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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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Q 1. 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 2. 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 3. 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 4. 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 5. 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 6. 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 7. 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 8. 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 9. 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