11 chapter reaction kinetics text book exercise

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CHAPTER11 REACTION KINETICS

TEXT BOOK EXERCISE

Q1. Multiple choice questions

(i) In a zero order reaction, the rate is independent of

(a) Temperature of reaction.

(b) Concentration of reactants.

(c) Concentration of products

(d) None of these

(ii) If the rate equation of a reaction 2A + B product is , rate

=k[A]2[B]. and A is present in large excess, then order of reaction

is

(a) 1 (b) 2

(c) 3 (d) none of these

(iii) The rate of reaction

(a) Increases as the reaction proceeds.

(b) Decreases as the reaction proceeds.

(c) Remains the same as the reaction proceeds.

(d) May decrease or increase as the reaction proceeds.

(iv) With increase of 10oC temperature, the rate of reaction doubles.

This increase in rate of reaction is due to:

(a) Decrease in activation energy of reaction.

(b) Decrease in the number of collisions between reactant

molecules.

(c) Increase in activation energy of reactants.

(d) Increase in number of effective collisions.

(v) The unit of the rate constant is the same as that of the rate of

reaction in

(a) First order reaction

(b) Second order reaction

(c) Zero order reaction

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(d) Third order reaction

Ans. (i) b (ii) a (iii) b (iv) d

(v) c

Q2. Fill in the blanks with suitable words. (i) The rate of an endothermic reaction_______wiht the

increase in temperature.

(ii) All radioactive disintegration nuclear is of ________order.

(iii) For a fast reaction the rate constant is relatively _______and

half-life is ________.

(iv) The second order reaction becomes ________if one of the

reactants is in large excess.

(v) Arrhenius equation can be used to find out_________of a

reaction.

An. (i) increases (ii) first (iii) large : small

(iv) First order (v) energy of activation

Q3. Indicate TRUE or FALSE as the case may be. (i) The half-life order reaction increases with temperature.

(ii) The reactions having zero activation energies are

instantaneous.

(iii) A catalyst makes a reaction more exothermic.

(iv) There is difference between rate law and the law of mass

action.

(v) The order of reaction is strictly determined by the

Stoichiometry of the balanced equation.

Ans. (i) False (ii) True (iii) False (iv) True

(v) False Q4. What is chemical kinetics? How do you compare chemical

kinetics with chemical equilibrium and thermodynamics?

Ans. Chemical thermodynamics can be used: (i) To predict whether or not a reaction will proceed to the

right, as written.

(ii) To predict the extent to which a reaction will proceed before

reaching a condition of equilibrium.

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(iii) To enable the amount of energy theoretically required by or

released during reactions to be calculated,

Thermodynamics cab tells us where the position of equilibrium lies

between the reactants and the products. It enables us to calculate the

equilibrium constants. It has the following limitations.

(i) It can only predict the us where possibility of a reaction but not

its success.

(ii) It cannot tell us about the rate at which a reaction will occur. It is

because time is not a thermodynamic variable.

(iii) It provides no information about the mechanism of the reaction.

This is because the change in any state function is path

independent.

Chemical kinetics can be used:

(i) To determinc the rate of reaction, that is the rate at which

products are formed or the rate at which reactions are used up in

the reaction.

(ii) To understand the factors that affects the rate of a reaction.

(iii) To find the mechanism of a reaction.

Example: When gaseous H2 and O2 are mixed, thermodynamics tell

us that H2O should be formed, since H2O is more energetically stable

than H2 and O2. It is a well known fact that at room temperature, a

mixture of H2 and O2 will not produce H2O. However, if the mixture is

sparked, water is produced with explosive violence, thereby proving the

prediction of thermodynamics. However, it does not give any indication

of how fast the reaction will proceed.

Q5. The fast of a chemical reaction with respect to products is written

with positive sing, but with respect to reactants is written with a

negative sing. Explain it with reference to the following

hypothetical reaction.

aA + bB cC + dD

Ans. The concentration of reactants decrease with time whereas the

concentration of products increases with time.

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The rate of disappearance of reactant is always negative. A

negative sign in the rate expression indicates that the

concentration of the reactant decreases with time.

For the general reaction:

aA + bB cC + dD

The rate of reaction is given by

Rate =- . = - . = + . = + .

Remember that in order to have a unique value of the reaction rate

(independent of the concentration term chosen), it is necessary to divide

each concentration change by its coefficient in the balanced equation for

the reaction.

Q6. What are instantaneous and average rates? Is it true that the

instantaneous rate of a reaction at the beginning of the reaction is

greater than average rate and becomes far less than the average rate

near the completion of reaction?

Q7. Differentiate between (i) Rate and rate constant of a reaction

Ans.

Rate of reaction Rate constant of

reaction

1. “The change

in concentration of

a reactant or

product per unit

time is called the

rate of reaction.”

2. The reaction

takes place in one

1. “The proportionality

constant in rate law

equation which relates

concentration to the rate

of reaction is called rate

constant of reaction.”

2. It is independent of

the concentration of the

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phase.

3. The

mechanism

involves the

formation of an

intermediate

substance.

reactant.

3. Its units depend

upon the order of reaction

and differs according to

order of reaction.

(ii) Homogeneous and heterogeneous catalyses

Ans.

Homogeneous catalyses Heterogeneous catalysis

1. “The process in which the

catalyst and

the reactants are in the same

phase is called homogeneous

catalysis.”

2. The reaction takes place in

one phase. The catalyst is

uniformly distributed and is

itself in the same phase.

3. The mechanism involves

the formation of an intermediate

substance.

1. “The process in which the

catalyst and the reaction are in

different phases is called

heterogeneous catalysis.”

2. The reaction takes place at

a phase boundary. The gaseous

molecules react at the surface of

the solid catalyst.

3. The mechanism involves

adsorption. Molecules are bound

to „active sites‟ on a solid

surface.

(iii) Fast step and the rate determining step

Ans.

Fast step Rate determining step

1. “A relatively rapid step of

the reaction mechanism is

called fast step.”

2. The overall rate of reaction

is independent if this step.

3. The number of molecules of

each reactant taking part in this

step does not appear in the rate

1. “The slowest step of the

reaction mechanism is called

rate determining step.”

2. The rate of this step

determines the overall rate of

reaction.

3. The number of molecules of

each reactant taking part in this

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equation. step appears in the rate equation.

Remember that except the slow step, all other steps of the reaction

mechanism are normally faster steps. The terms slow and fast are

relative. They don not necessarily imply that the reaction itself is slow or

fast.

(iv) Enthalpy change of reaction and energy of activation of reaction

Ans.

Enthalpy change of reaction Energy of activation of

reaction

1. “The heat change when the

reaction is carried out at

constant pressure is called

enthalpy change of reaction.”

2. It is the amount of heat

evolved or consumed in the

course of reaction.

3. It is given the symbol, H

1. “ The minimum amount of

energy , in addition to the

average kinetic energy. Which

the reactant molecules must

have for effective collisions is

called activation energy of

reaction.”

2. It is the minimum amount

of energy required to initiate a

reaction.

3. It is given the symbol, Ea

Q8. Justify the following statements

(i) Rate of chemical reaction is an ever changing parameter under

the given conditions.

Ans. The rate of chemical reaction change with time. It has a

maximum value at the beginning of the reaction, then gradually

decreases and finally becomes zero when the reactants are totally

converted into products.

(ii) The reaction rate decreases every moment but rate constant „k‟

of the reaction is constant quantity, under the given conditions.

Ans. The rate of chemical reaction is not uniform. It depends upon the

molar concentrations of reactants. The rate of reaction is directly

proportional to concentrations of reactants. Since the concentration

of reactants decreases every moment, so the rate of reaction, then

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gradually decreases with time. It has a maximum value at the start

of the reaction, then gradually decreases and finally becomes zero.

The constant, „k‟ of the reaction is independent of concentration.

So, it remains constant under the given conditions. However, it

changes with temperature.

(iii) 50% of a hypothetical first order reaction completes in one

hour. The remaining 50% needs more than one hour to complete.

Ans. Calculation of k:

For a first order reaction

T1/2 =

1 hour =

k=0693 hours-1

Calculation of time for remaining 50%

Completion of reaction:

For first order reaction rate equation:

K= log

t= log

Where [A]o is initial concentration and [A] is concentration at any time t.

[A]=

t =

t = =

t =5.6457 hours

Hence remaining 50% needs more than one hour to complete.

(iv) The radioactive decay is always a first order reaction.

Ans. Since the half-life of a first order reaction in constant and is

independent of initial concentration of the reaction, so the radio-

active decay is always a first order reaction.

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(v) The unit of rate constant of a second order reaction is dm3 mol

-

1s

-1 , but the unit of rate of reaction is mol dm

-3s

-1.

Ans. For second order reaction: A+B Products

Rate =k[A]2

Units of rate are: mol dm-3

s-1

.

Units of concentration of A: mol dm-3

Therefore, mol dm-3

s-1

=k(mol dm-3

)2

mol dm-3

s-1

=(mol2 dm

-6)

k= =mol-1

dm3s

-1

So, the units of rate constant of 2nd

order reaction are: dm3 mol

-

1s

-1

(vi) The sum of the coefficients of a balanced chemical equation is

not necessarily important to give the order of a reaction.

Ans. The order of a chemical reaction is the sum of the exponents of

the concentration terms in the rate equation. It cannot be written by

merely looking at the balanced chemical equation. Usually, the

powers of concentration terms in the rate equation are different

from coefficients of reactants in the balanced equation. The order

of a reaction can be determined only by experiment. For example,

for N2O5, the rate is proportional to [N2O5] and its power is one

although the coefficient of N2O5in the balanced equation is two.

N2O5 4NO2+O2

Rate = k [N2O5]

A reactant whose concentration does not affect the reaction rate

is not included in the rate equation. The total number of molecules

or atoms taking part in a reaction can never be zero or in fraction

while the order of reaction can be zero or in fraction.

(vii) The order of a reaction is obtained from the rate expression of a

reaction and the rate expression is obtained from the experiment.

Ans. The rate expression can be only being determined

experimentally. It expresses actual dependence of the rate on the

concentrations of the reactants. A reactant whose concentration

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does not affect the reaction rate is not included in the rate equation.

The order of reaction is the number of reacting molecules whose

concentrations alter as a result of a chemical reaction. Hence the

order of a reaction is obtained from the rate equation and the rate

expression is obtained from the experiment.

Q9. Explain that half-life method for measurement of the order of a

reaction can help us to measure the order of even those reactions

which have a fractional order.

Q10. A cure is obtained when a graph is plotted between time on x-

axis and concentration on y-axis. The measurement of the slopes of

various points give us the instantaneous rates of reaction. Explain

with suitable examples.

Q11. The rate determining step of a reaction is found out from the

mechanism of that reaction. Explain it with few examples.

Q12. Discuss the factors which influence the rates of chemical

reactions.

Q13. Explain the following facts about the reaction.

2NO(g) + 2H2(g) 2H2O(g) +N2(g)

(i) The changing concentrations of reactants change the

rates of this reaction.

(ii) Individual orders with respect to NO and H2 can be

measured.

(iii) The overall order can be evaluated by keeping the

concentration of one of the substances constant.

Q14. The collision frequency and the orientation of molecules are

necessary conditions for determining the proper rate of reaction.

Justify the statement.

Q15. How does Arrhenius equation help us to calculate the energy of

activation of a reaction?

Q16. Define the following terms and give examples.

(i) Homogeneous catalysis

(ii) Heterogeneous catalysis

(iii) Activation of a catalyst

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(iv) Auto-catalysis

(v) Catalytic poisoning

(vi) Enzyme catalysis.

Q17. Briefly describe the following with examples:

(i) Change of physical state of a catalyst at the end of

reaction.

Ans. There may be change in physical state such as the particle size or

change in the colour of the catalyst at the end of reaction. For

example, granular MnO2 used as a catalyst in the thermal

decomposition of KC1O3 is left as fine powder at the end of

reaction.

(ii) A very small amount of a catalyst may prove sufficient

to carry out a reaction.

Ans. Since a catalyst is not used up in the reaction, a very small

amount of catalyst is required. It can catalyses the reaction over

and over again. Sometimes a trace of a metal catalyst is required

to affect very large amount of reactants. For example, 1 mg of fine

platinum powder can convert 2.5 dm3 of H2 and 1.25 dm

3 of O2 to

water. Dry HC1 and NH3 combine in the presence of trace of

moisture to give dense white fume of NH4C1.

(iii) A finely divided catalyst may prove more effective.

Ans. A catalyst is more effective when it is present in a finely divided

form than it is used in bulk. With the increase of fine subdivision,

the free surface area is increased. As a result, the active sites on the

surface are increased. Consequently, the activity of the catalyst is

also enhanced. For example, a lump of platinum will have much

less catalytic activity than colloidal platinum. Finely divided nickel

is a better catalyst than lumps of solid nickel.

(iv) Equilibrium constant of a reversible reaction is not

changed in the presence of a catalyst.

Ans. A catalyst for the forward reaction is also a catalyst for the

reverse reaction. A catalyst speeds up the rate of both the forward

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and reverse reactions equally. So the equilibrium constant (k= )

for the reaction remains the same. A catalyst helps the equilibrium

to e established earlier. For example, the reaction of N2 and H2 to

from NH3 is very slow. In the presence of the catalyst, the

equilibrium

N2(g) + 3H2(g) 2NH2(g)

Is reached much sooner but the percentage yield remains

uncharged.

(iv) A catalyst is specific in its action.

Ans. A catalyst cans catalyst only a specific reaction. When a

particular catalyst works for one reaction it may not necessarily

work for any other reaction. It may increase the rate of one

reaction but not increase the rate of another reaction. For example,

MnO2 can catalyse the decomposition of KC1O3 but not that of

NH3. The decomposition of NH3 is catalysed by a hot tungsten

wire.

2KC1O3(g) 2KC1(s) + 3O2(g)

2NH3(g) N2(g) + 3H2(g)

Remember that transition metals catalyse reaction of different

types.

For example, the decomposition of aqueous solutions of

hydrogen peroxide (H2O2) is catalysed by MnO2 or colloidal

platinum.

2H2O2(l) 2H2O(l) + O2(g)

Q18. What are enzymes? Give examples in which they act as catalyst.

Mention the characteristics of enzyme catalysis.

Q19. In the reaction of NO and H2, it was observed that equimolecular

mixture of gases at 340.5 mm pressure was half changed in 102

seconds. In another experiment with an initial pressure of 288 mm

of Hg, the reaction was half completed in 140 seconds. Calculate

the order of reaction.

Ans. Initial pressure of reactants = Initial concentrations of reactants.

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Given: a1=340.5 mm : t1=102 seconds

A2=288 mm : t2=140 seconds

Order of reaction, n=?

Formula used:

n =1 +

n =1 + =1 + 1.89=2.89

n =3

Hence the reaction of third order.

Q20. A study of chemical kinetiecs of a reaction

A + B products

Gave the following data at 25oC. Calculate the rate law

Exp. [A]

[B] Rate

1 1.00 0.15

4.2 x 10-6

2 2.00 0.15

8.4 x 10 -6

3 1.00 0.2

5.6 x 10-6

Solution: Examination of the data shows that when the

concentration of B is kept constant 0.15. The concentration of A is

doubled form 1.00 to 2.00; the rate doubles from 4.2 x 10-6

tp 8.4 x 10-6

.

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Mathematically, the rate of reaction with respect to A is directly

proportional to the first power of the concentration of A. this is,

Rate [A]

Examination of the experiment 1 and 3 shows that when the

concentration of A is kept constant 1.00, the concentration of B is

increased from 0.15 to 0.2, the rate of the reaction with respect to B is

also increased form 4.1 x10-6

to 5.6 x10-6

. Mathematically, the rate of

reaction with respect to B is directly proportional to the first power of

the concentration of B. that is,

Rate [B]

Since the rate of a reaction is proportional to the product of

concentration of the reactants, so,

Rate [A][B]

Rate=k[A][B]

This equation is known as the rate law for the reaction.

Q21. Some reaction taking places around room temperature have

activation energies around 50kJ mol-1

.

(i) What is the value of the factor at 25oC.

Ans. Ea=50kJ mol-1

=50000 Jmol-1

R=8.3143 J K-1

mol-1

At 25oC, the factor = = =1.72 x10

-9

(ii) Calculate this factor at35oC, and 45

oC note the increase

in this factor for every 10oC rise in temperature.

Ans. At 25oC, the factor = = =3.314 x10

-9

At 25oC, the factor = = =6.12 x 10

-9

(iii) prove that for every 10oCrise in temperature, the factor

doubles and so rate constant also doubles.

Ans. At 25oC, the factor =1.72 x 10

-9

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At 35oC, the factor =3.314 x 10-9

At 45oC, the factor =6.12 x 10

-9

Since for every 10oC rise in temperature, the factor becomes

doubles and so rate constant also doubles.

Q22. H2 and 12 react to produce HI. Following data for rate constant at

various temperatures (K) have been collected.

Temperature(K) Rate constant(

cm3 mol

-1s

-1)(K)

500 6.814 x 10-4

550 2.64 x 10-2

600 0.56 x 10o

650 7.31 x 10o

700 66.67 x 10o

(i) Pot a graph between on x-axis and log k on the y-

axis.

(ii) Measure the slope of this straight line and calculate the

energy for actives this reaction.

Ans. 8326.32: 160.6 Jmol-1

.

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11 chapter reaction kinetics text book exercise

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