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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 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 2. 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 3. 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 4. 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 5. 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 6. 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 7. 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 8. 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 9. 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 10. 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 11. Units of rate constant of first and zero order reactions in terms of molarity {tex}\mathrm{M}{/tex} unit are respectively

{tex} \sec ^ { - 1 } , \operatorname { Msec } ^ { - 1 } {/tex}

B

{tex} \sec ^ { - 1 } , \mathrm { M } {/tex}

C

{tex} \mathrm { Msec } ^ { - 1 } , \mathrm { sec } ^ { - 1 } {/tex}

D

{tex} \mathrm { M } , \mathrm { sec } ^ { - 1 } {/tex}

Explanation


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Q 12. The rate law for a reaction between the substances {tex} A {/tex} and {tex} B {/tex} is given by rate {tex} = k [ A ] ^ { n } [ B ] ^ { m } {/tex}. On doubling the concentration of {tex} A {/tex} and halving the concentration of {tex} B , {/tex} the ratio of the new rate to the earlier rate of the reaction will be as

A

{tex} \frac { 1 } { 2 ^ { m + n } } {/tex}

B

{tex} ( m + n ) {/tex}

C

{tex} ( n - m ) {/tex}

{tex} 2^{( n - m )} {/tex}

Explanation


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Q 13. For the reaction {tex} A + 2 B \rightarrow C , {/tex} rate is given by {tex} R = [ A ] [ B ] ^ { 2 } {/tex} then the order of the reaction is

3

B

6

C

5

D

7

Explanation


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Q 14. For the reaction system: {tex}{ 2 \mathrm { NO } _ { ( \mathrm { g } ) } + \mathrm { O } _ { 2 ( \mathrm { g } ) } \rightarrow 2 \mathrm { NO } _ { 2 ( \mathrm { g } ) }}{/tex} volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to {tex} \mathrm { O } _ { 2 } {/tex} and second order with respect to {tex}\mathrm { NO } _ { 2 }{/tex} the rate of reaction will

A

diminish to one-fourth of its initial value

B

diminish to one-eighth of its initial value

increase to eight times of its initial value

D

increase to four times of its initial value.

Explanation


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Q 15. {tex} t _ { 1/4 } {/tex} can be taken as the time taken for the concentration of a reactant to drop to {tex} 3 / 4 {/tex} of its initial value. If the rate constant for a first order reaction is {tex} k , {/tex} the {tex} t _ { 1/4 } {/tex} can be written as

A

{tex} 0.10 / k {/tex}

{tex} 0.29 / k {/tex}

C

{tex} 0.69 / k {/tex}

D

{tex} 0.75 / k {/tex}

Explanation

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Q 16. The rate equation for the reaction {tex} 2 A + B \rightarrow C {/tex} is found to be : rate {tex} = k [ A ] [ B ] {/tex}. The correct statement in relation to this reaction is that the

A

unit of {tex} k {/tex} must be {tex} \mathrm { s } ^ { - 1 } {/tex}

B

{tex} t _ { 1 / 2 } {/tex} is a constant

C

rate of formation of {tex} C {/tex} is twice the rate of disappearance of {tex} A {/tex}

value of {tex} k {/tex} is independent of the initial concentrations of {tex} A {/tex} and {tex} B . {/tex}

Explanation


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Q 17. A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will be

A

remain unchanged

B

tripled

increased by a factor of {tex}4{/tex}

D

doubled

Explanation


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Q 18. In the respect of the equation {tex} k = A e ^ { - E _ { a } / R T } {/tex} in chemical kinetics, which one of the following statements is is correct?

A

{tex} k {/tex} is equilibrium constant

B

{tex} A {/tex} is adsorption factor

{tex} E _ { a } {/tex} is energy of activation

D

{tex} R {/tex} is Rydberg constant.

Explanation


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Q 19. The reaction {tex}\mathrm{ A + 2 B + C \rightarrow 2 D + E }{/tex} is found to be {tex} 1,2 {/tex} and zero order with respect to {tex} \mathrm{A , B} {/tex} and {tex} \mathrm C {/tex} respectively. What will be the final rate, if concentration of each reactant is doubled?

A

{tex}\mathrm{2\ times}{/tex}

B

{tex}\mathrm{4\ times}{/tex}

{tex}\mathrm{8\ times}{/tex}

D

{tex}\mathrm{16\ times}{/tex}

Explanation



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Q 20. Which of the following is an incorrect statement?

A

Half life of second order reaction decreases with increase in concentration of reactant

B

Half life of first order is independent of concentration of the reactant

C

The unit of rate constant of zero order is equal to unit of rate

The unit of frequency factor 'A' in Arrhenius equation is the unit of half life of the reaction

Explanation



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Q 21. For a hypothetical reaction {tex} x + y = A + B , {/tex} Rate {tex} = k [ x ] ^ { 5 / 2 } [ y ] ^ { - 1 / 2 } {/tex} on doubling the concentration of {tex} x {/tex} and {tex} y {/tex} the rate will become

A

2 times

4 times

C

8 times

D

Remains same

Explanation



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Q 22. The gaseous reaction {tex} A ( g ) \rightarrow 2 B ( g ) + C ( g ) {/tex} obeys first order kinetics. If the initial {tex} P = 90 \mathrm { mm } {/tex} and pressure after {tex}10{/tex} minutes {tex} = 180 \mathrm { mm } {/tex}. The velocity constant {tex} k {/tex} of the reaction is

A

{tex} 1.15 \times 10 ^ { + 3 } \mathrm { s } ^ { - 1 } {/tex}

B

{tex} 2.30 \times 10 ^ { + 3 } \mathrm { s } ^ { - 1 } {/tex}

C

{tex} 3.45 \times 10 ^ { - 3 } \mathrm { s } ^ { - 1 } {/tex}

{tex} 1.15 \times 10 ^ { - 3 } \mathrm { s } ^ { - 1 } {/tex}

Explanation



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Q 23. For a reaction {tex} 2 \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \longrightarrow 4 \mathrm { NO } _ { 2 } ( \mathrm { g } ) + \mathrm { O } _ { 2 } ( \mathrm { g } ) , {/tex} the rate and rate constant are {tex} 1.02 \times 10 ^ { - 4 } \mathrm { mol\ L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex} and {tex} 3.4 \times 10 ^ { - 5 } {/tex} {tex} \mathrm { s } ^ { - 1 } {/tex} respectively. The concentration of {tex} \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } {/tex} at this time will be

A

{tex} 1.732 \mathrm { mol } / \mathrm { L } {/tex}

{tex} 3 \mathrm { mol } / \mathrm { L } {/tex}

C

{tex} 1.02 \times 10 ^ { - 4 } \mathrm { mol } / \mathrm { L } {/tex}

D

{tex} 3.2 \times 10 ^ { 5 } \mathrm { mol } / \mathrm { L } {/tex}

Explanation

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Q 24. One gram atom of {tex} _ { 79 } \mathrm { Au } ^ { 198 } (\mathrm { t } _ { 1 / 2 } = 65 hours){/tex} decays by {tex}\beta{/tex} -emission to produce stable nuclide of {tex}\mathrm {Hg}{/tex}. How much {tex}\mathrm {Hg}{/tex} will be present after {tex}260{/tex} hours?

A

{tex} \frac { 1 } { 16 } \mathrm { g } {/tex} atom

B

{tex} \frac { 1 } { 32 } \mathrm { g } {/tex} atom

{tex} \frac { 15 } { 16 } \mathrm { g } {/tex} atom

D

{tex} \frac { 16 } { 15 } \mathrm { g } {/tex} atom

Explanation


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Q 25. {tex}1.0{/tex} g atom of an {tex} \alpha {/tex} -emiting isotope ({tex}\mathrm { t } _ { 1 } / 2 = 10 {/tex} days) is sealed in a container. The volume of gas accumulated in the container at STP after {tex}10{/tex} days would be

A

{tex} 1.0 \mathrm { L } {/tex}

B

{tex} 44.8 \mathrm { L } {/tex}

{tex} 11.2 \mathrm { L } {/tex}

D

{tex} 22.4 \mathrm { L } {/tex}

Explanation