Chemical Kinetics — Page 2 | JEE Chemistry

Q11

Consider the following nuclear reactions

{}_{92}^{238}M \to {}_Y^XN + 2{}_2^4He

{}_Y^XN \to {}_B^AL + 2{\beta ^ + }

The number of neutrons in the element L is

A 140

B 144

C 142

D 146

Correct Answer

Option B

Solution

We know that, $\alpha$ - decay formula is -

{ }_Z^A X \stackrel{-\alpha}{\rightarrow}{ }_{Z-2}^{A-4} Y

And the formula for $\beta$ is -

{ }_Z^A X \stackrel{-\beta}{\rightarrow}{ }_{Z-1}^A Y

Now for the given above question we see that, -

\begin{aligned} &{ }_{92}^{238} M \stackrel{-2 \alpha}{\rightarrow}{ }_{92-4}^{238-8} N \\\\ &{ }_Y^X X \stackrel{-2 \beta^{+}}{\rightarrow}{ }_{86}^{230} L \end{aligned}

Therefore we can now calculate the number of neutrons in $L$ , -

L=230-86

L=144

So, as we can see the no. of neutrons present in $\mathrm{L}$ is 144 .

Q12

The half – life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is

A 1.042 g

B 4.167 g

C 3.125 g

D 2.084 g

Correct Answer

Option C

Solution

{N_t} = {N_0}{\left( {{1 \over 2}} \right)^n}

where

n

is number of half life periods.

n = {{Total\,\,time} \over {half\,\,life}} = {{24} \over 4} = 6

$\therefore$

\,\,\,\,\,{N_t} = 200{\left( {{1 \over 2}} \right)^6} = 3.125g.

Q13

The photon of hard gamma radiation knocks a proton out of

{}_{12}^{24}Mg

nucleus to form

A the isotope of parent nucleus

B the isobar of parent nucleus

C the nuclide

{}_{11}^{23}Na

D the isobar of

{}_{11}^{23}Na

Correct Answer

Option C

Solution

{}_{12}M{g^{24}}\overset{\,}\longrightarrow {}_{11}N{a^{23}} + {}_1{H^1}.

Q14

t1/4 can be taken as the time taken for the concentration of a reactant to drop to

3 \over 4

of its initial value. If the rate constant for a first order reaction is K, the t1/4 can be written as

A 0.10 / K

B 0.29 / K

C 0.69 / K

D 0.75 / K

Correct Answer

Option B

Solution

{t_{1/4}} = {{2.303} \over K}\log {1 \over {3/4}}

= {{2.303} \over K}\log {4 \over 3}

= {{2.303} \over K}\left( {\log \,4 - \log 3} \right)

= {{2.303} \over K}\left( {2{{\log }^2} - \log 3} \right)

= {{2.303} \over K}\left( {2 \times 0.301 - 0.4771} \right)

= {{0.29} \over K}

Q15

A reaction involving two different reactants can never be

A Unimolecular reaction

B First order reaction

C second order reaction

D Bimolecular reaction

Correct Answer

Option A

Solution

The molecularity of reaction is the number of reactant molecules taking part in a single step of the reaction.

NOTE : The reaction involving two different reactant can never be unimolecular.

Q16

Hydrogen bomb is based on the principle of

A Nuclear fission

B Natural radioactivity

C Nuclear fusion

D Artificial radioactivity

Correct Answer

Option C

Solution

Option C Nuclear fusion Explanation : A hydrogen bomb, also known as a thermonuclear bomb, uses the principle of nuclear fusion.

In a fusion reaction, two lighter atomic nuclei combine to form a heavier nucleus, and a substantial amount of energy is released in the process.

In the case of a hydrogen bomb, isotopes of hydrogen (such as deuterium and tritium) fuse together to form helium, releasing a large amount of energy.

It's worth noting that a hydrogen bomb usually involves a two-stage process.

The first stage is a fission bomb (like those used in Hiroshima and Nagasaki) that creates the conditions necessary for the fusion reaction in the second stage.

Despite this, the majority of the energy in a hydrogen bomb comes from fusion, which is why it is categorized as a fusion weapon.

Q17

The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr: NO(g) + Br2 (g)

\leftrightharpoons

NOBr2 (g) NOBr2 (g) + NO (g)

\to

2NOBr (g) If the second step is the rate determining step, the order of the reaction with respect to NO(g) is

A 1

B 0

C 3

D 2

Correct Answer

Option D

Solution

\left( i \right)\,\,\,\,\,\,\,\,\,NO\left( g \right) + B{r_2}\left( g \right)\,\,\,\rightleftharpoons\,NOB{r_2}\left( g \right)

\left( {ii} \right)\,\,\,\,\,\,\,\,\,NOB{r_2}\left( g \right) + NO\left( g \right)\,\overset{\,}\longrightarrow 2NOBr\left( g \right)

Rate law equation

= k\left[ {NOB{r_2}} \right]\left[ {NO} \right]

But

NOB{r_2}

is intermediate and must not appear in the rate law equation from

1

st step

{K_C} = {{\left[ {NOB{r_2}} \right]} \over {\left[ {NO} \right]\left[ {B{r_2}} \right]}}

$\therefore$

\,\,\,\,\,\left[ {NOB{r_2}} \right] = {K_C}\left[ {NO} \right]\left[ {B{r_2}} \right]

$\therefore$

\,\,\,\,

Rate law equation

= k.{K_C}{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]

hence order of reaction is

2

w.r.t.

NO.

Q18

The energies of activation for forward and reverse reactions for A2 + B2

\leftrightharpoons

2AB are 180 kJ mol−1 and 200 kJ mol−1 respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol−1. The enthalpy change of the reaction ( A2 + B2

\to

2AB) in the presence of catalyst will be (in kJ mol−1)

A 300

B 120

C 200

D 20

Correct Answer

Option D

Solution

\Delta {H_R} = {E_f} - {E_b} = 180 - 200 = - 20kJ/mol

The nearest correct answer given in choices may be obtained by neglecting sign.

Q19

A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room?

A 1000 days

B 300 days

C 10 days

D 100 days

Correct Answer

Option D

Solution

Suppose activity of safe working

=A

Given

{A_0} = 10A

\lambda = {{0.693} \over {{t_{1/2}}}} = {{0.693} \over {30}}

{t_{{{1}/{2}}}} = {{2.303} \over \lambda }\log {{{A_0}} \over A}

= {{2.303} \over {0.693/30}}\log {{10A} \over A}

= {{2.303 \times 30} \over {0.693}} \times \log 10

= 100

days.

Q20

Which of the following nuclear reactions will generate an isotope?

A neutron particle emission

B positron emission

C

\alpha

-particle emission

D

\beta

-particle emission

Correct Answer

Option A

Solution

NOTE : Isotopes are atoms of same element having same atomic number but different atomic masses.

Neutron has atomic number

0

and atomic mass

1.

So loss of neutron will generate isotope.

e.g.,

{}_{92}{U^{238}} + {}_0{n^1} \to {}_{92}{U^{239}}

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