First Order reaction Questions in English

(C) Given that the initial concentration $[A]_0 = 0.08 \text{ mol dm}^{-3}$.
We are given the ratio $\frac{[A]_0}{[A]_t} = 5.00$.
To find the remaining concentration $[A]_t$,we rearrange the equation:
$[A]_t = \frac{[A]_0}{5.00} = \frac{0.08 \text{ mol dm}^{-3}}{5.00} = 0.016 \text{ mol dm}^{-3}$.
Thus,the concentration remaining after $40 \text{ minutes}$ is $0.016 \text{ mol dm}^{-3}$.

(B) The given reaction is a first-order reaction because the unit of the rate constant is $\text{minute}^{-1}$.
For a first-order reaction,the half-life $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 4.7672 \text{ minute}^{-1}$.
$t_{1/2} = \frac{0.693}{4.7672 \text{ minute}^{-1}} = 0.1454 \text{ minute}$.

(B) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given that $60 \%$ of the reactant is converted,the remaining concentration $[A]_t$ is $100 - 60 = 40 \%$ of the initial concentration $[A]_0$.
So,$[A]_0 = 100$ and $[A]_t = 40$.
Substituting the values: $k = \frac{2.303}{45} \log \frac{100}{40}$
$k = \frac{2.303}{45} \log(2.5)$
$k = \frac{2.303}{45} \times 0.3979$
$k \approx 0.02036 \ minute^{-1} \approx 0.0204 \ minute^{-1}$.

(B) For a first-order reaction, the half-life is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given the rate constant $k = 4.2 \times 10^{-2} \text{ day}^{-1}$.
Substituting the value of $k$:
$t_{1/2} = \frac{0.693}{4.2 \times 10^{-2}} = \frac{0.693}{0.042} = 16.5 \text{ days}$.

(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
For $90 \%$ completion,$[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$. Thus,$x = \frac{2.303}{k} \log_{10} \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{k} \log_{10} 10 = \frac{2.303}{k} \times 1$.
For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$. Thus,$t_{99.9 \%} = \frac{2.303}{k} \log_{10} \frac{[A]_0}{0.001[A]_0} = \frac{2.303}{k} \log_{10} 1000 = \frac{2.303}{k} \times 3$.
Comparing the two expressions,$t_{99.9 \%} = 3 \times \left( \frac{2.303}{k} \right) = 3x \ \text{minute}$.

(A) For a first-order reaction,the rate is given by $\text{Rate} = k[A]$.
Given $\text{Rate} = 0.00352 \ mol \ L^{-1} \ minute^{-1}$ and $[A] = 0.01 \ mol \ L^{-1}$.
Substituting these values: $0.00352 = k \times 0.01$.
Therefore,$k = \frac{0.00352}{0.01} = 0.352 \ minute^{-1}$.
The half-life period $(t_{1/2})$ for a first-order reaction is calculated as $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{0.352} \approx 1.969 \ minute$.

(B) For a first-order reaction,the half-life $(t_{1/2})$ is related to the rate constant $(k)$ by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 2.772 \times 10^{-3} \ s^{-1}$,we substitute this value into the equation:
$t_{1/2} = \frac{0.693}{2.772 \times 10^{-3} \ s^{-1}}$
$t_{1/2} = \frac{693}{2.772} \ s$
$t_{1/2} = 250 \ s$

(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
For $90 \%$ completion,$[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$. So,$k = \frac{2.303}{t} \log \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{t} \log(10) = \frac{2.303}{t}$.
Thus,$t = \frac{2.303}{k}$.
For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$. Let the time be $t'$.
$t' = \frac{2.303}{k} \log \frac{[A]_0}{0.001[A]_0} = \frac{2.303}{k} \log(1000) = \frac{2.303}{k} \times 3$.
Substituting $t = \frac{2.303}{k}$,we get $t' = 3t$.

(D) For a first-order reaction,the half-life is given by the formula: $t_{1/2} = \frac{0.693}{K}$.
Given the rate constant $K = 1 \times 10^{-3} \ sec^{-1}$.
Substituting the value: $t_{1/2} = \frac{0.693}{1 \times 10^{-3}} = 693 \ sec$.
To convert the time into minutes,divide by $60$: $t_{1/2} = \frac{693}{60} = 11.55 \ min$.

(A) For a first-order reaction,the integrated rate equation is $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$.
At half-life,$t = t_{1/2}$ and $[R] = \frac{[R]_0}{2}$.
Substituting these values: $k = \frac{2.303}{t_{1/2}} \log 2$.
Since $\log 2 \approx 0.3010$,we get $k = \frac{2.303 \times 0.3010}{t_{1/2}} = \frac{0.693}{t_{1/2}}$.
Therefore,the correct relation is $k \times t_{1/2} = 0.693$.

(A) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given $[A]_0 = 0.8 \ mol \ dm^{-3}$,$[A]_t = 0.2 \ mol \ dm^{-3}$,and $t = 12 \ hour$.
$k = \frac{2.303}{12} \log \frac{0.8}{0.2} = \frac{2.303}{12} \log 4 = \frac{2.303 \times 0.602}{12} \approx 0.1155 \ hour^{-1}$.
The half-life $t_{1/2}$ is given by $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{0.1155} = 6 \ hour$.
Alternatively,since the concentration decreases by a factor of $4$ $(0.8$ $\rightarrow 0.4$ $\rightarrow 0.2)$,it corresponds to $2$ half-lives. Thus,$2 \times t_{1/2} = 12 \ hour$,which implies $t_{1/2} = 6 \ hour$.

(A) For a first order reaction,the rate equation is: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
Here,$[A]_0 = 100$ and $[A]_t = 100 - 90 = 10$.
Substituting the values: $t = \frac{2.303}{k} \log \frac{100}{10}$
$t = \frac{2.303}{k} \log 10$
Since $\log 10 = 1$,we get $t = \frac{2.303}{k}$.

(A) For a first-order reaction,the rate constant $K$ is given by the formula:
$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given: $[A]_0 = 1.6 \ M$,$[A]_t = 0.4 \ M$,and $t = 12 \ hours$.
Substituting the values:
$K = \frac{2.303}{12} \log \frac{1.6}{0.4}$
$K = \frac{2.303}{12} \log 4$
Since $\log 4 = 2 \log 2 \approx 2 \times 0.3010 = 0.6020$:
$K = \frac{2.303 \times 0.6020}{12} \approx 0.1155 \ hour^{-1}$
Rounding to three decimal places,$K \approx 0.116 \ hour^{-1}$.

(D) For a first order reaction,the time $t$ required for completion is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $90 \%$ completion,$[A]_t = [A]_0 - 0.9[A]_0 = 0.1[A]_0$. Thus,$t = \frac{2.303}{k} \log \frac{[A]_0}{0.1[A]_0} = \frac{2.303}{k} \log 10 = \frac{2.303}{k}$.
For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$. Thus,$t_{99.9 \%} = \frac{2.303}{k} \log \frac{[A]_0}{0.001[A]_0} = \frac{2.303}{k} \log 10^3 = 3 \times \frac{2.303}{k}$.
Substituting $t$ into the equation,we get $t_{99.9 \%} = 3t$.

(A) For a first-order reaction,the half-life period $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given that the rate constant $k = 0.02 \ min^{-1}$,
$t_{1/2} = \frac{0.693}{0.02} \ min = 34.65 \ min$.

(C) For a first-order reaction,the integrated rate equation is: $\ln [A]_t = \ln [A]_0 - Kt$
This can be rearranged as: $\ln \frac{[A]_0}{[A]_t} = Kt$
Converting to base $10$: $\log_{10} \frac{[A]_0}{[A]_t} = \frac{K}{2.303} \cdot t$
Comparing this with the equation of a straight line $y = mx$,the slope $m = \frac{K}{2.303}$.
Given the slope is $1 \times 10^{-3}$,we have: $\frac{K}{2.303} = 1 \times 10^{-3}$
Therefore,$K = 2.303 \times 10^{-3}$.

(D) For a first-order reaction,the rate constant $K$ is given by the formula: $K = \frac{0.693}{t_{1/2}}$.
Given,$t_{1/2} = 2.5 \ hours$.
Convert the half-life into seconds: $t_{1/2} = 2.5 \times 60 \times 60 \ sec = 9000 \ sec$.
Now,substitute the value into the formula: $K = \frac{0.693}{9000 \ sec} = 7.7 \times 10^{-5} \ sec^{-1}$.

(A) The half-life period $t_{1/2}$ for a first-order reaction is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given the rate constant $k = 0.693 \times 10^{-2} \ min^{-1}$.
Substituting the value of $k$ in the formula:
$t_{1/2} = \frac{0.693}{0.693 \times 10^{-2}} \ min$
$t_{1/2} = 10^2 \ min = 100 \ min$
To convert the time into seconds,we multiply by $60 \ s/min$:
$t_{1/2} = 100 \times 60 \ s = 6000 \ s$.

(D) For a $1^{st}$ order reaction:
$k = \frac{2.303}{t} \log \frac{a_0}{a_t}$
Given:
$a_0 = 20 \ mmol$
$a_t = 10 \ mmol$
$t = 1.151 \ min$
Substituting the values:
$k = \frac{2.303}{1.151} \log \left( \frac{20}{10} \right)$
$k = \frac{2.303}{1.151} \times \log 2$
$k = \frac{2.303 \times 0.3010}{1.151}$
$k = 0.60 \ min^{-1}$
Alternatively,since the concentration reduces to half ($20 \ mmol$ to $10 \ mmol$),the time taken is the half-life $(t_{1/2})$:
$t_{1/2} = 1.151 \ min$
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1.151} \approx 0.60 \ min^{-1}$

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Substituting the given value,$k = \frac{0.693}{6.93} = 0.1 \ hour^{-1}$.
For $80 \%$ completion,the remaining concentration $[A]_t = [A]_0 - 0.80[A]_0 = 0.20[A]_0$.
The time $t$ is calculated using the formula $t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$.
$t = \frac{2.303}{0.1} \log_{10} \frac{100}{20} = 23.03 \times \log_{10} 5$.
Using $\log_{10} 5 \approx 0.699$,we get $t = 23.03 \times 0.699 \approx 16.10 \ hours$.

(A) For a first-order reaction,the rate law is given by:
$Rate = k[A]^1$
Comparing this with the equation of a straight line passing through the origin,$y = mx$,where $y = \text{Rate}$,$x = [A]$,and $m = \text{slope}$:
$Rate = k[A]$
Thus,the slope $m = k$.

(B) For a first order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given: $[A]_t = \frac{[A]_0}{8}$ and $t = 23.03 \ min$.
Substituting the values: $k = \frac{2.303}{23.03} \log \frac{[A]_0}{[A]_0/8} = 0.1 \log 8 = 0.1 \times 3 \log 2 = 0.3 \times 0.3010 = 0.0903 \ min^{-1}$.
The half-life period $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{0.0903} \approx 7.67 \ min \approx 7.7 \ min$.

(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t}$.
Given $[A]_0 = 1.0 \ M$,$[A]_t = 0.25 \ M$,and $t = 10 \ hours$.
$k = \frac{1}{10} \ln \frac{1.0}{0.25} = \frac{\ln 4}{10} \ h^{-1}$.
The half-life $t_{1/2}$ is given by $t_{1/2} = \frac{\ln 2}{k}$.
Substituting the value of $k$: $t_{1/2} = \frac{\ln 2}{(\ln 4) / 10} = \frac{10 \ln 2}{2 \ln 2} = 5 \ hours$.

(C) For a $1^{st}$ order reaction,the integrated rate equation is given by:
$2.303 \log \left(\frac{[A]_0}{[A]_t}\right) = kt$
Rearranging this equation,we get:
$\log \left(\frac{[A]_0}{[A]_t}\right) = \left(\frac{k}{2.303}\right) t$
Comparing this with the linear equation $y = mx + c$,where $y = \log \left(\frac{[A]_0}{[A]_t}\right)$,$x = t$,$m = \frac{k}{2.303}$,and $c$ is the intercept.
Since there is no constant term added,the intercept $c = 0$.

(D) The given rate constant $k = 2 \times 10^{-2} \ s^{-1}$ indicates a first-order reaction.
The integrated rate equation for a first-order reaction is $\ln \frac{[A]_0}{[A]_t} = kt$,which can be written as $2.303 \log \frac{[A]_0}{[A]_t} = kt$.
Given $[A]_0 = 1.0 \ mol \ dm^{-3}$,$t = 100 \ s$,and $k = 2 \times 10^{-2} \ s^{-1}$.
Substituting the values: $\log \frac{1}{[A]_t} = \frac{kt}{2.303} = \frac{2 \times 10^{-2} \times 100}{2.303} = \frac{2}{2.303} \approx 0.868$.

(D) For a first-order reaction,the rate constant $K$ is given by the formula:
$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given: $[A]_0 = 0.08 \ mol$,$[A]_t = 0.02 \ mol$,$t = 23.03 \ min$.
Substituting the values:
$K = \frac{2.303}{23.03} \log \frac{0.08}{0.02}$
$K = 0.1 \times \log 4$
Since $\log 4 = 2 \log 2 \approx 2 \times 0.3010 = 0.6020$:
$K = 0.1 \times 0.6020 = 0.0602 \ min^{-1}$.

(B) For the reaction: $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$
At $t = 0$: Pressure of $A = P_{i}$,$B = 0$,$C = 0$.
At $t = t$: Pressure of $A = P_{i} - x$,$B = x$,$C = x$.
Total pressure $P = (P_{i} - x) + x + x = P_{i} + x$.
Therefore,$x = P - P_{i}$.
The pressure of $A$ at time $t$ is $P_{A} = P_{i} - x = P_{i} - (P - P_{i}) = 2P_{i} - P$.
For a first-order reaction,the integrated rate law is $k = \frac{2.303}{t} \log_{10} \frac{P_{i}}{P_{A}}$.
Substituting $P_{A}$,we get $k = \frac{2.303}{t} \log_{10} \frac{P_{i}}{2P_{i} - P}$.

(A) For a $1^{st}$ order reaction,the half-life $(t_{1/2})$ is calculated using the rate constant $(k)$:
$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$
Given $k = 0.1989 \ min^{-1}$,
$t_{1/2} = \frac{0.693}{0.1989} \approx 3.48 \ min$.

(C) For a first-order reaction,the integrated rate equation is:
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
At half-life $(t_{1/2})$,the concentration of the reactant becomes half of its initial concentration,i.e.,$[A]_t = \frac{[A]_0}{2}$.
Substituting this in the equation:
$t_{1/2} = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0 / 2} = \frac{2.303}{k} \log 2$
Since $\log 2 \approx 0.3010$,we get:
$t_{1/2} = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k}$

(B) For a first-order reaction, the rate constant $k$ is given by:
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{20} \text{ min}^{-1}$
The time required for a first-order reaction is:
$t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]_t}\right)$
Given $[A]_t = \frac{[A]_0}{10}$, so $\frac{[A]_0}{[A]_t} = 10$.
$t = \frac{2.303 \times 20}{0.693} \log(10) = \frac{46.06}{0.693} \times 1 \approx 66.46 \text{ min}$.
The closest option provided is $66.56 \text{ min}$.

(A) For a first order reaction,the amount of reactant remaining after $87.5 \%$ conversion is $100 \% - 87.5 \% = 12.5 \%$.
Let the initial concentration $[A]_0 = 100$ and the concentration at time $t = 15 \ min$ be $[A]_t = 12.5$.
The number of half-lives $n$ can be calculated as $12.5 = 100 \times (1/2)^n$,which gives $(1/2)^n = 1/8$,so $n = 3$.
Since $t = n \times t_{1/2}$,we have $15 = 3 \times t_{1/2}$,which means $t_{1/2} = 5 \ min$.
The rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5} \ min^{-1}$.

(B) For a first-order reaction,the rate constant $k$ is given by:
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{6.0 \ h} = 0.1155 \ h^{-1}$
Given initial concentration $[A]_0 = 0.4 \ M$ and final concentration $[A]_t = 0.12 \ M$.
Using the integrated rate equation for a first-order reaction:
$t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$
$t = \frac{2.303}{0.1155 \ h^{-1}} \times \log_{10} \left( \frac{0.4}{0.12} \right)$
$t = \frac{2.303}{0.1155} \times \log_{10} (3.333)$
$t = 19.939 \times 0.5228 \approx 10.42 \ h$

(D) The rate constant $(k)$ of a reaction is a characteristic property that depends only on temperature and the nature of the reactants.
It is independent of the concentration of the reactants.
Therefore,even if the concentration of $A$ is reduced to half,the rate constant remains unchanged at $0.25 \ s^{-1}$.

(A) For a first order reaction,the rate law is given by: $\text{Rate} = k[N_{2}O_{5}]$
Given,$k = 6.2 \times 10^{-4} \ s^{-1}$ and $[N_{2}O_{5}] = 1.25 \ mol \ L^{-1}$.
Substituting the values:
$\text{Rate} = (6.2 \times 10^{-4} \ s^{-1}) \times (1.25 \ mol \ L^{-1})$
$\text{Rate} = 7.75 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$

(A) For a first-order reaction,the integrated rate equation is given by: $t = \frac{2.303}{k} \log \frac{[A]_{0}}{[A]_{t}}$
Given: $[A]_{0} = 100$,$[A]_{t} = 100 - 75 = 25$,and $k = 0.02232 \ min^{-1}$.
Substituting the values: $t = \frac{2.303}{0.02232} \log \frac{100}{25}$
$t = \frac{2.303}{0.02232} \log 4$
Since $\log 4 \approx 0.6021$,we get: $t = \frac{2.303 \times 0.6021}{0.02232} \approx 62.12 \ min$.

(B) For a first order reaction,the integrated rate equation is given by: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Rearranging this equation,we get: $\log_{10} [A]_t = -\frac{k}{2.303} t + \log_{10} [A]_0$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \log_{10} [A]_t$,$x = t$,and $c = \log_{10} [A]_0$,the slope $m$ is equal to $-k / 2.303$.

(C) The rate constant $k = 1 \times 10^{-2} \ s^{-1}$,initial concentration $[A]_0 = 20 \ g$,and final concentration $[A]_t = 5 \ g$.
For a first order reaction,the time $t$ is given by the formula:
$t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$
Substituting the values:
$t = \frac{2.303}{1 \times 10^{-2}} \log_{10} \frac{20}{5}$
$t = 2.303 \times 10^2 \times \log_{10}(4)$
Since $\log_{10}(4) \approx 0.602$:
$t = 230.3 \times 0.602 \approx 138.6 \ s$
Therefore,the correct option is $C$.

(D) The correct option is $(D)$.
For a first-order reaction,the rate constant $k$ is given by the formula:
$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$
Given:
Initial concentration $[A]_0 = 100$
Amount consumed $= 20 \%$,so remaining concentration $[A]_t = 100 - 20 = 80$
Time $t = 15 \ min$
Substituting the values:
$k = \frac{2.303}{15} \log_{10} \frac{100}{80}$
$k = \frac{2.303}{15} \log_{10} (1.25)$
$k = \frac{2.303}{15} \times 0.0969$
$k \approx 0.01487 \ min^{-1} = 1.487 \times 10^{-2} \ min^{-1}$
Thus,the value is approximately $1.48 \times 10^{-2} \ min^{-1}$.

(C) For a first order reaction,the rate equation is given by $t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$.
Given,rate constant $k = 0.00813 \ min^{-1}$.
Let the initial concentration $[A]_0 = 100$.
For $60 \%$ completion,the amount reacted is $60$,so the remaining concentration $[A]_t = 100 - 60 = 40$.
Substituting the values in the formula:
$t = \frac{2.303}{0.00813} \log_{10} \frac{100}{40}$
$t = \frac{2.303}{0.00813} \log_{10} (2.5)$
$t = \frac{2.303}{0.00813} \times 0.3979$
$t \approx 112.7 \ min$.

(B)
For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 3 \ min$,so $k = \frac{0.693}{3} = 0.231 \ min^{-1}$.
Let the initial concentration $[A]_0 = 100$.
Since the concentration is reduced by $90 \%$,the remaining concentration $[A]_t = 100 - 90 = 10$.
Using the first-order integrated rate equation: $t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$.
$t = \frac{2.303}{0.231} \log_{10} \frac{100}{10} = \frac{2.303}{0.231} \times 1 = 9.969 \ min$.

(C) The reaction is a first order reaction.
Since the concentration decreases from $0.2 \ M$ to $0.1 \ M$ (which is half of the initial concentration),the time taken is the half-life period,$t_{1/2} = 100 \ minutes$.
For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{0.693}{t_{1/2}}$.
Substituting the value: $k = \frac{0.693}{100 \ min} = 6.93 \times 10^{-3} \ min^{-1}$.

(B) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]_{t}}$.
Given that the reaction is $25 \%$ completed,if the initial concentration $[A]_{0} = 100$,then the remaining concentration $[A]_{t} = 100 - 25 = 75$.
The time taken $t = 40 \ min$.
Substituting these values into the formula:
$k = \frac{2.303}{40} \log_{10} \frac{100}{75}$.
Simplifying the fraction $\frac{100}{75}$ gives $\frac{4}{3}$.
Therefore,$k = \frac{2.303}{40} \log_{10} \frac{4}{3}$.

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Given that $75 \%$ is completed in $60 \ min$,the remaining concentration $[A]_t$ is $100 - 75 = 25 \%$ of $[A]_0$.
$k = \frac{2.303}{60} \log_{10} \frac{100}{25} = \frac{2.303}{60} \log_{10} 4 = \frac{2.303 \times 0.6020}{60} \approx 0.0231 \ min^{-1}$.
The time required for $50 \%$ completion is the half-life $t_{1/2}$.
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0231} = 30 \ min$.

(A) For a first-order reaction,$A \rightarrow \text{product}$,the rate is given by: $\text{Rate} = -\frac{d[A]}{dt} = k[A]$.
Rearranging and integrating the equation: $\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -\int_0^t k dt$.
This yields: $\ln \frac{[A]_t}{[A]_0} = -kt$.
Thus,the integrated rate equation is $k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t}$.
Alternatively,using base $10$ logarithms,$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Both options $A$ and $B$ represent the correct integrated rate equation for a first-order reaction.

(C) For a first order reaction,the rate law is given by $-\frac{d[A]}{dt} = k[A]$.
Integrating this expression from time $t=0$ (where $[A] = a$) to time $t$ (where $[A] = a-x$),we get:
$k = \frac{2.303}{t} \log \frac{a}{a-x}$
where '$a$' is the initial concentration and '$a-x$' is the concentration at time '$t$'.

(C) The integrated rate equation for a first-order reaction is given by $k = \frac{1}{t} \ln \frac{a}{a-x}$.
Here,$a$ is the initial concentration and $(a-x)$ is the concentration at time $t$.
Alternatively,using base $10$ logarithms,it is expressed as $k = \frac{2.303}{t} \log_{10} \frac{a}{a-x}$.

(B) The standard integrated rate equation for a first-order reaction is $k = \frac{2.303}{t} \log \frac{a}{a-x}$.
By using the logarithmic property $\log(\frac{x}{y}) = -\log(\frac{y}{x})$,we can rewrite the equation as $k = \frac{-2.303}{t} \log \frac{a-x}{a}$.
Therefore,option $B$ is a mathematically correct representation of the integrated rate equation for a first-order reaction.

(A) For a first order reaction,the half-life period $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Given the rate constant $k = 1.155 \times 10^{-3} \ s^{-1}$.
Substituting the value of $k$ in the formula:
$t_{1/2} = \frac{0.693}{1.155 \times 10^{-3}} \ s$.
$t_{1/2} = \frac{0.693}{1.155} \times 10^{3} \ s$.
$t_{1/2} = 0.6 \times 1000 \ s$.
$t_{1/2} = 600 \ s$.

(D) The half-life period $(t_{1/2})$ for a reaction of order $n$ is given by the relation: $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$,where $[A]_0$ is the initial concentration.
For a first order reaction,$n = 1$.
Substituting $n = 1$ in the expression: $t_{1/2} \propto \frac{1}{[A]_0^{1-1}} = \frac{1}{[A]_0^0} = \text{constant}$.
Therefore,for a first order reaction,the half-life period is independent of the initial concentration.

(A) For a first-order reaction,the rate constant is given by:
$k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right)$
Given $k = 0.02303 \ min^{-1}$ and $t = 60 \ min$.
Assuming initial concentration $[A]_0 = 100$,we need to find $[A]$.
$0.02303 = \frac{2.303}{60} \log \left( \frac{100}{[A]} \right)$
$0.02303 \times \frac{60}{2.303} = \log \left( \frac{100}{[A]} \right)$
$0.01 \times 60 = \log \left( \frac{100}{[A]} \right)$
$0.6 = \log \left( \frac{100}{[A]} \right)$
Since $\log(4) \approx 0.602$,we have $\frac{100}{[A]} = 4$,so $[A] = 25$.
Alternatively,using half-life:
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02303} \approx 30 \ min$.
After $60 \ min$ $(2 \times t_{1/2})$,the amount remaining is $(\frac{1}{2})^2 \times 100 \% = 25 \%$.