First Order reaction Questions in English

(B) For a first-order reaction,the rate law is given by: $\text{Rate} = k[R]^1$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \text{Rate}$,$x = [R]$,$m = \text{slope}$,and $c = 0$.
Thus,the slope of the graph of $\text{Rate}$ versus $[R]$ is equal to the rate constant $k$.

(A) The integrated rate law for a first-order reaction is given by:
$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$
Rearranging the equation:
$\log_{10} \frac{[A]_0}{[A]_t} = \frac{k}{2.303} t$
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} \frac{[A]_0}{[A]_t}$,$x = t$,$c = 0$,and $m$ is the slope:
The slope $m = \frac{k}{2.303}$.

(A) The integrated rate equation for a first-order reaction is given by: $\ln \frac{[R]_0}{[R]} = Kt$.
Converting to base $10$ logarithm: $\log \frac{[R]_0}{[R]} = \frac{Kt}{2.303}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{[R]_0}{[R]}$ and $x = t$,the slope $m$ is equal to $\frac{K}{2.303}$.
Therefore,the graph of $\log \frac{[R]_0}{[R]}$ versus $t$ gives a slope of $\frac{K}{2.303}$.

(B) For a first-order reaction,the integrated rate equation is given by:
$K = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$
Rearranging this equation to the form $y = mx + c$:
$\log \frac{[R]_0}{[R]} = \frac{K}{2.303} \times t$
Comparing this with the equation of a straight line $y = mx$,where $y = \log \frac{[R]_0}{[R]}$,$x = t$,and the slope $m = \frac{K}{2.303}$.
Therefore,the slope of the graph is $\frac{K}{2.303}$.

(C) For a first order reaction,the rate equation is given by $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$.
Given that the concentration reduces to $1/16^{th}$ of its initial value,$[R] = \frac{[R]_0}{16}$,so $\frac{[R]_0}{[R]} = 16$.
Substituting the values: $t = \frac{2.303}{k} \log(16)$.
$t = \frac{2.303}{60} \log(2^4) = \frac{2.303 \times 4 \times 0.3010}{60}$.
$t = \frac{2.303 \times 1.204}{60} \approx 0.0462 \text{ s}$.
Therefore,$t = 4.6 \times 10^{-2} \text{ s}$.

(B) For a first-order reaction,the integrated rate equation is $\log \frac{[R]_0}{[R]} = \frac{kt}{2.303}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{[R]_0}{[R]}$,$x = t$,$m = \frac{k}{2.303}$,and $c = 0$.
Since the y-intercept $c$ is $0$,the graph of $\log \frac{[R]_0}{[R]}$ versus $t$ passes through the origin.
Therefore,the correct option is $B$.

(A) For a first order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k}$.
Given $t_{1/2} = 40 \ minutes$.
Convert time to seconds: $t_{1/2} = 40 \times 60 \ s = 2400 \ s$.
Now,calculate the rate constant $k = \frac{0.693}{t_{1/2}}$.
$k = \frac{0.693}{2400} \ s^{-1} = 0.00028875 \ s^{-1}$.
Rounding to three significant figures,$k = 2.88 \times 10^{-4} \ s^{-1}$.

(B) For a first-order reaction,the integrated rate equation is $\log \frac{[R]_0}{[R]} = \frac{kt}{2.303}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{[R]_0}{[R]}$,$x = t$,$m = \frac{k}{2.303}$,and $c = 0$.
Since the intercept $c$ is $0$,the graph of $\log \frac{[R]_0}{[R]}$ versus $t$ passes through the origin.
Therefore,the correct option is $B$.

(B) For a $1^{st}$ 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 that the reaction is $75 \%$ complete,the remaining concentration $[A]_t = [A]_0 - 0.75 [A]_0 = 0.25 [A]_0$.
Substituting the values: $K = \frac{2.303}{20} \log \left( \frac{[A]_0}{0.25 [A]_0} \right)$.
$K = \frac{2.303}{20} \log (4) = \frac{2.303}{20} \times 0.6021 \approx \frac{1.386}{20} = 0.0693 \ s^{-1}$.
Alternatively,$t_{75\%} = 2 \times t_{1/2}$. Since $t_{1/2} = \frac{0.693}{K}$,then $20 = 2 \times \frac{0.693}{K}$,which gives $K = \frac{1.386}{20} = 0.0693 \ s^{-1}$.

(B) For a first-order reaction, the integrated rate equation is $t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]}\right)$.
Given $k = 2.303 \times 10^{-2} \text{ s}^{-1}$ and $[A] = \frac{[A]_0}{10}$.
Substituting these values, we get $t = \frac{2.303}{2.303 \times 10^{-2}} \log\left(\frac{[A]_0}{[A]_0 / 10}\right)$.
$t = \frac{1}{10^{-2}} \log(10) = 10^2 \times 1 = 100 \text{ s}$.

(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.
At half-life $(t = t_{1/2})$,the concentration $[A] = \frac{[A]_0}{2}$.
Substituting these values: $k = \frac{2.303}{t_{1/2}} \log \frac{[A]_0}{[A]_0/2} = \frac{2.303}{t_{1/2}} \log 2$.
Thus,$t_{1/2} = \frac{2.303 \times 0.3010}{k} = \frac{0.693}{k}$.
Since $t_{1/2}$ depends only on the rate constant $k$ and not on the initial concentration $[A]_0$,it is independent of the concentration.

(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: $[A]_0 = 0.05 \ M$,$[A]_t = 0.05 - 0.015 = 0.035 \ M$,and $t = 45 \ min$.
Substituting the values: $k = \frac{2.303}{45} \log \frac{0.05}{0.035} = \frac{2.303}{45} \log (1.4286) \approx 0.02673 \ min^{-1}$.
The half-life period is: $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02673} \approx 25.92 \ min$.

(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 5 \ min$,so $k = \frac{0.693}{5} \ min^{-1}$.
The time $t$ required for $99.9 \%$ completion is given by the formula $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$.
Substituting the values: $t = \frac{2.303}{0.693/5} \log \frac{[A]_0}{0.001[A]_0}$.
$t = \frac{2.303 \times 5}{0.693} \log(1000)$.
Since $\log(1000) = 3$ and $\frac{2.303}{0.693} \approx 3.32$,we have $t \approx 3.32 \times 5 \times 3 = 49.8 \approx 50 \ min$.
Thus,the correct option is $D$.

(A) For a first order reaction,the half-life $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Given $k = 5.5 \times 10^{-14} \ s^{-1}$.
Substituting the value of $k$ in the formula:
$t_{1/2} = \frac{0.693}{5.5 \times 10^{-14}} \ s$.
$t_{1/2} = 0.126 \times 10^{14} \ s$.
$t_{1/2} = 1.26 \times 10^{13} \ s$.

(B) For a reaction,the half-life $t_{1/2}$ is related to the initial concentration $[R]_0$ by the expression $t_{1/2} \propto [R]_0^{1-n}$,where $n$ is the order of the reaction.
If the plot of $t_{1/2}$ versus $[R]_0$ is a straight line parallel to the $x$-axis,it implies that $t_{1/2}$ is independent of the initial concentration $[R]_0$.
This occurs when $1-n = 0$,which means $n = 1$.
Therefore,the reaction is a first-order reaction.
The unit of the rate constant for a first-order reaction is $s^{-1}$.

(C) The number of half-lives $(n)$ is calculated as: $n = \frac{t}{t_{1/2}} = \frac{80 \ s}{20 \ s} = 4$.
The concentration of reactant remaining after $n$ half-lives is given by the formula: $[A_t] = [A_0] \times (\frac{1}{2})^n$.
Substituting the values: $[A_t] = 0.2 \ M \times (\frac{1}{2})^4 = 0.2 \times \frac{1}{16} = 0.0125 \ M$.

(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 45 \ min$,so $k = \frac{0.693}{45} \ min^{-1}$.
For $99.9 \%$ completion,the amount remaining $(a-x)$ is $100 - 99.9 = 0.1$.
The time $t$ required is $t = \frac{2.303}{k} \log \frac{a}{a-x}$.
Substituting the values: $t = \frac{2.303}{0.693 / 45} \log \frac{100}{0.1} = \frac{2.303 \times 45}{0.693} \log 1000$.
Since $\log 1000 = 3$,we have $t = \frac{2.303 \times 45 \times 3}{0.693} \approx 448.5 \ min$.
Converting to hours: $t = \frac{448.5}{60} \approx 7.475 \ hrs \approx 7.5 \ hours$.

(A) For a first order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$.
Given that $99 \%$ of the reaction is complete,the remaining concentration $[R]$ is $100 \% - 99 \% = 1 \%$ of the initial concentration $[R]_0$.
So,$[R] = 0.01 [R]_0 = \frac{[R]_0}{100}$.
Substituting these values into the equation:
$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]_0 / 100} = \frac{2.303}{k} \log(100)$.
Since $\log(100) = 2$,we get $t = \frac{2.303 \times 2}{k} = \frac{4.606}{k}$.

(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 $60 \%$ completion,$[A]_t = [A]_0 - 0.60[A]_0 = 0.40[A]_0$ and $t = 50 \ min$.
$k = \frac{2.303}{50} \log \frac{[A]_0}{0.40[A]_0} = \frac{2.303}{50} \log 2.5$.
For $93.6 \%$ completion,$[A]_t = [A]_0 - 0.936[A]_0 = 0.064[A]_0$.
$k = \frac{2.303}{t'} \log \frac{[A]_0}{0.064[A]_0} = \frac{2.303}{t'} \log \frac{1}{0.064} = \frac{2.303}{t'} \log 15.625$.
Equating the two expressions for $k$:
$\frac{2.303}{50} \log 2.5 = \frac{2.303}{t'} \log 15.625$.
$t' = 50 \times \frac{\log 15.625}{\log 2.5} = 50 \times \frac{\log (2.5)^3}{\log 2.5} = 50 \times 3 = 150 \ min$.

(A) For a $1^{st}$ order reaction,the amount remaining after $n$ half-lives is given by the formula: $[A] = [A]_0 \times (1/2)^n$.
Number of half-lives $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{240 \ minutes}{60 \ minutes} = 4$.
Percentage remaining $= (1/2)^n \times 100 = (1/2)^4 \times 100$.
Percentage remaining $= \frac{1}{16} \times 100 = 6.25 \%$.

(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.
Given that the concentration is reduced to $12.5 \%$ in $1 \ hr$ $(60 \ min)$,we have $[A]_0 = 100$ and $[A] = 12.5$.
$k = \frac{2.303}{60} \log \frac{100}{12.5} = \frac{2.303}{60} \log 8 = \frac{2.303}{60} \times 0.903 = 0.0346 \ min^{-1}$.
The half-life $t_{1/2}$ is calculated as $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0346} \approx 20 \ min$.
Alternatively,$12.5 \% = (1/2)^3$,which means $3$ half-lives have passed in $60 \ min$. Thus,$t_{1/2} = 60/3 = 20 \ min$.

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 10 \ min$,so $k = \frac{0.693}{10} = 0.0693 \ min^{-1}$.
After $20 \ min$ (which is $2 \times t_{1/2}$),the concentration $[A]$ will be reduced to $\frac{1}{4}$ of the initial concentration.
$[A] = \frac{12}{4} = 3 \ M$.
The rate of reaction is given by $Rate = k[A]$.
$Rate = 0.0693 \times 3 \ M \ min^{-1}$.

(A) For a first order reaction,the half-life $t_{1/2} = 25 \ min$.
The number of half-lives $n$ in $100 \ min$ is calculated as $n = \frac{100 \ min}{25 \ min} = 4$.
The amount of reactant remaining after $n$ half-lives is given by $\frac{A_0}{2^n}$.
Amount remaining $= \frac{100}{2^4} = \frac{100}{16} = 6.25 \%$.
The amount of reactant converted into product $= 100 \% - 6.25 \% = 93.75 \%$.

(B) For a first order reaction,the rate constant is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.
Case $I$: Given $x = 60 \%$,$t = 20 \ min$,so $a-x = 40$.
$k = \frac{2.303}{20} \log \frac{100}{40} = \frac{2.303}{20} \log 2.5$.
$k = \frac{2.303}{20} \times 0.3979 \approx 0.0458 \ min^{-1}$.
Case $II$: For $84 \%$ completion,$x = 84$,so $a-x = 16$.
$t = \frac{2.303}{k} \log \frac{100}{16} = \frac{2.303}{0.0458} \log 6.25$.
$t = \frac{2.303}{0.0458} \times 0.7959 \approx 40 \ min$.

(C) The reaction is $A(g) \longrightarrow B(g) + C(g)$.
Let the initial pressure of $A$ be $P_i$.
At $t = 0$: $P_A = P_i$,$P_B = 0$,$P_C = 0$. Total pressure $P_{total} = P_i = 180 \ mm \ of \ Hg$ (at completion).
At $t = 20 \ min$: $P_A = P_i - x$,$P_B = x$,$P_C = x$.
Total pressure $P_t = (P_i - x) + x + x = P_i + x = 100 \ mm \ of \ Hg$.
Since $P_i = 180 \ mm \ of \ Hg$,we have $180 + x = 100$,which is not possible as $x$ must be positive. Re-evaluating the data: The total pressure at completion is $2P_i = 180 \ mm \ of \ Hg$,so $P_i = 90 \ mm \ of \ Hg$.
At $t = 20 \ min$: $P_t = P_i + x = 100 \ mm \ of \ Hg$.
$90 + x = 100 \implies x = 10 \ mm \ of \ Hg$.
The partial pressure of $A$ at $20 \ min$ is $P_A = P_i - x = 90 - 10 = 80 \ mm \ of \ Hg$.

(A) For the reaction $2A_{(g)} \longrightarrow B_{(g)} + C_{(s)}$,let the initial pressure of $A$ be $P_0 = 2p$.
At $t = \infty$ (completion),$2A$ is fully consumed,so $P_{total} = P_B + P_C = p + p = 2p = 200 \ Pa$. Thus,$p = 100 \ Pa$ and $P_0 = 200 \ Pa$.
At $t = 10 \ min$,the pressure of $A$ is $2p - x$,$B$ is $x/2$,and $C$ is solid (negligible pressure).
$P_{total} = (2p - x) + x/2 = 2p - x/2 = 300 \ Pa$.
Substituting $2p = 200$,we get $200 - x/2 = 300$,which implies $x/2 = -100$. This suggests the reaction stoichiometry or data provided is inconsistent with standard gas laws.
Assuming the standard model where $P_{total} = P_0 + (n-1)x$,for $2A \rightarrow B + C$,$P_{total} = P_0 + (2-1)x = P_0 + x$.
At $t = \infty$,$P_{\infty} = P_0/2 = 200 \implies P_0 = 400 \ Pa$.
At $t = 10$,$P_t = P_0 + x/2 = 300 \implies 400 + x/2 = 300 \implies x/2 = -100$.
Given the options,the intended calculation is $k = \frac{2.303}{t} \log \frac{P_0}{P_A}$,where $P_A = 2(P_0 - P_t) = 2(400 - 300) = 200$.
$k = \frac{2.303}{10} \log \frac{400}{200} = 0.2303 \times 0.3010 \approx 0.0693 \ min^{-1}$.

(A) The reaction is $2 A_{(g)} \rightarrow B_{(g)} + C_{(s)}$.
Let the initial pressure of $A$ be $P_0$.
At $t = \infty$ (completion),only $B_{(g)}$ is present,so $P_B = P_0 / 2 = 200 \ Pa$,which means $P_0 = 400 \ Pa$.
At $t = 10 \ min$,let the pressure of $A$ reacted be $2x$.
Initial: $P_A = 400, P_B = 0, P_C = 0$.
At $t = 10$: $P_A = 400 - 2x, P_B = x, P_C = 0$ (since $C$ is solid).
Total pressure $P_t = (400 - 2x) + x = 400 - x = 300 \ Pa$.
Therefore,$x = 100 \ Pa$.
The partial pressure of $A$ at $t = 10 \ min$ is $P_A = 400 - 2(100) = 200 \ Pa$.
For a first-order reaction,$k = \frac{2.303}{t} \log \left( \frac{P_0}{P_A} \right)$.
$k = \frac{2.303}{10} \log \left( \frac{400}{200} \right) = \frac{2.303}{10} \log 2$.
$k = \frac{2.303 \times 0.3010}{10} \approx 0.0693 \ min^{-1}$.

(B) Given,half-life $t_{1/2} = 69.3 \ s$.
For a first-order reaction,the rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{69.3} = 10^{-2} \ s^{-1}$.
For first-order kinetics,$k = \frac{2.303}{t} \log \frac{P_0}{P_t}$,where $P_0$ is initial pressure and $P_t$ is pressure at time $t$.
$10^{-2} = \frac{2.303}{230} \log \frac{0.4}{P_t}$.
$1 = \log \frac{0.4}{P_t} \implies \frac{0.4}{P_t} = 10 \implies P_t = 0.04 \ atm$.
From the reaction $A_{(g)} \rightarrow P_{(g)} + Q_{(g)} + R_{(g)}$:
At $t = 0$,$P_A = 0.4 \ atm$,$P_P = 0, P_Q = 0, P_R = 0$.
At $t = 230 \ s$,$P_A = 0.04 \ atm$. The decrease in pressure of $A$ is $0.4 - 0.04 = 0.36 \ atm$.
According to stoichiometry,the pressure of products formed is $P_P = 0.36 \ atm, P_Q = 0.36 \ atm, P_R = 0.36 \ atm$.
Total pressure $P_{total} = P_A + P_P + P_Q + P_R = 0.04 + 0.36 + 0.36 + 0.36 = 1.12 \ atm$.

(D) For a first-order reaction,the rate law is given by: $\text{Rate} = k[A]$.
From the given data,we can find the rate constant $k$:
$0.2 = k \times 0.02 \implies k = \frac{0.2}{0.02} = 10 \ \text{min}^{-1}$.
Now,we find $x$ using the rate $0.4$:
$0.4 = 10 \times x \implies x = 0.04 \ M$.
Next,we find $y$ using the rate $1.0$:
$1.0 = 10 \times y \implies y = 0.10 \ M$.
The ratio $x : y$ is $0.04 : 0.10 = 4 : 10 = 2 : 5$.

(A) Given: $\Delta t = 25 \ min = 25 \times 60 \ s = 1500 \ s$.
Change in concentration $\Delta [R] = [R]_f - [R]_i = 0.02 - 0.03 = -0.01 \ mol \ L^{-1}$.
Rate $= -\frac{\Delta [R]}{\Delta t} = -\frac{-0.01}{1500} \ mol \ L^{-1} \ s^{-1}$.
Rate $= \frac{0.01}{1500} = \frac{1}{150000} = 6.667 \times 10^{-6} \ mol \ L^{-1} \ s^{-1}$.

(D) For a first order reaction,the rate constant $k$ is given by:
$k = \frac{2.303}{t} \log \frac{[X]_0}{[X]_t}$
Given $[X]_0 = 1.0 \ M$,$[X]_t = 0.25 \ M$,and $t = 40 \ min$:
$k = \frac{2.303}{40} \log \frac{1.0}{0.25} = \frac{2.303}{40} \log 4$
$k = \frac{2.303 \times 0.60}{40} = 0.034545 \ min^{-1}$
Now,the rate of reaction when $[X] = 0.1 \ M$ is:
$Rate = k[X] = 0.034545 \times 0.1$
$Rate = 0.0034545 \ mol \ L^{-1} \ min^{-1} = 3.45 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$

(C) For a first order reaction,the concentration after time $t$ is given by $[A]_t = [A]_0 \times (1/2)^n$,where $n$ is the number of half-lives.
Given that the concentration is reduced to $1/8$ of the initial concentration,we have $(1/2)^n = 1/8$.
Since $1/8 = (1/2)^3$,we find that $n = 3$.
This means $3$ half-lives have passed in $75 \ minutes$.
Therefore,$3 \times t_{1/2} = 75 \ minutes$.
$t_{1/2} = 75 / 3 = 25 \ minutes$.
The closest option is $25.1 \ minutes$.

(A) For a first order reaction,the time taken to complete a fraction $f$ is given by $t = \frac{2.303}{k} \log(\frac{1}{1-f})$.
For half-life $(t_{1/2})$,$f = 0.5$,so $t_{1/2} = \frac{2.303}{k} \log(2)$.
For $\frac{3}{4}$ th completion,$f = 0.75$,so $t_{3/4} = \frac{2.303}{k} \log(\frac{1}{1-0.75}) = \frac{2.303}{k} \log(4) = \frac{2.303}{k} \log(2^2) = 2 \times \frac{2.303}{k} \log(2)$.
Therefore,the ratio $\frac{t_{3/4}}{t_{1/2}} = \frac{2 \times \frac{2.303}{k} \log(2)}{\frac{2.303}{k} \log(2)} = 2$.

(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,$k = 4.606 \times 10^{-3} \ s^{-1}$ and $[A]_t = \frac{1}{10} [A]_0$,so $\frac{[A]_0}{[A]_t} = 10$.
Substituting the values: $4.606 \times 10^{-3} = \frac{2.303}{t} \log(10)$.
Since $\log(10) = 1$,we have $4.606 \times 10^{-3} = \frac{2.303}{t}$.
Solving for $t$: $t = \frac{2.303}{4.606 \times 10^{-3}} = \frac{1}{2} \times 10^3 = 500 \ s$.

(C) For a first order reaction,the integrated rate law is given by:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given:
$k = 2.303 \times 10^{-3} \ s^{-1}$
$[A]_0 = 4 \ g$
$[A]_t = 0.2 \ g$
Substituting the values:
$t = \frac{2.303}{2.303 \times 10^{-3}} \log \frac{4}{0.2}$
$t = \frac{1}{10^{-3}} \log 20$
$t = 1000 \times 1.301 = 1301 \ s$
Converting time to hours:
$t = \frac{1301}{3600} \approx 0.36 \ hours$
Thus,the correct option is $C$.

(A) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \left( \frac{a}{a-x} \right)$.
For $10 \%$ completion,$x = 0.1a$ and $t = 20 \text{ min}$.
$k = \frac{2.303}{20} \log \left( \frac{a}{0.9a} \right) = \frac{2.303}{20} \log \left( \frac{1}{0.9} \right)$.
For $19 \%$ completion,$x = 0.19a$ and $t = ?$.
$k = \frac{2.303}{t} \log \left( \frac{a}{0.81a} \right) = \frac{2.303}{t} \log \left( \frac{1}{0.81} \right) = \frac{2.303}{t} \log \left( \frac{1}{0.9^2} \right) = \frac{2.303}{t} \times 2 \log \left( \frac{1}{0.9} \right)$.
Equating the two expressions for $k$:
$\frac{2.303}{20} \log \left( \frac{1}{0.9} \right) = \frac{2.303}{t} \times 2 \log \left( \frac{1}{0.9} \right)$.
$\frac{1}{20} = \frac{2}{t} \implies t = 40 \text{ minutes}$.

(B) The time interval between $10:00 \ am$ and $11:30 \ am$ is $t = 90 \ min$.
At $10:00 \ am$,$20 \%$ is completed,so the remaining concentration is $80 \%$ of initial concentration $([A]_0)$.
At $11:30 \ am$,$20 \%$ is remaining,so $[A]_t = 0.20 [A]_0$.
The rate constant $k$ for a first-order reaction is given by $k = \frac{2.303}{t} \log \frac{[A]_{initial}}{[A]_{final}}$.
Substituting the values: $k = \frac{2.303}{90} \log \frac{0.80 [A]_0}{0.20 [A]_0} = \frac{2.303}{90} \log 4$.
Using $\log 4 \approx 0.602$,$k = \frac{2.303 \times 0.602}{90} \approx 0.0154 \ min^{-1}$.
The half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0154} \approx 45 \ min$.

(B) For a first order reaction,the rate equation is given by:
$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given: $K = 4.606 \times 10^{-3} \ s^{-1}$
If $90 \%$ of the initial quantity decomposes,then the remaining quantity $[A]_t = 100 \% - 90 \% = 10 \%$.
Let $[A]_0 = 100$. Then $[A]_t = 10$.
Substituting the values:
$t = \frac{2.303}{4.606 \times 10^{-3}} \log \frac{100}{10}$
$t = \frac{2.303}{4.606 \times 10^{-3}} \log(10)$
Since $\log(10) = 1$:
$t = \frac{2.303}{4.606 \times 10^{-3}} \times 1 = 0.5 \times 10^3 = 500 \ s$.

(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 $K = 3.3 \times 10^{-4} \ s^{-1}$ and the reaction is $90 \%$ complete,so $[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$.
Substituting the values:
$3.3 \times 10^{-4} = \frac{2.303}{t} \log \frac{[A]_0}{0.10[A]_0}$
$3.3 \times 10^{-4} = \frac{2.303}{t} \log(10)$
Since $\log(10) = 1$:
$t = \frac{2.303}{3.3 \times 10^{-4}} \ s$
$t \approx 0.6978 \times 10^4 \ s = 6978 \ s$
To convert time into minutes:
$t = \frac{6978}{60} \ min \approx 116.3 \ min$
Rounding to the nearest provided option,the answer is $116.67 \ min$.

(C) For the reaction $A_{(g)} \rightarrow B_{(g)} + 2C_{(g)}$,let the initial concentration of $A$ be $100 \ M$ at $t = 0$.
At $t = 24 \ min$,let the concentration of $A$ reacted be $x$.
Concentration of $A$ remaining $= 100 - x$.
Concentration of $B = x$ and concentration of $C = 2x$.
Total concentration of products $= x + 2x = 3x$.
Given the ratio of products to reactant is $1:3$,so $\frac{3x}{100 - x} = \frac{1}{3}$.
$9x = 100 - x \implies 10x = 100 \implies x = 10$.
Concentration of $A$ remaining $[A]_t = 100 - 10 = 90$.
Rate constant $k = \frac{2.303}{t} \log(\frac{[A]_0}{[A]_t}) = \frac{2.303}{24} \log(\frac{100}{90}) = \frac{2.303}{24} \log(1.11)$.
Using $\log 1.11 = 0.046$,$k = \frac{2.303 \times 0.046}{24} \approx 0.004415 \ min^{-1}$.
Half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.004415} \approx 157.19 \ min$,which is approximately $157.8 \ min$.

(D) For a first order reaction,the integrated rate equation is: $\ln(a-x) = -Kt + \ln a$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln(a-x)$,$x = \text{time}$,slope $m = -K$,and intercept $c = \ln a$.
From the given graph:
Intercept $c = -2.303 = \ln a$.
Therefore,$a = e^{-2.303} \approx 10^{-1} \ mol \ L^{-1}$.
Slope $m = -(10)^{-2} = -K$.
Therefore,$K = 10^{-2} \ s^{-1}$.
Thus,the rate constant is $10^{-2} \ s^{-1}$ and the initial concentration is $10^{-1} \ mol \ L^{-1}$.

(C) For a first order reaction,the integrated rate equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given that $93.75 \%$ of the reaction is complete,the remaining concentration is $[A]_t = [A]_0 - 0.9375[A]_0 = 0.0625[A]_0$.
Substituting this into the rate equation for $t = x$:
$k = \frac{2.303}{x} \log \frac{[A]_0}{0.0625[A]_0} = \frac{2.303}{x} \log(16) = \frac{2.303}{x} \log(2^4) = \frac{2.303 \times 4 \times \log(2)}{x} = \frac{4 \times 0.693}{x}$.
We know that the half-life $t_{1/2}$ is given by $k = \frac{0.693}{t_{1/2}}$.
Equating the two expressions for $k$:
$\frac{0.693}{t_{1/2}} = \frac{4 \times 0.693}{x}$.
Therefore,$t_{1/2} = \frac{x}{4}$ minutes.

(A) For a first order reaction,the rate constant $k$ is given by the formula:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$
Given values are:
$t = 5 \ min$
$[A]_0 = 0.6 \ mol \ L^{-1}$
$[A] = 0.2 \ mol \ L^{-1}$
Substituting the values in the formula:
$k = \frac{2.303}{5} \log \frac{0.6}{0.2}$
$k = \frac{2.303}{5} \log 3$
Using $\log 3 = 0.4771$:
$k = \frac{2.303 \times 0.4771}{5}$
$k = \frac{1.0988}{5} = 0.21976 \ min^{-1} \approx 0.219 \ min^{-1}$

(C) For a first-order reaction $A \longrightarrow B$:
The rate of formation of $B$ is given by $R = \frac{d[B]}{dt} = k[A]$.
For first-order kinetics,the concentration of reactant $A$ at time $t$ is $[A] = [A]_0 e^{-kt}$.
Substituting this into the rate expression: $R = k[A]_0 e^{-kt}$.
This equation shows that the rate $R$ decreases exponentially with time $t$. Therefore,the graph of $R$ vs $t$ is an exponential decay curve.

(B) For a first order reaction,the rate constant $k$ is given by the equation:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given that $80 \%$ of the reaction is completed,we have:
$[A]_0 = 100$
$[A]_t = 100 - 80 = 20$
Substituting these values into the equation:
$k = \frac{2.303}{t} \log \frac{100}{20}$
$k = \frac{2.303}{t} \log 5$
Since $\log 5 \approx 0.699$,we get:
$k = \frac{2.303 \times 0.699}{t}$
$k \approx \frac{1.609}{t}$
Therefore,the time $t$ required is:
$t \approx \frac{1.6}{k}$

(A) Given: Rate constant $k = 4.606 \times 10^{-3} \ s^{-1}$.
Initial concentration $[A]_0 = 400 \ g$.
Final concentration $[A]_t = 50 \ g$.
For a first order reaction,the formula is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Substituting the values: $t = \frac{2.303}{4.606 \times 10^{-3}} \log \frac{400}{50}$.
$t = \frac{2.303}{4.606 \times 10^{-3}} \log 8$.
$t = \frac{2.303}{4.606 \times 10^{-3}} \times 0.9030$.
$t = 0.5 \times 10^3 \times 0.9030 = 451.5 \ s$.
Converting to minutes: $t = \frac{451.5}{60} \ min = 7.525 \ min \approx 7.52 \ min$.

(C) For a $1^{\text{st}}$ order reaction,the integrated rate equation is given by:
$t = \frac{2.303}{k} \log \left(\frac{a}{a-x}\right)$
For the $\left(\frac{3}{4}\right)^{\text{th}}$ life,$x = \frac{3}{4}a$.
Substituting this value into the equation:
$t_{3/4} = \frac{2.303}{k} \log \left(\frac{a}{a - \frac{3}{4}a}\right)$
$t_{3/4} = \frac{2.303}{k} \log \left(\frac{a}{\frac{1}{4}a}\right)$
$t_{3/4} = \frac{2.303}{k} \log (4)$

(C) The unit of $k$ is $s^{-1}$,which indicates that it is a first-order reaction.
For a first-order reaction,the time required for a certain percentage of completion is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $t_{99.9}$,the remaining concentration is $0.1 \%$ of the initial concentration,so $t_{99.9} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log 1000 = \frac{2.303 \times 3}{k}$.
For $t_{50}$,the time is the half-life,$t_{50} = \frac{0.693}{k} = \frac{2.303 \times 0.301}{k}$.
The ratio $\frac{t_{99.9}}{t_{50}} = \frac{3 \times 2.303 / k}{0.301 \times 2.303 / k} = \frac{3}{0.301} \approx 10$.