1 MARK QUESTIONS

1. The gas phase decomposition of acetaldehyde CH3CHO→CH4+CO , follows the rate law. What are the units of its rate constant? Ans: atm1/2sec-1

2. State the order with respect to each reactant and overall reaction. H2O+3I-+2H+ 2H2O+ I3- Rate = k [H2O2]1[I-]1

Ans. Order of reaction= 1+1= 2

3. Give one example of pseudo first order reaction.

Ans. Acid Hydrolysis of an ester CH3COOC2H5+H2O H+¿→¿ CH3COOH+ C2H5OH

4. The conversion of molecules X to Y follows the second order of kinetics. If concentration of X is increased 3 times, how will it affect the rate of formation of Y?Ans .Rate = k [A]2=k [3A]2=k9[A]2 The rate of formation will become nine times.

5. The rate law for a reaction is Rate= K [A] [B]3/2. Can the reaction be an elementary process? Explain. Ans. No, an elementary process would have a rate law with orders equal to its molecularity and therefore must be in integral form.

6. What do you understand by ‘rate of reaction’? Ans: It is the change in conc of either the reactant or the product per unit time

7. Name the factors on which the rate of a particular reaction depends. Ans: Conc of reactant, temperature, surface area, presence of catalyst.

8. Why rate of reaction does not remain constant throughout? Ans:The rate of a reaction depends on the conc of the reactant and the conc of reactant decreases with passage of time

9. Define specific reaction rate or rate constant. Ans: It is the rate of a reaction when the conc of the reactant is unity.

10. What is half-life period of a reaction? Ans:It is the time required by the reaction to undergo 50% completion or the time by which the conc of the reactant reduces to half of its initial value

2 - MARK QUESTIONS

1. The rate of a particular reaction quadruples when the temperature changes from 293K to 313K. Calculate activation energy. Ans. K2/K1=4, T1= 293K, T2 = 313K

Log K2/K1= Ea

2.303R ¿¿

Log 4= Ea2.303 x 8.314 JK−1M−1

[313−293 ]313 X 293

Thus on calculating and substituting values we get, Ea = 52.86 kJmol-1

2. If the decomposition of nitrogen oxide as 2N2O5→ 4NO2+O2follows a first order kinetics. (i). Calculate the rate constant for a 0.05 M solution if the instantaneous rate is 1.5 x 10- molL-1s-1? Ans. Rate = K [N2O5]

K= Rate/ [N2O5] = 1.5 X10−60.05 =3.0X10-5

ii) What concentration of N2O5 would give a rate of 2.45 x 10-5mol l-1s-1?

Rate=K[N2O5]

[N2O5] = Rate/k=2.45 X10−53.0 X10−5 =0.82

3. Write the difference between order and molecularity of reaction.Ans. ORDER MOLECULARITY It is the sum of the powers of It is the number of reacting species undergoing concentration terms in the rate collision in a reaction simultaneously. law expression. It is determined experimentally It is a theoretical concept Order of reaction need not be a whole number. It is whole number only Order of reaction can be zero. It can’t be zero or fractional

4. Define Threshold energy and activation energy. How they are related? Ans. Threshold Energy: It is the minimum amount of energy which the reactant molecules must possess for the effective collision in forming the products. Activation Energy: It is the excess energy required by the reactants to undergo chemical reaction.

Activation energy = Threshold energy – Average kinetic energy of molecules. 5. a) Draw a schematic graph showing how the rate of a first order reaction changes in

concentration of reactants. Ans.

Variation of rate of first of first order reaction with concentration.

b) Rate of reaction is given by the equation Rate= k [A]2 [B]. What are the units of rate constant for this reaction?

Ans. Rate = k [A]2 [B],

k=mol L -1s-1 /(mol L-1)2(molL-1)

k= mol-2L2s-1

6. Can a reaction have zero kinetic energy?Ans. If Ea= 0, then according to Arrhenious equation, k= Ae-Ea/RT=Ae0=A, i.e. Rate constant = Collision frequency. This means every collision results into a chemical reaction which can not be true. Hence Ea can never be equal to zero.

7. Why equilibrium constant of a reaction does not change in the presence of a catalyst?Ans. The catalyst increases the rate of forward reaction as well as backward reaction to the same extant. Hence equilibrium is not disturbed but is attainted quickly.

8. What is the effect of adding catalyst on the free energy change (∆G) of a reaction?Ans. No change in ∆G occurs in the presence of catalyst (as free energy of the reactant and products remains the same).

3 MARKS QUESTIONS

1. Following reaction takes place in one step 2NO+ O2 → 2NO2. How will the rate of the reaction of the above reaction change if the volume of reaction vessel is diminished to1/3 of its original volume?

Ans. 2NO+O2 → 2NO2 dx/dt=k*[NO]2[O2] (Since it is one step) If the volume of reaction vessel is diminished to 1/3rd, concentration of both NO and O2 will become 3 times, the rate of reaction increased 27 times.

2. A reaction is first order in A and second order in B.(i) Write differential rate equation.(ii) How is the rate affected on increasing the concentration of B three times?(iii) How is the rate affected when the concentration of both A and B is doubled?

Ans. (i) dx/dt= k[A][B]2

(ii) Rate = kab2

If [B] is tripled, Rate =ka (3b)2= 9kab2=9 times.

(iii)If both [A] and [B] are doubled,Rate=k(2a)(2b)2=8kab2=8 times.

3. The half life period of a first order reaction is 60 minutes. What percentage will be left after 240 minutes?Ans. No of half lives = 240/60=4, i.e. n= 4Amount left after 4 half lives = [A]0/24 = [A]0/16 = 0.0625 of A0 = 6.25%

4. A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time will the reaction is 80% complete?Ans.

(i) For the first order reaction k= 2.303t log a/a-x

When x = 40100a = 0.4a =, t=50 minutes

K= 2.303/50 loga

a−0.4a

Or k= 2.30350 log

10.6 = 0.010216 min -1

(ii) T=? when reaction is 80% complete, i.e., x= 0.8aK= 0.010216 min-1

t=2.303k log

aa−x =

2.3030.010216min−1 log

aa−0.8a = 157.58 min.

5. What is the effect of temperature on the rate of reaction? Ans. With the increase in temperature the rate of reaction increases. This is because as per Maxwell/Boltzmann distribution curves the no of reactant molecules having energy equal to or greater than threshold energy increases. In most of the chemical reactions this becomes twice with every 100 rise in temperature.

6. State the role of activated complex in a chemical reaction.Ans. Activated complex is formed between the reactant and the catalyst. It provides the reaction an alternate path which requires less activation energy. Lesser the activation energy for a chemical reaction, faster is the reaction. Therefore the presence of catalyst increases the rate of reaction through the activated complex.

7. For a first order reaction, show that the time required for 99% completion of a first order reaction is twice the time required for the completion of 90%.

Ans. For the first order reaction, t= 2.303k log

aa−x

For 99% completion x= 99100 a= 0.99a

T99%= 2.303/klog a

a−0.99a = 2 x 2.303/k

For 90% completion, x= 90% of a = 0.90 a

T90%= 2.303k log

aa−0.90a =

2.303k log 10 = 2.303/k

Hence t99% = 2x t90%

Question bank with answer key

Q.1) Define (a) elementary step in a reaction ( b) rate of reaction.

Q.2) For zero order reaction, will the molecularity be zero?

Q.3) Why is the probability of reaction with molecularity higher than three is rare?

Q.4) With the help of diagram explain the role of activated complex in a reaction.

Q.5) A first order reaction takes 40 min for 30% decomposition calculate its t1/2 value.

Q.6) A first order reaction has the rate constant k = 5.5 x 10-14 s-1. Find its half life

Q.7) A first order reaction has a rate constant 1.15 x10-3 s-1. How long will 5g of this reactant take to reduce

to 3g?

Q.8) How rate constant is related to activation energy?

Q.9) Define activation energy of a reaction and order of reaction

Q.10 ) Explain the terms: (i) Rate of reaction (ii) Activation energy of a reaction

Q.11) Explain the terms: (i) Order of reaction (ii) molecularity of a reaction

Q.12) For the reaction C12H22O11 + H2O C6H12O6 + C6H12O6

Write (i) Rate of reaction expression (ii) Rate law equation (iii) molecularity (iv) order of reaction

Q.13) Illustrate graphically the effect of catalyst on activation energy

Q.14) Catalysts have no effect on equilibrium const. why?

Q15) What do you understand by the rate law and the rate const. of a reaction? Identify the order of reaction if the units of its rate const. are (i)mol/L/s (ii) L/mol/s

Q16) A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half? What is the unit of rate const. for such a reaction?

Q17) (a) Explain difference between the average rate and instantaneous rate of Reaction.

.

Q18) (a) Distinguish between molecularity and order of reaction

Q19) A reaction is of second order with respect to a reactant. How is the rate affected if the concentration of the reactant is

(i) doubled (ii) reduced to half?

Q20) (a) For a reaction A + B P , the rate law is given by, R = k [A]1/2[B]2

What is the order of this reaction?

( b ) A first order reaction is found to have a rate constant k=5.5 x 10 -14s-1 Find the half life of the reaction.

Q21) The rate of reaction becomes four times when the temperature changes from293 K to 313 K.Calculatethe energy of activation (Ea) of the reaction assuming that it does not change with temperature. [R = 8.314 J K -1, log4 = 0.6021]

Q22). For a chemical reaction R P, the variation in the concentration (R) vs. time (t) plot is given as

(i) Predict the order of the reaction. (ii) What is the slope of the curve?

Q.23) What will be the effect of temperature on rate constant?

ANSWERS OF QUESTION BANK CHAPTER 4 CHEMICAL KINETICS

ANS. 1 (a) Elementary step: Each step of a complex reaction is called an elementary step.

(b) Rate of reaction: It is the change in the concentration of any of the reactants or products per

unit time

Ans.2 Molecularity can not be zero or fraction

Ans. 3 Since probability of collision of more than three molecules simultaneously is least.

Ans. 4 Refer NCERT textbook fig. 4.11 page 115

Ans. 5 Its half life is 77.7 minute

Ans. 6 Its half life is 1.26 x 1013 s

Ans. 7 t= 4.438 x 102 s

Ans. 8 k = A e-Ea /RT

Ans. 9 Activation energy: The amount of energy which the reactants must absorb to pass over the

energy barrier between reactants and products is known as the activation energy.

Order of reaction :- The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.

If Rate = k [A]x [B]y Order = x + y

Ans. 10 (i) Rate of Reaction: It is defined as the decrease in the concentration of reactants per unit time

or increase in the concentration of products per unit time.

For a hypothetical reaction RP

Rate = rav = - ∆[R] / ∆t = + ∆ [P] / ∆t

(ii) Activation energy: The amount of energy which the reactants must absorb to pass over the

energy barrier between reactants and products is known as the activation energy.

Ans. 11 Order of reaction :- The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.

If Rate = k [A]x [B]y Order = x + y

Molecularity of areaction :- The number of reacting species taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction (values are limited from 1 to 3)

Ans. 12 (i)Rate = - ∆[C12H22O11]/ ∆t(ii) Rate = k’ [C12H22O11] [H2O]Here change in conc of water is very very small , hence it can be considerd constantK= k’ [H2O]Hence Rate = k[C12H22O11](iii) refer earlier questions

(iv) refer earlier questions

Ans. 13 Catalyst provides an alternate pathway by reducing activation energy between reactant and product. For fig. refer ncert book page no, 115 fig. 4.11

Ans .14 Catalyst catalyses the forward as well as backward reaction to the same extent so that the equilibrium state remains same but is reached earlier.

Ans. 15 Rate law- Rate law is an expression in which rate of reaction is expressed in terms of conc. Of reactants which are raised to some power which may or mat not be equal to their stoichiometric coefficients given in the balanced chemical equation.

Rate constant-rate constant is numerically equal to rate of reaction when molar concentration of all the reactants equal to 1 .

(b) (i) zero (ii) two or second order

Ans. 16 An experimentally determined expression which relates the rate of reaction with the concentration of

reactants is called rate law while the rate of reaction when concentration of each reactant is unity in a

rate law expression is called rate constant.

(i) Comparing power of mole in L–1 mol s–1 and (mol L–1)1–n s–1,

We get

1 = l – n n = 0 i.e., zero order reaction

(ii) Again comparing power of mole in L mol–1s–1 and (mol L–1) 1–n s–1, we get

–1 = 1 – n n = 2, i.e., second order reaction

Consider the reaction Products

As the reaction is of second order

Rate, r = k [R]2 ...(i)

If the concentration of the reactant reduced to half, then

Rate, r = k[R/2]2 = k [ R/4] …… (ii)

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

r’/r = k[R2/4] / k [R2] = 1/4

r’ = r.1/4 i. e. 1/4 th times of initial rate

The unit of rate constant is mol–1 L s–1.

Ans. 17

Average rate Instantaneous rate

Rate of change of conc. W.r.t. time measured over long time interval

Rate of change of conc. W.r.t. time measured over a very small time interval i. e. at a particular instant.

Rate avg = rav = - ∆[R] / ∆t = + ∆ [P] / ∆t Rate ins = rins = - d[R] / dt = + d [P] / dt = k [R]x

Ans. 18

Order of a reaction Molecularity of a reaction

1. It is experimental value It is theoritical value

2. Order is the sum of the exponents (powers ) of the concentration of reactant in rate law expression

Molecularuitry is the total number of atoms , ions or molecules of the reactant invonled in reaction

3. Order can be fractional Moleclarity will never be fraction or zero, it should be positive integers

Ans. 19 Rate = k[R]2

(i) If [R] is doubled, Rate = k[2R]2 = 4k[R]2 = 4 times of initial rate

(ii) If [R] is reduced to half

r’/r = k[R2/4] / k [R2] = 1/4

r’ = r.1/4 i. e. 1/4 th times of initial rate

Ans. 20 Order = sum of powers of conc. Terms in rate expression = ½ + 2 =2.5

Ans. 21 Ea = 52863.2177 J mol–1 or 52.863 kJ mol–1

Ans 22. The order of the reaction zero. Slope = -k

Ans. 23 Rate constant raised to double for every 10 0 rise of temperature.

HOTS FOR CHEMICAL KINETICS

1 MARK QUESTIONSQ. 1. In the reaction A — B, if the concentration of A is plotted against time, the nature of the

curve obtained will be as shown. What is the order of the reaction ?

Conc

entra

tion

T im eQ. 2. What is the effect of temperature on activation energy ?

Q. 3. Which will dissolve in water faster, powdered sugar or crystalline sugar and why ?

Q. 4. Which reaction will take place faster and why ?

500 °C

C (s) + ½ O2 (g) ————— CO (g)

1000 °C

C (s) + ½ O2 (g) ————— CO (g)

Q. 5. For a reaction A + H2O —— B; r = k [A]. What is its (i) Molecularity (ii) Order ?

2 MARKS QUESTIONS

Q. 1. A reaction : Reactant —— Product is represented by :

Conc. o freactan t

(R )

T im e ( t)

[R ] 0

(i) Predict the order of the reaction.

(ii) What does the slope of the graph represent ?

Q. 2. For a reaction, the activation energy is zero. What is the value of rate constant at 300 K if K = 1.6 × 106 s–1 at 280 K.

Q. 3. The slope of the line in the graph of log K is for a reaction is – 5841 K. Calculate Ea for the reaction.

3 MARKS QUESTIONS

K

Q. 1. Consider the reaction R —— P. The change in concentration of A with time is shown in the given plot :

Conc. [R ]

x

y

T im e ( t)(i) Predict the order of the reaction.(ii) Derive the expression for the time required for the completion of the reaction.

Q. 2. Answer the following questions on the basis of the given curve for a first order reaction :A —— P

T im e

[R ][R ]

0lo g

(i) What is the relation between slope of this line and rate constant ?(ii) Calculate the rate constant of the above reaction if the slope is 2 × 10–4 s–1.

Q. 3. For a certain chemical reaction variation in concentration in [R] VS time plot is given below. For this reaction write :

(i) What are the units of rate constant ?

1T

(ii) Give the relationship between k and t½.

(iii) What does the slope of the above line indicate ?Q. 4. Consider the following diagram representing potential energy plot and answer the

following questions :(i) What do ‘x’ and ‘y’ represent ?(ii) What does ‘z’ represent in this diagram ?(iii) Is the reaction endothermic or exothermic ?

Q. 5. Consider a plot between k vs where T is the temperature. On the basis of this plot, answer the following questions :

l n k

x

y

1T

(i) What is the slope in this line ?

(ii) What is the intercept of this line on the y-axis ?

(iii) What is the relation between K and T ?

Q. 6. Diagram given below shows a plot of potential energy Vs reaction co-ordination for a hypothetical reaction. Write answers to the following from the plot given :

1T

(a) Represent reactant, product and activated complex in terms of A, B and C ?(b) Is this reaction exothermic or endothermic ?(c) What will be the effect of a catalyst on Ea of the reaction ?

Q. 7. The rate of a first order reaction is 0.04 mol/h/s at 10 minutes and 0.03 mol/h/s at 20 minutes. Find the half life period of the reaction.

Q.8 For a decomposition reaction the values of rate const. k at two different temp. are given below

K1 = 2.15 x 10-8 L /mol/s at 650K K2= 2.39 x 10-7 L/mol/s at 700K Calculate Ea for this reaction (R = 8.314J/mol/K)

Q.9 The decomposition of PH3 proceeds according to the following equation.4PH3 P4 + 6H2 It is found that the reaction follows the following rate

equation Rate = k [PH3] The half life of PH3 is 39.9 s at120 0C

(i) How much time is required for 3/4th of PH3 to decompose? (ii) What fraction of the original sample of PH3 remain behind after one minute.

Q.10 The rate of reaction increases four times when the temperature

changes from 300K to 320 K Calculate Ea ,assuming that it does not change with temperature . ( R = 8.314J/K/mol)

Q.11 The following data were obtained during the first order thermal decomposition of SO2Cl2 at const. volume.

SO2Cl2 (g) SO2(g) + Cl2(g)

experiment Time (s) Total press. (atm)

1 0 0.5

2 100 0.6

Calculate the rate of reaction when total press. Is 0.65 atm.

A Y

C

Bx

Reaction Coordinate

P

O

T

E

N

T

I

Answer key for HOTSHOTS FOR CHEMICAL KINETICS

1 MARK QUESTIONSAns.1 First Order

Ans.2 There is no effect of temperature on activation energy.

Ans.3 Powdered sugar will dissolve in water faster as it has more surface area.

Ans.4 The second reaction is faster because increase in temperature increases the number of effective collisions and hence increase in rate.

Ans.5 Pseudo unimolecular rection order = 1

2 MARKS QUESTIONS

Ans.1 (i) Zero order

(ii) Slope = – k =

Ans.2 log = = = 0

= antilog (0) = 1 or K2 = K1 = 1.6 × 106 s–1

Ans.3 Slope = –

Ea = – 2.303 R × Slope

= – 2.303 × 8.314 × – 5841

= 1.118 × 105 g/mol

3 MARKS QUESTIONS

Ans.1 (i) Zero order(ii) For the reaction R —— P

d[R]dt

2

1

KK

a 2 1

1 2

E T T2.303R TT

2 1

1 2

T T02.303R TT

2

1

KK

aE2.303R

r = – = K [R]°– d [R] = K dtOn integration– [R] = Kt + CWhen t = 0 [R] = [R]0

On substitution– [R] = Kt – [R]0 [R] = – Kt + [R]0

Kt = [R]0 – [R] t = {[R]0 – [R]}

Ans.2 (i) Slope = (ii) Slope = 2 × 10–4 s–1

K = 2.303 × Slope= 2.303 × 2 × 10–4 s–1

= 4.606 × 10–4 s–1

Ans.3 (i) time–1 (s–1)

(ii) K = (iii) rate constant K of the reaction.

Ans.4 (i) ‘x’ represents Ea for forward reaction. ‘y’ represents Ea for backward reaction.

(ii) ‘z’ represents H, the enthalpy change for the reaction.

(iii) Exothermic reaction.

Ans.5 (i) Slope = –

(ii) Intercept = ln A

(iii) ln k or K = A e–Ea/RT

Ans.6 (a) A — Reactant

B — Product

d[R]dt

1K

K2.303

12

0.693t

aER

1T

C — Activated Complex

(b) Exothermic

(c) Catalyst will lower the activation energy for the reaction.

Ans.7 Rate = K C

r1 = K C1

r2 = KC2

K = log

When t = 10 min

K  = log = log  = 0.0287 min–1

t½ = =

= 24.14 min.

Q.4 The following data were obtained during the first order thermal decomposition of SO2Cl2 at const. volume.

SO2Cl2 (g) SO2(g) + Cl2(g)

experiment Time (s) Total press. (atm)

1 0 0.5

2 100 0.6

Calculate the rate of reaction when total press. Is 0.65 atm.

1 1

2 2

r (10 min) C 0.04r 20 min C 0.03

2.303t

1

2

CC

2.303t

0.040.03

2.30310

43

0.693K

0.6930.0287