Chemical Kinetics and Nuclear Chemistry

For a first order reaction, the integrated rate law is

$ t=\frac{2.303}{k}\log\left(\frac{[A]_0}{[A]_t}\right) $

1) Time to reach $\left(\frac{1}{8}\right)^{\text{th}}$ of initial concentration

Here, $[A]_t=\frac{[A]_0}{8}$

$ t_{1/8}=\frac{2.303}{k}\log\left(\frac{[A]_0}{[A]_0/8}\right) =\frac{2.303}{k}\log(8) $

Now, $\log(8)=\log(2^3)=3\log 2=3\times 0.3=0.9$

So,

$ t_{1/8}=\frac{2.303}{k}\times 0.9 $

2) Time to reach $\left(\frac{1}{10}\right)^{\text{th}}$ of initial concentration

Here, $[A]_t=\frac{[A]_0}{10}$

$ t_{1/10}=\frac{2.303}{k}\log\left(\frac{[A]_0}{[A]_0/10}\right) =\frac{2.303}{k}\log(10) $

And $\log(10)=1$, so

$ t_{1/10}=\frac{2.303}{k}\times 1 $

3) Required ratio

$ \frac{t_{1/8}}{t_{1/10}}=\frac{\frac{2.303}{k}\times 0.9}{\frac{2.303}{k}\times 1}=0.9 $

So,

$ \frac{t_{1/8}}{t_{1/10}}\times 10=0.9\times 10=9 $

Correct option: B (9)

Given :

$ \mathrm{A} \rightarrow \mathrm{D} ; \Delta_{\mathrm{r}} \mathrm{H}=(+) \mathrm{ve}=\mathrm{E}_{\mathrm{D}}-\mathrm{E}_{\mathrm{A}} \therefore \mathrm{E}_{\mathrm{D}}>\mathrm{E}_{\mathrm{A}} $

Mechanism

$ \begin{aligned} & \mathrm{A} \rightarrow \mathrm{~B} ; \Delta_{\mathrm{r}} \mathrm{H}=(+) \mathrm{ve} \Rightarrow \mathrm{E}_{\mathrm{B}}>\mathrm{E}_{\mathrm{A}} \\ & \mathrm{~B} \rightarrow \mathrm{C} ; \Delta_{\mathrm{r}} \mathrm{H}=(-) \mathrm{ve} \Rightarrow \mathrm{E}_{\mathrm{C}}<\mathrm{E}_{\mathrm{B}} \\ & \mathrm{C} \rightarrow \mathrm{D} ; \Delta_{\mathrm{r}} \mathrm{H}=(-) \mathrm{ve} \Rightarrow \mathrm{E}_{\mathrm{D}}<\mathrm{E}_{\mathrm{C}} \end{aligned} $

Arrhenius equation (NCERT):

$k = A\,e^{-\frac{E_a}{RT}}$

Given: catalyst lowers activation energy by $10\,\text{kJ mol}^{-1} = 10000\,\text{J mol}^{-1}$ and frequency factor $A$ is same.

So,

$ \frac{k_{\text{cat}}}{k_{\text{uncat}}} = \frac{A e^{-\frac{E_{a,\text{cat}}}{RT}}}{A e^{-\frac{E_{a,\text{uncat}}}{RT}}} = e^{\frac{E_{a,\text{uncat}}-E_{a,\text{cat}}}{RT}} = e^{\frac{10000}{RT}} $

At $27^\circ\text{C}$, $T = 300\,\text{K}$ and $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$:

$\frac{10000}{RT}=\frac{10000}{8.314\times 300}\approx 4.009$

So,

$\log_{10}\left(\frac{k_{\text{cat}}}{k_{\text{uncat}}}\right) = \log_{10}\left(e^{4.009}\right) = \frac{4.009}{2.303} \approx 1.741$

Correct option: A (1.741)

From the graph, in the first $10$ minutes the concentration of $\mathrm{A}$ decreases from $0.05$ to $0.04$.

So, amount of $\mathrm{A}$ dissociated in $10$ minutes $=0.05-0.04=0.01$.

For the reaction $\mathrm{A}\rightarrow \mathrm{nB}$, when $0.01$ of $\mathrm{A}$ reacts, $\mathrm{B}$ formed $=0.01\times n$.

From the graph, in the same time interval, concentration of $\mathrm{B}$ formed becomes $0.03$. So we equate $0.01\times n$ to $0.03$ and find $n$.

$\begin{aligned} & \mathrm{A} \rightarrow \mathrm{nB} \\ & 0.05 \quad 0 \\ & 0.04 \quad 0.01 \times \mathrm{n} \\ & 0.01 \times \mathrm{n}=0.03 \\ & \mathrm{n}=3\end{aligned}$

At $t=10$ minutes (i.e., at 10:10 am)

For reaction II, $5\ \mathrm{Br}^-$ are consumed when the reaction proceeds once. So, the rate of reaction is related to the rate of decrease of $\left[\mathrm{Br}^-\right]$ as:

$ \begin{aligned} & \text { Rate of reaction }=-\frac{1}{5} \frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}=\frac{1}{5} \times\left(2 \times 10^{-4}\right) \\ & =4 \times 10^{-5} \end{aligned} $

Now consider reaction I: $\mathrm{A} \rightarrow$ products.

At $t=10$ minutes, both reactions have the same rate. So, the rate of reaction I at this time is also $4 \times 10^{-5}\ \mathrm{mol\ L^{-1}\ min^{-1}}$.

Since reaction I is first order, its rate law is:

Rate of reaction $=\mathrm{k}[\mathrm{A}]$

Given $\left[\mathrm{A}\right]=10^{-2}\ \mathrm{mol\ L^{-1}}$ at 10:10 am, we write:

Rate of reaction $=4 \times 10^{-5}=\mathrm{k}[\mathrm{A}]$

So,

$ \mathrm{k}=\frac{4 \times 10^{-5}}{10^{-2}}=4 \times 10^{-3} \mathrm{~min}^{-1} . $

Arrhenius equation (NCERT) is

$k=Ae^{-E_a/RT}$

and in base-10 logarithm form,

$\log k=\log A-\frac{E_a}{2.303\,R}\left(\frac{1}{T}\right)$

Now check each statement:

A. $e^{-E_a/RT}$ corresponds to fraction of molecules having kinetic energy less than $E_a$.

Incorrect. This factor represents the fraction of molecules having energy equal to or greater than $E_a$ (able to cross the energy barrier).

B. At a given temperature, lower the $E_a$, faster is the reaction.

Correct. From $k=Ae^{-E_a/RT}$, smaller $E_a \Rightarrow$ larger $e^{-E_a/RT} \Rightarrow$ larger $k$.

C. Increase in temperature by about $10^\circ\text{C}$ doubles the rate of reaction.

Correct as a common empirical rule mentioned in NCERT: for many reactions, rate approximately doubles for a $10^\circ\text{C}$ rise.

D. Plot of $\log k$ vs $\frac{1}{T}$ gives a straight line with slope $=-\frac{E_a}{R}$.

Incorrect. For $\log$ (base 10), slope is

$-\frac{E_a}{2.303\,R}$

The slope $-\frac{E_a}{R}$ is for plotting $\ln k$ vs $\frac{1}{T}$.

So, correct statements are B and C only.

Correct option: Option C.

For a first order reaction, the rate law is:

Rate $= k[\mathrm{A}]$

This means the rate depends only on the concentration of $\mathrm{A}$, not on the total volume taken.

Run 1: $100\,\mathrm{mL}$ of $10\,\mathrm{M}$ $\mathrm{A}$ is used, so $[\mathrm{A}] = 10\,\mathrm{M}$.

Run 2: $200\,\mathrm{mL}$ of $10\,\mathrm{M}$ $\mathrm{A}$ is used, but the concentration is still $[\mathrm{A}] = 10\,\mathrm{M}$ (only volume increased, not concentration).

Run 3: $100\,\mathrm{mL}$ of $10\,\mathrm{M}$ $\mathrm{A}$ is mixed with $100\,\mathrm{mL}$ of water, so the solution gets diluted to double the volume.

So the new concentration becomes:

$[\mathrm{A}] = \dfrac{10 \times 100}{100+100} = 5\,\mathrm{M}$

Since Rate $= k[\mathrm{A}]$, the rates are:

Run 1: Rate $= k(10)$

Run 2: Rate $= k(10)$

Run 3: Rate $= k(5)$

So, the correct variation is:

$\text{Rate}_1 = \text{Rate}_2 > \text{Rate}_3$

Let initial pressure of $A$ be $P_0 = 1$ bar in a $1\ \text{L}$ closed vessel at constant $T$.

For the reaction:

$A(g) \rightarrow B(g) + C(g)$

Suppose $x$ bar (in pressure-equivalent terms) of $A$ decomposes.

Then after time $t$:

• Pressure of $A$ left $= (1-x)$ bar

• Pressure of $B$ formed $= x$ bar

• Pressure of $C$ formed $= x$ bar

So total pressure:

$P_\text{total} = (1-x) + x + x = 1 + x$

Given after $100$ min, $P_\text{total} = 1.5$ bar:

$1+x = 1.5 \Rightarrow x = 0.5$

So partial pressure of $A$ after $100$ min:

$P_A = 1 - x = 1 - 0.5 = 0.5\ \text{bar}$

For a first order reaction:

$k = \frac{2.303}{t}\log\left(\frac{[A]_0}{[A]_t}\right)$

Here, pressure is proportional to concentration, so:

$k = \frac{2.303}{100}\log\left(\frac{1}{0.5}\right)$

$= \frac{2.303}{100}\log 2$

Given $\log 2 = 0.3$:

$k = \frac{2.303}{100}\times 0.3 = 6.9\times 10^{-3}\ \text{min}^{-1}$

Correct option: B) $6.9 \times 10^{-3}\ \text{min}^{-1}$

For the reversible reaction $ R \rightleftharpoons P $, the equilibrium constant is given by:

$K_{eq} = \frac{[P]}{[R]} = \frac{x}{[R]_0 - x} = \frac{k_f}{k_b} = \frac{1}{4}$

Here, $x$ is the amount of $R$ that gets converted to $P$ at equilibrium. Using the above relation:

$\frac{x}{[R]_0 - x} = \frac{1}{4}$

Multiplying both sides by $( [R]_0 - x )$:

$4x = [R]_0 - x$

$\Rightarrow x = \frac{[R]_0}{5}$

This means that, out of the initial concentration $[R]_0$, one-fifth has been converted to $P$ at equilibrium.

The equilibrium fraction of $P$ is:

$\frac{[P]}{[R]_0} = \frac{x}{[R]_0} = \frac{[R]_0/5}{[R]_0} = \frac{1}{5} = 0.2$

The equilibrium fraction of $R$ is:

$\frac{[R]}{[R]_0} = \frac{[R]_0 - x}{[R]_0} = \frac{[R]_0 - [R]_0/5}{[R]_0} = \frac{4}{5} = 0.8$

At the start of the reaction (time = 0):

$ \begin{aligned} &\frac{[R]}{[R]_0} = \frac{[R]_0}{[R]_0} = 1 \\ &\frac{[P]}{[R]_0} = \frac{0}{[R]_0} = 0 \end{aligned} $

As the reaction proceeds, the value of $[R]/[R]_0$ decreases from 1 to 0.8, while $[P]/[R]_0$ increases from 0 to 0.2. The graph representing these changes corresponds to the correct option shown above.

For a first-order reaction, the integrated rate law is given by:

$\ln \frac{[A]}{[A]_0} = -kt$

where:

$[A]_0$ is the initial concentration,

$[A]$ is the concentration at time $t$,

$k$ is the rate constant.

Let's find the times $t_1$ and $t_2$ corresponding to when the concentration becomes $\frac{1}{4}[A]_0$ and $\frac{1}{8}[A]_0$ respectively.

For concentration $\frac{1}{4}[A]_0$:

Substitute into the integrated rate law:

$\ln\left(\frac{1}{4}\right) = -kt_1$

Recognizing that $\ln\left(\frac{1}{4}\right) = -\ln 4$, we have:

$-\ln 4 = -kt_1 \quad \Longrightarrow \quad t_1 = \frac{\ln 4}{k}$

For concentration $\frac{1}{8}[A]_0$:

Substitute into the integrated rate law:

$\ln\left(\frac{1}{8}\right) = -kt_2$

Recognizing that $\ln\left(\frac{1}{8}\right) = -\ln 8$, we have:

$-\ln 8 = -kt_2 \quad \Longrightarrow \quad t_2 = \frac{\ln 8}{k}$

Now, to find the ratio $\frac{t_1}{t_2}$:

$ \frac{t_1}{t_2} = \frac{\frac{\ln 4}{k}}{\frac{\ln 8}{k}} = \frac{\ln 4}{\ln 8} $

We can simplify further by expressing the logarithms in terms of $\ln 2$:

$\ln 4 = \ln (2^2) = 2\ln 2$

$\ln 8 = \ln (2^3) = 3\ln 2$

Thus,

$ \frac{t_1}{t_2} = \frac{2\ln 2}{3\ln 2} = \frac{2}{3} $

So, the ratio $\frac{t_1}{t_2}$ is $\frac{2}{3}$.

This corresponds to Option D.

$\begin{aligned} & \mathrm{P}_{\mathrm{t}}=\mathrm{P}^{\mathrm{o}}+\mathrm{x} \Rightarrow \mathrm{x}=\mathrm{P}_{\mathrm{t}}-\mathrm{P}^{\mathrm{o}}=\mathrm{P}_{\mathrm{t}}-\frac{\mathrm{P}_{\infty}}{2} \\ & \mathrm{P}_{\infty}=2 \mathrm{P}^{\mathrm{o}} \Rightarrow \mathrm{P}^{\mathrm{o}}=\frac{\mathrm{P}_{\infty}}{2} \\ & \mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{P}^{\mathrm{o}}}{\mathrm{P}^o-\mathrm{x}} \\ & \mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{P}_{\infty}}{2\left(\mathrm{P}_{\infty}-\mathrm{P}_{\mathrm{t}}\right)} \end{aligned}$

*Before applying medicine

$\frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{K}[\mathrm{A}]$ (First order growth) (Rate law)

$\frac{\mathrm{A}}{\mathrm{~A}_0}=\frac{\mathrm{N}}{\mathrm{~N}_0}=\mathrm{e}^{\mathrm{Kt}}$

*After applying medicine

$\mathrm{y=K_x{ }^2}$ Parabola

$\begin{aligned} &\begin{aligned} & \mathrm{P}_{\infty}=3 \mathrm{P}_0=240 \\ & \quad \mathrm{P}_0=80 \mathrm{~mm} \text { of } \mathrm{Hg} \\ & \mathrm{Kt}=\ln \left(\frac{\mathrm{P}_{\infty}-\mathrm{P}_0}{\mathrm{P}_{\infty}-\mathrm{Pt}}\right) \\ & \mathrm{K} \times 10=\ln \left(\frac{240-80}{240-160}\right) \\ & \mathrm{K}=\frac{\ln 2}{10}=0.0693 \mathrm{~min}^{-1} \end{aligned}\\ &\text { Option (3) is incorrect } \end{aligned}$

$\begin{aligned} & \mathrm{K}=\mathrm{A} \mathrm{e}^{-\mathrm{EaRT}} \\ & \operatorname{logk}=\log \mathrm{A}-\frac{\mathrm{Ea}}{2.303 \mathrm{RT}} \end{aligned}$

For graph between logk with $\frac{1}{\mathrm{~T}}$

$\mid \text { Slope of curve } \left\lvert\,=\frac{\mathrm{Ea}}{2.303 \mathrm{R}}\right.$

From given graph

Magnitude of slope $\Rightarrow(2)>(1)>(3)$

Hence $\mathrm{Ea}_2>\mathrm{Ea}_1>\mathrm{Ea}_3$

For zero order reaction

$\begin{aligned} & \text { Half life }=\frac{\mathrm{A}_{\mathrm{o}}}{2 \mathrm{k}} \\ & 60 \min =\frac{2}{2 \mathrm{k}} \\ & \mathrm{k}=\frac{1}{60} \mathrm{M} / \min \end{aligned}$

Now

$\begin{aligned} & A_t=A_o-k t \\ & t=\frac{A_o-A_t}{k} \\ &=\frac{0.5-0.25}{1 / 60} \\ & 0.25 \times 60 \\ & t=15 \mathrm{~min} \end{aligned}$

$\mathrm{A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{AB}$

$\begin{aligned} &\begin{aligned} & \mathrm{E}_{\mathrm{f}}=180 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{E}_{\mathrm{b}}=200 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta \mathrm{H}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}=-20 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\\ &\text { In presence of catalyst : }\\ &\begin{aligned} & \mathrm{E}_{\mathrm{f}}=180-100=80 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{E}_{\mathrm{b}}=200-100=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \end{aligned}$

$\mathrm{r}_1=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}$

Now $A$ is doubled \& B is halved in concentration

$\Rightarrow \mathrm{r}_2=\mathrm{k} 2^{\mathrm{n}}[\mathrm{~A}]^{\mathrm{n}} \cdot \frac{[\mathrm{~B}]^{\mathrm{m}}}{2^{\mathrm{m}}}$

Now $\frac{r_2}{r_1}=2^{(n-m)}$

Let’s examine each statement in the light of

$k = A\,e^{-\frac{E_a}{RT}}\,. $

A. “Arrhenius holds true only for an elementary homogeneous reaction.”

• False – it’s an empirical law and is routinely used (and found valid) for complex, composite or even heterogeneous processes.

B. “The unit of $A$ is the same as that of $k$.”

• True – since $e^{-E_a/RT}$ is dimensionless, $A$ must carry whatever units $k$ has.

C. “At a given temperature, a low activation energy means a fast reaction.”

• True – smaller $E_a$ makes $\exp(-E_a/RT)$ larger, hence a larger $k$.

D. “$A$ and $E_a$ as used in Arrhenius depend on temperature.”

• False – in the simple Arrhenius model both are treated as constants (over modest $T$-ranges).

E. “When $E_a \gg RT$, $A$ and $E_a$ become interdependent.”

• False – although in some data sets you observe a correlation (compensation effect), the basic Arrhenius form still treats them as independent parameters.

Therefore the only correct statements are B and C.

Answer: B and C only.

The initial conditions are:

$[A]_0 = 8[B]_0$

Half-life of A, $\left[\text{t}_{1/2}\right]_{A} = 10 \text{ min}$

Half-life of B, $\left[\text{t}_{1/2}\right]_{B} = 40 \text{ min}$

Both reactants follow first-order kinetics, and we want to find the time, $t$, when $[A]_t = [B]_t$.

Step-by-step Derivation:

Write the First-order Rate Equation:

For first-order reactions, the concentration $[X]_t$ at time $t$ is given by:

$ [X]_t = [X]_0 e^{-k_X t} $

where $k_X$ is the rate constant for substance $X$.

Express Half-life through Rate Constant:

For first-order reactions, the rate constant $k$ is related to the half-life $\text{t}_{1/2}$ by:

$ k = \frac{\ln 2}{\text{t}_{1/2}} $

Substituting the known half-lives:

$k_A = \frac{\ln 2}{10}$

$k_B = \frac{\ln 2}{40}$

Set Up the Equality from the Condition $[A]_t = [B]_t$:

$ [A]_0 e^{-k_A t} = [B]_0 e^{-k_B t} $

By inserting the initial condition $[A]_0 = 8[B]_0$, it becomes:

$ 8[B]_0 e^{-k_A t} = [B]_0 e^{-k_B t} $

Simplify and Solve for $t$:

Divide both sides by $[B]_0$:

$ 8 e^{-k_A t} = e^{-k_B t} $

Taking the natural logarithm of both sides:

$ \ln 8 = -k_A t + k_B t $

Replace $k_A$ and $k_B$ with their expressions:

$ \ln 8 = \left(\frac{\ln 2}{10} - \frac{\ln 2}{40}\right) t $

Simplify Further:

$ \ln 8 = \ln 2 \left(\frac{1}{10} - \frac{1}{40}\right) t $

Simplify $\left(\frac{1}{10} - \frac{1}{40}\right)$ to get:

$ \frac{1}{10} - \frac{1}{40} = \frac{4 - 1}{40} = \frac{3}{40} $

So,

$ \ln 8 = \ln 2 \times \frac{3}{40} \times t $

Final Expression for $t$:

$ t = \frac{\ln 8}{\ln 2 \times \frac{3}{40}} $

Calculate the value of $t$ using $\ln 8 = 3 \ln 2$:

$ t = \frac{3 \ln 2}{\ln 2 \times \frac{3}{40}} = \frac{40}{1} = 40 \text{ min} $

Therefore, after 40 minutes, the concentration of both reactants A and B will be the same.

$\mathrm{A} \rightarrow \mathrm{~B}$ Slow

$\Delta \mathrm{H}=+$ ve $\quad \mathrm{E}_{\mathrm{a}_1} \rightarrow \mathrm{High}$

$\mathrm{B} \rightarrow \mathrm{C}$ Fast

$\begin{aligned} & \Delta \mathrm{H}=-\mathrm{ve} \quad \mathrm{E}_{\mathrm{a}_2} \rightarrow \mathrm{Low} \\ & \mathrm{C} \rightarrow \mathrm{D} \\ & \Delta \mathrm{H}=-\mathrm{ve} \quad \mathrm{E}_{\mathrm{a}_3} \rightarrow \mathrm{Low} \end{aligned}$

Original concentration, $a = 16$ mg/mL

Concentration at time

t = 12 months, $a - x = 4$ mg/mL

For first order kinetics, the equation for late constant

$k = {{2.303} \over t}\log {a \over {a - x}}$

a $\to$ Initial concentration

$a-x\to$ Concentration at time, t

$ = {{2.303} \over {12\,months}}\log {{16\,mg/mL} \over {4\,mg/mL}}$

$ = {{2.303} \over {12}}\log 4$

$ = 0.1155\,month{s^{ - 1}}$

Drug is ineffective after 50% decomposition. So, the expiry time is after 50% decomposition. The time at which 50% decomposition occurs in a reaction is known as half life $({t_{1/2}})$. So, ${t_{1/2}}$ is the expiry time of the drug.

${t_{1/2}}$ formula is

${t_{1/2}} = {{0.693} \over k}$

Substitute k = 0.1155 months$^{-1}$,

${t_{1/2}} = {{0.613} \over {0.1155}}$ months$^{-1}$

= 6 months

Expiry time can also be calculate as,

$t = {{2.303} \over k}\log {a \over {a - x}}$

Initial is taken as 100%

The percent at expiry time is 50%

$ = {{2.303} \over {0.1155\,month{s^ - }}} \times \log {{100\% } \over {50\% }}$

$ = {{2.303} \over {0.1155}} \times \log {{100} \over {50}}$

$ = {{2.303} \over {0.1155}} \times 0.301$

$ = 6.002$

= 6 months

Correct answer is option (3) 6

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 $=k_2[A]\left[B_2\right]$

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

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

$\begin{aligned} & \mathrm{R}_1=\mathrm{K}[\mathrm{~A}]^1[\mathrm{~B}]^1 \\ & \mathrm{R}_1=\mathrm{K}\left[\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{~V}}\right]^1\left[\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{~V}}\right]^1 \\ & \mathrm{R}_2=\mathrm{K}\left[\frac{3 \mathrm{n}_{\mathrm{A}}}{\mathrm{~V}}\right]^1\left[\frac{3 \mathrm{n}_{\mathrm{B}}}{\mathrm{~V}}\right]^1 \\ & \mathrm{R}_2=9 \mathrm{R}_1 \end{aligned}$

Because no. of bacteria initial $=\mathrm{N}_0$ and No. of bacteria at any time $\mathrm{t}=\mathrm{N}$

Since bacterial growth is given as

$\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{\mathrm{Kt}}$

Where K = growth constant for bacterial growth

$\begin{array}{ll} \mathrm{t}_{1 / 2} \propto \frac{1}{\mathrm{~A}_0^{\mathrm{n}-1}} \\ \frac{\left(\mathrm{t}_{1 / 2}\right)_1}{\left(\mathrm{t}_{1 / 2}\right)_2}=\frac{\left(\mathrm{A}_0\right)_2^{\mathrm{n}-1}}{\left(\mathrm{~A}_0\right)_1^{\mathrm{n}-1}} & \\ \frac{200}{100}=\left(\frac{0.025}{0.100}\right)^{\mathrm{n}-1} & \mathrm{n}-1=-\frac{1}{2} \\ 2=\left(\frac{1}{4}\right)^{\mathrm{n}-1} & \mathrm{n}=\frac{1}{2}(\text { order }) \\ \Rightarrow \mathrm{t}_{1 / 2} \propto \sqrt{\mathrm{~A}_0} & \\ \frac{200}{\mathrm{t}_{1 / 2}}=\frac{(0.1)^{1 / 2}}{(1)^{1 / 2}} & \text { when } \mathrm{A}_0=1 \mathrm{M} \\ \mathrm{t}_{1 / 2}=200 \sqrt{10} \text { min } & \end{array}$

*Ist order kinetics have $t_{1 / 2}$ independent of their concentration. So upon changing the concentration $t_{1 / 2}$ should not change for first order reaction.

$\begin{aligned} & \frac{200}{t_{1 / 2}}=\frac{(0.1)^{1 / 2}}{(1.6)^{1 / 2}} \quad \text { when } \mathrm{A}_0=1.6 \mathrm{M} \\ & \mathrm{t}_{1 / 2}=800 \mathrm{~min} \end{aligned}$

Half life ( $\mathrm{t}_{1 / 2}$ ) of a first order reaction does not depend on initial concentration of the reactant $\left[\mathrm{R}_0\right]$. Therefore Statement I is true. Rate constant (k) of a first order reaction is given by

$ \begin{aligned} & \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{R}_0\right]}{[\mathrm{R}]} \\ & \log \frac{[\mathrm{R}]}{\left[\mathrm{R}_0\right]}=-\left(\frac{\mathrm{k}}{2.303}\right) \mathrm{t} \end{aligned} $

Plot of $\log \frac{[R]}{\left[R_0\right]}$ versus time is linear having

$ \text { slope }=-\frac{\mathrm{k}}{2.303} $

$\therefore$ Statement II is false.

$\begin{aligned} & \quad x=\frac{P_0}{2} \\ & P_{\text {total }}=P_0-\frac{P_0}{2}+P_0+\frac{P_0}{4}=\frac{7}{4} P_0 \\ & \text { Option (4) } \end{aligned}$

$\begin{aligned} &\text { For zero order reaction : } \mathrm{A} \rightarrow \mathrm{P}\\ &\begin{aligned} & \text { Rate }=\mathrm{k} \\ & {[\mathrm{~A}]_{\mathrm{t}}=\mathrm{a}_0-\mathrm{kt}} \end{aligned} \end{aligned}$

$E_1$ represents the activation energy for the backward reaction.

$E_1 + E_2$ is the activation energy for the forward reaction.

The product has more energy than the reactant, indicating that the product is less stable compared to the reactant.

This information is crucial in determining the relative stability of reactants and products and the required energy for the reactions to proceed in either direction.

Answer: Option B is not true.

Explanation:

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

$ t_{1/2} \;=\; \frac{\ln 2}{k} \;\;=\;\ln 2 \times \frac{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.

To determine the concentration of $ \mathrm{B} $ when the rate of formation of $ \mathrm{B} $ is set to zero, let's consider the reaction sequence:

$ \mathrm{A} \xrightarrow{\mathrm{K}_1} \mathrm{B} \xrightarrow{\mathrm{K}_2} \mathrm{C} $

The rate of formation of $ \mathrm{B} $ from $ \mathrm{A} $ is given by:

$ \text{Rate of formation of } \mathrm{B} = \mathrm{K}_1[\mathrm{A}] $

The rate of consumption of $ \mathrm{B} $ to form $ \mathrm{C} $ is given by:

$ \text{Rate of consumption of } \mathrm{B} = \mathrm{K}_2[\mathrm{B}] $

At steady state, where the rate of formation of $ \mathrm{B} $ is set to zero, the rate of formation of $ \mathrm{B} $ is equal to the rate of its consumption. Therefore, we have:

$ \mathrm{K}_1[\mathrm{A}] = \mathrm{K}_2[\mathrm{B}] $

Solving for $ [\mathrm{B}] $, we get:

$ [\mathrm{B}] = \frac{\mathrm{K}_1}{\mathrm{K}_2} [\mathrm{A}] $

Therefore, the correct concentration of $ \mathrm{B} $ is:

Option D: $ \left(\frac{\mathrm{K}_1}{\mathrm{K}_2}\right)[\mathrm{A}] $

$\begin{array}{llll} \mathrm{A} \rightarrow & \mathrm{B} & + & \mathrm{C} \\ \mathrm{P}_{\mathrm{i}} & 0 & & 0 \\ \mathrm{P}_{\mathrm{i}}-\mathrm{x} & \mathrm{x} & & \mathrm{x} \end{array}$

$\begin{aligned} & \mathrm{P_t=P_i+x} \\ & \mathrm{P_i-x=P_i-P_t+P_i} \\ & \mathrm{=2 P_i-P_t} \\ & \mathrm{K=\frac{2.303}{t} \log \frac{P_i}{2 P_i-P_t}} \end{aligned}$

rate = $\frac{k[x]}{X_s+[X]}$

Case-1 : $[X] > > {X_s};[X] + {X_s} \approx [X]$

rate $ = {{k[X]} \over {[X]}} = k$ (Zero order w.r.t. X)

I $\to$ P, S

Case 2 : $[X] < < {X_s};[X] + {X_s} \approx {X_s}$

$\therefore$ rate $ = {{k[X]} \over {{X_s}}} = k'[X]$ (1^st order w.r.t. X)

$\therefore$ I $\to$ Q, T

Case-3 : $[X] \approx {X_s}$

rate $ = {{k[X]} \over {{X_s} + [X]}}$

In this case curve-R given in List-II will match.

$\therefore$ I $\to$ P, Q, R, S, T (The graph of half-life should start from origin)

(II) rate $ = {{k[X]} \over {{X_s} + [X]}}$

$\because$ $[X] < < {X_s}$

$\therefore$ ${X_s} + [X] \approx {X_s}$

$\therefore$ Rate $ = {{k[X]} \over {{X_s}}} = k'[X]$ (1^st order w.r.t. X)

$\therefore$ II $\to$ Q, T

(III) rate $ = {{k[X]} \over {{X_s} + [X]}}$

$\because$ $[X] > > {X_s}$

$\therefore$ ${X_s} + [X] \approx [X]$

$\therefore$ rate $ = {{k[X]} \over {[X]}} = k$ (Zero order w.r.t. X)

$\therefore$ III $\to$ P, S

(IV) rate $ = {{k{{[X]}^2}} \over {{X_s} + [X]}}$

$\because$ $[X] > > {X_s}$

$\therefore$ ${X_s} + [X] \approx [X]$

$\therefore$ rate $ = {{k{{[X]}^2}} \over {[X]}} = k[X]$ (1^st order w.r.t. X)

$\therefore$ IV $\to$ Q, T