data above, we see that H2O is in large excess and its concentration has almost remained

constant in the reaction. So, if we consider [H2O] as a constant, then, the rate law may be rewritten as:

Rate = { K [H2O]} [CH

3 COO C

2H

5]

⇒ Rate = K ' [CH3 COO C2H5]

We see that this rate law is same as the one for a first order reaction. This is an example of a pseudo-first order

reactions. They are defined as: the reactions in which the molecularity of the reaction is 2 or more but they follow first order kinetics are called as pseudo first order reactions.


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Temperature: The A rrhenius Equat ion

The average kinetic energy of a collection of molecules is proportional to the absolute temperature. At a particular

temperature, T 1, a definite fraction of the reactant molecules have sufficient energy, Ea, to react to form product

molecules upon collision. At a higher temperature, T 2 a greater fraction of the molecules possess the necessary

activation energy, and the reaction proceeds at a faster rate. This is depicted in Figure 11.

From experimental observations, Svante Arrhenius developed the mathematical relationship among activation

energy, absolute temperature, ad the specific rate constant of a reaction, k , at that temperature. The relationship,

called the Arrhenius equation, is

a – E / RTAe=k

or, in logarithmic form,

a aE Eln ln A – or log log A – RT 2.303 RT= =k k

In this expression, A is a constant having the same units as the rate constant. It is proportional to the frequency of

collisions between reacting molecules.

Actually A (called frequency factor or pre-exponential factor) is equal to Z ρ i.e. A = Z ρ where Z is the number of collisions of the molecules per second in a unit volume and ρ is the steric factor. The necessity of introducing thefactor ρ in the Arrhenius equation is explained by the fact that the collisions even between active molecules(i.e. molecules with sufficient energy to bring about reaction) do not always result in a reaction, but only when the

molecules have a definite orientation. (Refer figure – 8 of section – 3 for an example). The factor ρ is proportionalto the ratio of the ratio of the number of ways of the mutual orientation of the molecules favourable for proceeding

of a reaction to the total number of possible ways of orientation: the greater this ratio, the more rapidly will areaction proceed. The steric factor ρ is usually much smaller than unity; it has an especially great influence on therate of reactions proceeding with the participation of complex molecules (for example, glucose and proteins),

when the total number of various possible orientations is very great, and the number of orientations favourable for

proceeding of a reaction is very limited.

R is the universal gas constant, expressed with the same energy units in its numerator as are used for Ea. For

instance, when Ea is known in J/mol, the value R = 8.314 J/mol ⋅K is appropriate. Here the unit “mol” is interpreted

as “mole of reaction,” as described the unit on thermodynamics. One important point is the following: The greater

the value of Ea, the smaller the value of and the slower the reaction rate (other factors being equal). This is because

fewer collisions take place with sufficient energy to get over a high energy barrier.

The Arrhenius equation predicts that increasing T results in faster reaction for the same Ea and concentrations.

F r a c t i o n o f m o l e c u l e s w i t h

a g i v e n k i n e t i c e n e r g y

Kinetic energy Ea

T 1

T 2

T T 2 1>

Figure 11: The effect of temperature on the number of molecules that have kinetic energies greaterE . At , a higher fraction of molecules posses at least E , the activation energy.than a 2 aT

The area between the distribution curveand the horizontal axis in Figure is proportionalto the total number of molecules present. The totalarea is the same at and . The shaded areas

represent the number of particles that exceed theenergy of activation, E .

T T 1 2

a

Section - 4 THE ARRHENIUS EQUATION


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a – E / RTa aE / RT – E / Rt e

increasesdecreases increases increases⇒ ⇒ ⇒ ⇒ ⇒If Reaction

increases speeds upk T

Let’s look at how the rate constant varies with temperature for a given single reaction. Assume that the

activation energy and the factor A do not vary with temperature. We can write the Arrhenius equation for

two different temperatures. Then we subtract one equation from the other, and rearrange the result to

obtain, in natural logarithm (ln) form,

a2

1 1 2

Ek 1 1ln –

k R

=

T T

In base-10 logarithm (log) form this equation is written as

a2

1 1 2

Ek 1 1log –

k 2.303R

=

T T

Let’s substitute some typical values into this equation. The activation energy for many reactions that occur

near room temperature is about 50 kJ/mol (or 12 kcal/mol). For such a reaction, a temperature increase

from 300 K to 310 K would result in

2

1

k 50,000 J / mol 1 1ln – 0.647

k (8.314 J /mol K) 300 K 310 K

= = ⋅

2

1

k 1.91 2k

= ≈

Chemists sometimes use the rule of thumb that near room temperature the rate of a reaction approximately

doubles with a 10° C rise in temperature. Such a “rule” must be used with care, however, because it

obviously depends on the activation energy.

The specific rate constant, k , for the following first-order reaction is 9.16 × 10 – 3 s – 1 at 0.0° C. The activation

energy of this reaction is 88.0 kJ/mol. Determine the value of k at 2.0° C.

2 5 2 3 N O NO NO→ +

Critical thinking

First we tabulate the values, remembering to convert temperature to the Kelvin scale.

E a = 88,000 J/mol R = 8.314 J/mol.K

k 1 = 9.16 × 10 –3 s –1 at T

1 = 0.0° C + 273 = 273 K

k 2 = ? at T

2 = 2.0° C + 273 = 275 K

We use these values in the "two-temperature" form of the Arrhenius equation.

Example – 13


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Solution:a2

1 1 2

Ek 1 1ln

k R T T

= −

2

– 3 –1

k 88,000 J / mol 1 1ln – 0.282

J9.16 10 s 273 K 275 K 8.314

mol K

= = × ⋅

Taking inverse (natural) logarithms of both sides.

2

– 3 –1

k 1.32

9.16 10 s=

×

( ) –3 –1 – 2 –12k 1.32 9.16 10 s 1.21 10 s= × = ×

We see that a very small temperature difference, only 2°C, causes an increase in the rate constant

(and hence in the reaction rate for the same concentrations) of about 32%. Such sensitivity of rate to

temperature change makes the control and measurement of temperature extremely important in chemical

reactions.

The gas-phase decomposition of ethyl iodide to give ethylene and hydrogen iodide is a first-order reaction.

2 5 2 4C H I C H HI → +At 600 K, the value of k is 1.60 × 10 – 5 s – 1. When the temperature is raised to 700 K, the value of k increases to

6.36 × 10 – 3 s – 1. What is the activation energy for this reaction?

Criti cal thinking

We know k at two different temperatures. We solve the two-temperature form of the Arrhenius equation

for E a and evaluate

Solution: –5 –1 –3 –11 1 2 21.60 10 s at T 600 K 6.36 10 s at T 700 K k k = × = = × =

aR 8.314 J / mol K E ?= ⋅ =

We arrange the Arrhenius equation for Ea.

2

a2 1a

1 1 2

1 2

R lnE 1 1

ln so ER T T 1

T T

k

k k

k

= − = 1 −

Example – 14


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Substituting,

( ) –3 –1

–5 –1

a – 4 –1

J 6.36 10 s J8.314 ln 8.314 5.98

mol K 1.60 10 s mol K E

2.38 10 K 1 1

– 600 K 700 K

× ⋅ × ⋅ = =

×

Ea = 2.09 × 105 J/mol ( i.e. 209 kJ/mol)

____________________________________________________________________________

The determination of Ea in the manner illustrated in the last Example may be subject to considerable error, because

it depends on the measurement of k at only two temperatures. Any error in either of these k values would greatly

affect the resulting value of Ea. A more reliable method that uses many measured values for the same reaction is

based on a graphical approach. Let us rearrange the single-temperature logarithmic form of the Arrhenius equation

and compare it with the equation for a straight line.

ln k

y

= – EaR

1

T+ ln A

= m x b+

The value of the collision frequency factor, A, is very nearly constant over moderate temperature changes. Thus, ln

A can be interpreted as the constant term in the equation (the intercept). The slope of the straight line obtained by

plotting ln k versus 1/T can be interpreted as – Ea/ R. This allows us to determine the value of the activation energy

from the slope (Figure 12).

Slope =

l n

k

1/T

EaR

Figure 12: A graphical method for determining activation energy, Ea. At each of several different temperatures,

the rate constant, k , is determined by methods such as those in Section 2. A plot of ln k versus 1/T gives astraight line with negative slope. The slope of this straight line is – Ea/R. Alternatively, a plot of log k versus

1/T gives a straight line whose slope is – Ea/2.303 R. Use of this graphical method is often desirable, because it

partially compensates for experimental errors in individual k and T values.

Please attempt the following problem before seeing its solution:

• The chemical reaction A→ B + C has a rate constant that obeys the Arrhenius equation. Predict whathappens to both the rate constant k and the rate of the reaction if the following were to occur:

a. a decrease in temperature.

b. an increase in the activation energy of the forward and reverese reactions.c. an increase in both activation energy and temperature.


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Solution: The Arrhenius equation is aE / RTAek −=

a. When the temperature is decreased, the rate constant, k , will also decrease. When k decreases,

the rate also decreases.

b. When the activation energy is increased, the rate constant, k , also decreases. When k decreases,the rate also decreases.

c. Since the activation energy is in the numerator and the temperature is in the denominator, you

cannot predict the effect without knowing the magnitude of the changes.

Catalysts

Catalysts are substances that can be added to reacting systems to increase the rate of reaction. They allow

reactions to occur via alternative pathways that increase reaction rates by lowering activation energies. The catalyst

causes unstable intermediates (called activated complexes) to appear as a new transition state which has a lower

energy as compared to the transition state corresponding to activated complexes of the uncatalysed reactions.

Since, the new transition state has lower energy, a much greater number of reacting species have the required

energy to reach this new transition state. This increases the rate of the reaction. Once this transition state is

attained, its decomposition leads to the formation of the products.

The activation energy is lowered in all catalysed reactions, as depicted in Figures 13 and 14. A catalyst does take

part in the reaction, but all of it is regenerated in later steps. Thus a catalyst does not appear in the balanced

equation for the reaction.

For constant T and the same concentrations,

a – E /RTa aE / RT – E / RT e

increasesdecreases decreases increases increases⇒ ⇒ ⇒ ⇒ ⇒aIf E Reaction

speeds up

k

Eforward

a

E

reversea

!E

Reaction coordinate for uncatalysed reaction

P o t e n t i

a l e n e r g y E

forwarda"

E

reverse

"a

!E

Reaction coordinate for catalysed reaction

P o t e n t i a l e n e r g y

E < E"a a

Figure 13: Potential energy diagrams showing the effect of a catalyst. The catalyst provides a different,

lower-energy mechanism for the formation of the products. A catalysed reaction typically occurs in several

steps, each with its own barrier, but the overall energy barrier is lower than the uncatalysed reaction. E∆ hasthe same value for each path. The value of E∆ depends only on the states of the reactants and products.


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Kinetic energy

F r i c t i o n o f m o l e c u l e s w i t h

a g i v e n k i n e t i c e n e r g y

Minimum kinetic energyfor uncatalysed reaction

Minimum kinetic energyfor catalysed reaction

Figure 14: When a catalyst is present, the energy barrier is lowered. Thus, more molecules possess the

minimum kinetic energy necessary for reaction. This is analogous to allowing more students to pass a course

by lowering the requirements.

Please attempt the following problem before seeing its solution:

• Considering the potential energy curves for two different reactions:

E n e r g y p e r m o l

E + F

G + H

Progress of reaction

E + F G + H

E n e r g y p e r m o l

A + B

C + D

Progress of reaction

A + B C + D

a. Which reaction has higher activation energy for the forward reaction?b. If both reactions were run at the same temperature and have the same orientation requirements to

react, which one would have the larger rate constant?

c. Are these reactions exothermic or endothermic?

Solution: a. Since the “hump” is larger, the A + B reaction has a higher activation energy.

b. Since the activation energy is lower, the E + F reaction would have the larger rate constant. Keep

in mind the inverse relationship between the activation energy, Ea, and the rate constant, k .

c. Since in both cases energy per mole of the reactants is greater than the products, both reactionsare exothermic.


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The activation energy of a reaction is 75.24 kJ/mol in the absence of a catalyst, and 50.14 kJ/mol with a catalyst.

How many times will the rate of the reaction grow in the presence of a catalyst if the reaction proceeds at 25° C?

Criti cal thinking

When the reaction is catalysed, the rate constant of the reaction increases and E a decreases. However

A remains constant. We can use this information to calculate the ratio of rate constants between

two alternative paths. This ratio shall give us the relative rates quite successfully.

Solution: Let the activation energy of the reaction without a catalyst be Ea, and with one, aE′ , let k and k ' be the

respective rate constants of the reaction. Using the Arrhenius equation, we find:

( )( )a a a

a

exp – E / RTE Eexp

exp – E / RT RT′ ′′ − = = k k

hence

a a a aE E E Ek k ln 2.303 log , and logk RT k 2.303RT

′ ′′ ′ ′− −= = =

k

k

Introducting the data of the example into the last equation, expressing the activation energy in joules,

and taking into account that

T = 298 K, we get:

( ) 375.24 – 50.14 10k 25.1 10log 4.40

k 2.30 8.314 298 2.30 8.314 298

3×′ ×= = =

× × × ×

We finally obtain k / k ′ = 2.5 × 104. Hence, a decrease in the activation energy by 25.1 kJ resulted inthe reaction rate growing 25000 times.

From the following data for the reaction between A and B, calculate (i) the order of the reaction with respect to

A and with respect to B, (ii) the rate constant at 300 K, (iii) the energy of activation, and (iv) the pre-exponential

factor.

[A]/mol L – 1

[B]/mol L – 1

Initial rate/mol L – 1

s at – 1

300 K 320 K

2.5 × 10 – 4

5.0 × 10 – 4

1.0 × 10 – 3

3.0 × 10 – 5

6.0 × 10 – 5

6.0 × 10 – 2

5.0 × 10 – 4

4.0 × 10 – 3

1.6 × 10 – 2

2.0 × 10 – 3

–

–

Example – 15

Example – 16


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Critical thinking

Sub-questions (i) and (ii) are left for you as an exercise. The answer are provided for cross-checking.

In sub-question (iii), we cannot calculate E a by the “one-temperature form” of the Arrhenius equation

as we don’t have A (called some times as pre-exponential factor). So, the other alternative is to see for the possibility of the two temperature form of Arrhenius equation to give us the solution. In such

a case we would need ratio of the rate constants of the same reaction at two different temperatures.

But so far in (i) and (ii) we have only got the value of rate constant at one particular temperature

(300 K). So, what do we do? Please go back to Example – 12 and note that in cases where ratio of

two quantities are involved we can substitute it with ratio of two other quantities which are directly

proportional to the ratio of the first two quantities.

Here, we know that the rate of a reaction at a particular temperature is proportional to the rate

constant at that particular temperature. Note also that the data given in table has two rates of

reaction at two different temperatures for first choice of concentrations of A and B. Can you use

this information to find your substitute for1

2

k k

?

Solution: The two temperature form of the Arrhenius equation is:

a2

1 1 2

Ek 1 1log

k 2.303 R T T

= −

... (i)

But, we know that,

[ ] [ ]2 1

2 2(rate) k A B= ... (ii)

and

[ ] [ ]2 1

1 1(rate) k A B= ... (iii)

Dividing (ii) by (iii) to get:

[ ] [ ]

[ ] [ ]

2 1

222 1

1 1

k A B(rate)

(rate) k A B

= ... (iv)

Hence, if we substitute A and B at the same instant, we get:

( )

( )22

11

rate k

rate k = (for the first choice of concentrations of A and B)

⇒ – 3

2

– 41

k 2.0 10

k 5.0 10

×=

× ... (v)


15/29



Substituting (v) in (i), we get,

( )( )

– 3a

– 4

E2.0 10 1 1log

2.303 8.314 J / K /mol 300 K 320 K 5.0 10

×= − ×

⇒( )( )( )( )

( )a

2.303 8.314 300 320 log 4E

320 300=

−

= 5.53 × 104 J/mol

= 55.33 kJ/mol.

(iv) To calculate the pre-exponential factor A, it is easier if we take logarithm of the Arrhenius equation.

Hence,

aElog A log k 2.303RT

= + ... (A)

Here, k = 2.67 × 108 (at 300 K) (You should have calculated it in sub-question (ii) above)

Ea = 5.53 × 104 J/mol

R = 8.314 J K – 1 mol – 1

T = 300 K

Substituting these values in (A) we get

( )( )( )( )

4 –18 5.53 10 J mollog A log 2.67 10

2.303 8.314 J / K / mol 300 K

×= × +

⇒ log A = 8.43 + 9.63 = 18.06

⇒ A = 1.145 × 108 M – 2 s – 1.

Answer (i) order w.r.t. A = 2 order w.r.t. B = 1

Answer (ii) k = 2.67 × 108 M – 2 s – 1.

A first-order reaction, A B→ , requires activation energy of 70 kJ mol – 1. When a 20% solution of A was kept at25° C for 20 minutes, 25% decomposition took place. What will be percent decomposition in the same time in a

30% solution maintained at 40° C? Assume that activation energy remains constant in this range of temperature.

Example – 17


16/29



Critical thinking

First of all we need to understand an important characteristic of first order reactions. Consider the

reactions:

(first order in A) (second order in A) (zero order in A)A B ...(i) and A C ...(ii) and A D ...(iii)→ → →

We have seen that the integrated rate equation for the reaction (i), (ii) and (iii) are:

[ ]

[ ]0

t

Aln akt

A= (first order) ... (iv)

[ ] [ ]t 0

1 1akt

A A− = (second order) ... (v)

[ ] [ ]0 t

A A akt− = (zero order) ... (vi)

Suppose we have two initial concentrations for A, say, 2M and 5M respectively. And we want to knowthe time after which a certain percentage of the reactants have reacted (say, 25%). Then in the first

order case, we will see:

2Mln akt

1.5M= and

5Mln akt

3.75M=

(1.5 M and 3.75M is what is left after 25% decomposition in the respective cases)

⇒ ln1.33 akt= and ln1.33 akt=

⇒ ( )ln 1.33

t

ak

= and ( )ln 1.33

t

ak

=

We see that, in both the cases, the time required is same and observing closely we can realise, that the

extent of reaction in a first order reaction is independent of the initial concentration.

You may use the same two concentrations in second order and zero order case to see that the extent of

reaction is dependent upon the initial concentration.

Zero order case is shown:

2M 1.5M akt− = and 5M 3.75M akt− =

⇒0.5M

tak

= and t1.25M

ak =

Since, the time required for 25% decomposition is different in both the cases, we can see that theextent of reaction in the same time period is different too and is dependent upon the initial concentration

in this case.

Solution: We know that the fraction of reactants reacted in a first order reaction is independent of initial concen-

tration. So, we need not worry about the terms 20% solution and 30% solution.

Now, it is given that 25% of A is decomposed in 20 minutes at 25° C. And we need to find the percent

decomposition in the same time at 40° C for the same reaction.

Quite obviously, the rate constants of this reaction at two temperatures are different. We can also see

that the rate constant for the first reaction can be calculated using the integrated rate equation and then

this can be used in the “two-temperature” form to get the rate constant of the reaction at 40ºC. This

rate constant at 40ºC can then be used in the integrated rate equation to obtain the required data.We illustrate the above methodology we have planned over here:


17/29



We know that,

0

t

[A] aktlog

[A] 2.303= ...(vii)

Here, at 25ºC,

a = 1

k 1 = ?

t = 20 min.

0

t

[A] 100

[A] 275= (as 25% decomposition has taken place)

Substituting these values in (vii), we get,

1100 1. k .(20 min)log75 2.303

=

⇒ k 1 = 1.44 × 10 –2 min –1

Now, letting k 2 be the rate constant of this reaction at 40ºC, we can use the two temperature form of

Arrhenius equation as:

a2

1 1 2

Ek 1 1log

k 2.303 R T T

= −

⇒3

22 1 1

k 70 10 J / mol 1 1log(1.44 10 ) 2.303 8.314 JK mol 298 K 313 K − − −

×= − × × Solving for k

2, we will get,

2 1

2k 5.56 10 min− −= ×

Finally,

t 2

0

[A] k log t

[A] 2.303

= − This remains 20 min. because we have been

asked to find the percentage decomposition at40ºC in the same time interval as the firstreaction (at 25ºC) during which 25%

decomposition took place.

2 15.56 10 min

2.303

− − ×= −

(20 min)

⇒ t0

[A] 330.33

[A] 100= =

i.e.,

percentage of A remaining at (t = 20 min) = 33%

percentage of A reacted till (t = 20 min) = (100 – 33) = 67%


18/29



Q. 1 For a first-order reaction

(a) the degree of dissociation is equal to (1 – e – kt

)(b) a plot of reciprocal concentration of the reactant versus time gives a straight line

(c) the time taken for the completion of 75% reaction is thrice that of t1/2

of the reaction

(d) the pre-exponential factor in the Arrhenius equation has the dimension of time T – 1

Q. 2 In the Arrhenius equation, k = A exp (– Ea/RT), A may be termed as the rate constant at............. .

Q. 3 A catalyst is a substance which

(a) increases the equilibrium concentration of the product

(b) changes the equilibrium constant of the reaction

(c) shortens the time to reach equilibrium

(d) supplies energy to the reaction

Q. 4 A catalyst

(a) increases the average kinetic energy of reacting molecules

(b) decrease the activation energy

(c) alters the reaction mechanism

(d) increases the frequency of collisions of reacting species

Q. 5 For an endothermic reaction where H∆ represents the enthalpy of the reaction, the minimum value for theenergy of activation will be

(a) less than H∆ (b) zero(c) more than H∆ (d) equal to H∆

TRUE or FALSE (Q. 6 - Q. 8)

Q. 6 Catalyst makes a reaction more exothermic.

Q. 7 Catalyst does not affect the energy of activation in a chemical reaction.

Q. 8 The rate of an exothermic reaction increases with increasing temperature.

Q. 9 In the Arrhenius equation for a certain reaction, the value of A and Ea (activation energy) are 4 × 1013 s –1

and 98.6 kJ mol – 1, respectively. If the reaction is of first-order, at what temperature will its half-life period

be ten minutes?

Q. 10 A first-order reaction is 50 percent completed in 30 min at 27° C and in 10 min at 47° C. Calculate the

reaction rate constant at 27° C and the energy of activation of the reaction.

Q. 11 The trans → cis isomerisation of 1, 2-dichloroethylene proceeds with Ea = 231 kJ/mol and

H∆ = 4.2 kJ/mol. What is Ea for the cis →trans isomerisation?

TRY YOURSELF - III


19/29



Q. 12 Will the value of the rate constant of a reaction change (a) when one catalyst is replaced with another one,

and (b) when the concentrations of the reactants and products are changed?

Q. 13 Does the heat effect of a reaction depend on its activation energy? Substantiate your answer.

Q. 14 For which reaction—the forward or the reverse one—is the activation energy greater if the forward

reaction proceeds with the liberation of heat?

Q. 15 How many times will the rate of a reaction proceeding at 298 K grow if its activation energy is lowered by

4 kJ/mol?

Q. 16 What is the activation energy of a reaction if its rate doubles when the temperature is raised from 290 K to

300 K?

Q. 17 What is the value of the activation energy for a reaction whose rate at 300 K is ten times greater than at

280 K?

Q. 18 Which of the following procedures will lead to a change in the rate constant of a reaction? (a) A change in

the pressure; (b) a change in the temperature; (c) a change in the volume of the reaction vessel; (d) the

introduction of a catalyst into the system; (e) a change in the concentration of the reactants and products.

Q. 19 The increase in a reaction rate with temperature is due chiefly to: (a) an increase in the average kinetic

energy of the molecules; (b) a growth in the number of active molecules; (c) a growth in the number of

collisions.

Q. 20 The first-order gaseous decomposition of N2O

4 into NO

2 has a k value of 4.5 × 103 s – 1 at 1° C and an

energy of activation of 58 kJ mol – 1. At what temperature its half-life would be 6.93 × 10 – 5 s?


20/29



Under certain conditions the reaction between sulphate 24(S O )

− ions and hydrogen ions is first order in each of

them, i.e., rate 24k[SO ]

−= [H+]. A student measured the initial rate of the reaction four times using the combinationsof solutions shown in the following table. Write down the actual concentrations of both ions in each experiment and

hence determine the value of the rate in terms of k .

_____________________________________________________________________________

Experiment Solution of sulphate ions Solution of hydrogen ionsConcentration volume Concentration volume

/mol dm –3 /cm3 /mol dm –3 /cm3

_____________________________________________________________________________

A 0.5 125 0.05 125

B 0.5 250 0.05 250

C 0.5 500 0.05 500

D 0.5 500 0.05 125

_____________________________________________________________________________

Criti cal thinking

(1) You will have to think carefully about the difference between concentration and volume.

(2) In the rate law for a reaction, the concentration used are in terms of the number of moles of each

reacting species w.r.t. the total concentration of the solution. Here the concentrations given are

of the reactants before they have been brought together for reaction. We have been given the

volume of each of the reactants’ solution. When they are mixed, the total volume will change and

hence, the concentration of the reacting species will change too. It is these new concentrations of

the reacting species that should be put in the rate law to get the respective rate of reaction.

Solution: As pointed out the key thing to remember is that the rate will change if the concentrations of the

reagents change.

In experiment A when the two solutions are mixed, the total volume becomes 250 cm3. This is twice

the volume of each of the separate solutions, so their concentrations are halved just as they are brought

together to react.

Hence, the concentration of 24SO − (0.5 m in a 125 cm3 solution) becomes

0.5M

2 in a 250 cm3

solution. And the concentration of H+ ions (0.05 M in a 125 cm3 solution) becomes0.05

2 in a 250 cm3

solution.

MISCELLANEOUS EXAMPLES

Example – 18


21/29



So, rate of the reaction in this case will be given by,

rate0.5 0.05

k 2 2

=

[substituting nto the given rate law]

rate = k × 6.25 × 10 –3

You are requested to provide the solution for experiments B, C and D yourself. The answers are given

here:

(B) rate = k × 6.25 × 10 –3

(C) rate = k × 6.25 × 10 –3

(D) rate = k × 4.00 × 10 –2

A certain compound A, was found the undergo two parallel first order rearrangements, forming B and C respectively

in the two arrangements. At 25° C, the first order rate constant for the formation of B was measured as1.0 × 10 – 5 s – 1 and for the formation of C as 2.0 × 10 – 6 s – 1. What is the percentage distribution of the rearrangement

products?

Critical thinking

What the question says is that the same compound A has tendency to form two different products

by two dif ferent paths and both these different reactions can occur simultaneously:

A k 1 =

1. 0 × 1 0

s – 5

– 1

k 2 = 2 .0 × 1

0 s

– 6 – 1

B

C Here, we can write the various rate laws as:

[ ][ ] [ ]1 2

d A – k A k A

dt= + ... (i)

[ ][ ]1

d Bk A

dt= ... (ii)

[ ][ ]2

d Ck A

dt= ... (iii)

Dividing (ii) by (i), we get,

[ ]

[ ]

[ ]

[ ] [ ]1

1 2

d B

k Adtd A k A k A

dt

=+

−

⇒[ ]

[ ]

[ ]1k Ad B

d A=

− ( ) [ ]1 2k k A+ Hence LHS gives us the fraction of amount of B formed to amount of A reacted and this is equal to

1

1 2

k

k k +. Can you see that you have all the information and analysis you need to get the answers for

this question?

Example – 19


22/29



Solution: A

k 1

k 2

B

C

percentage of B formed can be obtained by 1

1 2

k 100

k k

= × +

–5

–5 –6

1.0 10100

1.0 10 2.0 10

×= × × + ×

–5

–5 –5

1 10100

1.0 10 0.2 10

×= × × + × = 83.33 %

percentage of A formed can be obtained by2

1 2

k 100

k k

= × +

– 6

–5 –6

2.0 10100

1.0 10 2.0 10

×= × × + ×

–5

–5 –5

0.2 10100

1.0 10 0.2 10

×= ×

× + × 16.67%= .

Two reactions (i) A → products and (ii) B → products follow first-order kinetics. The rate of the reaction (i) isdoubled when the temperature is raised from 300 K to 310 K. The half-life for the reaction at 310 K is 30 min. At

the same temperature B decomposes twice as fast as A. If the energy of activation for the reaction (ii) is half that

of reaction (i), calculate the rate constant of the reaction (ii) at 300 K.

Critical thinking

The problem is actually very simple as all you have to do is certain calculations in a logical order.

Arrange all the data in a diagram and it will be immediately clear to you so as to how you can solve

this question.

In the solution to this problem, you will be provided with the rough draft of the data and then how

it can be reorganised to see for yourself so as to how solution can be obtained.

It is expected of you that you make the rough draft and its reorganisation yourself first and then

compare it with the one given here.

It is expected then that you provide the complete solution yourself. The answer is provided for

cross-checking.

Example – 20


23/29



Solution: Rough Draft (data collection from Question)

( ) ( )A 300K A 310 K 2k k = ... (i)

( )12

t for (i) at 310 K 30 min=... (ii)

( ) ( )B 310 K A 310 K k 2 k = ... (iii)

( ) ( )a ii a i1

E E2

= ... (iv)

( )B 300 K k ?= ... (v)

Reorganised draft

2k = k A (300 K) A (310 )K

and

30 min =0.693

k A (310 K)

get value of k at both

temperaturesA

get value of k

(using (iii))B (310 K)

Use in two temperatureform of Arrhenius Equation

to get E we have E

(using (iv))a (i) a (ii)#

Use both these informations in the two temperature

form of Arrhenius equation to get k B (300 K)

Answer ( ) ( )

–1 –1

B 300K B 310 K

1 0.0462 mink k 0.0327 min

2 2= = =


24/29



EXERCISE

Rate of Reaction and Rate Law

Q. 1 How will the rate of reaction ( ) ( ) ( )2 22 NO g O g 2NO g+ → change if the volume of the reactionvessel is deminished to one third of its initial value? (The rate law for this reaction involves all reactants and

their respective orders are same as the stoichiometric coefficients.)

Q. 2 One mole of gas A and two moles of gas B are introduced into one vessel, and two moles of gas A and one

mole of gas B into a second vessel having the same capacity. The temperature is the same in both vessels.

Will the rate of the reaction between gases A and B in these vessels differ if it is expressed by the equation

(a) rate1 = k

1[A] [B]; (b) rate

2 = k

2 [A]2[B]?

Q. 3 The reaction between substances A and B is expressed by the equation A + 2B →C. The initialconcentrations of the reactants are [A]

0 = 0.03 mol/l and [B]

0 = 0.05 mol/l. The rate constant of the

reaction is 0.4. Find the initial rate of the reaction and the rate after a certain time when the concentrationof substance A diminishes by 0.01 mol/l. (rate = k[A][B]2)

Q. 4 How will the rate of the reaction 2NO(g) + O2(g) →2NO

2(g) change if (a) the pressure in the system is

increased three times; (b) the volume of the system is diminished to one-third of its initial value; and (c) the

concentration of the NO is increased three times? (rate = [NO]2[O2] )

Q. 5 Rate of a reaction A + B → products, is given below as a function of different initial concentrations of Aand B:

[A] (mol/l) [B](mol/l) Initial rate (mol/l/min)

0.01 0.01 0.005

0.02 0.01 0.010

0.01 0.02 0.005

Determine the order of the reaction with respect to A and with respect to B. What is the half-life of A in the

reaction?

Q. 6 For the given reaction, A + B → products, following data were obtained.

[A] (mol/l) [B](mol/l) Initial rate (mol/l/min)

[A0] [B

0] R

0(mol L – 1 s – 1)

1. 0.1 0.2 0.05

2. 0.2 0.2 0.10

3. 0.1 0.1 0.05

(a) Write the rate law expression (b) Find the rate constant

Q. 7 The reaction, H2(g) + Br

2(g) →2 HBr(g) has a rate given by the rate law, Rate = k [H

2][Br

2]1/2. (a) What

are the units of the rate constant k ? (b) If the volume of the gas mixture is halved, by what factor is the rate

changed?

Q. 8 The reaction v1A + v

2B → products is first-order with respect to A and zero-order with respect to B. If

the reaction is started with [A]0 and [B]

0, the integrated rate expression of this reaction would be

(a)[ ]

[ ]

0

10

Aln k t

A x=

−(b)

[ ]

[ ]

0

110

A

ln k tA v x =− (c)

[ ]

[ ]

01 1

10

Aln v k t

A v x

=−

(d)[ ]

[ ]

0

1 110

Aln –v k t

A v x=

−


25/29



Q. 9 Rate of a reaction, A + B → products is given as a function of different initial concentrations of A and B.

[A]/mol L – 1 [B]/mol L – 1 r0/mol L – 1 min – 1

_____________ _________ _________ _________ _________ _____ _________ _________ _________ _

0.01 0.01 0.005 0.02 0.01 0.010

0.01 0.02 0.005

Determine the order of the reaction with respect to A and with respect to B. What is the half-life of A in the

reaction?

Q. 10 For a first order reaction A →B, the concentration of B at two instant is given: [B] t = 10 sec = 1.0 M and[B]

t = 25 sec = 4.0 M. Using this information can you calculate the instantaneous rate of the reaction at at least

one instant in the interval from t = 10 sec to t = 25 sec.

Q. 11 If the steady-state concentration of O3 in a polluted atmosphere is 2.0 × 10 – 8 mol/L, the rate of production

of O3 is 7.2 × 10 –13 M/hr, and O3 is destroyed by the reaction, 2 O3 →3 O2, what is the rate constant for

the reaction, assuming a rate law, Rate = – (1/2) ∆ [O3]/ ∆ t = k [O

3]2?

Q. 12 In the enzymatic fermentation of sugar, the sugar concentration decreased from 0.12 M to 0.06 M in

10 hours, and to 0.03 M in 20 hours. What is the order of the reaction? What is the rate constant k ?

Q. 13 The rate law for the reaction, Ce4+(aq) + Fe2+(aq) →Ce3+(aq) + Fe3+(aq), is: Rate = (1.0 × 103 M –1s –1)[Ce4+][Fe2+]. If 0.500 L of 0.0020 M Ce(SO

4)

2 is rapidly mixed with 0.500 L of 0.0020 M FeSO

4, how

long does it take for [Fe2+] to decrease to 1.0 × 10 – 4 M?

Integrated Rate Equation

Q. 14 The following statement(s) is (are) correct:

(a) A plot of log K p versus 1/T is linear

(b) A plot of log [X] versus time is linear for a first order reaction, X →P(c) A plot of p versus 1/T is linear at constant volume

(d) A plot of p versus 1/V is linear at constant temperature

Q. 15 At constant temperature and volume, X decomposes as

x

2X(g) 3Y(g) 2Z(g); P → + is the partial pressure of X.

Observation No. Time (in minute) Px (in mm of Hg)

1 0 800

2 100 400

3 200 200

(i) What is the order of reaction with respect to X?

(ii) Find the rate constant.

(iii) Find the time for 75% completion of the reaction

(iv) Find the total pressure when pressure of X is 700 mm of Hg.


26/29



Q. 16 While studying the decomposition of gaseous N2O

5 it is observed that a plot of logarithm of its partial

pressure versus time is linear. What kinetic parameters can be obtained from this observation?

Q. 17 The rate constant for the first-order decomposition of N2O

5(g) to NO

2(g) and O

2(g) is 7.48 × 10 – 3 s – 1 at

a given temperature. (a) Determine the length of time required for the total pressure in a system containing N

2O

5 at an initial pressure of 0.1 atm to rise to 0.145 atm. (b) Find the total pressure after 100 s of the

reaction.

The Arrhenius Equation.

Q. 18 The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25° C are

3.0 × 10 – 4s – 1, 104.4 kJ mol – 1 and 6.0 × 1014s – 1 respectively. The value of the rate constant as T → ∞is,

(a) 2.0 × 1018 s – 1 (b) 6.0 × 1014 s – 1 (c) infinity (d) 3.6 × 1030 s – 1

Q. 19 At 380° C, the half-life period for the first order decomposition of H2O2 is 360 min. The energy of activation of the reaction is 200 kJ mol – 1. Calculate the time required for 75% decomposition at 450° C.

Q. 20 The rate constant for the first order decomposition of a certain reaction is described by the equation

log(k/s – 1) = 14.34 – (1.25 × 104 K)/T

(i) What is the energy of activation of this reaction?

(ii) At what temperature will its half-life be 256 minutes?

Q. 21 The rate constant of a reaction is 1.5 × 107 s – 1 at 50° C and 4.5 × 107 s – 1 at 100° C. Evaluate the

Arrhenius parameters A and Ea.

Q. 22 A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in the presence of a

catalyst at the same rate, the temperature required is 400 K. Calculate the activation energy of the reaction

if the catalyst lowers the activation barrier by 20 kJ mol – 1.

Q. 23 The activation energy of a certain reaction is 15 kJ/mol. The reaction is exothermic, yielding 19 kJ/mol.

What is the activation energy of the reverse reaction?

Q. 24 The activation energy of the reaction O3(g) + NO(g) →O

2(g) + NO

2(g) is 10 kJ/mol. How many times

will the rate of the reaction change when the temperature is raised from 27 to 37° C?

Q. 25 The Ea of the reaction M + N →O + P is 80 kJ/mol. At 50° C, the products are formed at the rate of 0.15mol/L/min. What will be rate of formation of products at 100° C?

Q. 26 The time required for 10% completion of a first order reaction at 298 K is equal to that required for its

25% completion at 308 K. If the pre-exponential factor for the reaction is 3.56 × 109 s –1, calculate its rate

constant at 318 K and also the energy of activation.

Q. 27 A substance A was found to undergo two parallel first-order rearrangements A →B and A →C with rateconstants 1.26 × 10 – 4 s – 1 and 3.8 × 10 – 5 s – 1, respectively. What were the percentage distribution of B and

C?


27/29



TRY YOURSELF

[ ANSWERS KEY ]

TRY YOURSELF - I

1. (a) 2. (d) 3. (b)(c)

4. (d) 5. (c) 6. (d)

7. 16 times 8. [A]0 = 0.042 mol/l

[B]0 = 0.014 mol/l

9.

r a t e o f r e a c t i o n

zeroth order

second order

first order

concentration of reactant

10. rate = k [conc.] for a first order reaction. We can see that the slope is given by the

value of k here.

TRY YOURSELF - II

1. (i) 5.0 × 10 – 5 mol L – 1 s – 1 (ii) 4.175 × 10 – 5 mol L – 1 s – 1

2. (c) 3. (c)

4. (i) 5.26 % (ii) 4.62 × 105 seconds

5. 3.44 × 10 – 3 M min – 1

6. k = 5.21 × 10 – 3 min – 1 ( )

( )

2002.303 – log

(30 min) 233.8=


28/29



TRY YOURSELF - III

1. (A), (D) 2. very high temperature (or) zero activation energy

3. (c) 4. (b), (c) 5. (c)

6. False 7. False

8. False (The rate of reaction increases with increasing temperature because the no. of molecules

with Ea increases as the temperature increases. This has nothing to do with the reaction being

exothermic/endothermic)

9. 311 K ≈ 10. k 27° C = 0.0231 min – 1 11. 227 kJ/molE

a = 43.85 kJ/mol

12. (a) Yes (b) No 13. No 14. reverse one

15. 5 times 16. 49.9 kJ/mol 17. 80.3 kJ/mol

18. (b), (d) 19. (b) 20. 283 K


29/29


[ ANSWERS KEY ]

1. grow by 27 times. 2. (a) No, (b) Yes

3. 3 × 10 – 5, 7.2 × 10 – 6 4. (a) It will grow 27 times

(b) It will grow 27 times

(c) It will grow 9 times

5. 1, 0, 1.386 minutes 6. (a) R 0 = k[A

0]

(b) 0.5 sec – 1

7. (a) (mol/L) – 1/2 s – 1 8. (c)

(b) 2.8

9. (a) one, zero

(b) 1.386 minutes

10. Hint: Use mean value theorem (differential calculus).

11. 0.25 M – 1 s – 1 12. (a) first order (b) 6.9 × 10 – 2 hr – 1

13. 9 seconds 14. (A), (B), (D)

15. (i) 1 16. first order related

(ii) 6.93 × 10 – 3 min – 1

(iii) 200 min

(iv) 950 mmHg17. (a) 47.7 seconds 18. (b)

(b) 0.180 atm.

19. 20.4 minutes. 20. Ea = 2.39 kJ/mol; T = 669 K

21. A = 5.45 × 1010 s – 1; Ea = 22012.7 J/mol. 22. 100 kJ/mol

23. 34 kJ/mol. 24. 1.14 times

25. Hint: (rate)100° C

= (rate)50° C

100 C

50 C

k

k

°

°

26. Hint

a298 K

a298 K 308 K 308 K

gives E ?k use in two temperature2.303 100 2.303 100log t logfrom of Arrhenius eqn to get Ek 90 k 75 k

== = → →

k 318K

= ?

aE / RT

(pre exponential factor)

use k A.e−

↓−

=

4 1318 K k 9.43 10 s− −= ×

aE 76.5 kJ / mol=Q. 27 76.83% B and 23.17% C

EXERCISE

Chemical Kinetics Part - 2

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