First Order reaction Questions in English

(A) For a first order reaction,the integrated rate equation is given by:
$k = \frac{2.303}{t} \log \left(\frac{a}{a-x}\right)$
Rearranging the equation:
$kt = 2.303 \log a - 2.303 \log (a-x)$
$2.303 \log (a-x) = 2.303 \log a - kt$
$\log (a-x) = \log a - \frac{kt}{2.303}$
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(a-x)$,$x = t$,and $c = \log a$,the slope $m$ is equal to $-\frac{k}{2.303}$.

(A) For a $1^{st}$ order reaction,the rate constant $K$ is given by:
$K = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]_t} \right)$
Given $t = 570 \ s$ and $[A]_t = 32 \%$ of $[A]_0$,so $\frac{[A]_0}{[A]_t} = \frac{100}{32} = 3.125$.
$K = \frac{2.303}{570} \log_{10} (3.125)$
Since $\log_{10} (3.125) = \log_{10} (100/32) = \log_{10} 100 - \log_{10} 32 = 2 - 5 \log_{10} 2 = 2 - 5(0.301) = 2 - 1.505 = 0.495$.
$K = \frac{2.303 \times 0.495}{570} \approx \frac{1.14}{570} = 0.002 \ s^{-1} = 2 \times 10^{-3} \ s^{-1}$.
Rounding to the nearest integer,the value is $2$.

(B) Given $(t_{1/2})_A = 54 \, min$ and $(t_{1/2})_B = 18 \, min$.
Starting with equimolar concentrations,let $[A]_0 = [B]_0 = x$.
For first order kinetics,the concentration at time $t$ is given by $[A]_t = [A]_0 \times (1/2)^n$,where $n = t / t_{1/2}$.
We want $[A]_t = 16 \times [B]_t$.
Substituting the expressions: $x \times (1/2)^{t/54} = 16 \times x \times (1/2)^{t/18}$.
$(1/2)^{t/54} = 2^4 \times (1/2)^{t/18}$.
$(1/2)^{t/54} = (1/2)^{-4} \times (1/2)^{t/18}$.
$(1/2)^{t/54} = (1/2)^{(t/18) - 4}$.
Equating the exponents: $t/54 = (t/18) - 4$.
$4 = t/18 - t/54 = (3t - t) / 54 = 2t / 54 = t / 27$.
$t = 4 \times 27 = 108 \, min$.

(D) For a first-order reaction,the time required for $n \%$ completion is given by $t = \frac{1}{k} \ln \frac{[A]_0}{[A]_t}$.
Given $t_{1/2} = 1 \, \text{min}$,we know $k = \frac{\ln 2}{t_{1/2}} = \frac{0.69}{1} = 0.69 \, \text{min}^{-1}$.
For $99.9 \, \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$.
$t_{99.9} = \frac{1}{k} \ln \frac{[A]_0}{0.001[A]_0} = \frac{1}{k} \ln 1000 = \frac{1}{k} \ln 10^3 = \frac{3 \ln 10}{k}$.
Substituting the values: $t_{99.9} = \frac{3 \times 2.3}{0.69} = \frac{6.9}{0.69} = 10 \, \text{min}$.

(C) The reaction is a first order process.
The integrated rate equation for a first order reaction is given by:
$Kt = \ln \frac{[A]_{0}}{[A]_{t}}$
Given that the reaction proceeds $40\%$ to completion,the remaining concentration $[A]_{t}$ is $60\%$ of the initial concentration $[A]_{0}$.
So,$[A]_{t} = 0.60 [A]_{0}$ or $\frac{[A]_{0}}{[A]_{t}} = \frac{100}{60} = \frac{5}{3}$.
Substituting the values:
$3.3 \times 10^{-4} \ s^{-1} \times t = \ln \left(\frac{100}{60}\right)$
$3.3 \times 10^{-4} \times t = \ln(1.6667)$
$3.3 \times 10^{-4} \times t = 0.5108$
$t = \frac{0.5108}{3.3 \times 10^{-4}} \ s$
$t = 1547.95 \ s$
To convert the time into minutes:
$t = \frac{1547.95}{60} \ min$
$t = 25.799 \ min$
Rounding off to the nearest integer,we get $26 \ min$.

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3.33 \ h} = \frac{0.693}{10/3 \ h} = 0.2079 \ h^{-1}$.
The integrated rate law for a first order reaction is $\ln \frac{[A]_0}{[A]_t} = kt$,which can be written as $\log_{10} \frac{[A]_0}{[A]_t} = \frac{kt}{2.303}$.
Given that the fraction of sucrose remaining is $f = \frac{[A]_t}{[A]_0}$,we have $\frac{1}{f} = \frac{[A]_0}{[A]_t}$.
Substituting the values: $\log_{10} (\frac{1}{f}) = \frac{0.2079 \times 9}{2.303} = \frac{0.693 \times 3 \times 9}{10 \times 2.303} = \frac{0.693}{2.303} \times 2.7 = 0.3009 \times 2.7 = 0.81243$.
Thus,$\log_{10} (\frac{1}{f}) = 81.24 \times 10^{-2}$.
Rounding to the nearest integer,we get $81$.

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
For $50 \% $ completion,$t_{50 \%} = \frac{2.303}{k} \log \frac{100}{50} = \frac{2.303}{k} \log 2$.
For $75 \% $ completion,$t_{75 \%} = \frac{2.303}{k} \log \frac{100}{25} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} \log 2^2 = 2 \times \frac{2.303}{k} \log 2$.
Therefore,the ratio $\frac{t_{75 \%}}{t_{50 \%}} = \frac{2 \times (\frac{2.303}{k} \log 2)}{\frac{2.303}{k} \log 2} = 2$.

(A) Since the unit of the rate constant is $min^{-1}$,it follows first-order kinetics.
The integrated rate equation is $K \times t = 2.303 \log(A_0 / A_t)$.
Given that $10 \%$ of the virus is inactivated in $1 \ min$,we have $A_0 = 100$ and $A_t = 100 - 10 = 90$.
Substituting the values: $K \times 1 = 2.303 \times \log(100 / 90)$.
$K = 2.303 \times (\log 10 - \log 9) = 2.303 \times (1 - 2 \log 3)$.
$K = 2.303 \times (1 - 2 \times 0.477) = 2.303 \times (1 - 0.954) = 2.303 \times 0.046 = 0.105938$.
$K = 105.938 \times 10^{-3} \ min^{-1}$.
Rounding to the nearest integer,we get $106 \times 10^{-3} \ min^{-1}$.

(C) For a first order reaction,the rate constant $K$ is given by:
$K = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]_{t}}$
Given:
$[A]_{0} = 50 \ mol \ L^{-1}$
$[A]_{t} = 10 \ mol \ L^{-1}$
$t = 120 \ min$
Substituting the values:
$K = \frac{2.303}{120} \log \frac{50}{10}$
$K = \frac{2.303}{120} \times \log 5$
$K = \frac{2.303}{120} \times 0.6989$
$K \approx 0.013413 \ min^{-1}$
$K \approx 1.34 \times 10^{-2} \ min^{-1}$
Comparing with $X \times 10^{-2} \ min^{-1}$,we get $X = 1.34$. Rounding to the nearest integer,$X = 1$.

(C) For a first order reaction,the rate constant $k$ is given by:
$k = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]_{t}}$
Given:
$[A]_{0} = 2.40 \times 10^{-2} \ mol \ L^{-1}$
$[A]_{t} = 1.60 \times 10^{-2} \ mol \ L^{-1}$
$t = 1 \ hour = 60 \ min$
Substituting the values:
$k = \frac{2.303}{60} \log \left( \frac{2.40 \times 10^{-2}}{1.60 \times 10^{-2}} \right)$
$k = \frac{2.303}{60} \log (1.5)$
$k = \frac{2.303}{60} \times 0.1761$
$k \approx 0.00676 \ min^{-1} = 6.76 \times 10^{-3} \ min^{-1}$
Rounding to the nearest integer,we get $7 \times 10^{-3} \ min^{-1}$.

(NONE) For the reaction $A \rightarrow 2B$:
At $t=0$,$[A]_0 = 1 \ mol$ and $[B] = 0$.
At $t=100 \ min$,let the amount of $A$ reacted be $x$. Then $[A]_t = 1-x$ and $[B] = 2x$.
Given $2x = 0.2 \ mol$,so $x = 0.1 \ mol$.
Thus,$[A]_t = 1 - 0.1 = 0.9 \ mol$.
The rate constant $k$ is given by $k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t} = \frac{1}{100} \ln \frac{1}{0.9} = \frac{1}{100} \ln(1.111)$.
Using $\ln(1.111) \approx 0.105$,$k \approx 0.00105 \ min^{-1}$.
The half-life $t_{1/2} = \frac{\ln 2}{k} = \frac{0.69}{0.00105} \approx 657 \ min$.

(A) For a first order reaction,the rate constant $K$ is given by the formula:
$K = \frac{2.303}{t} \log \left(\frac{[A]_0}{[A]_t}\right)$
Given:
Initial concentration $[A]_0 = 0.1 \, M$
Concentration after time $t$,$[A]_t = 0.001 \, M$
Time $t = 5 \, \min$
Substituting the values:
$K = \frac{2.303}{5} \log \left(\frac{0.1}{0.001}\right)$
$K = \frac{2.303}{5} \log(100)$
Since $\log(100) = 2$:
$K = \frac{2.303 \times 2}{5} = \frac{4.606}{5} = 0.9212 \, \min^{-1}$

(C) For a first order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.
The time $t$ required for a reaction to complete $90 \%$ is given by $t_{90\%} = \frac{2.303}{K} \log \left( \frac{100}{100 - 90} \right) = \frac{2.303}{K} \log 10 = \frac{2.303}{K}$.
Substituting $K = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{t_{1/2}}$ into the equation:
$t_{90\%} = \frac{2.303}{0.693} \times t_{1/2}$.
Since $2.303 \approx \ln 10$ and $0.693 \approx \ln 2$,we have $t_{90\%} = \frac{\ln 10}{\ln 2} \times t_{1/2} = \frac{2.303}{2.303 \times 0.3010} \times t_{1/2} = \frac{1}{0.3010} \times t_{1/2} \approx 3.32 \times t_{1/2}$.
Therefore,$x = 3.32$.

(C) The rate constants for first-order reactions are given by $k = \frac{\ln 2}{t_{1/2}}$.
$k_{A} = \frac{\ln 2}{100} \, s^{-1}$ and $k_{B} = \frac{\ln 2}{50} \, s^{-1}$.
The concentration at time $t$ is given by $[A]_t = [A]_0 e^{-k_A t}$ and $[B]_t = [B]_0 e^{-k_B t}$.
Given $[A]_0 = [B]_0$,we set $[A]_t = 4[B]_t$.
$[A]_0 e^{-k_A t} = 4 [A]_0 e^{-k_B t}$.
$e^{(k_B - k_A)t} = 4$.
Taking natural log on both sides: $(k_B - k_A)t = \ln 4 = 2 \ln 2$.
$(\frac{\ln 2}{50} - \frac{\ln 2}{100})t = 2 \ln 2$.
$(\frac{2 \ln 2 - \ln 2}{100})t = 2 \ln 2$.
$(\frac{\ln 2}{100})t = 2 \ln 2$.
$t = 200 \, s$.

(B) For a first order reaction,the time $t$ required for completion is given by $t = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_t}\right)$.
For $67 \, \%$ completion,$[A]_t = [A]_0 - 0.67[A]_0 = 0.33[A]_0 \approx \frac{1}{3}[A]_0$.
Thus,$t_{67 \, \%} = \frac{1}{k} \ln \left(\frac{1}{1/3}\right) = \frac{\ln 3}{k}$.
The half-life is $t_{1/2} = \frac{\ln 2}{k}$.
Therefore,$\frac{t_{67 \, \%}}{t_{1/2}} = \frac{\ln 3}{\ln 2} \approx \frac{1.0986}{0.6931} \approx 1.585$.
Given $t_{67 \, \%} = (x \times 10^{-1}) \times t_{1/2}$,we have $x \times 10^{-1} = 1.585$,which implies $x = 15.85$.
Rounding to the nearest integer,$x = 16$.

(C) For a first order reaction,the integrated rate equation is $\ln(P/P_0) = -kt$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln(P/P_0)$ and $x = t$,the slope $m = -k$.
From the given graph,the slope is $-3.465 \times 10^4 \ s^{-1}$.
Therefore,$-k = -3.465 \times 10^4 \ s^{-1}$,which gives $k = 3.465 \times 10^4 \ s^{-1}$.
Given $k = x \times 10^4 \ s^{-1}$,we have $x = 3.465 \approx 3$ (rounding to the nearest integer as per options).

(D) The half-life for a reaction of order $n$ is given by the relation $t_{1/2} \propto \frac{1}{P_0^{n-1}}$,where $P_0$ is the initial pressure.
Given: $(t_{1/2})_1 = 240 \ s = 4 \ min$ at $P_1 = 500 \ Torr$.
Given: $(t_{1/2})_2 = 4 \ min$ at $P_2 = 250 \ Torr$.
Since the half-life remains constant $(4 \ min)$ despite the change in initial pressure,the half-life is independent of the initial pressure.
For $t_{1/2}$ to be independent of $P_0$,the exponent $(n-1)$ must be $0$.
Therefore,$n-1 = 0$,which implies $n = 1$.
The reaction is of the $1^{st}$ order.

(C) Since the half-life is independent of the initial concentration,it is a $1^{st}$ order reaction.
$k = \frac{0.693}{T_{1/2}} = \frac{0.693}{200} = 3.465 \times 10^{-3} \, s^{-1}$.
For $80\%$ decomposition,the remaining concentration is $20\%$ of the initial concentration $(A = 0.2 A_{0})$.
The rate equation is $t = \frac{2.303}{k} \log \frac{A_{0}}{A}$.
$t = \frac{2.303}{3.465 \times 10^{-3}} \log \frac{A_{0}}{0.2 A_{0}} = \frac{2.303}{3.465 \times 10^{-3}} \log 5$.
Using $\log 5 = 0.70$ and $\frac{2.303}{k} = \frac{200}{0.693} \approx 288.66$.
$t = 288.66 \times 0.70 \approx 202.06 \times 2.303 \approx 466.67 \, s \approx 467 \, s$.

(B) For a first-order reaction,the half-life $t_{1/2}$ is given as $70 \ min$.
Converting time to seconds: $t_{1/2} = 70 \times 60 \ s = 4200 \ s$.
The rate constant $k$ is calculated as:
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{4200} \ s^{-1}$.
$k = \frac{693}{4200000} \ s^{-1} = \frac{693}{42} \times 10^{-6} \ s^{-1}$.
$k = 165 \times 10^{-6} \ s^{-1}$.
Thus,the value of $x$ is $165$.

(C) For a first-order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 0.3010 \ min$,so $K = \frac{0.693}{0.3010} \approx 2.303 \ min^{-1}$.
The integrated rate equation is $\ln \frac{[A]_0}{[A]_t} = Kt$,which can be written as $\log \frac{[A]_0}{[A]_t} = \frac{Kt}{2.303}$.
Substituting the values: $\log \frac{[A]_0}{[A]_t} = \frac{2.303 \times 2.0}{2.303} = 2.0$.
Therefore,$\frac{[A]_0}{[A]_t} = 10^2 = 100$.

(D) For a $1^{st}$ order reaction,the half-life $t_{1/2}$ is given as $1 \, h$. This means $50 \, \%$ of the reaction is completed in $1 \, h$.
For $87.5 \, \%$ completion,the amount remaining is $100 \, \% - 87.5 \, \% = 12.5 \, \%$.
We can relate the number of half-lives $(n)$ to the fraction remaining: $\frac{[A]_t}{[A]_0} = (\frac{1}{2})^n$.
Here,$\frac{12.5}{100} = \frac{1}{8} = (\frac{1}{2})^3$.
Thus,$n = 3$ half-lives are required.
Total time $t = n \times t_{1/2} = 3 \times 1 \, h = 3 \, h$.

(B)
As given in the question,the concentration of a substance becomes one-half of its original value,which refers to the half-life period $(t_{1/2})$.
The half-life period is the time required to reduce the initial concentration to half of its value.
For a $zero$ order reaction,$t_{1/2} = \frac{[A]_0}{2k}$.
For a $first$ order reaction,$t_{1/2} = \frac{0.693}{k}$.
For a $second$ order reaction,$t_{1/2} = \frac{1}{k[A]_0}$.
For a $third$ order reaction,$t_{1/2} = \frac{1}{2k[A]_0^2}$.
Here,$[A]_0$ is the initial concentration.
Since the half-life of a $first$ order reaction is independent of the initial concentration,the correct option is $(b)$.

(B) For a first order reaction $R \longrightarrow P$,the rate law is given by: $\text{Rate} = -\frac{d[R]}{dt} = k[R]$.
Rearranging the terms,we get: $\frac{d[R]}{[R]} = -k dt$.
Integrating both sides: $\int \frac{d[R]}{[R]} = -\int k dt \Rightarrow \ln[R] = -kt + C$.
At $t = 0$,$[R] = [R_0]$,so $\ln[R_0] = C$.
Substituting $C$ back into the equation: $\ln[R] = -kt + \ln[R_0]$.
Rearranging: $\ln \frac{[R]}{[R_0]} = -kt$.
Taking the exponential of both sides: $\frac{[R]}{[R_0]} = e^{-kt}$.
Therefore,the concentration at time $t$ is: $[R] = [R_0] e^{-kt}$.

(B)
For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 30 \, min$,so $k = \frac{0.693}{30} \, min^{-1}$.
The time $t$ required for $75 \, \%$ completion is given by the formula $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $75 \, \%$ completion,$[A]_t = [A]_0 - 0.75[A]_0 = 0.25[A]_0$.
Thus,$t = \frac{2.303}{0.693/30} \log \frac{[A]_0}{0.25[A]_0} = \frac{2.303 \times 30}{0.693} \log 4$.
Since $\frac{2.303}{0.693} \approx \frac{1}{0.301} \approx 3.32$,we have $t = 30 \times 2 = 60 \, min$.
Alternatively,$75 \, \%$ completion corresponds to $2 \times t_{1/2} = 2 \times 30 = 60 \, min$.

(B) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k} = 30 \ min$.
The time required for $75 \%$ completion $(T_{75\%})$ is the time taken for two half-lives to pass.
$T_{75\%} = 2 \times t_{1/2} = 2 \times 30 \ min = 60 \ min$.
Alternatively,using the formula $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$:
$t = \frac{2.303}{k} \log \frac{100}{100-75} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} \times 2 \log 2$.
Since $t_{1/2} = \frac{2.303 \log 2}{k} = 30 \ min$,then $t = 2 \times 30 = 60 \ min$.

(D) $A.$ Incorrect. $A$ first-order reaction theoretically takes infinite time to complete.
$B.$ Incorrect. $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4.6 \times 10^{-3}} \approx 150.65 \ s$.
$C.$ Incorrect. $t_{10\%} = \frac{1}{k} \ln(\frac{100}{90}) = \frac{2.303}{k} \log(1.11) \approx \frac{0.105}{k}$. $t_{90\%} = \frac{1}{k} \ln(\frac{100}{10}) = \frac{2.303}{k} \log(10) \approx \frac{2.303}{k}$. Thus,$t_{90\%} = 22 \times t_{10\%}$ (approx),not $25$.
$D.$ Correct. For first order,$[A]_t = [A]_0 e^{-kt}$. Degree of dissociation $\alpha = \frac{[A]_0 - [A]_t}{[A]_0} = 1 - e^{-kt}$.
$E.$ Correct. For a first-order reaction,rate $= k[A]^1$. Units of rate are $mol \ L^{-1} \ s^{-1}$ and units of $[A]$ are $mol \ L^{-1}$,so units of $k$ are $s^{-1}$. They are not the same.
Wait,re-evaluating $E$: Rate unit is $M \ s^{-1}$,$k$ unit is $s^{-1}$. They are different. So $E$ is incorrect.
Only statement $D$ is correct. The number of correct statements is $1$.

(B) For a first order reaction,the time $t$ is given by:
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
Given:
$k = 2.011 \times 10^{-3} \ s^{-1}$
$[A]_0 = 7 \ g$
$[A]_t = 2 \ g$
$\log 7 = 0.845$
$\log 2 = 0.301$
Substituting the values:
$t = \frac{2.303}{2.011 \times 10^{-3}} \log \left( \frac{7}{2} \right)$
$t = \frac{2.303}{2.011 \times 10^{-3}} (\log 7 - \log 2)$
$t = \frac{2.303}{2.011 \times 10^{-3}} (0.845 - 0.301)$
$t = \frac{2.303 \times 0.544}{2.011} \times 10^3$
$t = \frac{1.252832}{2.011} \times 1000$
$t \approx 622.989 \ s$
The nearest integer is $623$.

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{1}{t} \ln \frac{a_0}{a_t}$.
For $60 \%$ decomposition,$a_t = a_0 - 0.6 a_0 = 0.4 a_0$ and $t_1 = 540 \ s$.
$k = \frac{1}{540} \ln \frac{a_0}{0.4 a_0} = \frac{1}{540} \ln \frac{10}{4} = \frac{1}{540} (\ln 10 - \ln 4) = \frac{1}{540} (2.3 - 2 \times 0.3 \times 2.3 / 2.3) = \frac{1}{540} (2.3 - 1.38) = \frac{0.92}{540}$.
For $90 \%$ decomposition,$a_t = a_0 - 0.9 a_0 = 0.1 a_0$.
$t_2 = \frac{1}{k} \ln \frac{a_0}{0.1 a_0} = \frac{1}{k} \ln 10 = \frac{1}{k} \times 2.3$.
Substituting $k$: $t_2 = \frac{540}{0.92} \times 2.3 = \frac{540 \times 2.3}{0.92} = \frac{1242}{0.92} = 1350 \ s$.

(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given $[A]_t = \frac{[A]_0}{32}$ and $k = 20 \, min^{-1}$.
Substituting the values: $20 = \frac{2.303}{t} \log \frac{[A]_0}{[A]_0 / 32} = \frac{2.303}{t} \log 32$.
Since $32 = 2^5$,$\log 32 = 5 \log 2 = 5 \times 0.3010 = 1.505$.
$t = \frac{2.303 \times 1.505}{20} = \frac{3.466}{20} = 0.1733 \, min$.
Converting to $10^{-2} \, min$: $0.1733 \, min = 17.33 \times 10^{-2} \, min$.
The nearest integer is $17$.

(B) For a first-order reaction,the number of half-lives $(n)$ is calculated as $n = \frac{t}{t_{1/2}} = \frac{8000}{2000} = 4$.
After $n$ half-lives,the remaining concentration $[A]_t$ is given by $[A]_t = \frac{[A]_0}{2^n}$.
Substituting the values: $0.02 = \frac{[A]_0}{2^4}$.
$0.02 = \frac{[A]_0}{16}$.
$[A]_0 = 0.02 \times 16 = 0.32 \, M$.

(B) For parallel first-order reactions,the effective rate constant $k_{eff}$ is the sum of individual rate constants: $k_{eff} = k_1 + k_2$.
Since $k = \frac{\ln 2}{t_{1/2}}$,we have $\frac{\ln 2}{t_{eff}} = \frac{\ln 2}{t_1} + \frac{\ln 2}{t_2}$.
This simplifies to $\frac{1}{t_{eff}} = \frac{1}{t_1} + \frac{1}{t_2}$.
Given $t_1 = 12 \ min$ and $t_2 = 3 \ min$,we calculate $\frac{1}{t_{eff}} = \frac{1}{12} + \frac{1}{3} = \frac{1+4}{12} = \frac{5}{12}$.
Thus,$t_{eff} = \frac{12}{5} \ min = 2.4 \ min$.
The nearest integer is $2 \ min$.

(C) For a first order reaction,the time taken for $50 \%$ completion is the half-life,$t_{50} = t_{1/2}$.
At $87.5 \%$ completion,the remaining amount of reactant is $A_t = A_0 - 0.875 A_0 = 0.125 A_0 = \frac{A_0}{8}$.
Using the half-life concept:
$A_0$ $\xrightarrow{t_{1/2}} \frac{A_0}{2}$ $\xrightarrow{t_{1/2}} \frac{A_0}{4}$ $\xrightarrow{t_{1/2}} \frac{A_0}{8}$.
This shows that $t_{87.5} = 3 \times t_{1/2}$.
Since $t_{50} = t_{1/2}$,we have $t_{87.5} = 3 \times t_{50}$.
Therefore,$x = 3$.

(D) For the reaction $A ( g ) \rightarrow 2 B ( g ) + C ( g )$,let the initial pressure of $A$ be $P_0 = 800 \ mm \ Hg$.
At $t = 10 \ min$,the total pressure $P_t = P_0 - x + 2x + x = P_0 + 2x = 1600 \ mm \ Hg$.
Substituting $P_0 = 800$,we get $800 + 2x = 1600$,so $2x = 800$,which means $x = 400 \ mm \ Hg$.
The pressure of $A$ remaining at $t = 10 \ min$ is $P_0 - x = 800 - 400 = 400 \ mm \ Hg$.
Since the initial pressure was $800 \ mm \ Hg$ and it became $400 \ mm \ Hg$ in $10 \ min$,the half-life $t_{1/2} = 10 \ min$.
After $30 \ min$ $(3 \times t_{1/2})$,the pressure of $A$ remaining is $P_A = P_0 \times (1/2)^3 = 800 \times (1/8) = 100 \ mm \ Hg$.
The amount of $A$ reacted is $800 - 100 = 700 \ mm \ Hg$.
According to stoichiometry,$A \rightarrow 2B + C$,the pressure of $B$ produced is $2 \times 700 = 1400 \ mm \ Hg$ and $C$ produced is $700 \ mm \ Hg$.
The total pressure at $t = 30 \ min$ is $P_A + P_B + P_C = 100 + 1400 + 700 = 2200 \ mm \ Hg$.

(C) For a first order reaction,the time $t$ is given by $t = \frac{2.303}{k} \log \left( \frac{a}{a-x} \right)$.
For $99.9 \%$ completion,$x = 0.999a$,so $a-x = 0.001a = \frac{a}{1000}$.
Thus,$t_{99.9 \%} = \frac{2.303}{k} \log \left( \frac{a}{a/1000} \right) = \frac{2.303}{k} \log(10^3) = \frac{2.303 \times 3}{k}$.
The half-life is $t_{1/2} = \frac{0.693}{k} = \frac{2.303 \times 0.301}{k}$.
Taking the ratio: $\frac{t_{99.9 \%}}{t_{1/2}} = \frac{2.303 \times 3 / k}{2.303 \times 0.301 / k} \approx \frac{3}{0.3} = 10$.

(A) For a first-order reaction,the rate $r = k[A]_t = k[A]_0 e^{-kt}$.
Given $r_1 = 0.04 \ mol \ L^{-1} \ s^{-1}$ at $t_1 = 10 \ min$ and $r_2 = 0.03 \ mol \ L^{-1} \ s^{-1}$ at $t_2 = 20 \ min$.
$\frac{r_1}{r_2} = \frac{k[A]_0 e^{-kt_1}}{k[A]_0 e^{-kt_2}} = e^{k(t_2 - t_1)}$.
$\frac{0.04}{0.03} = e^{k(20 - 10)} = e^{10k}$.
$\frac{4}{3} = e^{10k} \implies 10k = \ln(\frac{4}{3}) = 2.303 \log(\frac{4}{3})$.
$10k = 2.303 \times (2 \log 2 - \log 3) = 2.303 \times (2 \times 0.3010 - 0.4771) = 2.303 \times (0.6020 - 0.4771) = 2.303 \times 0.1249 \approx 0.2876$.
$k = 0.02876 \ min^{-1}$.
Half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02876} \approx 24.09 \ min$.
Thus,the half-life is approximately $24 \ minutes$.

(A) For the reaction $A(g) \rightarrow B(g) + C(g)$:

Time $t=0$	$P_i$	$0$	$0$
Time $t$	$P_i - x$	$x$	$x$

Total pressure at time $t$ is $P_t = (P_i - x) + x + x = P_i + x$.
Therefore,$x = P_t - P_i$.
The partial pressure of reactant $A$ at time $t$ is $P_A = P_i - x = P_i - (P_t - P_i) = 2 P_i - P_t$.
Substituting into the first-order rate equation $k = \frac{2.303}{t} \log \frac{P_i}{P_A}$:
$k = \frac{2.303}{t} \log \frac{P_i}{2 P_i - P_t}$.

(B) The rate law $r = k[A]$ indicates that the reaction is of the first order.
For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k} = 120 \ \text{min}$.
Thus,$k = \frac{0.693}{120} \ \text{min}^{-1}$.
For $90 \%$ decomposition,the amount remaining is $100 - 90 = 10 \%$.
The time $t$ is given by the formula $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right)$.
Substituting the values: $t = \frac{2.303}{(0.693 / 120)} \log \left( \frac{100}{10} \right)$.
$t = \frac{2.303 \times 120}{0.693} \times \log(10)$.
Since $\frac{2.303}{0.693} \approx \frac{1}{0.301} \approx 3.32$,we get $t = 120 \times 3.32 \approx 399 \ \text{minutes}$.

(B) For the reaction $A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$:
Let the initial pressure of $A$ be $P_0 = 0.1 \ atm$.
At time $t = 115 \ s$,let the decrease in pressure of $A$ be $x$.
Then,the pressures are: $P_A = P_0 - x$,$P_B = 2x$,and $P_C = x$.
The total pressure $P_t = (P_0 - x) + 2x + x = P_0 + 2x$.
Given $P_t = 0.28 \ atm$ at $t = 115 \ s$,we have $0.1 + 2x = 0.28$,which gives $2x = 0.18$,so $x = 0.09 \ atm$.
The pressure of $A$ at $t = 115 \ s$ is $P_A = 0.1 - 0.09 = 0.01 \ atm$.
For a first-order reaction,the rate constant $k = \frac{1}{t} \ln \frac{P_0}{P_A} = \frac{1}{115} \ln \frac{0.1}{0.01} = \frac{1}{115} \ln(10)$.
Using $\ln(10) \approx 2.303$,$k = \frac{2.303}{115} \approx 0.02002 \ s^{-1} = 2.002 \times 10^{-2} \ s^{-1}$.
The nearest integer is $2$.

(A) Given the reaction $A + B \rightarrow C$ and the rate law $\text{rate} = k[A]^{1/2}[B]^{1/2}$.
Since the initial concentrations $[A]_0 = [B]_0 = 1 \ M$,at any time $t$,$[A] = [B]$.
Substituting this into the rate law: $\text{rate} = k[A]^{1/2}[A]^{1/2} = k[A]$.
This confirms the reaction follows first-order kinetics with respect to $A$.
The integrated rate equation for a first-order reaction is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given $k = 4.6 \times 10^{-2} \ s^{-1}$,$[A]_0 = 1 \ M$,and $[A]_t = 0.1 \ M$.
Substituting the values: $4.6 \times 10^{-2} = \frac{2.303}{t} \log \frac{1}{0.1}$.
$4.6 \times 10^{-2} = \frac{2.303}{t} \times 1$.
$t = \frac{2.303}{4.6 \times 10^{-2}} \approx 50.06 \ s$.
Rounding to the nearest integer,$t = 50 \ s$.

(C) For a first order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \left(\frac{[A]_0}{[A]_t}\right)$.
For $99.9 \%$ completion,$[A]_t = 0.1 \% \text{ of } [A]_0$,so $[A]_t = 0.001 [A]_0$. Thus,$t_{99.9 \%} = \frac{2.303}{K} \log \left(\frac{100}{0.1}\right) = \frac{2.303}{K} \log(10^3) = \frac{2.303}{K} \times 3$.
For $90 \%$ completion,$[A]_t = 10 \% \text{ of } [A]_0$,so $[A]_t = 0.1 [A]_0$. Thus,$t_{90 \%} = \frac{2.303}{K} \log \left(\frac{100}{10}\right) = \frac{2.303}{K} \log(10) = \frac{2.303}{K} \times 1$.
Taking the ratio,$\frac{t_{99.9 \%}}{t_{90 \%}} = \frac{3}{1} = 3$.
Therefore,the time required for $99.9 \%$ completion is $3$ times the time required for $90 \%$ completion.

(D) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{\ln 2}{k}$.
Given $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{k_2}{k_1} = \frac{5}{2}$.
For a first-order reaction,the time $t$ to complete a fraction $x$ is $t = \frac{1}{k} \ln \frac{1}{1-x}$.
For Reaction $1$,$t_1 = \frac{1}{k_1} \ln \frac{1}{1 - 2/3} = \frac{1}{k_1} \ln 3$.
For Reaction $2$,$t_2 = \frac{1}{k_2} \ln \frac{1}{1 - 4/5} = \frac{1}{k_2} \ln 5$.
Taking the ratio: $\frac{t_1}{t_2} = \frac{k_2}{k_1} \times \frac{\ln 3}{\ln 5} = \frac{5}{2} \times \frac{\log_{10} 3}{\log_{10} 5}$.
Substituting the given values: $\frac{t_1}{t_2} = \frac{5}{2} \times \frac{0.477}{0.699} = 2.5 \times 0.6824 = 1.706$.
Rounding to the nearest integer,$1.706 \approx 1.7 = 17 \times 10^{-1}$.

(A) For the reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$,let the initial pressure of $A$ be $P_0$.
At $t = 0$: $P_A = P_0, P_B = 0, P_C = 0, P_{total} = P_0$.
At $t = 23 \ s$: $P_A = P_0 - x, P_B = 2x, P_C = x, P_{total} = P_0 + 2x = 200 \ torr$.
At $t = \infty$: $P_A = 0, P_B = 2P_0, P_C = P_0, P_{total} = 3P_0 = 300 \ torr$.
Thus,$P_0 = 100 \ torr$.
Substituting $P_0$ in $P_{total} = P_0 + 2x = 200$,we get $100 + 2x = 200$,so $x = 50 \ torr$.
The pressure of $A$ at $t = 23 \ s$ is $P_A = P_0 - x = 100 - 50 = 50 \ torr$.
For a first-order reaction,$k = \frac{2.303}{t} \log \frac{P_0}{P_A} = \frac{2.303}{23} \log \frac{100}{50} = \frac{2.303}{23} \log(2) = \frac{2.303 \times 0.301}{23} \approx 0.0301 \ s^{-1} = 3.01 \times 10^{-2} \ s^{-1}$.
The nearest integer is $3$.

(A) For a first-order reaction,the concentration at time $t$ is given by $C_t = C_0 e^{-kt}$,which shows exponential decay. Thus,statement $A$ is correct.
The half-life is $t_{1/2} = \frac{\ln 2}{k}$. Since the rate constant $k$ increases with temperature according to the Arrhenius equation,$t_{1/2}$ decreases as temperature increases. Thus,statement $B$ is correct.
For a first-order reaction,$t_{1/2} = \frac{0.693}{k}$,which is independent of the initial concentration $C_0$. Thus,statement $C$ is incorrect.
After $n$ half-lives,the remaining reactant is $C_t = \frac{C_0}{2^n}$. For $n = 8$,$C_t = \frac{C_0}{2^8} = \frac{C_0}{256}$.
Percentage completion = $\frac{C_0 - C_t}{C_0} \times 100 = \frac{C_0 - \frac{C_0}{256}}{C_0} \times 100 = (1 - \frac{1}{256}) \times 100 \approx 99.6 \%$. Thus,statement $D$ is correct.
Therefore,statements $A, B,$ and $D$ are correct.

(B) For a first-order reaction,the rate constant $K$ is given by $K = \frac{1}{t} \ln \left( \frac{C_0}{C_t} \right)$.
For $t_{1/8}$,the concentration $C_t = C_0 / 8$,so $K t_{1/8} = \ln(8) = 3 \ln(2)$.
For $t_{1/10}$,the concentration $C_t = C_0 / 10$,so $K t_{1/10} = \ln(10)$.
Taking the ratio: $\frac{t_{1/8}}{t_{1/10}} = \frac{3 \ln(2)}{\ln(10)} = 3 \log_{10} 2$.
Given $\log_{10} 2 = 0.3$,we have $\frac{t_{1/8}}{t_{1/10}} = 3 \times 0.3 = 0.9$.
Therefore,$\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9$.

(A) For the reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$:
At $t=0$,pressure of $A$ is $P_0$,and total pressure is $P_0$.
At time $t$,let the pressure of $A$ reacted be $x$. Then $P_A = P_0 - x$,$P_B = 2x$,and $P_C = x$.
Total pressure $P_t = (P_0 - x) + 2x + x = P_0 + 2x$.
Thus,$x = \frac{P_t - P_0}{2}$.
Partial pressure of $A$ at time $t$ is $P_A = P_0 - \frac{P_t - P_0}{2} = \frac{3P_0 - P_t}{2}$.
For a first order reaction,$k = \frac{1}{t} \ln \left( \frac{P_0}{P_A} \right) = \frac{1}{t} \ln \left( \frac{P_0}{(3P_0 - P_t)/2} \right) = \frac{1}{t} \ln \left( \frac{2P_0}{3P_0 - P_t} \right)$.
Rearranging gives $\ln(3P_0 - P_t) = \ln(2P_0) - kt$. This represents a straight line with a negative slope $(-k)$,matching graph $A$.
For a first order reaction,the rate constant $k$ is independent of the initial concentration $[A]_0$,matching graph $D$.
Also,$t_{1/3}$ (time for $A$ to become $1/3$ of its initial value) is $\frac{\ln 3}{k}$,which is independent of $[A]_0$.

(B) For the reaction $2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)}$ at constant $V$ and $T$:
At $t = 0$,$P_{N_2O_5} = 1 \ atm$,$P_{N_2O_4} = 0$,$P_{O_2} = 0$.
At $t = Y \times 10^3 \ s$,let the pressure of $O_2$ formed be $P$. Then $P_{N_2O_5} = 1 - 2P$,$P_{N_2O_4} = 2P$,and $P_{O_2} = P$.
The total pressure $P_T = (1 - 2P) + 2P + P = 1 + P = 1.45 \ atm$.
Thus,$P = 0.45 \ atm$.
The initial pressure of $N_2O_5$ is $P_0 = 1 \ atm$ and the pressure at time $t$ is $P_t = 1 - 2P = 1 - 2(0.45) = 0.1 \ atm$.
For a first-order reaction,$k = \frac{2.303}{t} \log \left(\frac{P_0}{P_t}\right)$.
Substituting the values: $5 \times 10^{-4} = \frac{2.303}{Y \times 10^3} \log \left(\frac{1}{0.1}\right)$.
$5 \times 10^{-4} = \frac{2.303}{Y \times 10^3} \times 1$.
$0.5 = \frac{2.303}{Y} \implies Y = \frac{2.303}{0.5} = 4.606$.
Given the options provided,there is a discrepancy in the calculation or the provided options. Based on standard calculation,$Y = 4.606$.

(A) The reaction is a $SN^1$ reaction,which follows first-order kinetics.
For a first-order reaction:
$1$. The half-life $t_{1/2} = \frac{0.693}{k}$ is independent of the initial concentration $[P]_0$. Thus,the plot of $t_{1/2}$ vs $[P]_0$ is a horizontal line.
$2$. The rate of reaction is given by $rate = k[P]$. The initial rate is $rate_0 = k[P]_0$. Thus,the plot of initial rate vs $[P]_0$ is a straight line passing through the origin.
$3$. The concentration of product $Q$ at time $t$ is $[Q] = [P]_0 - [P] = [P]_0(1 - e^{-kt})$. Thus,$\frac{[Q]}{[P]_0} = 1 - e^{-kt}$. This plot is an exponential growth curve.
$4$. For first-order kinetics,$\ln(\frac{[P]}{[P]_0}) = -kt$. The plot of $\ln(\frac{[P]}{[P]_0})$ vs time is a straight line with a negative slope $-k$ passing through the origin.

(D) The reaction is $2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)}$.
For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{P_0}{P_t}$,where $P_0$ is the initial pressure of $N_2O_5$ and $P_t$ is the pressure at time $t$.
Given $k = 4.606 \times 10^{-2} \ s^{-1}$ and $t = 100 \ s$.
$4.606 \times 10^{-2} = \frac{2.303}{100} \log \frac{0.6}{P_{N_2O_5}}$
$2 = \log \frac{0.6}{P_{N_2O_5}} \Rightarrow \frac{0.6}{P_{N_2O_5}} = 10^2 = 100$.
$P_{N_2O_5} = \frac{0.6}{100} = 0.006 \ atm$.
Let $x$ be the pressure of $N_2O_5$ decomposed. Then $0.6 - x = 0.006 \Rightarrow x = 0.594 \ atm$.
The total pressure $P_{total} = (0.6 - x) + x + \frac{x}{2} = 0.6 + \frac{x}{2}$.
$P_{total} = 0.6 + \frac{0.594}{2} = 0.6 + 0.297 = 0.897 \ atm$.
$P_{total} = 897 \times 10^{-3} \ atm$.
Rounding to the nearest integer,$X = 897$.

(C) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{\ln 2}{k}$. Since this expression is independent of the initial concentration $[R]_0$,the graph of $t_{1/2}$ versus $[R]_0$ is a horizontal line. Thus,Statement $I$ is true.
For a first-order reaction,the integrated rate equation is $\ln \frac{[R]_0}{[R]} = kt$,which can be written as $\log \frac{[R]_0}{[R]} = \frac{k}{2.303} \times t$. This is an equation of a straight line $y = mx$ where $y = \log \frac{[R]_0}{[R]}$,$x = t$,and the slope $m = \frac{k}{2.303}$. Thus,Statement $II$ is true.

(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given $[A]_0 = 16 \ mg/mL$,$[A]_t = 4 \ mg/mL$,and $t = 12 \ months$.
$k = \frac{2.303}{12} \log \frac{16}{4} = \frac{2.303}{12} \log 4 = \frac{2.303 \times 0.602}{12} \approx 0.1155 \ month^{-1}$.
The drug becomes ineffective after $50 \%$ decomposition,meaning $50 \%$ remains. Thus,$[A]_t = 0.5 \times [A]_0$.
The time taken for $50 \%$ decomposition is the half-life $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{0.1155} = 6 \ months$.