Chemical Kinetics Important Question Paper

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Questions:

Generic 2

9 x 2 = 18

Write two differences between 'order of reaction' and 'molecularity of reaction'.

For a reaction A \(\longrightarrow \) B, the rate of reaction becomes twenty seven times when the concentration of A is increased three ttimes. what is the order of the reaction?

In the reaction aA +bB \(\longrightarrow\) products, if concentration of A is doubled (keeping B constant) the initial rate becomes four times and if B is doubled (keeping A constant), the rate becomes double. What is the rate law equation and order of reaction ?

Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of the one example .

The rate constant for a second order reaction is k = \(\frac {2.303}{t(a-b)} log \frac {b(a-x)}{(b-x)}\)where a and b are initial concentrations of the two reactants A and B involved. If one of the reactants is present in excess, it becomes pseudo unimolecular. Explain how ?

Why molecularity is applicable only for elementary reactions and order is applicable for elementary as well as complex reactions?

Show that for a first order reaction, the time required for half the change (half-life period) is independent of initial concentration.

A first order reactions has rate constant k = 5.5 \(\times\) 10^-14 s^-1. Find the half life of the reaction.

t_1/2 of the reaction increases with increase in initial concentration. What is the order of reaction?

Generic 5

4 x 5 = 20

10.

In summers, due to higher temperature reaction become fast and use of refrigerator is must. Activation energy plays a role in deciding the speed of a reaction.
Answer the questions:
(i) Acook cries less on cutting onion kept in refrigerator. Why?
(ii) In refrigerator where should you store meat and why?
(iii) Diamond is forever. Comment.

11.

At constant temperature and volume, X decomposes as 2 X (g) \(\longrightarrow\) 3 Y (g) + 2 Z (g). P_x is the partial pressure of X.

Observation No.	Time (in minutes)	P_x (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 time for 75% completion of the reaction.
(iii) Find the total pressure when pressure of X is 700 mm of Hg.

12.

Hydrogen peroxide, H₂O₂(aq) decomposes to H₂O(l) and O₂(g) in a reaction that is of first order in H₂O₂ and has a rate constant, k = 1.06 \(\times\) 10^-3 min^-1.
(i) How long will it take 15% of a sample of H₂O₂ to decompose?
(ii) How long will it take 85% of a simple of H₂O₂ to decompose?

13.

(a) Express clearly what you understand by 'rate expression' and 'rate constant' of a reaction.
(b) Nitrogen pentoxide decomposes according to the equation
2N₂O₅(g) \(\to\) 4 NO₂ (g) + O₂ (g)
This first order reaction was allowed to proceed at 40^oC and the data given below were collected:

[N₂O₅] (M)	Time (min)
0.400	0.00
0.289	20.00
0.209	40.00
0.151	60.00
0.109	80.00

(i) Calculate the rate constant for the reaction. Include units with you answer.
(ii) Calculate the initial rate of reaction.
(iii) After how many minutes will [N₂O₅] be equal to 0.350 M?

Multiple Choice Question

13 x 1 = 13

14.

In a hypothetical reaction X \(\longrightarrow\) Y, the activation energy for the forward and the backward reaction are 15 and 9 kJ mol^-1 respectively. The potential energy of X is 10 kJ mol^-1. Then

(a)

Threshold energy of the reaction is 25 kJ

(b)

The potential energy of Y is 16 kJ

(c)

Heat of reaction is 6 kJ

(d)

The reaction is endothermic

15.

A reactant (A) forms two products:
A \(\xrightarrow { { k }_{ 1 } } \) B, Activation Energy \({ E }_{ { a }_{ 1 } }\)
A \(\xrightarrow { { k }_{ 1 } } \) C, Activation Energy \({ E }_{ { a }_{ 2 } }\)
If \({ E }_{ { a }_{ 2 } }\) = 2\({ E }_{ { a }_{ 1 } }\) , then k₁and k₂ are related as

(a)

k₁ = 2k₂ \({ e }^{ { Ea }/_{ 2 }RT }\)

(b)

k₂= k₁ \({ e }^{ { Ea }/_{ 2 }RT }\)

(c)

k₂= k₁ \({ e }^{ { Ea }/_{ 2 }RT }\)

(d)

k₁ = Ak₂\({ e }^{ { Ea }/_{ 2 }RT }\)

16.

Kinetics of the reaction A (g) \(\longrightarrow\) 2 B (g) + C (g) is followed by measuring the total pressure at different times. It is given that
Initial pressure of A = 0.5 atm.
Total pressure of A after 2 hours = 0.7 atm
Rate constant of the reaction = 1 \(\times 10 ^{-3}s^{-1}\)
What is the rate of reaction \(-\frac {d[A]}{dt}\) when the total pressure is 0.7 atm?

(a)

2.0 \(\times 10^{-4}M s^{-1}\)

(b)

4.0\(\times 10^{-4}M s^{-1}\)

(c)

5.0\(\times 10^{-4}M s^{-1}\)

(d)

7.0\(\times 10^{-4}M s^{-1}\)

17.

The initial rates of reaction 3 A + 2 B + C \(\longrightarrow\) Product, at different initial concentration are given below :

Initial rate , Ms^-1	[A]₀, M	[B]₀, M	[C]₀, M
5.0 \(\times 10 ^{-3}\)	0.010	0.005	0.010
5.0 [A]₀, M	0.010	0.005	0.015
1.0 \(\times 10 ^{-2}\)	0.010	0.010	0.010
1.25\(\times 10 ^{-3}\)	0.005	0.005	0.010

The order with respect to the reactant A,B and C are respectively.

(a)

3,2,0

(b)

3,2,1

(c)

2,2,0

(d)

2,2,1

(e)

2,1,0

18.

The rate constant of the reaction A \(\longrightarrow\) B is 0.6 \(\times 10^3\) mole per litre per second. If the concentration of A is 5 M, then concentration of B after 20 minutes is

(a)

0.36 M

(b)

0.72 M

(c)

1.08 M

(d)

3.60 M

19.

The rate of a gaseous reaction is generally expressed in terms of \(\frac {dP}{dt}\). If it were expressed in terms of change in number of moles per unit time \((\frac {dn}{dt})\) or in terms of change in molar concentration per unit time \((\frac {dC}{dt}),\) which of the following relationship will hold good ?

(a)

\(\frac { dC }{ dt } =\frac { dn }{ dt } =\frac { dP }{ dt } \)

(b)

\(\frac { dC }{ dt } =\frac{1}{V}(\frac { dn }{ dt }) =\frac{1}{RT}(\frac { dP }{ dt }) \)

(c)

\(\frac { dC }{ dt } =\frac { dn }{ dt } =\frac{1}{RT}(\frac { dP }{ dt }) \)

(d)

None of these

20.

During decomposition of an activated complex

(a)

energy is always realeased

(b)

energy is always absorbed

(c)

energy is not change

(d)

reaction may be formed

21.

Which of the following statements are applicable to a balanced chemical equation of an elementary reaction ?

(a)

Order is same as molecularity

(b)

Order is less than the molecularity

(c)

Order is grater than the molecularity

(d)

Molecularity can never be zero.

22.

Which of the following statement is incorrect about the collision theory of chemical reaction ?

(a)

If considers reactions molecules or atmos to be hard spheres and ignores their structural features

(b)

Number of effective collisions determines the rate of reaction

(c)

Collision at atoms or molecules possessing sufficient threshold energy results into the product information

(d)

Molecules should collide with sufficient threshold energy and proper orientation for the collision to be effective.

23.

Consider the Arrhenius equation given below and mark the correct option. k = Ae^-E_a/RT

(a)

Rate constant increases exponentially with increasing activation energy and decreasing temperature

(b)

Rate constant decreases exponentially with increasing activation energy and decreasing temperature

(c)

Rate constant increases exponentially with decreasing activation energy and decreasing temperature

(d)

Rate constant increases exponentially with decreasing activation energy and increasing temperature

24.

The reaction A \(\longrightarrow\) B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 9.9 mole of A to produce 0.675 mole of B ?

(a)

1 hour

(b)

0.5 hour

(c)

0.25 hour

(d)

2 hours

25.

The rate of the reaction 2 NO + CI₂ \(\rightarrow\) 2NOCI is given by the rate equation : rate = k [NO]₂ [CI₂]. The value of the rate constant can be increased by

(a)

increasing the temperature

(b)

increasing the concentration of NO

(c)

increasing the concentration of CI₂

(d)

doing all of these

26.

Time required for 100 percent completion of a zero order reaction is

(a)

\(\frac {2k}{a}\)

(b)

\(\frac {a}{2k}\)

(c)

\(\frac {a}{k}\)

(d)

a k

Generic 3

9 x 3 = 27

27.

Rate constant 'k' of a reaction varies with temperature 'T' according to the equation \(logk=logA-\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ T } \right) \) where E_a is the activation energy. When a graph is plotted for log k vs, 1/T, a straight line with a slope of - 4250 K is obtained. Calculate 'E_a' for the reaction. (R = 8.314 K^-1mol^-1).

28.

For the first order thermal decomposition reaction, the following data were obtained:
\(C_2H_5Cl_{(g)}\rightarrow C_2H_4{(g)}+HCl{(g)}\)

Time/sec	Total pressure/atm
0	0.30
300	0.50

Calculate the rate constant.
(Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)

29.

For a reaction
\({ N }_{ 2 }+{ 3H }_{ 2 }\longrightarrow { 2NH }_{ 3 }\)
the rate of reaction measured as \(\frac { \Delta \left[ { NH }_{ 3 } \right] }{ \Delta t } \) was found to be \(2.4\times { 10 }^{ -4 } \ mol{ L }^{ -1 }{ S }^{ -1 }\) . calculate the rate of the reaction expressed in terms of
(i) \({ N }_{ 2 }\)
(ii) \({ H }_{ 2 }\)

30.

Two reactions,
(i) A \(\longrightarrow\) Products
(ii) B \(\longrightarrow\) Products, follow first order kinetics.
The rate of reaction
(i) is doubled when temperature is raised from 300 K to 310 K. The half life for this reaction at 310 K is 30 minutes. 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
(iii), calculate the rate constant of reaction (ii) at 300 K.

31.

The rate of a particular reaction doubles when temperature changes from 27^oC to 37⁰C. Calculate the energy of activation of such a reaction.

32.

The decomposition of Cl₂O₇ at 400K in the gas phase to Cl₂ and O₂ is a first order reaction.
(i) After 50 seconds at 400K,the pressure of Cl₂O₇ falls from 0.062 to 0.044 atm. Calculate the rate constant
(ii) Calculate the pressure of Cl₂O₇ after 100 sec of decomposition at this temperature.

33.

From the following data,show that the decomposition of hydrogen peroxide is a reaction of the first order:

t	0	10	20
x	46.1	29.8	19.3

where t is the time in minutes and x is the volume of standard KMnO₄ solution in cm³ required for titrating the same volume of the reaction mixture.

34.

The following rate data were obtained at 303K for the following reaction:2A + B\(\longrightarrow\)C + D

Experiment	[A]/mol	[B]/mol	Initial rate of fornmation of D/mol L^-1 min^-1
I	0.1	0.1	6.0\(\times\)10^-3
II	0.3	0.2	7.2\(\times\)10^-2
III	0.3	0.4	2.88\(\times\)10^-1
IV	0.4	0.1	2.4\(\times\)10^-2

What is the rate law? What is the order with respect to each reactant and the overall order?Also calculate the rate constant and write its units.

35.

For the reaction, 4NH₃(g) + 5O₂(g)\(\longrightarrow\)4NO(g) + 6H₂O(g), if the rate expression in terms of disappearance of NH₃ is \(-{\Delta [NH_3]\over \Delta t}\), write the rate expression in terms of concentration of O₂ and H₂O.

Fill up / 1 Marks

1 x 1 = 1

36.

The overall rate of a reaction depends upon the .............. .

Match the following

1 x 1 = 1

37.

e^-E_a/RT

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Answers Key:

Generic 2

9 x 2 = 18

Order of reaction	Molecularity of reaction
(i) It is the sum of powers to which concentration terms are raised in rate law or rate equation	(i) It is the sum of number of molecules which takes part in a chemical reaction.
(ii) It is determined experimentally, can be in fraction and even zero.	(ii) It is determined theoretically and is always a whole number

\(=k[A]^2[B], \text { order }=2+1=3\)

Thermodynamic feasibility\(\left( -ve\ \Delta G \right) \) cannot alone decide the rate of reaction because the reaction may be feasible but its rate may be very very slow because of very high activation energy. For example, conversion of diamond to graphite is highly feasible but this reaction is very slow because it has high activation energy.

Suppose B is in excess so that b>> a or x. Neglecting a and x in comparison to b, the equation reduced to k b = k' = \(\frac {2.303}{t} log \frac {a}{(a-x)}\) which is same as for reactions of 1st order.

Complex reaction proceeds through several elementary reactions. Molecularity of each elementary reaction may be different, therefore, molecularity of complex reaction can't be determined. Order of complex reaction is determined by slowest step in mechanism (involving elementary reactions).

Half-life is time in which 50% of the reaction is complete.
At \({ t }_{ { 1 }/{ 2 }\prime }\quad [A]={ { [A] }_{ 0 } }/{ 2 }\)
\( k=\frac { 2.303 }{ { t }_{ { 1 }/{ 2 } } } \log { \frac { { [A] }_{ 0 } }{ { { [A] }_{ 0 } }/{ 2 } } } \)
\(=\frac { 2.303 }{ { t }_{ { 1 }/{ 2 } } } \log { 2 } \)
\(=\frac { 2.303 }{ { t }_{ { 1 }/{ 2 } } } \times 0.3010\)
\(=\frac { 0.693 }{ { t }_{ { 1 }/{ 2 } } }\)
\({ t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ k } \)
It shows that half-life of first order reaction is independent of initial concentration.

\({ t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ k } \)
\(=\frac { 0.693 }{ 5.5\times { 10 }^{ -14 }{ s }^{ -1 } }\)
\(=1.26\times { 10 }^{ 13 }s\)

Zero order reaction.

Generic 5

4 x 5 = 20

10.

(i) Due to lower temperature, less vapours are formed which cause less tears.
(ii) Meat should be stored in the coldest part of the refrigerator. This slows down the growth of microorganisms which are responsible for causing spoilage.
(iii) This statement is not correct as conversion of diamond to graphite is spontaneous thermodynamically. But process is so slow that this will take thousands of years.

11.

(i) As pressure of X is changing with time, it cannot be a zero order reaction. Let us now check it for 1st order.
At t = 100 min, \(k={2.303\over 100}log{P_0\over P_t}={2.303\over 100}log {800\over 400}=6.932\times 10^{-3}min^{-1}\)
At t = 200 min, \(k={2.303\over 200}log{800\over 200}={2.303\over 800}log4=6.932\times10^{-3}min^{-1}\)
As k comes out to be constant, hence it is a reaction of 1st order
(ii)\(t_{75./.}={2.303\over k}log{100\over 100-75}={2.303\over 6.932\times10-3min}log4=200min\)
(iii) 2x (g)⟶ 3 y (g) + 2z (g)
Initial Pressure 800mm 0 0
Pressure after time t 800 - 2p 3 p 2 p
When pressure of X is 700 mm, 800 - 2 p = 700 or p = 50 mm
Total pressure = (800 - 2 p) + 3 p + 2 p = 800 + 3 p = 800 + 3 x 50 = 950 mm.

12.

(i) \(k=\frac { 2.303 }{ t } \log { \frac { { \left[ A \right] }_{ 0 } }{ \left[ A \right] } } \)
\(t=\frac { 2.303 }{ 1.06\times { 10 }^{ -3 } } \log { \frac { { \left[ A \right] }_{ 0 } }{ \frac { 85 }{ 100 } { \left[ A \right] }_{ 0 } } }\)
\(=\frac { 2.303 }{ 1.06\times { 10 }^{ -3 } } \times \left[ \log { 20 } -\log { 17 } \right] \)
\(=\frac { 2.303 }{ 1.06\times { 10 }^{ -3 } } \times \left[ 1.3010-1.2304 \right] \)
\(=\frac { 2.303\times 0.0706 }{ 1.06\times { 10 }^{ -3 } } \times \frac { 0.1626\times { 10 }^{ 3 } }{ 1.06 } \)
\(=1.534\times { 10 }^{ 2 }=153.4 \ min\)
(ii) \(t=\frac { 2.303 }{ 1.06\times { 10 }^{ -3 } } \log { \frac { { \left[ A \right] }_{ 0 } }{ \frac { 15 }{ 100 } { \left[ A \right] }_{ 0 } } }\)
\(=\frac { 2.303 }{ 1.06\times { 10 }^{ -3 } } \times \left[ \log { 20 } -\log { 3 } \right] \)
\(=\frac { 2.303 }{ 1.06\times { 10 }^{ -3 } } \times \left[ 1.3010-0.4771 \right] \)
\(=\frac { 2.303 }{ 1.06\times { 10 }^{ -3 } } \times 0.8239\)
\(=\frac { 1.897 }{ 1.06 } \times 1000=1790.0\)

13.

(a) Rate expression is a way of expressing rate of reaction, e.g.
N₂(g) + 3H₂(g)⇾ 2NH₃(g)
\({-d[N_2]\over dt}={-{1\over 3}}{d[H_2]\over dt}=+{1\over 2}{d[NH_3]\over dt}\)
Rate constant is defined as equal torate of reaction when molar cone. of reactants is unity. Its unit depends upon order of reaction.
(b) (i)\(k={2.303\over t}log{[R]_0\over [R]}={2.303\over 20}log{0.400\over 0.289}={2.303\over 20}[10 0.400 - 10 0.289]\)
\(={2.303\over 20}=[\bar1.6021-\bar1.4609]={2.303\over 20}\times1.1412\)
\(={0.3521\over 20}=0.016285\ min^{-1}\)
\(k={2.303\over 40}log{0.400\over 0.209}=={2.303\over 40}[log0.400-log0.209={2.303\over 40}\times0.2820\)
\(={0.6494\over 40}=0.01623\ min^{-1}\)
\(k={2.303\over 60}[log 0.400 - log 0.151] ={2.303\over 60}[\bar1.6021- \bar1.1790] ={2.303\over 60}\times 0.4231\)
\(={0.9739\over 60}=0.01623 \ min^{-1}\)
\(k={0.01625+0.01623+001623\over 3}\)
= 0.016236 min^-1
(ii) Initial rate = k[N₂O₅]
= 0.016236 x 0.4 = 0.00649 mol L^-1 S^-1
(iii) \(k={2.303\over t}log{[R]_0\over [R]}\Rightarrow k={2.303\over t}log{0.4\over 0.35}\)
\(t={2.303\over 1.6236\times10^{-12}}(log40-log35)={2.303\over 1.625\times10^{12}}(1.6021-1.5441)\)
\(t={2.303\times10^2\times0.0580\over 1.6236}={13.3574\over 1.6236}=8.227min.\)

Multiple Choice Question

13 x 1 = 13

14.

15.

(d)
k₁ = Ak₂\({ e }^{ { Ea }/_{ 2 }RT }\)

16.

(b)
4.0\(\times 10^{-4}M s^{-1}\)

17.

(e)
2,1,0

18.

(b)
0.72 M

19.

(b)
\(\frac { dC }{ dt } =\frac{1}{V}(\frac { dn }{ dt }) =\frac{1}{RT}(\frac { dP }{ dt }) \)

20.

Activated complex has higher energy. When it decomposes. energy is always realeased and it may give products or reactants back.

21.

(b)
Order is less than the molecularity

22.

(c) is incorrect because formation of product depends not only on energy but also on proper orientation at the time of colllision.

23.

(d)
Rate constant increases exponentially with decreasing activation energy and increasing temperature

24.

(a) : The fraction of a reacted in each case is same (0.2/0.8 = 1/4), (0.9 - 0.675)/0.90 = 0.225/0.90 = 1/4). Hence, time taken is same.

25.

(a) : The rate of constant of a reaction depends only on temperature and does not depend upon concentrations of the reactants.

26.

Generic 3

9 x 3 = 27

27.

Given, slope = -4250K,
R = 8.314 JK-1mol^-1
From equation
log K = \(logA-\frac { { E }_{ a } }{ 2.303RT } \)
Comparing with straight line euqation, y = mx + c
\(-\frac { { E }_{ a } }{ 2.303 } =-4250\)
⇒ E_a = 2.303 x 8.314 x 4250
= 81.37 KJ mol^-1.

28.

\(P_0=0.30 atm\) \(P_t=0.50atm\) t = 300s
Rate constant, k = \(\frac{2.303}{t}log{\frac{P_0}{2P_0-P_t}}\)
\(=\frac{2.303}{300}log\frac{0.30}{2\times0.30-0.50}\)
\(=\frac{2.303}{300}log\frac{0.30}{0.60-0.50}\)
\(=\frac{2.303}{300}log\frac{0.30}{0.10}\)
\(=\frac{2.303}{300}log 3\)
\(=\frac{2.303}{300} \times 0.4771\)
\(=\frac{1.099}{300}\)
\(=0.0036 s^{ -1}=3.66\times10^{ -3}s^{ -1}\)

29.

\(\frac { \Delta \left[ { N }_{ 2 } \right] }{ \Delta t } =\frac { 3 }{ 2 } \frac { \Delta \left[ { NH }_{ 3 } \right] }{ \Delta t }\)
\(=\frac { 1 }{ 2 } \times 2.4\times { 10 }^{ -4 }\)
\(=1.2\times { 10 }^{ -4 } \ mol \ { L }^{ -1 }{ S }^{ -1 }\)
\(-\frac { \Delta \left[ { H }_{ 2 } \right] }{ \Delta t } =\frac { 3 }{ 2 } \frac { \Delta \left[ { NH }_{ 3 } \right] }{ \Delta t } \)
\(=\frac { 3 }{ 2 } \times 2.4\times { 10 }^{ -4 }\)
\(=3.6\times { 10 }^{ -4 } \ mol \ { L }^{ -1 }{ S }^{ -1 }\)

30.

Calculation of activation energy of reaction (i)
T₁= 300 K, T₂ = 310 K, k₁ = k, k₂ = 2 k
\(log{k_2\over k_1}={E_A\over 2.303E}\left(T_2-T_2\over T_1T_2\right ),ie.,\ log2={E_a\over 2.303\times8.314}\times{10\over 300\times310}\ or\ E_a=53.60kJmol^{-1}\)
Calculation of rate constant of reaction (i) at 310 K
\(k={0.693\over t_{1/2}}={0.693\over 30\ min}=2.31\times10^{-2}min^{-1}\)
Rate constant of reaction (ii) at 310 K = 2 x 2·31x 10^-2 min^-1 = 4·62 x 10^-2 min^-1
Energy 0f ac tiva tion 0f reac tion (ii) =\({53.60kJ\ mol^{-1}\over 2}=26.80kJ\ mol^{-1}\)
Aim. To calculate k for reaction (ii) at 300 K
\(log{4.62\times10^{-2}\over k_{300}}={26.80\over 2.303\times8.314\times10^{-3}}\times{10\over 300\times310}=0.0151\)
or \({4.62\times10^{-2}\over k_{300k}}=Antilog\ or\ 0151 = 1.035\ or \ k_{300k}={4.62\times10^{-2}\over 1.035}=4.46\times10^{-2}min^{-1}\)