jee main chemistry important questions | Chemical Kinetics jee main questions FREE

Q 1 :

Time required for completion of 99.9% of a First order reaction is ______ times of half life ( $t_{1 / 2}$ ) of the reaction. [2024]

(10)

For first order reactions, rate constant $k$ is given by:

$k = \frac{2.303}{t} \log \frac{a_{o}}{a_{t}}$

When reaction is 99.9% complete, 0.1% reactants are left i.e.

$a_{t} = \frac{0.1}{100} a_{o},$ $\frac{a_{o}}{a_{t}} = 1000$ , thus

$k = \frac{2.303}{t_{99.9 %}} \log 1000 = \frac{2.303}{t_{99.9 %}} \times 3$ (I)

When reaction is 50% complete, 50% reactants are left i.e.

$a_{t} = \frac{50}{100} a_{o},$ $\frac{a_{o}}{a_{t}} = 2$ , thus

$k = \frac{2.303}{t_{50 %}} \log 2 = \frac{2.303}{t_{50 %}} \times 0.301$ (II)

Equate I and II

$\frac{2.303}{t_{99.9 %}} \times 3 = \frac{2.303}{t_{50 %}} \times 0.301$

$t_{99.9 %} \approx 10 \times t_{50 %}$

Q 2 :

The half-life of radioisotope bromine -82 is 36 hours. The fraction which remains after one day is ______ $\times 10^{- 2}$ .

(Given antilog 0.2006 = 1.587) [2024]

(63)

For first order reaction,

$k = \frac{2.303}{t} \log_{10} \frac{a_{o}}{a_{t}}$

At half life time i.e 36 hrs:

$k = \frac{2.303}{36} \log_{10} \frac{a_{o}}{a_{o} / 2}$

$k = \frac{2.303}{36} \log_{10} 2$ (i)

At after one day i.e 24 hrs:

$k = \frac{2.303}{24} \log_{10} \frac{a_{o}}{a_{(t = 24)}}$ (ii)

Equating (i) and (ii)

$\frac{2.303}{24} \log_{10} \frac{a_{o}}{a_{(t = 24)}} = \frac{2.303}{36} \log_{10} 2$

$\log_{10} \frac{a_{o}}{a_{(t = 24)}} = \frac{24}{36} \log_{10} 2$

$\log_{10} \frac{a_{o}}{a_{(t = 24)}} = \frac{24}{36} \times 0.301 = 0.20066$

$\frac{a_{o}}{a_{(t = 24)}} = antilog (0.2006) = 1.587$

$\frac{a_{(t = 24)}}{a_{o}} = \frac{1}{1.587} = 0.63 = 63 \times 10^{- 2}$

Thus the fraction which remains after one day is $63 \times 10^{- 2}$ .

Q 3 :

The rate of First order reaction is 0.04 mol $L^{- 1}$ $s^{- 1}$ at 10 minutes and 0.03 mol $L^{- 1}$ $s^{- 1}$ at 20 minutes after initiation. Half life of the reaction is ______ minutes. (Given log2 = 0.3010, log3 = 0.4771) [2024]

(24)

For a first-order reaction $A \to$ products, the rate of the reaction is given by: $r = k {[A]}_{t} = k {[A]}_{0} e^{- kt}$

Substituting value of rate at t = 10 min and t = 20 min:

at, t = 10 min :

$0.04 mol L^{- 1} s^{- 1} = k {[A]}_{0} e^{- 10 k}$ (I)

at, t = 20 min :

$0.03 mol L^{- 1} s^{- 1} = k {[A]}_{0} e^{- 20 k}$ (II)

$I \div II$

$\frac{4}{3} = \frac{e^{- 10 k}}{e^{- 20 k}}$

$e^{10 k} = \frac{4}{3}$

$\ln$ $e^{10 k} = \ln \frac{4}{3}$

$10 k$ $\ln$ $e = 2.303 \log_{10} \frac{4}{3}$

$10 k = 2.303 \log_{10} \frac{4}{3}$ $(\ln$ $e = 1)$

$k = \frac{2.303}{10} \log_{10} \frac{4}{3}$ (III)

Also for a first-order reaction,

$t_{1 / 2} = \frac{2.303 \log_{10} 2}{k}$ or $k = \frac{2.303 \log_{10} 2}{t_{1 / 2}}$ (IV)

Equating III and IV:

$\frac{2.303 \log_{10} 2}{t_{1 / 2}} = \frac{2.303}{10} \log_{10} \frac{4}{3}$

$\frac{\log_{10} 2}{t_{1 / 2}} = \frac{1}{10} (\log_{10} 4 - \log_{10} 3)$

$\frac{\log_{10} 2}{t_{1 / 2}} = \frac{1}{10} (2 \log_{10} 2 - \log_{10} 3)$

$\frac{0.3010}{t_{1 / 2}} = \frac{1}{10} (2 \times 0.3010 - 0.4771)$

$t_{1 / 2} = \frac{0.3010}{0.01249} min = 24.1 min$

Q 4 :

$r = k [A]$ for a reaction. 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is ______ minutes. [2024]

(399)

From rate expression it is clear that the reaction is a first order reaction. As 50% of the reactants decompose in 120 min, $t_{1 / 2} = 120$ min. For first order reactions, the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{a_{o}}{a_{t}}$

When reaction is 90% complete, 10% reactants are left i.e. $a_{t} = \frac{10}{100} a_{o}, \frac{a_{o}}{a_{t}} = 10$

$\Rightarrow k = \frac{2.303}{t_{90 %}} \log 10$ (I)

When reaction is 50% complete, 50% reactants are left i.e. $a_{t} = \frac{50}{100} a_{o}, \frac{a_{o}}{a_{t}} = 2$

$\Rightarrow k = \frac{2.303}{t_{50 %}} \log 2$ (II)

Equating I and II

$\frac{2.303}{t_{90 %}} \log 10 = \frac{2.303}{t_{50 %}} \log 2$

$\frac{1}{t_{90 %}} \times 1 = \frac{1}{120} \times 0.301$

$t_{90 %} = 398.67 min \approx 399 min$

Q 5 :

The ratio of $\frac{{}^{14}C}{{}^{12}C}$ in a piece of wood is $\frac{1}{8}$ part that of atmosphere. If half life of ${}^{14}C$ is 5730 years, the age of wood sample is ______ years. [2024]

(17190)

${}^{14}C$ is radioactive and undergoes first order decay. In a piece of wood amount of ${}^{14}C$ decreases because of radioactive decay but amount of ${}^{12}C$ remains same. So if ratio of $\frac{{}^{14}C}{{}^{12}C}$ in a piece of wood is $\frac{1}{8}$ , this implies ratio of amount ${}^{14}C$ remained in the wood to the initial amount is $\frac{1}{8}$ .

In first order reaction equal percentage of reaction is completed in equal time interval. Thus concentration gets halved after every $t_{1 / 2}$ .

$1 \overset{t_{1 / 2}}{\to} \frac{1}{2} \overset{t_{1 / 2}}{\to} \frac{1}{4} \overset{t_{1 / 2}}{\to} \frac{1}{8}$

To get to ${(\frac{1}{8})}^{th}$ value of initial amount, it takes $3 t_{1 / 2}$ time i.e.

$3 \times 5730 years = 17190 years$

Q 6 :

Consider the following reaction, the rate expression of which is given below

$A + B \to C$

$rate = k {[A]}^{1 / 2} {[B]}^{1 / 2}$

The reaction is initiated by taking 1 M concentration of A and B each. If the rate constant (k) is $4.6 \times 10^{- 2}$ $s^{- 1}$ , then the time taken for A to become 0.1 M is _____ sec. (nearest integer) [2024]

(50)

$A$ $+$ $B$ $\to$ $C$

$t = 0$ $a_{o}$ $a_{o}$ 0

$t = t$ $a_{o} - x$ $a_{o} - x$ $x$

$r = k {[A]}^{1 / 2} {[B]}^{1 / 2} = k {(a_{o} - x)}^{1 / 2} {(a_{o} - x)}^{1 / 2}$

$\frac{d x}{d t} = k (a_{o} - x)$

$\frac{d x}{(a_{o} - x)} = k d t$

Integrate both sides

$\int_{0}^{x} \frac{dx}{(a_{o} - x)} = \int_{0}^{t} k dt$

${[\frac{\ln (a_{o} - x)}{- 1}]}_{0}^{x} = {[k t]}_{0}^{t}$

$[\frac{\ln (a_{o} - x)}{- 1} - \frac{\ln (a_{o})}{- 1}] = k t$

$\ln (\frac{a_{o}}{a_{o} - x}) = k t$

$2.303 \log_{10} (\frac{a_{o}}{a_{o} - x}) = k t$

$In the question a_{0} = 1 M and (a_{0} - x) = 0.1 M$

$2.303 \log_{10} (\frac{1}{0.1}) = 4.6 \times 10^{- 2} t$

$2.303 \log_{10} 10 = 4.6 \times 10^{- 2} t$

$2.303 \times 1 = 4.6 \times 10^{- 2} t$

$t = 50 s$

Q 7 :

Time required for 99.9% completion of a first order reaction is ________ times the time required for completion of 90% reaction. (nearest integer) [2024]

(3)

Rate constant of a first order reaction is given as: $k = \frac{2.303}{t} \log \frac{a_{o}}{a_{t}}$

When reaction is 90% complete, then only 10% reactants remain i.e. at $t_{90 %}$ , $a_{t} = 0.1$ $a_{0}$

$k = \frac{2.303}{t_{90 %}} \log \frac{a_{0}}{0.1 a_{0}} = \frac{2.303}{t_{90 %}} \log 10 = \frac{2.303}{t_{90 %}} \times 1 = \frac{2.303}{t_{90 %}}$ ...(i)

When reaction is 99.9% complete, then only 0.1% reactants remain i.e. $t_{99.9 %},$ $a_{t} = 0.001$ $a_{0}$

$k = \frac{2.303}{t_{99.9 %}} \log \frac{a_{0}}{0.001 a_{0}} = \frac{2.303}{t_{99.9 %}} \log 10^{3} = \frac{2.303}{t_{99.9 %}} \times 3 \log 10 = \frac{2.303}{t_{99.9 %}} \times 3$ ...(ii)

Equate I and II

$\frac{2.303}{t_{90 %}} = \frac{2.303}{t_{99.9 %}} \times 3$

$t_{99.9 %} = t_{90 %} \times 3$

Q 8 :

Consider the two different first order reactions given below

$A + B \to C (Reaction 1)$

$P \to Q (Reaction 2)$

The ratio of the half life of Reaction 1 : Reaction 2 is 5 : 2.

If $t_{1}$ and $t_{2}$ represent the time taken to complete $2 / 3^{r d}$ and $4 / 5^{t h}$ of reaction 1 and reaction 2, respectively, then the value of the ratio $t_{1} : t_{2}$ is _______ $\times 10^{- 1}$ (nearest integer).

[Given: $\log_{10}$ (3) = 0.477 and $\log_{10}$ (5) = 0.699] [2024]

(17)

For a first order reaction, rate constant $k$ is related to initial conc. of reactants ( $r_{o}$ ) and conc. of reactants at time $t$ ( $r_{t}$ ) as:

$k = \frac{2.303}{t} \log_{10} \frac{r_{o}}{r_{t}}$

Reaction 1: When reaction is $2 / 3^{r d}$ complete, then $1 / 3^{r d}$ of reactants remain, $r_{t} = \frac{1}{3} r_{o}$

$k_{1} = \frac{2.303}{t_{1}} \log_{10} \frac{r_{o}}{(\frac{1}{3}) r_{o}}$

$t_{1} = \frac{2.303}{k_{1}} \log_{10} 3$ ...(i)

Reaction 2: When reaction is $4 / 5^{t h}$ complete, then $1 / 5^{t h}$ of reactants remain, $r_{t} = \frac{1}{5} r_{o}$

$k_{2} = \frac{2.303}{t_{2}} \log_{10} \frac{r_{o}}{(\frac{1}{5}) r_{o}}$

$t_{2} = \frac{2.303}{k_{2}} \log_{10} 5$ ...(ii)

$(i) \div (ii)$

$\frac{t_{1}}{t_{2}} = \frac{k_{2}}{k_{1}} \frac{\log_{10} 3}{\log_{10} 5}$

For first order reaction, $t_{1 / 2} = \frac{0.693}{k}$

So, $\frac{t_{1}}{t_{2}} = \frac{0.693 / {(t_{1 / 2})}_{2}}{0.693 / {(t_{1 / 2})}_{1}} \frac{\log_{10} 3}{\log_{10} 5}$

$\frac{t_{1}}{t_{2}} = \frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} \frac{\log_{10} 3}{\log_{10} 5} = \frac{5 \times 0.477}{2 \times 0.699} = 1.7 = 17 \times 10^{- 1}$

Q 9 :

Consider the following reaction: [2024]

$A + B \to C$

The time taken for A to become 1/4th of its initial concentration is twice the time taken to become 1/2 of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is _______.

(1)

For first order reaction, Equal percentage of reaction is completed in equal time interval. Since for the given reaction, the time for A to become 1/4th of its initial concentration is twice the time taken to become 1/2 of the same, the reaction is first order w.r.t.A.

$a_{o} \overset{t_{1 / 2}}{\to} \frac{a_{o}}{2} \overset{t_{1 / 2}}{\to} \frac{a_{o}}{4}$

For a zero order reaction, concentration of reactants decreases linearly with time.

So reaction is zero order w.r.t.B.

Rate law expression is: $r = k {[A]}^{1} {[B]}^{0}$

Order = 1 + 0 = 1.

Q 10 :

Given below are two statements: [2024]

Statement I: The rate law for the reaction $A + B \to C$ is $rate (r) = k {[A]}^{2} [B] .$ When the concentration of both A and B is doubled, the reaction rate is increased "x" times.

Statement II:

The figure is showing "the variation in concentration against time plot" for a "y" order reaction.

The value of x + y is _______.

(8)

Statement I: $r_{1} = k {[A_{1}]}^{2} [B_{1}]$

$r_{2} = k {[A_{2}]}^{2} [B_{2}] = k {[2 A_{1}]}^{2} [2 B_{1}]$

$\frac{r_{2}}{r_{1}} = \frac{k {[2 A_{1}]}^{2} [2 B_{1}]}{k {[A_{1}]}^{2} [B_{1}]} = 8 = x$

Statement II: For zero order reaction: $[R] = {[R]}_{0} - k t, y = 0$

$x + y = 8$

Q 11 :

${[A]}_{0} / {mol L}^{- 1}$ $t_{1 / 2} / min$

0.100 200

0.025 100

For a given reaction $R \to P$ , $t_{1 / 2}$ is related to ${[A]}_{0}$ as given in the table.
Given: log 2 = 0.30

Which of the following is true?

(A) The order of the reaction is 1/2.

(B) If ${[A]}_{0}$ is 1 M, then $t_{1 / 2}$ is $200 \sqrt{10}$ min.

(C) The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.

(D) $t_{1 / 2}$ is 800 min for ${[A]}_{0}$ = 1.6 M.

Choose the correct answer from the options given below: [2025]

(C) and (D) Only
(A) and (B) Only
(A), (B) and (D) Only
(A) and (C) Only

1

2

3

4

(3)

(A) For $n^{th}$ order reaction, $t_{1 / 2}$ is related to initial concentration $(a_{0})$ as:

$t_{1 / 2} \propto a_{0}^{1 - n}$

$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{{(a_{0})}_{1}}{{(a_{0})}_{2}})}^{1 - n}$

$\frac{200}{100} = {(\frac{0.10}{0.025})}^{1 - n}$

$2 = 4^{1 - n}$

$n = \frac{1}{2}$

(B) $\frac{{(t_{1 / 2})}_{for a_{0} = 1 M}}{{(t_{1 / 2})}_{1}} = {(\frac{1}{{(a_{0})}_{1}})}^{1 - n}$

$\frac{{(t_{1 / 2})}_{for a_{0} = 1 M}}{200} = {(\frac{1}{0.1})}^{1 - \frac{1}{2}}$

${(t_{1 / 2})}_{for a_{0} = 1 M} = 200 \sqrt{10} min$

(D) $\frac{{(t_{1 / 2})}_{for a_{0} = 1.6 M}}{{(t_{1 / 2})}_{1}} = {(\frac{1.6}{{(a_{0})}_{1}})}^{1 - n}$

$\frac{{(t_{1 / 2})}_{for a_{0} = 1.6 M}}{200} = {(\frac{1.6}{0.1})}^{1 - \frac{1}{2}}$

${(t_{1 / 2})}_{for a_{0} = 1.6 M} = 800 min$

Q 12 :

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Where N – Number of Bacteria at any time, $N_{0}$ – Initial number of Bacteria. [2025]

1

2

3

4

(1)

Initial number of bacteria = $N_{0}$

Number of bacteria at time t = N

Rate of increase of bacteria = $\frac{d N}{d t}$

As bacteria grow by first order:

$\frac{d N}{d t} = k N$

$\frac{d N}{N} = k dt$

Integrate both sides:

$\int_{N_{0}}^{N} \frac{d N}{N} = \int_{0}^{t} k dt$

${[\ln N]}_{N_{0}}^{N} = k {[t]}_{0}^{t}$

$N = N_{0} e^{k t}$

This is exponential growth.

Q 13 :

Drug X becomes ineffective after 50% decomposition. The original concentration of drug in a bottle was 16 mg/mL which becomes 4 mg/mL in 12 months. The expiry time of the drug in months is _______.

Assume that the decomposition of the drug follows first order kinetics. [2025]

6
12
2
3

1

2

3

4

(1)

After every $t_{1 / 2}$ , concentration of reactants in a first order reaction becomes half.

$16 \frac{mg}{mL} \overset{6 months}{\to} 8 \frac{mg}{mL} \overset{6 months}{\to} 4 \frac{mg}{mL}$

So in 6 months, the drug becomes ineffective.

Q 14 :

In a reaction A + B → C, initial concentrations of A and B are related as ${[A]}_{0} = 8 {[B]}_{0}$ . The half-lives of A and B are 10 min and 40 min, respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same? [2025]

60 min
80 min
20 min
40 min

1

2

3

4

(4)

${[A]}_{0}$

$8 M \overset{10 min}{\to} 4 M \overset{10 min}{\to} 2 M \overset{10 min}{\to} 1 M \overset{10 min}{\to} 0.5 M$

${[B]}_{0}$

$1 M \overset{40 min}{\to} 0.5 M$

Q 15 :

Half life of zero order reaction A → product is 1 hour, when initial concentration of reaction is 2.0 mol $L^{- 1}$ . The time required to decrease concentration of A from 0.50 to 0.25 mol $L^{- 1}$ is: [2025]

0.5 hour
4 hour
15 min
60 min

1

2

3

4

(3)

Half-life $(t_{1 / 2})$ of a zero order reaction is related to initial concentration $(a_{0})$ by:

$t_{1 / 2} = \frac{a_{0}}{2 k}$

Thus, rate constant $(k)$ is:
$k = \frac{a_{0}}{2 t_{1 / 2}} = \frac{2 mol/L}{2 \times 1 hour} = \frac{2 mol/L}{2 \times 60 min} = \frac{1}{60} {mol L}^{- 1} {min}^{- 1}$

Concentration at any time $t$ $(a_{t})$ is given by:

$a_{t} = a_{0} - k t,$ $0.25 = 0.5 - (\frac{1}{60}) t, t = 15 min.$

Q 16 :

Reaction A(g) → 2B(g) + C(g) is a first order reaction. It was started with pure A.

t / min Pressure of system at time t / mm Hg

10 160

$\infty$ 240

Which of the following options is incorrect? [2025]

Initial pressure of A is 80 mm Hg.
The reaction never goes to completion.
Rate constant of the reaction is 1.693 $\min^{- 1}$ .
Partial pressure of A after 10 minutes is 40 mm Hg.

1

2

3

4

(3)

$\begin{matrix} A(g) & ⟶ & 2 B(g) & + & C(g) \\ t =0 (pressure) & p ° & 0 & 0 \\ t =10 min & p ° - p & 2 p & p & Total= p ° + 2 p = 160 - (I) \\ t = \infty & 0 & 2 p ° & p ° & Total= 3 p ° = 240 - (II) \end{matrix}$

From equations (I) and (II):

$p ° = 80 mmHg$

$p = 40 mmHg$

Thus, initial pressure of A $(p °) = 80 mmHg$ .

Pressure of A after 10 min

$p ° - p = (80 - 40) mm Hg = 40 mmHg$

First-order reactions never go to completion.

Rate constant:

$k = \frac{2.303}{t} \log \frac{p °}{p ° - p}$

$= \frac{2.303}{10} \log \frac{80}{80 - 40}$

$= \frac{2.303}{10} \log 2$

$= \frac{2.303}{10} \times 0.3 = 0.06909 {min}^{- 1}$

Q 17 :

A person’s wound was exposed to some bacteria and then bacterial growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay (r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the ‘before’ and ‘after’ situation of the application of the medicine?

Given: N = Number of bacteria, t = time, bacterial growth follows first order kinetics. [2025]

1

2

3

4

(2)

Before applying medicine:

Number of bacteria at any instant t = N

Rate of increase of bacteria = $\frac{d N}{d t}$

As bacterial growth is first order reaction:

$\frac{d N}{d t} = k N$

$\frac{d N}{N} = k dt$

Integrating both sides:

$\int_{N_{0}}^{N} \frac{d N}{N} = \int_{0}^{t} k dt$

${[\ln N]}_{N_{0}}^{N} = k {[t]}_{0}^{t}$

$\ln \frac{N}{N_{0}} = k t$

$\frac{N}{N_{0}} = e^{k t}$

After applying medicine:

Rate of decrease of bacteria $(r) = - \frac{d N}{d t}$
As per given data:

$r = - \frac{d N}{d t} = k {[N]}^{2}$

Q 18 :

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_{1}$ and $t_{2}$ (s), respectively. The ratio $t_{1} / t_{2}$ will: [2025]

$\frac{4}{3}$
$\frac{3}{2}$
$\frac{3}{4}$
$\frac{2}{3}$

1

2

3

4

(4)

$\frac{t_{1}}{t_{2}} = \frac{2 t_{1 / 2}}{3 t_{1 / 2}} = \frac{2}{3}$

Q 19 :

For the reaction A → products.

The concentration of A at 10 minutes is ______ $\times 10^{- 3}$ mol $L^{- 1}$ (nearest integer).

The reaction was started with 2.5 mol $L^{- 1}$ of A. [2025]

(2435)

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

$For zero order reaction: t_{1 / 2} = \frac{a_{0}}{2 k}$

Thus, slope of $t_{1 / 2}$ vs $a_{0}$ is $\frac{1}{2 k}$ .

$As per given in graph, \frac{1}{2 k} = 76.92$

$k = \frac{1}{2 \times 76.92}$

Also, for zero-order reaction, $a_{t} = a_{0} - k t$

$a_{t = 10} = 2.5 - \frac{1}{2 \times 76.92} \times 10 = 2435 \times 10^{- 3} mol/L$

Q 20 :

For the thermal decomposition of $N_{2} O_{5} (g)$ at constant volume, the following table can be formed, for the reaction mentioned below:

$2 N_{2} O_{5} (g) \to 2 N_{2} O_{4} (g) + O_{2} (g)$

Sr. No. Times Total Pressure (atm)

1 0 0.6

2 100 $' x'$

$x$ = _____ $\times 10^{- 3}$ atm (nearest integer)

Given: Rate constant for the reaction is $4.606 \times 10^{- 2} s^{- 1}$ . [2025]

(897)

$\begin{matrix} 2 N_{2} O_{5} (g) & ⟶ & 2 N_{2} O_{4} (g) & + & O_{2} (g) \\ t = 0 & p ° & 0 & 0 \\ t = 100 s & p ° - 2 p & 2 p & p \end{matrix}$

Total pressure at $t = 0$ is $p ° + 0 + 0 = p ° = 0.6$

Total pressure at $t = 100 s$ is $(p ° - 2 p) + 2 p + p = p ° + p = x$

From above two equations, $p = x - 0.6$

For first order reaction, rate constant is given as:

$k = \frac{2.303}{t} \log \frac{P_{N_{2} O_{5}} (t = 0)}{P_{N_{2} O_{5}} (t = t)}$

$4.606 \times 10^{- 2} = \frac{2.303}{100} \log \frac{0.6}{0.6 - 2 (x - 0.6)}$

$4.606 \times 10^{- 2} = \frac{2.303}{100} \log \frac{0.6}{1.8 - 2 x}$

$\log \frac{0.6}{1.8 - 2 x} = 2$

$\frac{0.6}{1.8 - 2 x} = 10^{2} = 100$

$1.8 - 2 x = 0.006$

$x = 0.897 atm = 897 \times 10^{- 3} atm$

Q 21 :

$A \to B$

The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is 191.48 kJ ${mol}^{- 1}$ and the frequency factor is $10^{20}$ , the time required for 50% molecules of A to become B is ______ picoseconds (nearest integer).

[R = 8.314 J $K^{- 1} {mol}^{- 1}$ ] [2025]

(69)

Activation energy $E_{a} = 191.48 {kJ mol}^{- 1} = 191480 {J mol}^{- 1}$

Frequency factor $(A) = 10^{20}$

Temperature $(T) = 1000 K$

Gas constant R = 8.314 $J/mol K$

By Arrhenius equation,

$= A e^{\frac{- E_{a}}{R T}} = 10^{20} e^{\frac{- 191480}{8.314 \times 1000}}$

$= 10^{20} e^{- 23.03} = 10^{20} e^{- \ln 10 \times 10}$

$= 10^{20} e^{- \ln 10^{10}}$

$= \frac{10^{20}}{e^{\ln 10^{10}}} = \frac{10^{20}}{10^{10}} = 10^{10} s^{- 1}$

$t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{10^{10}} s = 69.3 \times 10^{- 12} s$

$= 69.3 picosecond \approx 69 picosecond$

Q 22 :

A student has studied the decomposition of a gas ${AB}_{3}$ at 25°C. He obtained the following data.

$p$ (mm Hg) 50 100 200 400

relative $t_{1 / 2}$ (s) 4 2 1 0.5

The order of the reaction is: [2023]

1
0.5
0 (zero)
2

1

2

3

4

(4)

$T_{\frac{1}{2}} \propto {(C_{0})}^{1 - n}$

$\frac{{(T_{\frac{1}{2}})}_{1st}}{{(T_{\frac{1}{2}})}_{2nd}} = {(\frac{P_{1}}{P_{2}})}^{1 - n}$

$\Rightarrow \frac{4}{2} = {(\frac{50}{100})}^{1 - n} \Rightarrow 2 = {(\frac{1}{2})}^{1 - n}$

$2 = {(2)}^{n - 1}$

$n - 1 = 1 \Rightarrow n = 2$

$Order = 2$

Q 23 :

The graph which represents the following reaction is:

${(C_{6} H_{5})}_{3} C - Cl \to_{Pyridine}^{{OH}^{-}} {(C_{6} H_{5})}_{3} C - OH$ [2023]

1

2

3

4

(4)

Q 24 :

For the first order reaction $A \to B$ , the half-life is 30 min. The time taken for 75% completion of the reaction is ______ min. (Nearest integer)

Given: log 2 = 0.3010

log 3 = 0.4771

log 5 = 0.6989 [2023]

(60)

$t_{75 %} = 2 t_{1 / 2} [For 1st order reaction]$

$t_{75 %} = 2 \times 30 = 60 min .$

Q 25 :

A first order reaction has the rate constant, $k = 4.6 \times 10^{- 3} s^{- 1}$ . The number of correct statement/s from the following is/are ______. Given: log 3 = 0.48 [2023]

A. Reaction completes in 1000 s.

B. The reaction has a half-life of 500 s.

C. The time required for 10% completion is 25 times the time required for 90% completion.

D. The degree of dissociation is equal to $(1 - e^{- kt})$ .

E. The rate and the rate constant have the same unit.

(2)

(C & D)

Statement 4 is correct

$C_{t} = C_{0} e^{- k t}$

$C_{0} \to C_{0} - C_{t}$

$1 \to \frac{C_{0} - C_{t}}{C_{0}}$

Q 26 :

If compound A reacts with B following first order kinetics with rate constant $2.011 \times 10^{- 3} s^{- 1}$ . The time taken by A (in seconds) to reduce from 7 g to 2 g will be ______. (Nearest integer) [log 5 = 0.698, log 7 = 0.845, log 2 = 0.301] [2023]

(623)

$K = \frac{1}{t} \ln (\frac{a}{a - x})$

$K = \frac{1}{t} \ln (\frac{W_{0}}{W_{t}})$

$t = \frac{1}{K} \ln (\frac{7}{2})$

$t = \frac{2.303}{K} [\log 7 - \log 2]$

$= \frac{2.303}{K} [0.845 - 0.301]$

$= \frac{2.303}{2.011 \times 10^{- 3}} \times 0.544 = 622.989 sec$

Q 27 :

An organic compound undergoes first order decomposition. If the time taken for the 60% decomposition is 540 s, then the time required for 90% decomposition will be is ______ s. (Nearest integer).

(Given: ln 10 = 2.3; log 2 = 0.3) [2023]

(1350)

First order reaction

$t = \frac{1}{k} \ln \frac{C_{0}}{C_{t}}$

$540 = \frac{1}{k} \ln \frac{100}{40}$ $. . . (1)$

$t = \frac{1}{k} \ln \frac{100}{10}$ $. . . (2)$

$From (1) and (2)$

$\frac{540}{t} = \frac{\ln \frac{100}{40}}{\ln \frac{100}{10}} = \frac{\log 100 - \log 40}{\log 100 - \log 10}$

$\frac{540}{t} = \frac{2 - 1.6}{2 - 1} = 0.4$

$t = \frac{540}{0.4} = 1350 s$

Q 28 :

The rate constant for a first order reaction is 20 $\min^{- 1}$ . The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is ______ $\times 10^{- 2}$ min. (Nearest integer)

(Given: ln 10 = 2.303, log 2 = 0.3010) [2023]

(17)

$t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{20} = 0.03465 min$

${(\frac{1}{2})}^{n} = (\frac{1}{32}) = {(\frac{1}{2})}^{5}$

$No. of half-life = 5$

$Time required = 0.03465 \times 5 = 0.17325 min$

$= 17.325 \times 10^{- 2} min$

Q 29 :

A and B are two substances undergoing radioactive decay in a container. The half life of A is 15 min and that of B is 5 min. If the initial concentration of B is 4 times that of A and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? ______ min. [2023]

(15)

$t_{1 / 2 (A)} = 15 min, t_{1 / 2 (B)} = 5 min$

${[B]}_{initial} = 4 {[A]}_{initial}$

${[B]}_{final} = {[B]}_{initial} {(\frac{1}{2})}^{t / 15}$

${[A]}_{final} = {[A]}_{_{initial}} {(\frac{1}{2})}^{t / 5}$

${[B]}_{final} = {[A]}_{final}$

$t = 15$

Q 30 :

$A \to B$

The above reaction is of zero order. Half-life of this reaction is 50 min. The time taken for the concentration of A to reduce to one-fourth of its initial value is _______ min. (Nearest integer) [2023]

(75)

$Zero order$

$t_{1 / 2} = 50$

$t_{3 / 4} = 1.5$ $t_{1 / 2} = 75$

Topic Question Set

Q 1 :

Time required for completion of 99.9% of a First order reaction is ______ times of half life ( $t_{1 / 2}$ ) of the reaction. [2024]

Q 2 :

The half-life of radioisotope bromine -82 is 36 hours. The fraction which remains after one day is ______ $\times 10^{- 2}$ .

(Given antilog 0.2006 = 1.587) [2024]

Q 3 :

The rate of First order reaction is 0.04 mol $L^{- 1}$ $s^{- 1}$ at 10 minutes and 0.03 mol $L^{- 1}$ $s^{- 1}$ at 20 minutes after initiation. Half life of the reaction is ______ minutes. (Given log2 = 0.3010, log3 = 0.4771) [2024]

Q 4 :

$r = k [A]$ for a reaction. 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is ______ minutes. [2024]

Q 5 :

The ratio of $\frac{{}^{14}C}{{}^{12}C}$ in a piece of wood is $\frac{1}{8}$ part that of atmosphere. If half life of ${}^{14}C$ is 5730 years, the age of wood sample is ______ years. [2024]

Q 7 :

Time required for 99.9% completion of a first order reaction is ________ times the time required for completion of 90% reaction. (nearest integer) [2024]

Q 12 :

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Where N – Number of Bacteria at any time, $N_{0}$ – Initial number of Bacteria. [2025]

Q 13 :

Drug X becomes ineffective after 50% decomposition. The original concentration of drug in a bottle was 16 mg/mL which becomes 4 mg/mL in 12 months. The expiry time of the drug in months is _______.

Assume that the decomposition of the drug follows first order kinetics. [2025]

Q 15 :

Half life of zero order reaction A → product is 1 hour, when initial concentration of reaction is 2.0 mol $L^{- 1}$ . The time required to decrease concentration of A from 0.50 to 0.25 mol $L^{- 1}$ is: [2025]

Q 16 :

Reaction A(g) → 2B(g) + C(g) is a first order reaction. It was started with pure A.

t / min Pressure of system at time t / mm Hg

10 160

$\infty$ 240

Which of the following options is incorrect? [2025]

Q 18 :

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_{1}$ and $t_{2}$ (s), respectively. The ratio $t_{1} / t_{2}$ will: [2025]

Q 19 :

For the reaction A → products.

The concentration of A at 10 minutes is ______ $\times 10^{- 3}$ mol $L^{- 1}$ (nearest integer).

The reaction was started with 2.5 mol $L^{- 1}$ of A. [2025]

Q 22 :

A student has studied the decomposition of a gas ${AB}_{3}$ at 25°C. He obtained the following data.

$p$ (mm Hg) 50 100 200 400

relative $t_{1 / 2}$ (s) 4 2 1 0.5

The order of the reaction is: [2023]

Q 23 :

The graph which represents the following reaction is:

${(C_{6} H_{5})}_{3} C - Cl \to_{Pyridine}^{{OH}^{-}} {(C_{6} H_{5})}_{3} C - OH$ [2023]

Q 24 :

For the first order reaction $A \to B$ , the half-life is 30 min. The time taken for 75% completion of the reaction is ______ min. (Nearest integer)

Given: log 2 = 0.3010

log 3 = 0.4771

log 5 = 0.6989 [2023]

Q 26 :

If compound A reacts with B following first order kinetics with rate constant $2.011 \times 10^{- 3} s^{- 1}$ . The time taken by A (in seconds) to reduce from 7 g to 2 g will be ______. (Nearest integer) [log 5 = 0.698, log 7 = 0.845, log 2 = 0.301] [2023]

Q 27 :

An organic compound undergoes first order decomposition. If the time taken for the 60% decomposition is 540 s, then the time required for 90% decomposition will be is ______ s. (Nearest integer).

(Given: ln 10 = 2.3; log 2 = 0.3) [2023]

Q 28 :

The rate constant for a first order reaction is 20 $\min^{- 1}$ . The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is ______ $\times 10^{- 2}$ min. (Nearest integer)

(Given: ln 10 = 2.303, log 2 = 0.3010) [2023]

Q 30 :

$A \to B$

The above reaction is of zero order. Half-life of this reaction is 50 min. The time taken for the concentration of A to reduce to one-fourth of its initial value is _______ min. (Nearest integer) [2023]

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${[A]}_{0} / {mol L}^{- 1}$	$t_{1 / 2} / min$
0.100	200
0.025	100

t / min	Pressure of system at time t / mm Hg
10	160
$\infty$	240

Sr. No.	Times	Total Pressure (atm)
1	0	0.6
2	100	$' x'$

Topic Question Set

Q 1 : Time required for completion of 99.9% of a First order reaction is ______ times of half life (t1/2) of the reaction. [2024]

Q 2 : The half-life of radioisotope bromine -82 is 36 hours. The fraction which remains after one day is ______ ×10-2. (Given antilog 0.2006 = 1.587) [2024]

Q 3 : The rate of First order reaction is 0.04 mol L-1 s-1 at 10 minutes and 0.03 mol L-1 s-1 at 20 minutes after initiation. Half life of the reaction is ______ minutes. (Given log2 = 0.3010, log3 = 0.4771) [2024]

Q 4 : r=k[A] for a reaction. 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is ______ minutes. [2024]

Q 5 : The ratio of C14C12 in a piece of wood is 18 part that of atmosphere. If half life of C14 is 5730 years, the age of wood sample is ______ years. [2024]

Q 7 : Time required for 99.9% completion of a first order reaction is ________ times the time required for completion of 90% reaction. (nearest integer) [2024]

Q 12 : For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth? Where N – Number of Bacteria at any time, N0 – Initial number of Bacteria. [2025]

Q 13 : Drug X becomes ineffective after 50% decomposition. The original concentration of drug in a bottle was 16 mg/mL which becomes 4 mg/mL in 12 months. The expiry time of the drug in months is _______. Assume that the decomposition of the drug follows first order kinetics. [2025]

Q 15 : Half life of zero order reaction A → product is 1 hour, when initial concentration of reaction is 2.0 mol L-1. The time required to decrease concentration of A from 0.50 to 0.25 mol L-1 is: [2025]

Q 16 : Reaction A(g) → 2B(g) + C(g) is a first order reaction. It was started with pure A. t / min Pressure of system at time t / mm Hg 10 160 ∞ 240 Which of the following options is incorrect? [2025]

Q 18 : In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are t1 and t2(s), respectively. The ratio t1/t2 will: [2025]

Q 19 : For the reaction A → products. The concentration of A at 10 minutes is ______ ×10-3 mol L-1 (nearest integer). The reaction was started with 2.5 mol L-1 of A. [2025]

Q 22 : A student has studied the decomposition of a gas AB3 at 25°C. He obtained the following data. p(mm Hg) 50 100 200 400 relative t1/2(s) 4 2 1 0.5 The order of the reaction is: [2023]

Q 23 : The graph which represents the following reaction is: (C6H5)3C-Cl→PyridineOH-(C6H5)3C-OH [2023]

Q 24 : For the first order reaction A→B, the half-life is 30 min. The time taken for 75% completion of the reaction is ______ min. (Nearest integer) Given: log 2 = 0.3010 log 3 = 0.4771 log 5 = 0.6989 [2023]

Q 26 : If compound A reacts with B following first order kinetics with rate constant 2.011×10-3s-1. The time taken by A (in seconds) to reduce from 7 g to 2 g will be ______. (Nearest integer) [log 5 = 0.698, log 7 = 0.845, log 2 = 0.301] [2023]

Q 27 : An organic compound undergoes first order decomposition. If the time taken for the 60% decomposition is 540 s, then the time required for 90% decomposition will be is ______ s. (Nearest integer). (Given: ln 10 = 2.3; log 2 = 0.3) [2023]

Q 28 : The rate constant for a first order reaction is 20 min-1. The time required for the initial concentration of the reactant to reduce to its 132 level is ______ ×10-2 min. (Nearest integer) (Given: ln 10 = 2.303, log 2 = 0.3010) [2023]

Q 30 : A→B The above reaction is of zero order. Half-life of this reaction is 50 min. The time taken for the concentration of A to reduce to one-fourth of its initial value is _______ min. (Nearest integer) [2023]

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Q 1 :

Time required for completion of 99.9% of a First order reaction is ______ times of half life ( $t_{1 / 2}$ ) of the reaction. [2024]

Q 2 :

The half-life of radioisotope bromine -82 is 36 hours. The fraction which remains after one day is ______ $\times 10^{- 2}$ .

(Given antilog 0.2006 = 1.587) [2024]

Q 3 :

The rate of First order reaction is 0.04 mol $L^{- 1}$ $s^{- 1}$ at 10 minutes and 0.03 mol $L^{- 1}$ $s^{- 1}$ at 20 minutes after initiation. Half life of the reaction is ______ minutes. (Given log2 = 0.3010, log3 = 0.4771) [2024]

Q 4 :

$r = k [A]$ for a reaction. 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is ______ minutes. [2024]

Q 5 :

The ratio of $\frac{{}^{14}C}{{}^{12}C}$ in a piece of wood is $\frac{1}{8}$ part that of atmosphere. If half life of ${}^{14}C$ is 5730 years, the age of wood sample is ______ years. [2024]

Q 7 :

Time required for 99.9% completion of a first order reaction is ________ times the time required for completion of 90% reaction. (nearest integer) [2024]

Q 12 :

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Where N – Number of Bacteria at any time, $N_{0}$ – Initial number of Bacteria. [2025]

Q 13 :

Drug X becomes ineffective after 50% decomposition. The original concentration of drug in a bottle was 16 mg/mL which becomes 4 mg/mL in 12 months. The expiry time of the drug in months is _______.

Assume that the decomposition of the drug follows first order kinetics. [2025]

Q 15 :

Half life of zero order reaction A → product is 1 hour, when initial concentration of reaction is 2.0 mol $L^{- 1}$ . The time required to decrease concentration of A from 0.50 to 0.25 mol $L^{- 1}$ is: [2025]

Q 16 :

Reaction A(g) → 2B(g) + C(g) is a first order reaction. It was started with pure A.

t / min Pressure of system at time t / mm Hg

10 160

$\infty$ 240

Which of the following options is incorrect? [2025]

Q 18 :

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_{1}$ and $t_{2}$ (s), respectively. The ratio $t_{1} / t_{2}$ will: [2025]

Q 19 :

For the reaction A → products.

The concentration of A at 10 minutes is ______ $\times 10^{- 3}$ mol $L^{- 1}$ (nearest integer).

The reaction was started with 2.5 mol $L^{- 1}$ of A. [2025]

Q 22 :

A student has studied the decomposition of a gas ${AB}_{3}$ at 25°C. He obtained the following data.

$p$ (mm Hg) 50 100 200 400

relative $t_{1 / 2}$ (s) 4 2 1 0.5

The order of the reaction is: [2023]

Q 23 :

The graph which represents the following reaction is:

${(C_{6} H_{5})}_{3} C - Cl \to_{Pyridine}^{{OH}^{-}} {(C_{6} H_{5})}_{3} C - OH$ [2023]

Q 24 :

For the first order reaction $A \to B$ , the half-life is 30 min. The time taken for 75% completion of the reaction is ______ min. (Nearest integer)

Given: log 2 = 0.3010

log 3 = 0.4771

log 5 = 0.6989 [2023]

Q 26 :

If compound A reacts with B following first order kinetics with rate constant $2.011 \times 10^{- 3} s^{- 1}$ . The time taken by A (in seconds) to reduce from 7 g to 2 g will be ______. (Nearest integer) [log 5 = 0.698, log 7 = 0.845, log 2 = 0.301] [2023]

Q 27 :

An organic compound undergoes first order decomposition. If the time taken for the 60% decomposition is 540 s, then the time required for 90% decomposition will be is ______ s. (Nearest integer).

(Given: ln 10 = 2.3; log 2 = 0.3) [2023]

Q 28 :

The rate constant for a first order reaction is 20 $\min^{- 1}$ . The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is ______ $\times 10^{- 2}$ min. (Nearest integer)

(Given: ln 10 = 2.303, log 2 = 0.3010) [2023]

Q 30 :

$A \to B$

The above reaction is of zero order. Half-life of this reaction is 50 min. The time taken for the concentration of A to reduce to one-fourth of its initial value is _______ min. (Nearest integer) [2023]