Chemical Kinetics NEET Previous Year Questions with Answers

Q 1 :
If the rate constant of a reaction is 0.03 $s^{- 1}$ , how much time does it take for 7.2 mol $L^{- 1}$ concentration of the reactant to get reduced to 0.9 mol $L^{- 1}$ ?

(Given: log 2 = 0.301) [2025]

210 s
21.0 s
69.3 s
23.1 s

1

2

3

4

(3)

Rate constant, $k = 0.03$ $s^{- 1}$

For $1^{st}$ order reaction, $k = \frac{2.303}{t} \log \frac{a}{a - x}$

So, $t = \frac{2.303}{0.03} \log \frac{7.2}{0.9} = \frac{2.303}{0.03} \log 8 = \frac{2.303}{0.03} \times 3 \log 2$

$= \frac{2.303}{0.03} \times 3 \times 0.301$ $[Given, \log 2 = 0.301]$

$= 69.32$ $s$

Q 2 :
If the half-life $(t_{1 / 2})$ for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to [2025]

5 minutes
10 minutes
2 minutes
4 minutes

1

2

3

4

(2)

For a first order reaction,

$t_{1 / 2} = \frac{0.693}{k}$

or,
$k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{1} = 0.693$ ${min}^{- 1}$ $(∵ t_{1 / 2} = 1 minute)$

For 99.9% completion of reaction, $k = \frac{2.303}{t} \log \frac{a}{a - x}$

or, $t = \frac{2.303}{k} \log \frac{a}{a - x} = \frac{2.303}{0.693} \log \frac{100}{100 - 99.9}$

or, $t = \frac{2.303}{0.693} \times 3 = 9.97$ $min \approx 10$ $min$

Q 3 :
For a first order reaction A $\to$ Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in $\min^{- 1}$ is [2022]

1.3818
0.9212
0.4606
0.2303

1

2

3

4

(2)

For a first order reaction

$k = \frac{2.303}{t} \log \frac{a}{a - x}$

$k = \frac{2.303}{5} \log \frac{0.1}{0.001}$

$k = \frac{2.303}{5} \log 10^{2} = \frac{2.303 \times 2}{5} = 0.9212$ ${min}^{- 1}$

Q 4 :
The given graph is a representation of kinetics of a reaction.

[IMAGE 1]

The y and x axes for zero and first order reactions, respectively, are [2022]

zero order (y = concentration and x = time), first order (y = $t_{1 / 2}$ and x = concentration)
zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)
zero order (y = rate and x = concentration), first order (y = $t_{1 / 2}$ and x = concentration)
zero order (y = rate and x = concentration), first order (y = rate and x = $t_{1 / 2}$ )

1

2

3

4

(3)

[IMAGE 2]

Q 5 :
The rate constant for a first order reaction is $4.606 \times 10^{- 3}$ $s^{- 1}$ . The time required to reduce 2.0 g of the reactant to 0.2 g is [2020]

100 s
200 s
500 s
1000 s

1

2

3

4

(3)

For a first order reaction,

$k = \frac{2.303}{t} \log \frac{{[R]}_{0}}{[R]}$

$t = \frac{2.303}{4.606 \times 10^{- 3} s^{- 1}} \log (\frac{2}{0.2}) = \frac{2.303 \times 10^{3}}{4.606} = 500$ $s$

Q 6 :
If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by [2019]

$t = 2.303 / k$
$t = 0.693 / k$
$t = 6.909 / k$
$t = 4.606 / k$

1

2

3

4

(4)

For $1^{st}$ order reaction,

$t = \frac{2.303}{k} \log \frac{a}{a - x} = \frac{2.303}{k} \log \frac{100}{100 - 99}$

$= \frac{2.303}{k} \log 10^{2} = \frac{2.303}{k} \times 2 \times \log 10 = \frac{4.606}{k}$

Q 7 :
A first order reaction has a rate constant of $2.303 \times 10^{- 3}$ $s^{- 1}$ . The time required for 40 g of this reactant to reduce to 10 g will be

[Given that $\log_{10}$ 2 = 0.3010] [2019]

230.3 s
301 s
2000 s
602 s

1

2

3

4

(4)

For a first order reaction, $k = \frac{2.303}{t} \log \frac{{[A]}_{0}}{{[A]}_{t}}$

$2.303 \times 10^{- 3} = \frac{2.303}{t} \log \frac{40}{10}$

$t = \frac{1}{10^{- 3}} \log 2^{2} = \frac{2}{10^{- 3}} \log 2 = \frac{2}{10^{- 3}} \times 0.3010 = 602$ $s$

Q 8 :
The correct difference between first and second order reactions is that [2018]

the rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
the half-life of a first-order reaction does not depend on ${[A]}_{0}$ ; the half-life of a second-order reaction does depend on ${[A]}_{0}$
a first-order reaction can be catalysed; a second-order reaction cannot be catalysed
the rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations.

1

2

3

4

(2)

For the first order reaction, $t_{1 / 2} = \frac{0.693}{k}$ which is independent of initial concentration ${[A]}_{0}$ .

For second order reaction, $t_{1 / 2} = \frac{1}{k {[A]}_{0}}$

Half-life depends on initial concentration of reactant.

Q 9 :
When initial concentration of the reactant is doubled, the half-life period of a zero order reaction [2018]

is halved
is doubled
is tripled
remains unchanged

1

2

3

4

(2)

${(t_{1 / 2})}_{zero} = \frac{{[A]}_{0}}{2 k}$

As the half-life of a zero order reaction is directly proportional to initial concentration.

$∴$ If ${[A]}_{0}$ = doubled then, $t_{1 / 2}$ = doubled.

Q 10 :
A first order reaction has a specific reaction rate of $10^{- 2}$ $\sec^{- 1}$ . How much time will it take for 20 g of the reactant to reduce to 5 g? [2017]

138.6 sec
346.5 sec
693.0 sec
238.6 sec

1

2

3

4

(1)

For a first order reaction,

$k = \frac{2.303}{t} \log \frac{{[A]}_{0}}{{[A]}_{t}} or 10^{- 2} = \frac{2.303}{t} \log \frac{20}{5}$

$10^{- 2} = \frac{2.303 \times 0.6020}{t} or t = 138.6$ $sec$

Topic Question Set

Q 1 :
If the rate constant of a reaction is 0.03 $s^{- 1}$ , how much time does it take for 7.2 mol $L^{- 1}$ concentration of the reactant to get reduced to 0.9 mol $L^{- 1}$ ?

(Given: log 2 = 0.301) [2025]

Q 2 :
If the half-life $(t_{1 / 2})$ for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to [2025]

Q 3 :
For a first order reaction A $\to$ Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in $\min^{- 1}$ is [2022]

Q 4 :
The given graph is a representation of kinetics of a reaction.

[IMAGE 1]

The y and x axes for zero and first order reactions, respectively, are [2022]

Q 5 :
The rate constant for a first order reaction is $4.606 \times 10^{- 3}$ $s^{- 1}$ . The time required to reduce 2.0 g of the reactant to 0.2 g is [2020]

Q 6 :
If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by [2019]

Q 7 :
A first order reaction has a rate constant of $2.303 \times 10^{- 3}$ $s^{- 1}$ . The time required for 40 g of this reactant to reduce to 10 g will be

[Given that $\log_{10}$ 2 = 0.3010] [2019]

Q 8 :
The correct difference between first and second order reactions is that [2018]

Q 9 :
When initial concentration of the reactant is doubled, the half-life period of a zero order reaction [2018]

Q 10 :
A first order reaction has a specific reaction rate of $10^{- 2}$ $\sec^{- 1}$ . How much time will it take for 20 g of the reactant to reduce to 5 g? [2017]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : If the rate constant of a reaction is 0.03 s-1, how much time does it take for 7.2 mol L-1 concentration of the reactant to get reduced to 0.9 mol L-1? (Given: log 2 = 0.301) [2025]

Q 2 : If the half-life (t1/2) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to [2025]

Q 3 : For a first order reaction A → Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min-1 is [2022]

Q 4 : The given graph is a representation of kinetics of a reaction. [IMAGE 1] The y and x axes for zero and first order reactions, respectively, are [2022]

Q 5 : The rate constant for a first order reaction is 4.606×10-3 s-1. The time required to reduce 2.0 g of the reactant to 0.2 g is [2020]

Q 6 : If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by [2019]

Q 7 : A first order reaction has a rate constant of 2.303×10-3 s-1. The time required for 40 g of this reactant to reduce to 10 g will be [Given that log10 2 = 0.3010] [2019]

Q 8 : The correct difference between first and second order reactions is that [2018]

Q 9 : When initial concentration of the reactant is doubled, the half-life period of a zero order reaction [2018]

Q 10 : A first order reaction has a specific reaction rate of 10-2 sec-1. How much time will it take for 20 g of the reactant to reduce to 5 g? [2017]

Want Us to Email you About Special Offers & Updates?

Q 1 :
If the rate constant of a reaction is 0.03 $s^{- 1}$ , how much time does it take for 7.2 mol $L^{- 1}$ concentration of the reactant to get reduced to 0.9 mol $L^{- 1}$ ?

(Given: log 2 = 0.301) [2025]

Q 2 :
If the half-life $(t_{1 / 2})$ for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to [2025]

Q 3 :
For a first order reaction A $\to$ Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in $\min^{- 1}$ is [2022]

Q 4 :
The given graph is a representation of kinetics of a reaction.

[IMAGE 1]

The y and x axes for zero and first order reactions, respectively, are [2022]

Q 5 :
The rate constant for a first order reaction is $4.606 \times 10^{- 3}$ $s^{- 1}$ . The time required to reduce 2.0 g of the reactant to 0.2 g is [2020]

Q 6 :
If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by [2019]

Q 7 :
A first order reaction has a rate constant of $2.303 \times 10^{- 3}$ $s^{- 1}$ . The time required for 40 g of this reactant to reduce to 10 g will be

[Given that $\log_{10}$ 2 = 0.3010] [2019]

Q 8 :
The correct difference between first and second order reactions is that [2018]

Q 9 :
When initial concentration of the reactant is doubled, the half-life period of a zero order reaction [2018]

Q 10 :
A first order reaction has a specific reaction rate of $10^{- 2}$ $\sec^{- 1}$ . How much time will it take for 20 g of the reactant to reduce to 5 g? [2017]