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

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Chemical Kinetics

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Q 1. During the kinetic study of the reaction, {tex} 2 A + B \rightarrow C + D , {/tex} following results were obtained

Based on the above data which one of the following is correct?

A

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

B

Rate {tex} = k [ A ] [ B ] {/tex}

C

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

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

Explanation





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Q 2. In the reaction,
{tex} \ \mathrm{ Br O } _ {3 ( a q ) } ^ { - } + 5 \mathrm { Br } _ { ( a q ) } ^ { - } + 6 \mathrm { H } ^ { + } \rightarrow 3 \mathrm { Br } _ { 2 ( \ell ) } + 3 \mathrm { H } _ { 2 } \mathrm { O } _ { ( l ) } {/tex}
The rate of appearance of bromine {tex}\mathrm{(Br_{2} )}{/tex} is related to rate of disappearance of bromide ions as

A

{tex} \frac { d \left[ \mathrm { Br } _ { 2 } \right] } { d t } = - \frac { 5 } { 3 } \frac { d \left[ \mathrm { Br } ^ { - } \right] } { d t } {/tex}

B

{tex} \frac { d \left[ \mathrm { Br } _ { 2 } \right] } { d t } = \frac { 5 } { 3 } \frac { d \left[ \mathrm { Br } ^ { - } \right] } { d t } {/tex}

C

{tex} \frac { d \left[ \mathrm { Br } _ { 2 } \right] } { d t } = \frac { 3 } { 5 } \frac { d \left[ \mathrm { Br } ^ { - } \right] } { d t } {/tex}

{tex} \frac { d \left[ \mathrm { Br } _ { 2 } \right] } { d t } = - \frac { 3 } { 5 } \frac { d \left[ \mathrm { Br } ^ { - } \right] } { d t } {/tex}

Explanation


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Q 3. For the reaction {tex} A + B \rightarrow {/tex} products, it is observed that
(1) on doubling the initial concentration of {tex} A {/tex} only, the rate of reaction is also doubled and
(ii) on doubling the initial concentration of
both {tex} A {/tex} and {tex} B , {/tex} there is a change by a factor of 8 in the rate of the reaction. The rate of this reaction is given by

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

B

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

C

rate {tex} = k [ A ] [ B ] {/tex}

D

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

Explanation

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Q 4. The bromination of acetone that occurs in acid solution is represented by this equation.
{tex} \mathrm { CH } _ { 3 } \mathrm { COCH } _ { 3 ( a ) } + \mathrm { Br } _ { 2 ( a q ) } \rightarrow {/tex} {tex} \mathrm { CH } _ { 3 } \mathrm { COCH } _ { 2 } \mathrm { Br } _ { ( a q ) } + \mathrm { H } _ { ( a q ) } ^ { + } + \mathrm { Br } _ { ( a q ) } ^ { - } ( a q ) {/tex}
These kinetic data were obtained for given reaction concentrations.


Based on these data, the rate equation is

A

Rate {tex} = k \left[ \mathrm { CH } _ { 3 } \mathrm { COCH } _ { 3 } \right] \left[ \mathrm { Br } _ { 2 } \right] \left[ \mathrm { H } ^ { + } \right] ^ { 2 } {/tex}

B

Rate {tex} = k \left[ \mathrm { CH } _ { 3 } \mathrm { COCH } _ { 3 } \right] \left[ \mathrm { Br } _ { 2 } \right] \left[ \mathrm { H } ^ { + } \right] {/tex}

Rate {tex} = k \left[ \mathrm { CH } _ { 3 } \mathrm { COCH } _ { 3 } \right] [ \mathrm { H ^+} ] {/tex}

D

Rate {tex} = k \left[ \mathrm { CH } _ { 3 } \mathrm { COCH } _ { 3 } \right] \left[ \mathrm { Br } _ { 2 } \right] {/tex}

Explanation


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Q 5. In a first-order reaction {tex} A \rightarrow B , {/tex} if {tex} k {/tex} is rate constant and initial concentration of the reactant {tex} A {/tex} is 0.5{tex}\ \mathrm { M } {/tex}, then the half-life is

A

{tex} \frac { \log 2 } { k } {/tex}

B

{tex} \frac { \log 2 } { k \sqrt { 0.5 } } {/tex}

{tex} \frac { \ln 2 } { k } {/tex}

D

{tex} \frac { 0.693 } { 0.5 k } {/tex}

Explanation

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Q 6. For a first order reaction {tex} A \rightarrow B {/tex} the reaction rate at reactant concentration of {tex}0.01\ \mathrm { M } {/tex} is found to be {tex} 2.0 \times 10 ^ { - 5 } \mathrm { mol }\ \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } . {/tex} The half- life period of the reaction is

A

30{tex}\ \mathrm { s } {/tex}

B

220{tex}\ \mathrm { s } {/tex}

C

300{tex}\ \mathrm { s } {/tex}

347{tex}\ \mathrm { s } {/tex}

Explanation


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Q 7. The reaction {tex} A \rightarrow B {/tex} follows first order kinetics. The time taken for 0.8 mole of {tex} A {/tex} to produce 0.6 mole of {tex} B {/tex} is 1 hour. What is the time taken for conversion of 0.9 mole of {tex} A {/tex} to produce 0.675 mole of {tex} B ? {/tex}

1 hour

B

0.5 hour

C

0.25 hour

D

2 hours

Explanation


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Q 8. When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is

A

second

B

more than zero but less than first

C

Zero

first

Explanation

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Q 9. A reaction is 50% complete in 2 hours and 75% complete in 4 hours. The order of reaction is

1

B

2

C

3

D

0

Explanation

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Q 10. In a zero-order reaction, for every {tex} 10 ^ { \circ } \mathrm { C } {/tex} rise of temperature, the rate is doubled. If the temperature is increased from {tex} 10 ^ { \circ } \mathrm { C } {/tex} to {tex} 100 ^ { \circ } \mathrm { C } , {/tex} the rate of the reaction will become

A

256 times

512 times

C

64 times

D

128 times

Explanation

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Q 11. The half-life of a substance in a certain enzyme- catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L{tex} ^ { - 1 } {/tex} to 0.04 mg L {tex} ^ { - 1 } {/tex} is

A

414{tex}\ \mathrm { s } {/tex}

B

552{tex}\ \mathrm { s } {/tex}

690{tex}\ \mathrm { s }{/tex}

D

276{tex}\ \mathrm { s } {/tex}

Explanation

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Q 12. For the reaction {tex} \mathrm { N } _ { 2 } \mathrm { O } _ { 5 ( \mathrm { g } ) } \rightarrow 2 \mathrm { NO } _ { 2 ( \mathrm { g } ) } + 1 / 2 \mathrm { O } _ { 2 ( \mathrm { g } ) } {/tex} the value of rate of disappearance of {tex} \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } {/tex} is given as {tex} 6.25 \times 10 ^ { - 3 } \mathrm { mol } \,\mathrm { L } ^ { 1 - 1 } \mathrm { s } ^ { - 1 } {/tex} . The rate of formation of {tex} \mathrm { NO } _ { 2 } {/tex} and {tex} \mathrm { O } _ { 2 } {/tex} is given respectively as

A

{tex} 6.25 \times 10 ^ { - 3 } \mathrm { mol }\ \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex} and
{tex} 6.25 \times 10 ^ { - 3 } \mathrm { mol }\ \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex}

{tex} 1.25 \times 10 ^ { - 2 } \mathrm { mol }\ \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex} and
{tex} 3.125 \times 10 ^ { - 3 } \mathrm { mol }\ \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex}

C

{tex} 6.25 \times 10 ^ { - 3 } \mathrm { mol }\ \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex} and
{tex} 3.125 \times 10 ^ { - 3 } \mathrm { mol }\ \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex}

D

{tex} 1.25 \times 10 ^ { - 2 } \mathrm { mol }\ \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex} and
{tex} 6.25 \times 10 ^ { - 3 } \mathrm { mol }\ \mathrm { L } ^ { - 1 } \mathrm { s } ^ { - 1 } {/tex}

Explanation



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Q 13. Half-life period of a first order reaction is 1386 seconds. The specific rate constant of the reaction is

A

{tex} 0.5 \times 10 ^ { - 2 } \mathrm { s } ^ { - 1 } {/tex}

{tex} 0.5 \times 10 ^ { - 3 } \mathrm { s } ^ { - 1 } {/tex}

C

{tex} 5.0 \times 10 ^ { - 2 } \mathrm { s } ^ { - 1 } {/tex}

D

{tex} 5.0 \times 10 ^ { - 3 } \mathrm { s } ^ { - 1 } {/tex}

Explanation

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Q 14. The reaction of hydrogen and iodine monochloride is given as:
{tex} \mathrm { H } _ { 2 ( g ) } + 2 \mathrm { ICl } _ { ( g ) } \rightarrow 2 \mathrm { HCl } _ { ( g ) } + \mathrm { I } _ { 2 ( g ) } {/tex}
This reaction is of first order with respect to {tex} \mathrm { H } _ { 2 ( g ) } {/tex} and {tex} \mathrm { ICl } _ { ( g ) } , {/tex} following mechanisms were proposed.
{tex}Mechanism\ A {/tex}:
{tex}\mathrm{H_{2(g)}+2ICl_(g)\rightarrow 2HCl_(g)+I_{2(g)}}{/tex}
{tex}Mechanism\ B{/tex}:
{tex}\mathrm{H_{2(g)}+ICl_(g)\rightarrow HCl_(g)+HI_{2(g)};}{/tex} slow
{tex}\mathrm{HI_{(g)}+ICl_(g)\rightarrow HCl_(g)+I_{2(g)};}{/tex} fast
Which of the above mechanism(s) can be consistent with the given information about the reaction?

A

{tex} A {/tex} and {tex} B {/tex} both

B

Neither {tex} A {/tex} nor {tex} B {/tex}

C

{tex} A {/tex} only

{tex} B {/tex} only

Explanation


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Q 15. The temperature dependence of rate constant {tex}(k){/tex} of a chemical reaction is written in terms of Arrhenius equation, {tex} k = A \cdot e ^ { - E ^ { * }/ R T } {/tex}. Activation energy {tex} \left( E ^ { * } \right) {/tex} of the reaction can be calculated by plotting

A

{tex} k\ v s\ T {/tex}

B

{tex} k\ v s\ \frac { 1 } { \log T } {/tex}

{tex} \log k\ v s\ \frac { 1 } { T } {/tex}

D

{tex} \log k\ v s\ \frac { 1 } { \log T } {/tex}

Explanation

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Q 16. The experimental data for the reaction,
{tex} 2 A + B _ { 2 } \rightarrow 2 A B {/tex} is

{tex} \begin{array} { c c c c } { \text { Experiment } } & { [ A ] } & { \left[ B _ { 2 } \right] } & {\left. \text { Rate (mole } s ^ { - 1 } \right) } \\ { 1 } & { 0.50 } & { 0.50 } & { 1.6 \times 10 ^ { - 4 } } \\ { 2 } & { 0.50 } & { 1.00 } & { 3.2 \times 10 ^ { - 4 } } \\ { 3 } & { 1.00 } & { 1.00 } & { 3.2 \times 10 ^ { - 4 } } \end{array} {/tex}
The rate equation for the above data is

A

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

B

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

Rate {tex} = k \left[ B _ { 2 } \right] {/tex}

D

Rate {tex} = k \left[ B _ { 2 } \right] ^ { 2 } {/tex}

Explanation




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Q 17. The plot that represents the zero order reaction is:

A

B

C

Explanation

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Q 18. The rate equation for a reaction,{tex} \mathrm { N } _ { 2 } \mathrm { O } \longrightarrow \mathrm { N } _ { 2 } + 1 / 2 \mathrm { O } _ { 2 } {/tex} . Rate {tex} = \mathrm { k } \left[ \mathrm { N } _ { 2 } \mathrm { O } \right] ^ { 0 } = \mathrm { k } {/tex}. If the initial concentration of the reactant is a {tex} \ \mathrm { mol \ Lit}^{-1} {/tex} , the half-life period of the reaction is

{tex} t _ { \frac { 1 } { 2 } } = \frac { a } { 2 k } {/tex}

B

{tex} - t _ { \frac { 1 } { 2 } } = k a {/tex}

C

{tex} t _ { \frac { 1 } { 2 } } = \frac { a } { k } {/tex}

D

{tex} t _ { \frac { 1 } { 2 } } = \frac { k } { a } {/tex}

Explanation

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Q 19. In a first order reaction, reactant concentration C varies with time t as

A

1/ C increases linearly with t

log C decreases linearly with t

C

C decreases with 1/ t

D

log C decreases with 1/ t

Explanation

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Q 20. The rate law for a reaction between the substances A and B is given by rate = k[A]n [B]m. On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

A

B

(m + n)

C

(n - m)

2(n - m)

Explanation

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Q 21. For the First order reaction with half life is 150 seconds, the time taken for the concentration of the reactant to fall from m/10 to m/100 will be approximately

A

600 s

B

900 s

500 s

D

1500 s

Explanation

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Q 22. Mathematical expression for t1/4 i.e. when (1/4)th reaction is over following first order kinetics can be given by

A

B

D

Explanation

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Q 23. Which of the following is the fast reaction ?

A

B

C

Explanation