Chemical Kinetics and Nuclear Chemistry

(A) For a first-order reaction, the concentration of product at time $ t $ is given by:

$ [\mathrm{P}]_t = [\mathrm{R}]_0 (1 - e^{-kt}) $

This equation shows that as time increases, the concentration of the product $ [\mathrm{P}]_t $ gradually increases and finally becomes equal to the initial concentration of the reactant $ [\mathrm{R}]_0 $. The curve starts from zero and rises exponentially to a maximum value.

(B) For a first-order reaction, the rate law is:

$ \text{Rate} = -\frac{d[\mathrm{R}]}{dt} = k[\mathrm{R}] $

Rearranging, we get:

$ \frac{d[\mathrm{R}]}{dt} = -k[\mathrm{R}] $

Taking logarithms, we can express it as a straight-line equation similar to $ y = -mx $. Hence, a graph of $\ln [\mathrm{R}]$ (or $\log [\mathrm{R}]$) versus time $ t $ gives a straight line with a slope equal to $-k$.

(C) The rate of formation of the product is equal to the rate of disappearance of the reactant:

$ \frac{d[\mathrm{P}]}{dt} = -\frac{d[\mathrm{R}]}{dt} = k[\mathrm{R}] $

Substitute $[\mathrm{R}] = [\mathrm{R}]_0 e^{-kt}$:

$ \frac{d[\mathrm{P}]}{dt} = k[\mathrm{R}]_0 e^{-kt} $

This means the rate of product formation decreases exponentially with time.

(D) For a given temperature, the rate constant $ k $ remains constant. It does not change with concentration or time.

As the concentration of reactant becomes half at t = 50 s. So, half-time of reaction is 50 s.

Given,

${{dP} \over {dt}} = k{[X]^1}$

$ \Rightarrow - {1 \over 2}{{dX} \over {dt}} = {{dP} \over {dt}} = k{[X]^1}$

$ - {{dX} \over {dt}} = 2k{[X]^1}$

${{ - dX} \over {[X]}} = 2k\,dt$

$ - \int_{{X_0}}^X {{{dX} \over X} = 2k\,\int_0^t {dt} } $

$ - [\ln X]_{{X_0}}^X = 2k[t]_0^t$

$ - \ln X + \ln {X_0} = 2kt$

$\ln {{{X_0}} \over X} = 2kt$

At t_1/2, $X = {{{X_0}} \over 2}$

$\ln {{2{X_0}} \over {{X_0}}} = 2k{t_{1/2}}$

$k = {{\ln 2} \over {2{t_{1/2}}}} = {{0.693} \over {100}} = 6.93 \times {10^{ - 3}}{s^{ - 1}}$

At t = 50 s, $ - {{dx} \over {dt}} = 2k{[X]^1};[X] = 1$ mol

So, ${{ - dX} \over {dt}} = 2 \times 6.93 \times {10^{ - 3}} \times 1 = 13.86 \times {10^{ - 3}}$ mol L^-1 s^-1

At t = 100 s, $ - {1 \over 2}{{dX} \over {dt}} = {{ - dY} \over {dt}} \Rightarrow {{ - dY} \over {dt}} = k{[Y]^1};[Y] = {1 \over 2}$

$ - {{dY} \over {dt}} = 6.93 \times {10^{ - 3}} \times {1 \over 2} = 3.46 \times {10^{ - 3}}$ mol L^-1 s^-1

So, options (b), (c) and (d) are correct.

In decay sequence,



X₁ particle will deflect towards negatively charged plate due to presence of positive charge on $\alpha $- particles.

Hence, options (a, b, c) are correct.

Total pressure of gaseous mixture at beginning $(T=0)=\mathrm{P}_0$

Total pressure of gaseous mixture after time $t$ $(\mathrm{T}=t)=\mathrm{P}_t$

Initial concentration of $A=[A]_0$

Temperature of the reaction mixture (T) $=300 \mathrm{~K}$

For the first order reaction,


Total pressure after time $(\mathrm{T}=t)=\mathrm{P}_0-\mathrm{P}+2 \mathrm{P} +\mathrm{P}$

$ \begin{aligned} \mathrm{P}_t=\mathrm{P}_0+2 \mathrm{P} \\\\ \mathrm{P} =\frac{\mathrm{P}_t-\mathrm{P}_0}{2} \end{aligned} $

According to integrated rate law for the first order reaction:

$ \begin{aligned} &t =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\mathrm{P}_t} ~~~~...(i)\\\\ &t =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\mathrm{P}_0-\mathrm{P}} \\\\ &t =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\mathrm{P}_0-\frac{\left(\mathrm{P}_t-\mathrm{P}_0\right)}{2}} \\\\ &t =\frac{2.303}{k} \log \frac{2 \mathrm{P}_0}{2 \mathrm{P}_0-\mathrm{P}_t+\mathrm{P}_0} \\\\ &t =\frac{2.303}{k} \log \frac{2 \mathrm{P}_0}{3 \mathrm{P}_0-\mathrm{P}_t} \\\\ &t =\frac{1}{k} \ln \frac{2 \mathrm{P}_0}{3 \mathrm{P}_0-\mathrm{P}_t} \\\\ &\ln \left(3 \mathrm{P}_0-\mathrm{P}_t\right) =\ln \left(2 \mathrm{P}_0-\mathrm{P}_t\right)-k t ~~~~...(ii) \end{aligned} $

Comparing equation (ii) with equation of straight option line

$ \begin{aligned} & \ln \left(3 \mathrm{P}_0-\mathrm{P}_t\right) =\ln 2 \mathrm{P}_0-k t \\\\ &y =c+m x \end{aligned} $

Plot of $\ln \left(3 \mathrm{P}_0-\mathrm{P}_t\right)$ verses time $(t)$ is represented as:


Option (A) is correct.

(ii) For the first order reaction,


$ \mathrm{P}_t=\frac{\mathrm{P}_0}{3}+\frac{4 \mathrm{P}_0}{3}+\frac{2 \mathrm{P}_0}{3}=\frac{7 \mathrm{P}_0}{3} $

Substituting the value of $P_t$ in $e q$ (i)

$ \begin{aligned} t_{1 / 3} & =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\frac{7 \mathrm{P}_0}{3}} \\\\ t_{1 / 3} & =\frac{2.303}{k} \log \frac{3}{7} \end{aligned} $

$\mathrm{P}_{t_3}$ is independent of concentration of (A).

Option (B) is not correct.

(iii) At any time $t$, rate constant $(k)$ is independent of concentration of [A].

For example:

At $t_{y_3}$ :

$ k=\frac{2.303}{t_{1 / 3}} \times \log \frac{3}{7} $

Hence, Option (D) is correct.

Arrhenius equation is

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

where, A = Frequency factor

Taking into account orientation factor,

$k = P{Z_{AB}}{e^{ - {E_a}/RT}}$

where, P = steric factor, Z_AB = collision frequency

The value of steric factor lies between 0 and 1 predicted by Arrhenius equation. Thus, the experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation.

The activation energy of the reaction does not depend upon the value of the steric factor.

If P is very small, then catalyst is required to carry out the reaction at measurable rate.

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.

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.

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.