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Explore popular questions from Chemical Kinetics for JEE Main. This collection covers Chemical Kinetics previous year JEE Main questions hand picked by experienced teachers.

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Chemical Kinetics

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Q 1. In a reaction involving the synthesis of ammonia by Haber's process, {tex} \mathrm { N } _ { 2 } + 3 \mathrm { H } _ { 2 } \rightleftharpoons 2 \mathrm { NH } _ { 3 } , {/tex} the rate of reaction was measured {tex} \mathrm { as } = 2.5 \times 10 ^ { - 4 } \mathrm { mol } \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } . {/tex} The rate of change of concentration of {tex} \mathrm { H } _ { 2 } {/tex} will be

A

{tex} 1.25 \times 10 ^ { - 4 } \mathrm { mol } \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex}

B

{tex} 2.50 \times 10 ^ { - 4 } \mathrm { mol } \,\,\mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex}

{tex} 7.5 \times 10 ^ { - 4 } \mathrm { mol }\,\, \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex}

D

{tex} 5.0 \times 10 ^ { - 4 } \mathrm { mol } \,\,\mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex}

Explanation



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Q 2. In the formation of sulphur trioxide by contact process, {tex} 2 \mathrm { SO } _ { 2 } + \mathrm { O } _ { 2 } \rightleftharpoons 2 \mathrm { SO } _ { 3 } {/tex} , the rate of reaction was measured as = {tex}\mathrm{2.5 \times10^{-4} \,mol\,\,L^{-1} sec^{-1}.}{/tex} the rate of reaction expressed on terms of {tex}\mathrm{SO_3}{/tex} will be

A

{tex} - 1.25 \times 10 ^ { - 4 } \mathrm { mol } \mathrm { L } ^ { - 2 } \mathrm { sec } ^ { - 1 } {/tex}

B

{tex} 50 \times 10 ^ { - 4 } \mathrm { mol } \mathrm { L } ^ { - 1 } \mathrm { sec } ^ { - 1 } {/tex}

C

{tex} - 3.75 \times 10 ^ { - 4 } \mathrm { mol } \mathrm { L } ^ { - 1 } \mathrm { sec } ^ { - 1 } {/tex}

{tex} 5.00 \times 10 ^ { - 4 } \mathrm { mol }\,\, \mathrm { L } ^ { - 1 } \mathrm { sec } ^ { - 1 } {/tex}

Explanation

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Q 3. The rate constant of a reaction is equal to rate of reaction:

A

When concentrations of reactants do not change with time.

B

When concentrations of all reactants and products are equal.

C

At time {tex} t = 0 {/tex}

When concentrations of all reactants are unity.

Explanation





Rate constant is equal to rate of reaction when the product of the molar conc. of reactants is unity.

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Q 4. If the concentration of the reactants in the reaction {tex} 2A {/tex} {tex} + B \rightarrow C + D {/tex} is increased by three folds, the rate of the reaction will be increased by

27 times

B

9 times

C

64 times

D

01 times

Explanation

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Q 5. The rate of change in concentration of {tex} C {/tex} in the reaction {tex} 2 A + B \rightarrow 2 C + 3 D {/tex} was reported as {tex} 1.0 \mathrm { mol } \mathrm { L } ^ { - 1 } \mathrm { sec } ^ { - 1 } . {/tex} Calculate the reaction rate:

A

0.05 mol {tex} \mathrm { L } ^ { - 1 } \mathrm { sec } ^ { - 1 } {/tex}

B

0.01{tex} \mathrm { mol } \mathrm { L } ^ { - 1 } \mathrm { sec } ^ { - 1 } {/tex}

0.5{tex} \mathrm { mol } \mathrm { L } ^ { - 1 } \mathrm { sec } ^ { - 1 } {/tex}

D

None of these

Explanation

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Q 6. In a first order reaction, the initial concentration of the reactant was {tex} \mathrm { M } / 10 . {/tex} After 8 minutes 20 seconds the concentration becomes {tex} \mathrm { M } / 100 . {/tex} What is the rate constant?

A

{tex} 5 \times 10 ^ { - 3 } \mathrm { sec } ^ { - 1 } {/tex}

B

{tex} 2.303 \times 10 ^ { - 5 } \mathrm { sec } ^ { - 1 } {/tex}

C

{tex} 2.303 \times 10 ^ { - 4 } \mathrm { sec } ^ { - 1 } {/tex}

{tex} 4.606 \times 10 ^ { - 3 } \mathrm { sec } ^ { - 1 } {/tex}

Explanation



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Q 7. Substance A reacts according to a first order rate law with {tex} K = 5.0 \times 10 ^ { - 5 } \mathrm { s } ^ { - 1 } . {/tex} If the initial concentration of {tex} \mathrm { A } {/tex} is {tex} 1.0 \mathrm { M } , {/tex} the initial rate is

A

{tex} 1 \times 10 ^ { - 5 } \mathrm { Ms } ^ { - 1 } {/tex}

{tex} 5.0 \times 10 ^ { - 5 } \mathrm { Ms } ^ { - 1 } {/tex}

C

{tex} 1 \times 10 ^ { - 4 } \mathrm { Ms } ^ { - 1 } {/tex}

D

{tex} 5.0 \times 10 ^ { - 4 } \mathrm { Ms } ^ { - 1 } {/tex}

Explanation

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Q 8. The mechanism of the reaction is  
The rate constant of the reaction is

A

{tex} K _ { 2 } {/tex}

B

{tex} K _ { 2 } K _ { 1 } \left( K _ { - 1 } \right) {/tex}

C

{tex} K _ { 2 } K _ { 1 } {/tex}

{tex} K _ { 2 } \left( \frac { K _ { 1 } } { K _ { - 1 } } \right) {/tex}

Explanation



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Q 9. Dinitrogen pentaoxide decomposes as 2{tex} \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \rightarrow 4 \mathrm { NO } _ { 2 } {/tex} {tex} + \mathrm { O } _ { 2 } . {/tex} The rate can be given in three ways

{tex} \frac { - d \left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] } { d t } = {/tex}{tex} K _ { 1 } \left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] , \frac { d \left[ \mathrm { NO } _ { 2 } \right] } { d t } = K _ { 2 } \left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] , \frac { d \left[ \mathrm { NO } _ { 2 } \right] } { d t } = K _ { 3 } \left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] {/tex}

The relation between the rate constants {tex} K _ { 1 } , K _ { 2 } {/tex} and {tex} K _ { 3 } {/tex} is

{tex} K _ { 2 } = 2 K _ { 1 } {/tex} and {tex} K _ { 3 } = {K _ { 1 } / 2} {/tex}

B

{tex} K _ { 1 } = 2 K _ { 2 } {/tex} and {tex} K _ { 3 } = 2 K _ { 1 } {/tex}

C

{tex} K _ { 1 } = K _ { 2 } = K _ { 3 } {/tex}

D

{tex} K _ { 1 } = 2 K _ { 2 } = 3 K _ { 3 } {/tex}

Explanation

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Q 10. For a reaction,,{tex}\frac{d[x]}{dt}{/tex}is equal to

A

{tex} k _ { 1 } ( a - x ) - k _ { 2 } ( a - x ) {/tex}

B

{tex} k _ { 2 } ( a - x ) - k _ { 1 } ( a - x ) {/tex}

{tex} k _ { 1 } ( a - x ) + k _ { 2 } ( a - x ) {/tex}

D

{tex} - k _ { 1 } ( a - x ) - k _ { 2 } ( a - x ) {/tex}

Explanation

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Q 11. The mechanism of the reaction : {tex} A + 2 B + C \rightarrow D {/tex} is
{tex} (step 1) \quad (fast) equilibrium {tex} A + B \rightleftharpoons X {/tex}
{tex} \begin{array} { l l } { ( \operatorname { step } 2 ) } & { ( \text { slow } ) X + C \rightarrow Y } \\ { ( \text { step } 3 ) } & { \text { (fast) } Y + B \rightarrow D } \end{array} {/tex}
Which rate law is correct?

A

{tex} r = k [ C ] {/tex}

B

{tex} r = k [ A ] [ B ] ^ { 2 } [ C ] {/tex}

{tex} r = k [ A ] [ B ] [ C ] {/tex}

D

{tex} r = k [ D ] {/tex}

Explanation

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Q 12. Mechanism of the reaction {tex} 2 \mathrm { NO } + \mathrm { Cl } _ { 2 } \rightarrow 2 \mathrm { NOCl } {/tex} may be written as
2{tex} \mathrm { NO } \stackrel { K } { \rightleftharpoons } ( \mathrm { NO } ) _ { 2 } \cdots \cdots ( \text { fast } ) {/tex}
{tex} ( \mathrm { NO } ) _ { 2 } + \mathrm { Cl } _ { 2 } \stackrel { k } { \longrightarrow } 2 \mathrm { NOCl } \ldots \ldots . {/tex} (slow)
Rate equation would be

A

{tex} k K \left[ ( \mathrm { NO } ) ^ { 4 } \right] \left[ \mathrm { Cl } _ { 2 } \right] {/tex}

{tex} k K [ \mathrm { NO } ] ^ { 2 } \left[ \mathrm { Cl } _ { 2 } \right] {/tex}

C

{tex} k K \left[ \mathrm { Cl } _ { 2 } \right] {/tex}

D

{tex} k K [ \mathrm { NO } ] \cdot \left[ \mathrm { Cl } _ { 2 } \right] {/tex}

Explanation

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Q 13. The half-life of decomposition of {tex}\mathrm { N _ { 2 } O } _ { 5 } {/tex} is a first order reaction represented by {tex} \mathrm N _ { 2 } \mathrm { O } _ { 5 } \rightarrow \mathrm { N } _ { 2 } \mathrm { O } _ { 4 } + 1 / 2 \mathrm { O } _ { 2 } {/tex} After 15 minutes, the volume of {tex} \mathrm { O } _ { 2 } {/tex} , produced is 9{tex} \mathrm { mL } {/tex} and at the end of the reaction 35{tex} \mathrm { mL } {/tex} . The rate constant is equal to

{tex} \frac { 1 } { 15 } \log _ { e } \frac { 35 } { 26 } {/tex}

B

{tex} \frac { 1 } { 15 } \log _ { e } \frac { 44 } { 26 } {/tex}

C

{tex} \frac { 1 } { 15 } \log _ { e } \frac { 35 } { 36 } {/tex}

D

None of the foregoing

Explanation

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Q 14. 99{tex} \% {/tex} at a first order reaction was completed in 32{tex} \mathrm { min } {/tex} . When will 99.9{tex} \% {/tex} of the reaction complete?

48{tex} \mathrm { min } {/tex}

B

46{tex} \mathrm { min } {/tex}

C

50{tex} \mathrm { min } {/tex}

D

45{tex} \mathrm { min } {/tex}

Explanation


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Q 15. {tex} T _ { 0.5 } = {/tex} constant, confirms the first order of the reaction as one {tex} a ^ { 2 } T _ { 0.5 } = {/tex} constant confirms that the reaction is of

A

Zero order

B

First order

C

Second order

Third order

Explanation

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Q 16. The half-life period for a reaction at initial concentration of 0.5 and 0.1 moles {tex}\mathrm{L^{-1}}{/tex} are 200 second 100 second respectively. The order of the reaction is

A

0

B

1

2

D

3

Explanation

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Q 17. The slope of the line for the graph of {tex} \log\mathrm K {/tex} versus {tex} \frac { 1 } { T } {/tex} for the reaction, {tex} \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \rightarrow 2 \mathrm { NO } _ { 2 } + \frac { 1 } { 2 } \mathrm { O } _ { 2 } {/tex} is {tex} - 5000 . {/tex} Calculate the energy of activation of the reaction {tex} - \left( \mathrm { KJ } \mathrm { K } ^ { - 1 } \right. {/tex} {tex} \mathrm { mol } ^ { - 1 } ) {/tex}

95.7

B

9.57

C

957

D

None

Explanation



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Q 18. For a reaction, the rate constant is expressed as, {tex} \mathrm { k } = {/tex} A.e {tex} ^ { - 40000 / \mathrm { T } } . {/tex} The energy of the activation is

A

40000{tex} \mathrm { cal } {/tex}

B

88000{tex} \mathrm { cal } {/tex}

80000{tex} \mathrm { cal } {/tex}

D

8000{tex} \mathrm { cal } {/tex}

Explanation

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Q 19. A reaction takes place in three steps. The rate constants are {tex} k _ { 1 } , k _ { 2 } {/tex} and {tex} k _ { 3 } . {/tex} The over all rate constant {tex} k = \frac { k _ { 1 } k _ { 3 } } { k _ { 2 } } . {/tex} If (energy of activation) {tex} E _ { 1 } , E _ { 2 } {/tex} and {tex} E _ { 3 } {/tex} are {tex} 60,30 {/tex} and {tex}10{/tex} kJ. The overall energy of activation is:

80

B

30

C

400

D

60

Explanation

{tex}Ae^{{-60-10+30}/RT}{/tex}={tex}Ae^{{-40}/RT}{/tex}

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Q 20. For a chemical reaction {tex} A + 3 B \longrightarrow {/tex} Product It was observed that rate of reaction increases nine times when concentration of {tex} B {/tex} increased three times by keeping concentration of {tex} A {/tex} as constant. On doubling concentration of both rate increases eight times. Differential rate equation can be given as

A

{tex} r = k [ A ] [ B ] ^ { 3 } {/tex}

{tex} r = k [ A ] [ B ] ^ { 2 } {/tex}

C

{tex} r = k [ A ] ^ { 2 } [ B ] {/tex}

D

{tex} r = k [ A ] ^ { 2 } [ B ] ^ { 1 / 3 } {/tex}

Explanation