Chemical Kinetics — Page 8 | NEET Chemistry

Q71

Which of the following statement is not true for radioactive decay?

A Half life is

\ln 2

times of

\dfrac{1}{\text{ rate constant }}

.

B Decay constant increases with increase in temperature.

C Decay constant does not depend upon temperature.

D Amount of radioactive substance remained after three half lives is

\dfrac{1}{8}

th of original amount.

Correct Answer

Option B

Solution

Answer: Option B is not true.

Explanation: Option A states that the half‐life $t_{1/2}$ is $\ln 2$ times $\dfrac{1}{k}$ , where $k$ is the decay (rate) constant.

$t_{1/2} \;=\; \dfrac{\ln 2}{k} \;\;=\;\ln 2 \times \dfrac{1}{k}.$ This is correct.

Option B states that the decay constant $k$ increases with an increase in temperature.

However, for radioactive (nuclear) decay processes, the decay constant is essentially independent of external factors such as temperature and pressure.

Hence, saying that it increases with temperature is not true.

Option C states that the decay constant does not depend upon temperature, which is true for radioactive decay.

Option D states that the amount of radioactive substance left after three half‐lives is $\tfrac{1}{8}$ of the original.

Since one half‐life leaves $\tfrac{1}{2}$ of the sample, three half‐lives leave $\left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$ .

This is correct.

Therefore, the statement that is not true is Option B.

Q72

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

A remain unchanged

B triple

C increase by a factor of 4

D double

Correct Answer

Option C

Solution

Since the reaction is

2

nd order w.r.t CO. Thus, rate law is given as.

r = k{\left[ {CO} \right]^2}

Let initial concentration of

CO

is a i.e.

\left[ {CO} \right] = a

$\therefore$

\,\,\,\,{r_1} = k{\left( a \right)^2} = k{a^2}

when concentration becomes doubled, i.e.

\left[ {CO} \right] = 2a

$\therefore$

\,\,\,\,\,{r_2} = k{\left( {2a} \right)^2} = 4k{a^2}

$\therefore$

\,\,\,\,\,{r_2} = 4{r_1}

So, the rate of reaction becomes

4

times.

Q73

The reaction

A_2 + B_2 \rightarrow 2AB

follows the mechanism:

A_2 \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} A + A

(fast)

A + B_2 \xrightarrow{k_2} AB + B

(slow)

A + B \rightarrow AB

(fast)The overall order of the reaction is:

A 3

B 2.5

C 1.5

D 2

Correct Answer

Option C

Solution

Reaction given:

A_2+B_2 \rightarrow 2 A B

Mechanism:

A_2\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{k_{-1}}^{k_1}} A+A

(fast)

\begin{aligned} & A+B_2 \xrightarrow{k_2} A B+B \text{ (slow) } \\ & A+B \longrightarrow A B \text{ (fast) } \end{aligned}

The rate determining step in a mechanism is the slow step.

So, late $=k_2[A]\left[B_2\right]$ $A$ is the intermediate; to determine the concentration of $A$ , the equilibrium is considered.

A_2\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{k_{-1}}^{k_1}} A+A

\begin{aligned} \frac{k_1}{k_{-1}} & =\frac{[A][A]}{\left[A_2\right]} \\ & =\frac{[A]^2}{\left[A_2\right]} \end{aligned}

\begin{aligned} S_{0,}[A]^2 & =\frac{k_1}{k_{-1}}\left[A_2\right] \\ {[A] } & =\left\{\frac{k_1}{k_{-1}}\left[A_2\right]\right\}^{1 / 2} \end{aligned}

Substitute $[A]$ in the late, late $=k_2[A]\left[B_2\right]$

\begin{aligned} & =k_2\left(\frac{k_1}{k_{-1}}\right)^{1 / 2}\left[A_2\right]^{1 / 2}\left[B_2\right] \\ & =k_2\left(\frac{k_1}{k_{-1}}\right)^{1 / 2}\left[A_2\right]^{1 / 2}\left[B_2\right]^1 \end{aligned}

\begin{aligned} \text{ Overall rate } & =\frac{1}{2}+1 \\ & =\frac{3}{2} \\ & =1.5 \end{aligned}

Q74

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 (m + n)

B (n - m)

C 2( n - m)

D

{1 \over {{2^{(m + n)}}}}

Correct Answer

Option C

Solution

Rat{e_1} = k{\left[ A \right]^n}{\left[ B \right]^m};

Rat{e_2} = k{\left[ {2A} \right]^n}{\left[ {{{1}/{2}}B} \right]^m}

$\therefore$

\,\,\,\,{{Rat{e_2}} \over {Rat{e_1}}} = {{k{{\left[ {2A} \right]}^n}{{\left[ {{{1}/{2}}B} \right]}^m}} \over {k{{\left[ A \right]}^n}{{\left[ B \right]}^m}}}

= {\left[ 2 \right]^n}{\left[ {{{1}/{2}}} \right]^m} = {2^n}{.2^{ - m}} = {2^{n - m}}

Q75

In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is

A 30 minutes

B 60 minutes

C 7.5 minutes

D 15 minutes

Correct Answer

Option A

Solution

As the concentration of reactant decreases from

0.8

to

0.4

in

15

minutes hence the

{t_{1/2}}

is

15

minutes. To fall the concentration from

0.1

to

0.025

we need two half lives i.e.,

30

minutes.

Q76

At

30^{\circ} \mathrm{C}

, the half life for the decomposition of

\mathrm{AB}_{2}

is

200 \mathrm{~s}

and is independent of the initial concentration of

\mathrm{AB}_{2}

. The time required for

80 \%

of the

\mathrm{AB}_{2}

to decompose is Given:

\log 2=0.30

\quad \log 3=0.48

A 200 s

B 323 s

C 467 s

D 532 s

Correct Answer

Option C

Solution

Since, half-life is independent of the initial concentration of

A B_{2}

. Hence, the reaction is "First Order".

k=\frac{2.303 \log 2}{t_{1 / 2}}

\frac{2.303 \log 2}{t_{1 / 2}}=\frac{2.303}{t} \log \frac{100}{(100-80)}

\frac{2.303 \times 0.3}{200}=\frac{2.303}{t} \log 5

t=467 \mathrm{~s}

Q77

The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2=0.301) :

A 230.3 minutes

B 23.03 minutes

C 46.06 minutes

D 460.6 minutes

Correct Answer

Option C

Solution

For first order reaction

k = {{2.303} \over t}\log {{100} \over {100 - 99}}

{{0.693} \over {6.93}} = {{2.303} \over t}\log {{100} \over 1}

{{0.693} \over {6.93}} = {{2.303 \times 2} \over t}

\Rightarrow t = 46.06\min

Q78

Consider the reaction, 2A + B

\to

products. When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

A L mol-1 s-1

B no unit

C mol L-1 s-1

D s-1

Correct Answer

Option A

Solution

Rate = k [A]x[B]y When [B] is doubled, keeping [A] constant half-life of the reaction does not change.

For a first order reaction

{t_{1/2}} = {{0.693} \over K}\,\,

i.e. for a first order reaction

{t_{1/2}}

does not depend up on the concentration.

Hence the reaction is first order with respect to B.

Now when [A] is doubled, keeping [B] constant, the rate also doubles.

Hence the reaction is first order with respect to A.

$\therefore$

\,\,\,\,\,

Rate = k [A]1[B]1 Order of reaction

=1+1=2

Now for a nth order reaction, unit of rate constant is (L)n–1 (mol)1–n s–1 when n = 2, unit of rate constant is

L\,\,mo{l^{ - 1}}\,{\sec ^{ - 1}}.

Q79

Isotope(s) of hydrogen which emits low energy

\beta

-

particles with t1/2 value > 12 years is/are

A Protium

B Tritium

C Deuterium

D Deuterium and Tritium

Correct Answer

Option B

Solution

_1^1

H and

_1^2

H are stable while

_1^3

H is radioactive.

Q80

A student has studied the decomposition of a gas AB

_3

at 25

^\circ

C. He obtained the following data. .tg .tg p (mm Hg) 50 100 200 400 relative t

_{1/2}

(s) 4 2 1 0.5 The order of the reaction is

A 2

B 0.5

C 1

D 0 (zero)

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{t}_{1 / 2} \propto\left(\mathrm{P}_{\mathrm{o}}\right)^{1-\mathrm{n}} \\\\ & \frac{\left(\mathrm{t}_{1 / 2}\right)_1}{\left(\mathrm{t}_{1 / 2}\right)_2}=\frac{\left(\mathrm{P}_o\right)_1^{1-\mathrm{n}}}{\left(\mathrm{P}_{o_2}\right)_2^{1-\mathrm{n}}} \\\\ & \Rightarrow\left(\frac{4}{2}\right)=\left(\frac{50}{100}\right)^{1-\mathrm{n}} \\\\ & \Rightarrow 2=\left(\frac{1}{2}\right)^{1-\mathrm{n}} \\\\ & \Rightarrow 2=(2)^{\mathrm{n}-1} \\\\ & \Rightarrow \mathrm{n}-1=1 \\\\ & \Rightarrow \mathrm{n}=2 \end{aligned}