Chemical Kinetics and Nuclear Chemistry

We have

${O_3}(g) + C{l^ \bullet }(g) \to {O_2}(g) + Cl{O^ \bullet }(g)$ ...... (1)

$Cl{O^ \bullet }(g) + {O^ \bullet }(g) \to {O_2}(g) + C{l^ \bullet }(g)$ ...... (2)

On adding Eq. (1) and Eq. (2), we get the required equation for overall reaction

${O_3}(g) + {O^ \bullet }(g) \to 2{O_2}(g)$

Therefore,

${k_{overall}} = {k_1} \times {k_2}$

$ = 5.2 \times {10^9} \times 2.6 \times {10^{10}}$

$ = 1.4 \times {10^{20}}$ L mol^$-$1 s^$-$1

${H_2}{O_2}\left( {aq} \right) \to {H_2}O\left( {aq} \right) + {1 \over 2}{O_2}\left( g \right)$

For a first order reaction

$k = {{2.303} \over t}\log {a \over {\left( {a - x} \right)}}$

Given $a = 0.5,\left( {a - x} \right) = 0.125,\,t = 50\,\,$ min

$\therefore$ $\,\,\,\,\,k = {{2.303} \over {50}}\log {{0.5} \over {0.125}}$

$ = 2.78 \times {10^{ - 2}}\,{\min ^{ - 1}}$

$r = k\left[ {{H_2}{O_2}} \right] = 2.78 \times {10^{ - 2}} \times 0.05$

$ = 1.386 \times {10^{ - 3}}\,\,mol\,{\min ^{ - 1}}$

Now

$ - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}} = {{d\left[ {{H_2}O} \right]} \over {dt}} = {{2d\left[ {{O_2}} \right]} \over {dt}}$

$\therefore$ $\,\,\,\,\,{{2d\left[ {{O_2}} \right]} \over {dt}} - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}$

$\therefore$ $\,\,\,\,\,$ ${{d\left[ {{O_2}} \right]} \over {dt}} = {1 \over 2} \times {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}$

$ = {{1.386 \times {{10}^{ - 3}}} \over 2} = 6.93 \times {10^{ - 4}}\,mol\,{\min ^{ - 1}}$

Plot of Z (number of protons) verses N (number of neutrons) is represented as :


(a) For a stable nucleus $\mathrm{N} / \mathrm{Z}\left(\frac{\text { no. of neutrons }}{\text { no. of protons }}\right)$ should be equal to 1.

This happens for elements till atomic number $(Z=20)$. For such nucleus number of proton are equal to number of neutrons.

(b) For heavier nucleides with atomic number $(\mathrm{Z})$ greater than $20(Z>20)$, the number of neutron are more than proton as seen by curved (green) appearance in the plot.

(c) For nucleids with $\mathrm{N} / \mathrm{Z}<1$, the number of neutrons are less than the protons making the nucleus unstable. This is due to increase in coulombic repulsions between positively charged protons.

Such nuclei become stable by K-electron capture and positron emission.

K-electron capture :

$ { }_1^1 p+{ }_{-1}^0 e \rightarrow{ }_0^1 n $

The K-electron capture, proton is converted to the neutron as increases the $\mathrm{N} / \mathrm{Z}$ ratio.

$\beta^{ \pm}$decay (positron emission) :

$ { }_1^1 p \rightarrow{ }\underset{neutron}{_0^1 n}+\underset{positron}{e^{+}}+\underset{electron \,nutrino}{v_e} $

Proton decays to a neutron, positron and electron nutrino. This decreases the number of protons and increases the number of neutrons, thereby increasing $\mathrm{N} / \mathrm{Z}$ ratio.

Options (B) and (D) are correct.

The Arrhenius equation is a mathematical model of the temperature dependence of reaction rates. It is expressed as follows:

$ k=\mathrm{Ae}^{-\mathrm{E}_a / \mathrm{RT}} $

where :

- k is the rate constant

- A is the pre-exponential factor (also known as the frequency factor)

- E_a is the activation energy

- R is the universal gas constant

- T is the temperature (in Kelvin)

- e is Euler's number, a mathematical constant approximately equal to 2.71828

The options provided correlate to the Arrhenius equation as follows:

Option A : Incorrect. A high activation energy usually implies a slower reaction, not a fast one. The activation energy is the energy barrier that needs to be overcome for a reaction to occur. The higher the activation energy, the fewer the number of molecules that have enough energy to surpass this barrier and react, resulting in a slower reaction rate.


Option B : Correct. Temperature $\mathrm{T}_2$ is greater than $\mathrm{T}_1$; hence, the fraction of molecules (shaded in green) having energy greater than activation energy increases with increase in temperature result, rate of reaction increases.

Higher temperature, more number of collisions (between extents) having energy greater than activation energy happen increases speed of reaction.



Option C : Correct. A reaction at higher activation energy makes the reaction rate slow. An increase in temperature causes reactant molecules to cross the high energy barrier; i.e., $E_a$ and form activated complex.

Option D : Correct. The pre-exponential factor (A) takes into account the probability of proper orientation of reactant molecules to form the product and the frequency at which reactant molecules collide to form the product.

$ \mathrm{A}=\mathrm{Z}_{\mathrm{AB}} \times \mathrm{P} $

$Z_{A B}$ : Collision frequency or frequency with which A collides with B.

P : Probability of reactant molecules oriented in right orientation with respect to each other to form the product.

Option (D), where pre-exponential factor affects the rate reaction irrespective of the energy they possess.

Reactions of higher order $\left( { > 3} \right)$ are very rate due to very less chances of many molecules to undergo effective collisions.

The equilibrium yield of ammonia gets lowered with the increase of temperature. However, at higher temperature the initial rate of forward reaction would be greater than at lower temperature that is why the percentage yield of NH₃ would be more initially.

Let rate of reaction $ = {{d\left[ C \right]} \over t} = k{\left[ A \right]^x}{\left[ B \right]^y}$

Now from the given data

$1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y}\,\,\,\,\,...\left( i \right)$

$1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y}\,\,\,\,\,...\left( {ii} \right)$

$2.4 \times {10^{ - 3}} = k{\left[ {0.2} \right]^x}{\left[ {0.1} \right]^y}\,\,\,\,...\left( {iii} \right)$

Dividing equation $(i)$ by $(ii)$

$ \Rightarrow \,\,\,\,{{1.2 \times {{10}^{ - 3}}} \over {1.2 \times {{10}^{ - 3}}}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}}}$

We find, $y=0$

Now dividing equation $(i)$ by $(iii)$

$ \Rightarrow \,\,\,\,{{1.2 \times {{10}^{ - 3}}} \over {2.4 \times {{10}^{ - 3}}}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.2} \right]}^x}{{\left[ {0.1} \right]}^y}}}$

We find, $x=1$

Hence ${{d\left[ C \right]} \over {dt}} = k{\left[ A \right]^1}{\left[ B \right]^0}$

The rate law for an elementary reaction where the concentration of reactant M impacts the rate of the reaction is expressed as:

$\text{Rate} = k[M]^n$

where k is the rate constant, M is the concentration of the reactant, and n is the order of the reaction with respect to M.

In the scenario described, when the concentration of M is doubled, the rate of disappearance of M increases by a factor of 8. Let’s denote the initial concentration of M by [M] and the doubled concentration by 2[M]. According to the rate law, the initial rate would be:

$\text{Rate}_1 = k[M]^n$

Upon doubling M:

$\text{Rate}_2 = k(2[M])^n = k2^n[M]^n$

It is given that this rate is 8 times the initial rate:

$ \text{Rate}_2 = 8 \times \text{Rate}_1 $

$ k2^n[M]^n = 8k[M]^n $

Cancelling out the common terms and simplifying the equation, we have:

$ 2^n = 8 $

To find n, solve for n in:

$ 2^n = 2^3 $

Hence, n = 3, indicating that the order of the reaction with respect to M is 3. The correct option is therefore:

Option B: 3

Using Graham's law of diffusion, we have that for identical conditions, rate of diffusion varies inversely with square root of mass of the gases. So,

${x \over {24 - x}} = {\left( {{{40} \over {10}}} \right)^{1/2}} \Rightarrow x = 16$

Collision frequency of X is greater than Y.

Activation energy can be calculated from the equation

${{\log \,{k_2}} \over {\log \,{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$

given ${{{k_2}} \over {{k_1}}} = 2\,\,{T_2} = 310\,K\,\,{T_1} = 300\,K$

$ = \log 2 = {{ - {E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {310}} - {1 \over {300}}} \right)$

${E_a} = 53598.6J/mol$

$ = 53.6kJ/mol.$

This is a first-order reaction because t_75% = 2 $\times$ t_50%. The graph shows that the order with respect to Q is 0, so we can write the rate expression as

Rate = k[P]¹[Q]⁰

The nuclear transmutation of beryllium (Be) is given as:

$ _4^9 \mathrm{Be} + X \to _4^8 \mathrm{Be} + Y $

We examine the possible reactions to determine the values of X and Y:

(A) If $ X = _0^0\gamma $ and $ Y = _0^1n $

The reaction is:

$ _4^9 \mathrm{Be} + _0^0\gamma \longrightarrow _4^8 \mathrm{Be} + _0^1n $

Using the conditions of:

(i) Mass balance:

Mass of reactants $ = 9 + 0 = 9 $

Mass of products $ = 8 + 1 = 9 $

(ii) Charge balance:

Charge of reactants $ = 4 + 0 = 4 $

Charge of products $ = 4 + 0 = 4 $

Both mass and charge are balanced. Therefore, Option (A) is correct.

(B) If $ X = p = _1^1H $ and $ Y = D = _1^2H $

The reaction is:

$ _4^9 \mathrm{Be} + _1^1H \longrightarrow _4^8 \mathrm{Be} + _1^2H $

Using the conditions of:

(i) Mass balance:

Mass of reactants: $ 9 + 1 = 10 $

Mass of products $ = 8 + 2 = 10 $

(ii) Charge balance:

Charge of reactants $ = 4 + 1 = 5 $

Charge on products $ = 4 + 1 = 5 $

Both mass and charge are balanced. Therefore, Option (B) is correct.

(C) If $ X = _0^1n $ and $ Y = D = _1^2H $

The reaction is:

$ _4^9 \mathrm{Be} + _0^1n \longrightarrow _4^8 \mathrm{Be} + _1^2H $

Using the conditions of:

(i) Mass balance:

Mass of reactants $ = 9 + 1 = 10 $

Mass of products $ = 8 + 2 = 10 $

(ii) Charge balance:

Charge on reactants $ = 4 + 0 = 4 $

Charge on products $ = 4 + 1 = 5 $

Charge on reactants $ \neq $ Charge on products. Since, charge is not balanced, Option (C) is incorrect.

(D) If $ X = _0^0\gamma $ and $ Y = p = _1^1H $

The reaction is:

$ _4^9 \mathrm{Be} + _0^0\gamma \longrightarrow _4^8 \mathrm{Be} + _1^1H $

Using the conditions of:

(i) Mass balance:

Mass of reactants: $ 9 + 0 = 9 $

Mass of products: $ 8 + 1 = 9 $

(ii) Charge balance:

Charge on products: $ 4 + 1 = 5 $

Charge on reactants: $ 4 + 0 = 4 $

Charge on reactants $ \neq $ Charge on products. Since the charge is not balanced, Option (D) is incorrect.

For a first order reaction

$k = {{2.0303} \over t}\,\log \,{a \over {a - x}}$

$ = {{2.303} \over {40}}\log {{0.1} \over {0.025}}$

$ = {{2.303} \over {40}}\log 4$

$ = {{2.303 \times 0.6020} \over {40}}$

$ = 3.47 \times {10^{ - 2}}$

$R = K{\left( A \right)^1} = 3.47 \times {10^{ - 2}} \times 0.01$

$ = 3.47 \times {10^{ - 4}}$

Since for every ${10^ \circ }C$ rise in temperature rate doubles for ${50^ \circ }C$ rise in temp increase in reaction rate $ = {2^5} = 32$ times

Bombardment of aluminum by an $\alpha$-particle leads to its artificial disintegration in two ways, as shown in the reactions below. The resultant products X, Y, and Z are identified as follows :

Reaction (i)

Disintegration of aluminum into silicon :

$^{4}_{2} \text{He} + ^{27}_{13} \text{Al} \rightarrow ^{30}_{14} \text{Si} + ^{A}_{Z} X$

Conservation conditions :

(a) Charge balance :

$ 2 + 13 = 14 + Z \implies Z = 1 $

(b) Mass balance :

$ 4 + 27 = 30 + A \implies A = 1 $

The particle $A_Z X$ is $^1_1 \text{H}$, a proton (an isotope of hydrogen).

Reaction (ii)

Disintegration of aluminum into phosphorus :

$^{4}_{2} \text{He} + ^{27}_{13} \text{Al} \rightarrow ^{30}_{15} \text{P} + ^{A}_{Z}Y$

Conservation conditions :

(a) Charge balance :

$ 2 + 13 = 15 + Z \implies Z = 0 $

(b) Mass balance :

$ 4 + 27 = 30 + A \implies A = 1 $

The particle $Y$ has one unit mass and no charge, which is a neutron $^1_0 n$.

Reaction (iii)

Disintegration of phosphorus into silicon :

$^{30}_{15}\text{P} \longrightarrow \, ^{30}_{14}\text{Si} + ^{A}_{Z}\text{Z} $

Conservation conditions :

(a) Mass balance :

$ 30 = 30 + A \implies A = 0 $

(b) Charge balance :

$ 15 = 14 + Z \implies Z = 1 $

The particle has zero mass but one unit of positive charge. Hence, the particle is a positron $\left(^{0}_{+1}\beta\right)$.

Summary

The particles are :

X = $^1_1 \text{H}$ (proton)

Y = $^1_0 n$ (neutron)

Z = $^{0}_{+1}\beta$ (positron)

The given chemical equation represents a first-order reaction:

$2N_2O_5 (g) \to 4NO_2 (g) + O_2 (g)$

In first-order reactions, the rate of reaction is directly proportional to the concentration of the reactant. The rate law can be expressed as:

$Rate = k[N_2O_5]$

where k is the rate constant.

The integrated rate equation for a first-order reaction is:

$\ln [N_2O_5] = -kt + \ln [N_2O_5]_0$

where [N_2O_5]_0 is the initial concentration and [N_2O_5] is the concentration at time t.

Rearranging gives:

$[N_2O_5] = [N_2O_5]_0 e^{-kt}$

Now, let's analyze each option:

Option A: "the concentration of the reactant decreases exponentially with time".

This matches the expression $[N_2O_5] = [N_2O_5]_0 e^{-kt}$ which shows an exponential decline in the concentration of the reactant. Therefore, option A is correct.

Option B: "the half-life of the reaction decreases with increasing temperature".

In first-order reactions, the half-life is determined by the equation $t_{1/2} = \frac{0.693}{k}.$ Increases in temperature typically increase the rate constant k, thus decreasing the half-life. However, it is not directly related to the half-life itself but rather to the rate constant via the Arrhenius equation. Nonetheless, the essence of the statement is correct as increasing temperature generally decreases the half-life.

Option C: "the half-life of the reaction depends on the initial concentration of the reactant".

The half-life for a first-order reaction given by $t_{1/2} = \frac{0.693}{k}$ is independent of the initial concentration [N_2O_5]_0. Therefore, option C is incorrect.

Option D: "the reaction proceeds to 99.6% completion in eight half-life durations".

Completion to a certain level can be found using the formula for decay over multiple half-lives, given as $[N_2O_5] = [N_2O_5]_0 \times \left(\frac{1}{2}\right)^n$ where n is the number of half-lives. For eight half-lives, we have:

$Final\, Concentration = [N_2O_5]_0 \times \left(\frac{1}{2}\right)^8 = [N_2O_5]_0 \times \frac{1}{256}$

The reaction proceeds to $\frac{255}{256} \approx 99.6\%$ completion. Thus, option D is correct.

In summary, options A, B, and D are correct, while option C is incorrect.

For the reaction

$A\,\, \to \,\,$ Product ; given ${t_{1/2}} = 1\,\,$ hour

for a zero order reaction

${t_{completion}}\,\, = {{\left[ {{A_0}} \right]} \over k} = {{initial\,\,conc.} \over {rate\,\,cons\tan t}}$

$\therefore$ $\,\,\,\,\,{t_{1/2}} = {{\left[ {{A_0}} \right]} \over {2K}}$

or $\,\,\,\,\,k = {{\left[ {{A_0}} \right]} \over {2{t_{1/2}}}} = {2 \over {2 \times 1}} = 1\,\,mol\,li{t^{ - 1}}\,h{r^{ - 1}}$

Further for a zero order reaction

$k = {{dx} \over {dt}} = {{change\,\,\,in\,\,\,concentration} \over {time}}$

$I = {{0.50 - 0.25} \over {time}}\,\,\,$

$\therefore$ $\,\,\,\,\,$ time $ = 0.25\,\,hr.$

Since the slow step is the rate determining step hence-

if we consider option $(1)$ we find

Rate $ = k\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]$

Now if we consider option $(2)$ we find

Rate $ = k\left[ {C{l_2}} \right]\left[ {H{S^ - }} \right]\,\,\,\,\,...\left( 1 \right)$

From equation $(i)$

$k = {{\left[ {{H^ + }} \right]\left[ {H{S^ - }} \right]} \over {{H_2}S}}$ $\,\,\,\,\,$ or

$\left[ {H{S^ - }} \right] = {{k\left[ {{H_2}S} \right]} \over {{H^ + }}}$

Substituting this value in equation $(1)$ we find

Rate $ = k\left[ {C{l_2}} \right]K{{\left[ {{H_2}S} \right]} \over {{H^ + }}}$

$ = k'{{\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]} \over {\left[ {{H^ + }} \right]}}$

hence only, mechanism $(1)$ is consistent with the given rate equation.

According to the Arrhenius equation

$k = A{e^{ - {E_a}/RT}}$

where, k = rate constant, E_a = activation energy and T = temperature

From the expression, we have that as temperature increases, the rate constant ($k$) increases exponentially.

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 $

Given that $\log k = 6 - {{2000} \over T}$

According to Arrhenius equation

$k = A{e^{ - {E_a}/RT}}$

$\log K = \log A - {E \over {2.303RT}}$

Comparing with the given equation and solving, we get

$A = 1.0 \times {10^6}\,{s^{ - 1}}$

${E_a} = 38.3$ kJ/mol

The rates of reactions for the reaction

${1 \over 2}A \to 2B$

can be written either as

$ - 2{d \over {dt}}\left[ A \right]\,\,$ with respect to $'A'$

or $\,\,\,\,\,$ ${1 \over 2}{d \over {dt}}\left[ B \right]\,\,\,\,$ with respect to $'B'$

From the above, we have

$ - 2{d \over {dt}}\left[ A \right] = {1 \over 2}{d \over {dt}}\left[ B \right]$

or $\,\,\,\,\,$ $ - {d \over {dt}}\left[ A \right] = {1 \over 4}{d \over {dt}}\left[ B \right]$

i.e., correct answer is (b)

Rate constant for first order kinetics:

${k_1} = {{0.693} \over {{t_{1/2}}}} = {{0.693} \over {40\,{s^{ - 1}}}}$ ..... (i)

Rate constant for zero order kinetics:

${k_0} = {{{A_0}} \over {2{t_{1/2}}}} = {{1.386} \over {2 \times 20}}$ mol dm$^{-3}$ s$^{-1}$ ..... (ii)

Divide (i) by (ii)

${{{k_1}} \over {{k_0}}} = {{0.693} \over {40}} \times {{40} \over {1.386}} = {{0.693} \over {1.386}}$ mol$^{-1}$ dm$^{-3}$ = 0.5 mol$^{-1}$ dm$^{3}$

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

Suppose activity of safe working $=A$

Given ${A_0} = 10A$

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

${t_{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 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.

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. $e.g.,$
$${}_{92}{U^{238}} + {}_0{n^1} \to {}_{92}{U^{239}}$$

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

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

The rate law expression is

$\mathrm{R}=k[\mathrm{G}]^{a}[\mathrm{H}]^{b} $ ...... (i)

The rate of reaction increases by a factor of 2 on doubling the concentration of '$\mathrm{G}$'.

The rate law expression becomes

$\mathrm{R}^{\prime}=2 \mathrm{R}=k 2^{a}[\mathrm{G}]^{a}[\mathrm{H}]^{b}$ ..... (ii)

Divided equation (ii) with eqn. (i)

$\begin{aligned}\frac{\mathrm{R}^{\prime}}{\mathrm{R}} & =\frac{2 \mathrm{R}}{\mathrm{R}}=\frac{k 2^{a}[\mathrm{G}]^{a}[\mathrm{H}]^{b}}{k 2[\mathrm{G}]^{a}[\mathrm{H}]^{b}} \\2 & =2^{a} \\a & =1\end{aligned}$

The rate of reaction increases by a factor of 8 on doubling the concentration of $\mathrm{G}$ and '$\mathrm{H}$'.

The rate law expression becomes

Rate $=\mathrm{R}^{\prime \prime} k 2^{a}[\mathrm{G}]^{a} 2^{b}[\mathrm{H}]^{b}$ ..... (iii)

Divide equation (iii) with equation (i)

$\begin{aligned}\frac{8 \mathrm{R}}{\mathrm{R}} & =\frac{k 2^{a}[\mathrm{G}]^{a} 2^{b}[\mathrm{H}]^{b}}{k[\mathrm{G}]^{a}[\mathrm{H}]^{b}}\end{aligned}$

$8=2^a2^b$

or $b=2$

The overall order of reaction is

$a+b=1+2=3$

In Arrhenius equation $K = A\,{e^{ - E/RT}},\,\,E$ is the energy of activation, which is required by the colliding molecules to react resulting in the formation of products.

${}_{92}^{238}U\buildrel { - \alpha } \over \longrightarrow {}_{90}^{234}Th\buildrel {\, - 2\beta } \over \longrightarrow {}_{92}^{234}U$

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

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

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. $NO.$

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.

The option A states that due to higher rate of circulation of ${ }^{14} \mathrm{C}$ from atmosphere in the form of carbon dioxide, the carbon content remain constant in living organism. The living organisms absorb ${ }^{14} \mathrm{C}$ from atmosphere during photosynthesis and the content remain constant till the organism is alive. After the death of living organism, the absorption decreases as carbon

dioxide absorption decreases. But, carbon content refers to the content of any isotope of carbon. Hence, the option A is incorrect.

The option B states that the method of carbon dating is used for calculating age of earth crust and rocks. The carbon dating method involves use of ${ }^{14} \mathrm{C}$ to determine the age of organic material present in any substance. Rocks and earth crust contain organic material in the form of remains of dead plants and animals, but also contains inorganic material such as minerals. The age of remains of dead organism is an organic material and the age of this organic material can be calculated using carbon dating. But, carbon dating is not useful for determining the age of minerals present in the rocks and Earth crust. Hence, option B is incorrect.

The option C states radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organism. The cosmic rays are ionizing radiation released from radioactive material. The rate of absorption of ${ }^{14} \mathrm{C}$ equals rate of disintegration of ${ }^{14} \mathrm{C}$. The ratio of ${ }^{14} \mathrm{C}$ to ${ }^{12} \mathrm{C}$ in living matter is $1: 10^{12}$ which is constant. Hence, option C is correct.

The option D states that carbon dating cannot be used to determine concentration of ${ }^{14} \mathrm{C}$ in dead beings. This is not true as carbon dating technique is used for determining age of organic material in fossils, rocks and dead organisms. Hence, option D is incorrect.

The carbon dating technique is used for determining the age of organic material including fossils. The carbon dating used ${ }^{14} \mathrm{C}$ isotope of carbon, based on the capturing of neutron in the upper atmosphere.

The half-life period for ${ }^{14} \mathrm{C}$ is 5770 years. The 6 years is not a meaningful time period for determination of age of fossil as there is vast difference between 5,770 years and 6 years. Hence, option A is incorrect.

The option B is correct as half-life period of ${ }^{14} \mathrm{C}$ is 5,770 years which is closer to 6,000 years. Hence, option B is correct.

The accuracy of carbon dating decreases if the fossils are more than 30,000 years older, Hence, the accuracy will be negligible in case the age of fossil is 60,000 years. Hence, option C is incorrect.

The option D is incorrect as carbon dating is effective in determining age of fossils aged below 30,000 years.

As stated in question, a nuclear explosion increases the concentration of $\mathrm{C}^{14}$ in nearby areas. Therefore, there is no increase in concentration of $\mathrm{C}^{14}$ in faraway areas.

The concentration of $C^{14}$ is $C_1$ in nearby areas whereas, $C_2$ in faraway areas. The $T_1$ denotes the age of the fossil at nearby areas and $\mathrm{T}_2$ becomes the age of the fossil in far away area.

The radioactive decay is first order reaction.

The decay constant, $\lambda$ can be given for first order reaction as follows:

$ \lambda=\frac{1}{\mathrm{~T}_1-\mathrm{T}_2} \ln \frac{\mathrm{C}_1}{\mathrm{C}_2} $

The age of fossil will be lesser in nearby areas as compared to that of faraway areas because the rate of absorption of $\mathrm{C}^{14}$ is higher in nearby areas as compared to that of far away areas. Thus, the amount of $\mathrm{C}^{14}$ increases in nearby area as compared to that of faraway areas and the time required for decay of $\mathrm{C}^{14}$ in nearby areas will be higher than that of faraway areas.

The chemical reaction between

$\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CD}_3$ and alc. KOH is given below.

Note: The bond between C and deuterium is much strong than that of C and H , hence, elimination takes place from $\mathrm{C}-\mathrm{H}$ bonds only. This chemical reaction undergoes E2 mechanism as two substituents, H and Br atoms are removed simultaneously from the compound to form alkene.

Hence, option A in column I matches with option Q from column II.

(B) The strength of C-D bonds is greater than that of $\mathrm{C}-\mathrm{H}$ bonds. Thus, breaking of $\mathrm{C}-\mathrm{D}$ will require more energy than that of $\mathrm{C}-\mathrm{H}$ bonds. More the amount of energy required to break the bonds, higher is the stability of the compound. Higher is the stability of the compound, lower is the reactivity and vice-versa.

Hence, $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CH}_3$ will have higher reactivity than that of $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CD}_3$. Hence, $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CH}_3$ will react faster than that of $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CD}_3$ through E 2 mechanism. The mechanism is E2 as two substituents are eliminated in a single step from the compound.

Hence, option B in column I matches with option Q in column II.

(C) The chemical reaction between $\mathrm{Ph}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}$is a two-step process as shown below.

The step (i) is a fast step in which elimination of proton leads to the formation of a carbanion. The step (i) is reversible and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD}$ along with

ethanol is used as a solvent. This is first order reaction as reactant will incorporate deuterium (D). The step (ii) is a slow step in which final product is formed by elimination of bromide ion. The reaction is unimolecular elimination reaction and the reactant for the reaction is a carbanion produced during the first step. The carbanion is a conjugate base of the reactant. Hence, it is E1 cb reaction mechanism.

Hence, option C in column I matches with options R and S in column II.

(D) The chemical reactions for both

$\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br}$ and $\mathrm{PhCD}_2-\mathrm{CH}_2-\mathrm{Br}$ are shown below:

E1 elimination in $\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br}$, to form an alkene

Step 1:

$ \mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br} \xrightarrow{\text { Slow }} \mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}} \mathrm{H}_2+\mathrm{Br}^{-} $

Step 2:

$ \mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}} \mathrm{H}_2+\mathrm{Br}^{-} \xrightarrow{\text { Fast }} \mathrm{PhCH}=\mathrm{CH}_2+\mathrm{H}^{+}$

E1 elimination in $\mathrm{PhCD}_2-\mathrm{CH}_2-\mathrm{Br}$, to form an alkene

Step 1:

$ \mathrm{PhCD}_2-\mathrm{CH}_2-\mathrm{Br} \xrightarrow{\text { Slow }} \mathrm{PhCD}_2-\stackrel{+}{\mathrm{C}} \mathrm{H}_2+\mathrm{Br}^{-} $

$ \begin{aligned} &\text { Step 2: }\\ &\mathrm{PhCD}_2-\stackrel{+}{\mathrm{C}} \mathrm{H}_2+\mathrm{Br}^{-} \xrightarrow{\text { Fast }}\mathrm{PhCD}=\mathrm{CHD}+\mathrm{H}^{+}\end{aligned} $

For both chemical reactions, the first step is slow step and hence, rate determining step.

For $\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br}$ reaction,

Rate of reaction $\propto \mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br}$

For $\mathrm{PhCD}_2-\mathrm{CH}_2-\mathrm{Br}$ reaction,

Rate of reaction $\propto \mathrm{PhCD}_2-\mathrm{CH}_2-\mathrm{Br}$

Both chemical reactions are of first order. Both chemical reactions involve elimination of bromide ion and a proton and hence, it undergoes E1 mechanism

Hence, option D in column I matches with options P and S in column II.

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.

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.

The graph show that reaction is exothermic.

$\log \,k = {{ - \Delta H} \over {RT}} + 1$

For exothermic reaction $\Delta H < 0$

$\therefore$ $\,\,\,\,\,log\,\,k\,\,Vs{1 \over T}\,\,$ would be negative straight line with positive slope.

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

$ = {{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}$

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