First Order reaction Questions in English

(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 $k$:
$t_{1/2} = \frac{0.693}{5.5 \times 10^{-14} \ s^{-1}}$
$t_{1/2} = 0.126 \times 10^{14} \ s = 1.26 \times 10^{13} \ s$.

For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$.
When the reaction is $99.9 \%$ complete,the remaining concentration is $[R] = [R]_0 - 0.999[R]_0 = 0.001[R]_0 = 10^{-3}[R]_0$.
Substituting this into the rate equation:
$k = \frac{2.303}{t_{99.9}} \log \frac{[R]_0}{10^{-3}[R]_0} = \frac{2.303}{t_{99.9}} \log 10^3 = \frac{2.303 \times 3}{t_{99.9}} = \frac{6.909}{t_{99.9}}$.
Thus,$t_{99.9} = \frac{6.909}{k}$.
For the half-life of a first order reaction,$t_{1/2} = \frac{0.693}{k}$.
Taking the ratio:
$\frac{t_{99.9}}{t_{1/2}} = \frac{6.909 / k}{0.693 / k} \approx 10$.
Therefore,$t_{99.9} = 10 \times t_{1/2}$.

(N/A) For a first order reaction,the integrated rate equation is given by:
$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$
Given:
Initial amount $[R]_0 = 5 \, g$
Final amount $[R] = 3 \, g$
Rate constant $k = 1.15 \times 10^{-3} \, s^{-1}$
Substituting the values:
$t = \frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3}$
$t = \frac{2.303}{1.15 \times 10^{-3}} \times (\log 5 - \log 3)$
$t = \frac{2.303}{1.15 \times 10^{-3}} \times (0.6989 - 0.4771)$
$t = \frac{2.303}{1.15 \times 10^{-3}} \times 0.2218$
$t \approx 444.38 \, s$
Thus,the time taken is approximately $444 \, s$.

(N/A) For a $1^{st}$ order reaction,the half-life is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given that $t_{1/2} = 60 \ min$,we can rearrange the formula to solve for the rate constant $k$:
$k = \frac{0.693}{t_{1/2}}$
Substituting the given value:
$k = \frac{0.693}{60} \ min^{-1}$
$k = 0.01155 \ min^{-1}$
Expressing in scientific notation:
$k = 1.155 \times 10^{-2} \ min^{-1}$

$(i) \ \text{Half-life}, t_{1/2} = \frac{0.693}{k} = \frac{0.693}{200 \ s^{-1}} = 3.465 \times 10^{-3} \ s \approx 3.47 \times 10^{-3} \ s$
$(ii) \ \text{Half-life}, t_{1/2} = \frac{0.693}{k} = \frac{0.693}{2 \ min^{-1}} = 0.3465 \ min \approx 0.35 \ min$
$(iii) \ \text{Half-life}, t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4 \ years^{-1}} = 0.17325 \ years \approx 0.173 \ years$

(N/A) The radioactive decay follows first-order kinetics.
Given: $t_{1/2} = 5730 \text{ years}$,$[R]_0 = 100$,$[R] = 80$.
First,calculate the decay constant $k$:
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \text{ years}^{-1}$.
Using the first-order integrated rate equation:
$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$
$t = \frac{2.303}{0.693 / 5730} \times \log \left( \frac{100}{80} \right)$
$t = \frac{2.303 \times 5730}{0.693} \times \log(1.25)$
$t \approx 19039.5 \times 0.0969 \approx 1845 \text{ years}$.
Thus,the age of the sample is approximately $1845 \text{ years}$.

(N/A) $(i)$ Plot $[N_2O_5]$ vs $t$ shows a curve indicating a first-order reaction.
$(ii)$ Initial concentration $[N_2O_5]_0 = 1.63 \times 10^{-2} \, mol \, L^{-1}$. Half-life corresponds to the time when concentration becomes half,i.e.,$0.815 \times 10^{-2} \, mol \, L^{-1}$. From the graph,$t_{1/2} \approx 1450 \, s$.
$(iii)$ Plot $\log[N_2O_5]$ vs $t$ gives a straight line.
$(iv)$ Since the plot of $\log[N_2O_5]$ vs $t$ is a straight line,the reaction is of the first order. The rate law is: $\text{Rate} = k[N_2O_5]$.
$(v)$ Slope of $\log[N_2O_5]$ vs $t$ graph $= \frac{-k}{2.303}$.
Using points $(0, -1.79)$ and $(3200, -2.46)$:
$\text{Slope} = \frac{-2.46 - (-1.79)}{3200 - 0} = \frac{-0.67}{3200} = -2.09 \times 10^{-4} \, s^{-1}$.
$k = -\text{Slope} \times 2.303 = 2.09 \times 10^{-4} \times 2.303 \approx 4.82 \times 10^{-4} \, s^{-1}$.
$(vi)$ $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4.82 \times 10^{-4}} \approx 1438 \, s$. This is in close agreement with the value obtained in $(ii)$.

(A) For a first order reaction,the time $t$ is given by the formula:
$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$
Given $k = 60 \ s^{-1}$ and $[R] = \frac{[R]_0}{16}$,so $\frac{[R]_0}{[R]} = 16$.
Substituting the values:
$t = \frac{2.303}{60} \log(16)$
$t = \frac{2.303}{60} \times 1.204$
$t \approx 0.0462 \ s = 4.62 \times 10^{-2} \ s$.
Thus,the required time is $4.62 \times 10^{-2} \ s$.

For a first order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $99 \%$ completion,$[A]_t = [A]_0 - 0.99[A]_0 = 0.01[A]_0$. Thus,$t_{99\%} = \frac{2.303}{k} \log \frac{[A]_0}{0.01[A]_0} = \frac{2.303}{k} \log 100 = \frac{2.303}{k} \times 2$.
For $90 \%$ completion,$[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$. Thus,$t_{90\%} = \frac{2.303}{k} \log \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{k} \log 10 = \frac{2.303}{k} \times 1$.
Comparing the two,$t_{99\%} = 2 \times t_{90\%}$.
Therefore,the time required for $99 \%$ completion is twice the time required for $90 \%$ completion.

(N/A) For a first order reaction,the rate constant $k$ is given by:
$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$
Given $t = 40 \ min$ and $[R] = [R]_0 - 0.30[R]_0 = 0.70[R]_0$,
$k = \frac{2.303}{40} \log \frac{100}{70} = \frac{2.303}{40} \log(1.4286)$
$k = \frac{2.303}{40} \times 0.1549 = 8.918 \times 10^{-3} \ min^{-1}$
Now,the half-life $t_{1/2}$ is calculated as:
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.918 \times 10^{-3}} \ min$
$t_{1/2} \approx 77.7 \ min$

(N/A) The decomposition reaction is: $(CH_3)_2CHN=NCH(CH_3)_2(g) \rightarrow C_6H_{14}(g) + N_2(g)$.
Let $P_0$ be the initial pressure of azoisopropane at $t=0$. At time $t$,let $p$ be the decrease in pressure of azoisopropane.
Total pressure $P_t = (P_0 - p) + p + p = P_0 + p$.
Thus,$p = P_t - P_0$.
The pressure of azoisopropane at time $t$ is $P_0 - p = P_0 - (P_t - P_0) = 2P_0 - P_t$.
For a first-order reaction: $k = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}$.
At $t = 360 \ s$: $k = \frac{2.303}{360} \log \frac{35.0}{2(35.0) - 54.0} = \frac{2.303}{360} \log \frac{35.0}{16.0} \approx 2.175 \times 10^{-3} \ s^{-1}$.
At $t = 720 \ s$: $k = \frac{2.303}{720} \log \frac{35.0}{2(35.0) - 63.0} = \frac{2.303}{720} \log \frac{35.0}{7.0} = \frac{2.303}{720} \log 5 \approx 2.235 \times 10^{-3} \ s^{-1}$.
Average $k = \frac{2.175 \times 10^{-3} + 2.235 \times 10^{-3}}{2} = 2.21 \times 10^{-3} \ s^{-1}$.

(D) The thermal decomposition of $SO_{2}Cl_{2}$ at a constant volume is represented by the following equation:

Time Phase	Reaction: $SO_{2}Cl_{2(g)} \longrightarrow SO_{2(g)} + Cl_{2(g)}$
At $t=0$	$P_{0} \longrightarrow 0 + 0$
At $t=t$	$(P_{0} - p) \longrightarrow p + p$

After time $t$,the total pressure $P_{t} = (P_{0} - p) + p + p = P_{0} + p$.
Thus,$p = P_{t} - P_{0}$.
The pressure of $SO_{2}Cl_{2}$ at time $t$ is $P_{SO_{2}Cl_{2}} = P_{0} - p = P_{0} - (P_{t} - P_{0}) = 2P_{0} - P_{t}$.
For a first order reaction,$k = \frac{2.303}{t} \log \frac{P_{0}}{2P_{0} - P_{t}}$.
At $t = 100 \ s$ and $P_{t} = 0.6 \ atm$,$k = \frac{2.303}{100} \log \frac{0.5}{2(0.5) - 0.6} = \frac{2.303}{100} \log \frac{0.5}{0.4} = 0.02303 \times \log(1.25) \approx 2.231 \times 10^{-3} \ s^{-1}$.
When $P_{t} = 0.65 \ atm$,the pressure of $SO_{2}Cl_{2}$ is $P_{SO_{2}Cl_{2}} = 2P_{0} - P_{t} = 2(0.5) - 0.65 = 1.0 - 0.65 = 0.35 \ atm$.
The rate of reaction is $Rate = k \times P_{SO_{2}Cl_{2}} = (2.231 \times 10^{-3} \ s^{-1}) \times (0.35 \ atm) = 7.81 \times 10^{-4} \ atm \ s^{-1}$.

Given: $k = 2.0 \times 10^{-2} \ s^{-1}$,$t = 100 \ s$,$[A]_{0} = 1.0 \ mol \ L^{-1}$.
Since the unit of $k$ is $s^{-1}$,the reaction is a first-order reaction.
For a first-order reaction,the integrated rate equation is:
$k = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]}$
Substituting the values:
$2.0 \times 10^{-2} = \frac{2.303}{100} \log \frac{1.0}{[A]}$
$\log \frac{1.0}{[A]} = \frac{2.0 \times 10^{-2} \times 100}{2.303} = \frac{2}{2.303} \approx 0.8684$
$\frac{1.0}{[A]} = \text{antilog}(0.8684) \approx 7.385$
$[A] = \frac{1.0}{7.385} \approx 0.135 \ mol \ L^{-1}$.
Alternatively,using $[A] = [A]_{0} e^{-kt}$:
$[A] = 1.0 \times e^{-(2.0 \times 10^{-2} \times 100)} = 1.0 \times e^{-2} \approx 1.0 \times 0.1353 = 0.1353 \ mol \ L^{-1}$.

(N/A) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 3.00 \ h$,we have $k = \frac{0.693}{3.00 \ h} = 0.231 \ h^{-1}$.
The integrated rate law for a first order reaction is $\ln \frac{[R]_0}{[R]} = kt$ or $\log \frac{[R]_0}{[R]} = \frac{kt}{2.303}$.
Substituting the values: $\log \frac{[R]_0}{[R]} = \frac{0.231 \ h^{-1} \times 8 \ h}{2.303} = \frac{1.848}{2.303} \approx 0.8024$.
Taking the antilog,$\frac{[R]_0}{[R]} = 10^{0.8024} \approx 6.3445$.
The fraction remaining is $\frac{[R]}{[R]_0} = \frac{1}{6.3445} \approx 0.1576$.
Thus,the fraction of sucrose remaining after $8 \ h$ is approximately $0.158$.

(N/A) First-order reaction: $A$ reaction in which the rate of reaction is proportional to the first power of the concentration of the reactant $R$ is called a first-order reaction.
The rate of a first-order reaction $\propto [R]^1$.
For the reaction $R \to P$,the differential rate expression is:
$Rate = -\frac{d[R]}{dt} = k[R]$
$\therefore \frac{d[R]}{[R]} = -k dt \dots (i)$
Integrating this equation on both sides:
$\int \frac{d[R]}{[R]} = -\int k dt$
$\ln [R] = -kt + I \dots (ii)$
Here,$I$ is the constant of integration.
When $t = 0$,$[R] = [R]_0$,where $[R]_0$ is the initial concentration of the reactant. Substituting these values into equation $(ii)$:
$\ln [R]_0 = -k(0) + I \implies I = \ln [R]_0 \dots (iii)$
Substituting $I = \ln [R]_0$ into equation $(ii)$:
$\ln [R] = -kt + \ln [R]_0 \dots (iv)$
Rearranging the terms:
$kt = \ln [R]_0 - \ln [R]$
$kt = \ln \frac{[R]_0}{[R]}$
$k = \frac{1}{t} \ln \frac{[R]_0}{[R]} \dots (v)$
Converting to base $10$ logarithm:
$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \dots (vi)$
Taking the antilog of equation $(iv)$,we get:
$[R] = [R]_0 e^{-kt} \dots (vii)$

(N/A) The integrated rate equations for a first-order reaction are as follows:
$1. \ln [R] = -k(t) + \ln [R]_0$
$2. \log [R] = -\frac{k}{2.303}(t) + \log [R]_0$
These equations follow the straight-line form $y = mx + c$. Thus,the plots of $\ln [R]$ versus $t$ and $\log [R]$ versus $t$ are straight lines with a negative slope and a positive intercept on the $Y$-axis.
- For the plot of $\ln [R]$ vs $t$: The slope is $-k$ and the intercept is $\ln [R]_0$.
- For the plot of $\log [R]$ vs $t$: The slope is $-\frac{k}{2.303}$ and the intercept is $\log [R]_0$.
Additionally,based on the integrated rate equation $\log \frac{[R]_0}{[R]} = \frac{k}{2.303}(t)$,the plot of $\log \frac{[R]_0}{[R]}$ versus $t$ is a straight line passing through the origin $(0, 0)$ with a slope of $\frac{k}{2.303}$.

(N/A) The integrated rate equation for a first-order reaction is:
$\ln [R] = -kt + \ln [R]_0$ $\quad \dots (I)$
At time $t_1$,the concentration is $[R]_1$:
$\ln [R]_1 = -kt_1 + \ln [R]_0$ $\quad \dots (II)$
At time $t_2$,the concentration is $[R]_2$:
$\ln [R]_2 = -kt_2 + \ln [R]_0$ $\quad \dots (III)$
Subtracting equation $(III)$ from equation $(II)$:
$\ln [R]_1 - \ln [R]_2 = (-kt_1 + \ln [R]_0) - (-kt_2 + \ln [R]_0)$
$\ln [R]_1 - \ln [R]_2 = -kt_1 + kt_2$
$\ln \frac{[R]_1}{[R]_2} = k(t_2 - t_1)$
Converting to base $10$ logarithm:
$2.303 \log \frac{[R]_1}{[R]_2} = k(t_2 - t_1)$
$\log \frac{[R]_1}{[R]_2} = \frac{k}{2.303}(t_2 - t_1)$

(N/A) The following are examples of first-order reactions:
$1$. Hydrogenation of ethene:
$C_{2}H_{4(g)} + H_{2(g)} \rightarrow C_{2}H_{6(g)}$
$Rate = k[C_{2}H_{4}]$
$2$. All natural and artificial radioactive decay processes follow first-order kinetics.
$3$. Decomposition of $N_{2}O_{5}$:
$2N_{2}O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$
$Rate = k[N_{2}O_{5}]$
$4$. Decomposition of $SO_{2}Cl_{2}$:
$SO_{2}Cl_{2(g)} \rightarrow SO_{2(g)} + Cl_{2(g)}$
$Rate = k[SO_{2}Cl_{2}]$
$5$. Acid-catalyzed hydrolysis of sucrose (pseudo first-order):
$C_{12}H_{22}O_{11} + H_{2}O \xrightarrow{H^+} C_{6}H_{12}O_{6} + C_{6}H_{12}O_{6}$

(N/A) The reaction is: $A_{(g)} \to B_{(g)} + C_{(g)}$
| Time | $A_{(g)}$ | $B_{(g)}$ | $C_{(g)}$ |
| :--- | :--- | :--- | :--- |
| $t = 0$ | $p_i \ atm$ | $0 \ atm$ | $0 \ atm$ |
| $t = t$ | $(p_i - x) \ atm$ | $x \ atm$ | $x \ atm$ |
Here,$p_i$ is the initial pressure of $A$ at $t = 0$,and $x$ is the decrease in pressure of $A$ at time $t$.
The total pressure $p_t$ at time $t$ is given by the sum of partial pressures:
$p_t = (p_i - x) + x + x = p_i + x$
From this,we can express $x$ in terms of $p_t$ and $p_i$:
$x = p_t - p_i$
The partial pressure of $A$ at time $t$ is:
$p_A = p_i - x = p_i - (p_t - p_i) = 2p_i - p_t$
For a first-order reaction,the integrated rate equation is:
$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$
Substituting the partial pressures for concentrations:
$k = \frac{2.303}{t} \log \frac{p_i}{p_A} = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t}$

(N/A) For a first-order reaction,the rate constant $(k)$ is given by the equation: $k = \frac{2.303}{t} \log \frac{[R]_{0}}{[R]}$ ...$(i)$
At half-life,$t = t_{1/2}$ and the concentration of the reactant $[R] = \frac{[R]_{0}}{2}$.
Substituting these values into equation $(i)$:
$k = \frac{2.303}{t_{1/2}} \log \frac{[R]_{0}}{[R]_{0}/2}$
$k = \frac{2.303}{t_{1/2}} \log 2$
Since $\log 2 \approx 0.3010$:
$k = \frac{2.303 \times 0.3010}{t_{1/2}}$
$k = \frac{0.693}{t_{1/2}}$
Therefore,$t_{1/2} = \frac{0.693}{k}$.
Conclusion: For a first-order reaction,the half-life is independent of the initial concentration of the reactant.

(N/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:
Initial concentration $[A]_0 = 1.24 \times 10^{-2} \ mol \ L^{-1}$
Final concentration $[A]_t = 0.20 \times 10^{-2} \ mol \ L^{-1}$
Time $t = 60 \ min$
Substituting the values:
$k = \frac{2.303}{60} \log \frac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}}$
$k = \frac{2.303}{60} \log(6.2)$
$k = \frac{2.303}{60} \times 0.7924$
$k \approx 0.0304 \ min^{-1}$

(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: $k = 1.20 \times 10^{-3} \, s^{-1}$,$[A]_0 = 5 \, g$,and $[A]_t = 3 \, g$.
Substituting the values: $1.20 \times 10^{-3} = \frac{2.303}{t} \log \frac{5}{3}$.
$t = \frac{2.303}{1.20 \times 10^{-3}} \times \log(1.666)$.
$t = \frac{2.303}{1.20 \times 10^{-3}} \times 0.2218$.
$t \approx 425.6 \, s \approx 426 \, s$.

(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,$t_{1/2} = 60 \ min$.
Substituting the value in the formula: $60 = \frac{0.693}{k}$.
Therefore,$k = \frac{0.693}{60} \ min^{-1}$.
$k = 0.01155 \ min^{-1}$.

(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,$k = 70 \, s^{-1}$ and $[A]_t = \frac{1}{18} [A]_0$,which implies $\frac{[A]_0}{[A]_t} = 18$.
Substituting the values: $t = \frac{2.303}{70} \log(18)$.
Since $\log(18) \approx 1.255$,we get $t = \frac{2.303 \times 1.255}{70}$.
$t = \frac{2.890}{70} \approx 0.0413 \, s$.

(N/A) $(i)$ Initial rate $= k[N_2O_5]_0 = (5.0 \times 10^{-4} \ s^{-1}) \times (0.25 \ mol \ L^{-1}) = 1.25 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
$(ii)$ Half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{5.0 \times 10^{-4} \ s^{-1}} = 1386 \ s$.
$(iii)$ For $75\%$ completion,$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} = \frac{2.303}{5.0 \times 10^{-4}} \log \frac{100}{25} = 4606 \times 0.6021 \approx 2773 \ s$.
$(iv)$ After $t = 30 \ min = 1800 \ s$,$[N_2O_5] = [N_2O_5]_0 e^{-kt} = 0.25 \times e^{-(5.0 \times 10^{-4} \times 1800)} = 0.25 \times e^{-0.9} \approx 0.25 \times 0.4066 \approx 0.1016 \ mol \ L^{-1}$.
Amount reacted $= 0.25 - 0.1016 = 0.1484 \ mol \ L^{-1}$.
Since $1 \ mol \ N_2O_5$ gives $2 \ mol \ NO_2$,$[NO_2] = 2 \times 0.1484 = 0.2968 \ mol \ L^{-1} \approx 0.30 \ mol \ L^{-1}$.

(B) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Substituting the given values: $k = \frac{2.303}{45 \, min} \log \frac{0.80}{0.06}$
$k = \frac{2.303}{45} \log(13.33) \approx \frac{2.303}{45} \times 1.1248 \approx 0.05756 \, min^{-1}$
The half-life $(t_{1/2})$ is calculated as: $t_{1/2} = \frac{0.693}{k}$
$t_{1/2} = \frac{0.693}{0.05756} \approx 12.04 \, min$.
Note: Based on standard calculation,the closest option is $12.50 \, min$.

(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 $t = 20 \ min$ for $15\%$ completion,$[A]_t = 100 - 15 = 85\%$ of $[A]_0$.
$k = \frac{2.303}{20} \log \frac{100}{85} = \frac{2.303}{20} \log(1.1765) \approx \frac{2.303 \times 0.0706}{20} \approx 0.00813 \ min^{-1}$.
Now,for $75\%$ completion,$[A]_t = 100 - 75 = 25\%$ of $[A]_0$.
$t = \frac{2.303}{k} \log \frac{100}{25} = \frac{2.303}{0.00813} \log(4) \approx \frac{2.303 \times 0.6020}{0.00813} \approx 170.58 \ min$.

(A) For a first-order reaction,the integrated rate equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given: $t = 100 \ min$,$[A]_t = \frac{1}{8} [A]_0$,so $\frac{[A]_0}{[A]_t} = 8$.
Substituting the values: $k = \frac{2.303}{100} \log(8) = \frac{2.303}{100} \times 0.903 = 2.08 \times 10^{-2} \ min^{-1}$.
The half-life time is calculated as $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{2.08 \times 10^{-2}} \approx 33.3 \ min$.

(A) For the reaction $N_2O_{5(soln)} \to 2NO_{2(soln)} + \frac{1}{2}O_{2(g)}$,let the initial concentration of $N_2O_5$ be $[A]_0 = 0.25 \ mol \ L^{-1}$.
Let $x$ be the amount of $N_2O_5$ dissociated at time $t$.
According to stoichiometry,$[NO_2] = 2x = 0.20 \ mol \ L^{-1}$,so $x = 0.10 \ mol \ L^{-1}$.
The concentration of $N_2O_5$ remaining at time $t$ is $[A]_t = [A]_0 - x = 0.25 - 0.10 = 0.15 \ mol \ L^{-1}$.
For a first-order reaction,$k = \frac{2.303}{t} \log(\frac{[A]_0}{[A]_t})$.
Substituting the values: $5.0 \times 10^{-4} = \frac{2.303}{t} \log(\frac{0.25}{0.15})$.
$t = \frac{2.303}{5.0 \times 10^{-4}} \log(1.666) \approx 4606 \times 0.2218 \approx 1022 \ s$.

(A) 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: $k = 5 \times 10^{-4} \ s^{-1}$,$[A]_0 = 0.2 \ mol \ L^{-1}$,and $[A]_t = 25 \% \text{ of } [A]_0 = 0.25 \times [A]_0$.
Substituting the values: $5 \times 10^{-4} = \frac{2.303}{t} \log \frac{[A]_0}{0.25[A]_0}$.
$5 \times 10^{-4} = \frac{2.303}{t} \log(4)$.
Since $\log(4) \approx 0.6021$,we have $t = \frac{2.303 \times 0.6021}{5 \times 10^{-4}}$.
$t = \frac{1.3866}{5 \times 10^{-4}} = 2773.2 \ s \approx 2773 \ s$.

(B) For a first order reaction,the half-life $t_{1/2}$ is $16 \ min$.
The time required for $n$ half-lives is given by $t = n \times t_{1/2}$.
For $87.5 \%$ completion,the remaining amount is $100 \% - 87.5 \% = 12.5 \%$.
Since $12.5 \% = (1/2)^3 \times 100 \%$,the reaction has undergone $3$ half-lives $(n=3)$.
Therefore,$t = 3 \times 16 \ min = 48 \ min$.

(A) For a first order reaction,the rate law is given by $Rate = k[R]$.
The half-life period $(t_{1/2})$ for a first order reaction is derived as $t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$.
Since $t_{1/2}$ is independent of the initial concentration $[R]_0$,it can be expressed as $t_{1/2} \propto [R]_0^0$,which means $t_{1/2}$ is proportional to the initial concentration raised to the power of $0$.
Both statements $(i)$ and $(ii)$ are correct. Therefore,the correct option is $(A)$.

(B) For a first order reaction,the half-life $(t_{1/2})$ is given by the formula $t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$.
Statement $1$: $t_{1/2} = \frac{0.693}{2k}$ is False because the correct expression is $\frac{0.693}{k}$.
Statement $2$: $t_{1/2} \propto k$ is False because $t_{1/2}$ is inversely proportional to the rate constant $(t_{1/2} \propto \frac{1}{k})$.
Therefore,both statements are False $(F, F)$.

(A) For a first-order reaction $R \rightarrow P$,the rate law is given by $\text{Rate} = k[R]^1$.
Since the rate of disappearance of reactant $R$ is defined as $-\frac{d[R]}{dt}$,we have $\text{Rate} = -\frac{d[R]}{dt} = k[R]$.
Therefore,Statement $I$ is True $(T)$.
Statement $II$ suggests $\text{Rate} = -k[R]$,which is incorrect because the rate of a reaction must always be a positive quantity,and the rate constant $k$ is positive. Thus,Statement $II$ is False $(F)$.

(A) For a first-order reaction,the integrated rate law is given by:
$[R] = [R]_{0} e^{-kt}$
Taking the natural logarithm on both sides:
$\ln [R] = \ln ([R]_{0} e^{-kt})$
Using the property $\ln(ab) = \ln a + \ln b$:
$\ln [R] = \ln [R]_{0} + \ln(e^{-kt})$
Since $\ln(e^x) = x$:
$\ln [R] = -kt + \ln [R]_{0}$
Comparing this with the given statements:
Statement $(i)$ is $\ln [R] = -kt + \ln [R]_{0}$,which is $True$ $(T)$.
Statement $(ii)$ is $\ln [R] = +kt + \ln [R]_{0}$,which is $False$ $(F)$.
Therefore,the correct option is $i-T, ii-F$.

(A) For a first order reaction $R \rightarrow P$,the integrated rate law is given by:
$k = \frac{1}{t} \ln \frac{[R]_0}{[R]}$
where $[R]_0$ is the initial concentration and $[R]$ is the concentration at time $t$.
Statement $I$ matches this formula,so it is True $(T)$.
Statement $II$ is $k = \frac{1}{t} \ln \frac{[R]}{[R]_0}$,which is the reciprocal of the correct expression,so it is False $(F)$.
Therefore,the correct option is $I-T, II-F$.

(D) For a first-order reaction,the integrated rate equation is given by $k = \frac{1}{t} \ln \frac{[R]_0}{[R]_t}$.
At two different times $t_1$ and $t_2$,we have:
$k = \frac{1}{t_1} \ln \frac{[R]_0}{[R]_1} \implies \ln [R]_1 = \ln [R]_0 - kt_1$
$k = \frac{1}{t_2} \ln \frac{[R]_0}{[R]_2} \implies \ln [R]_2 = \ln [R]_0 - kt_2$
Subtracting the two equations: $\ln [R]_1 - \ln [R]_2 = -kt_1 - (-kt_2) = k(t_2 - t_1)$.
Thus,$k = \frac{1}{t_2 - t_1} \ln \frac{[R]_1}{[R]_2} = -\frac{1}{t_1 - t_2} \ln \frac{[R]_1}{[R]_2} = \frac{1}{t_1 - t_2} \ln \frac{[R]_2}{[R]_1}$.
Comparing this with the given statements:
Statement $I$: $k = \frac{1}{t_1 - t_2} \ln \frac{[R]_1}{[R]_2}$ is False because the denominator should be $(t_2 - t_1)$ or the expression should be $-\frac{1}{t_1 - t_2} \ln \frac{[R]_1}{[R]_2}$.
Statement $II$: $k = -\frac{1}{t_1 - t_2} \ln \frac{[R]_2}{[R]_1} = \frac{1}{t_2 - t_1} \ln \frac{[R]_2}{[R]_1}$,which is also False based on the standard derivation.
Therefore,both statements are False.

(A) 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:
$\frac{kt}{2.303} = \log \frac{[R]_0}{[R]}$
Since $\log \frac{[R]_0}{[R]} = -\log \frac{[R]}{[R]_0}$,we can write:
$\frac{kt}{2.303} = -\log \frac{[R]}{[R]_0}$
Or,$\log \frac{[R]}{[R]_0} = -\frac{kt}{2.303}$
Therefore,statement $I$ is $True$ $(T)$ and statement $II$ is $False$ $(F)$.

(N/A) For a pseudo first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{C_{0}}{C}$.
At $t = 30 \ min$: $k_{1} = \frac{2.303}{30} \log \frac{0.850}{0.800} = 0.07677 \times 0.02633 = 2.02 \times 10^{-3} \ min^{-1}$.
At $t = 60 \ min$: $k_{2} = \frac{2.303}{60} \log \frac{0.850}{0.754} = 0.03838 \times 0.0520 = 1.996 \times 10^{-3} \ min^{-1}$.
At $t = 90 \ min$: $k_{3} = \frac{2.303}{90} \log \frac{0.850}{0.710} = 0.02559 \times 0.0781 = 2.00 \times 10^{-3} \ min^{-1}$.
Since $k$ is constant,it follows a pseudo first order reaction. The average $k \approx 2.00 \times 10^{-3} \ min^{-1}$.
The rate law is $Rate = k^{\prime} [Ester][H_{2}O]$. Since $[H_{2}O]$ is constant,$k = k^{\prime} [H_{2}O]$.
$k^{\prime} = \frac{k}{[H_{2}O]} = \frac{2.00 \times 10^{-3} \ min^{-1}}{55 \ mol \ L^{-1}} = 3.636 \times 10^{-5} \ L \ mol^{-1} \ min^{-1}$.

(N/A) Average rate $= -\frac{\Delta[R]}{\Delta t} = -\frac{0.173 - 0.312}{60 - 30} = -\frac{-0.139}{30} = 4.63 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
$(b)$ For a first order reaction,$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$.
Using data at $t = 30 \ s$: $k = \frac{2.303}{30} \log \frac{0.551}{0.312} = 0.07677 \times \log(1.766) = 0.07677 \times 0.247 = 1.896 \times 10^{-2} \ s^{-1}$.
Using data at $t = 60 \ s$: $k = \frac{2.303}{60} \log \frac{0.551}{0.173} = 0.03838 \times \log(3.185) = 0.03838 \times 0.503 = 1.930 \times 10^{-2} \ s^{-1}$.
Taking the average,$k \approx 1.9 \times 10^{-2} \ s^{-1}$.

(N/A) For a pseudo first-order reaction,the rate constant $k'$ is given by the formula:
$k' = \frac{2.303}{t} \log \frac{C_0}{C_t}$
Given $C_0 = 0.850 \text{ mol L}^{-1}$.
At $t = 30 \text{ min}$,$C_t = 0.800 \text{ mol L}^{-1}$:
$k'_1 = \frac{2.303}{30} \log \frac{0.850}{0.800} = 0.07677 \times \log(1.0625) \approx 0.07677 \times 0.0263 = 0.00202 \text{ min}^{-1}$.
At $t = 60 \text{ min}$,$C_t = 0.754 \text{ mol L}^{-1}$:
$k'_2 = \frac{2.303}{60} \log \frac{0.850}{0.754} = 0.03838 \times \log(1.1273) \approx 0.03838 \times 0.0520 = 0.00200 \text{ min}^{-1}$.
At $t = 90 \text{ min}$,$C_t = 0.710 \text{ mol L}^{-1}$:
$k'_3 = \frac{2.303}{90} \log \frac{0.850}{0.710} = 0.02559 \times \log(1.1972) \approx 0.02559 \times 0.0781 = 0.00200 \text{ min}^{-1}$.
Since the values of $k'$ are nearly constant,the reaction follows pseudo first-order kinetics.
The average value of $k' \approx 2.00 \times 10^{-3} \text{ min}^{-1}$.

(N/A) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{V_{\infty} - V_0}{V_{\infty} - V_t}$.
Here,$V_0 = 24.40 \, mL$,$V_{\infty} = 47.50 \, mL$,and $V_t$ is the volume at time $t$.
$1$. At $t = 20 \, min$: $k = \frac{2.303}{20} \log \frac{47.50 - 24.40}{47.50 - 25.82} = \frac{2.303}{20} \log \frac{23.10}{21.68} \approx 0.00308 \, min^{-1}$.
$2$. At $t = 75 \, min$: $k = \frac{2.303}{75} \log \frac{47.50 - 24.40}{47.50 - 29.35} = \frac{2.303}{75} \log \frac{23.10}{18.15} \approx 0.00315 \, min^{-1}$.
$3$. At $t = 120 \, min$: $k = \frac{2.303}{120} \log \frac{47.50 - 24.40}{47.50 - 31.75} = \frac{2.303}{120} \log \frac{23.10}{15.75} \approx 0.00318 \, min^{-1}$.
Since the value of $k$ remains approximately constant at different time intervals,the reaction follows first-order kinetics.

(T, T, T) True $(T)$: The hydrolysis of an ester follows the reaction: $RCOOR' + H_2O \rightarrow RCOOH + R'OH$. This produces a carboxylic acid and an alcohol.
$(b)$ True $(T)$: In the hydrolysis of an ester,water is taken in large excess. Therefore,its concentration remains effectively constant throughout the reaction.
$(c)$ True $(T)$: Since water is in large excess,its concentration $[H_2O]$ does not change significantly during the reaction,which is why the reaction follows pseudo-first-order kinetics.

(N/A) The hydrolysis of sucrose $(C_{12}H_{22}O_{11})$ in the presence of an acid catalyst $(H^{+})$ produces glucose and fructose. The reaction is as follows:
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^{+}} C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{fructose})$

(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 $75 \%$ completion,$[A]_t = [A]_0 - 0.75[A]_0 = 0.25[A]_0$,so $k = \frac{2.303}{90} \log \frac{1}{0.25} = \frac{2.303}{90} \log 4 = \frac{2.303 \times 2 \log 2}{90} = \frac{2.303 \times 0.6}{90} \ min^{-1}$.
For $60 \%$ completion,$[A]_t = [A]_0 - 0.60[A]_0 = 0.40[A]_0$,so $t = \frac{2.303}{k} \log \frac{1}{0.4} = \frac{2.303}{k} \log 2.5$.
Substituting $k$: $t = \frac{2.303 \times 90}{2.303 \times 0.6} \times 0.4 = \frac{90 \times 0.4}{0.6} = \frac{36}{0.6} = 60 \ minutes$.

(D) For a first-order reaction,the concentration at time $t$ is given by $[C]_t = [C]_0 e^{-kt},$ where $k = \frac{\ln 2}{t_{1/2}}.$
Given $[A]_0 = [B]_0,$ we want to find $t$ such that $[A]_t = 4[B]_t.$
Substituting the expressions: $[A]_0 e^{-(\ln 2 / 300)t} = 4[B]_0 e^{-(\ln 2 / 180)t}.$
Since $[A]_0 = [B]_0,$ we have $e^{-(\ln 2 / 300)t} = 4 e^{-(\ln 2 / 180)t}.$
Rearranging gives $e^{(\frac{\ln 2}{180} - \frac{\ln 2}{300})t} = 4.$
Taking the natural logarithm on both sides: $(\frac{\ln 2}{180} - \frac{\ln 2}{300})t = \ln 4 = 2 \ln 2.$
Dividing by $\ln 2$: $(\frac{1}{180} - \frac{1}{300})t = 2.$
Solving for $t$: $(\frac{300 - 180}{180 \times 300})t = 2 \Rightarrow \frac{120}{54000}t = 2.$
$t = \frac{2 \times 54000}{120} = \frac{108000}{120} = 900 \ s.$

(D) For a first order reaction,the rate equation is given by: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$
Given: $k = 4.606 \times 10^{-3} \ s^{-1}$,$[A]_0 = 2.0 \ g$,$[A]_t = 0.2 \ g$
Substituting the values: $4.606 \times 10^{-3} = \frac{2.303}{t} \log_{10} \frac{2.0}{0.2}$
$4.606 \times 10^{-3} = \frac{2.303}{t} \log_{10} (10)$
Since $\log_{10} (10) = 1$,we have: $4.606 \times 10^{-3} = \frac{2.303}{t}$
$t = \frac{2.303}{4.606 \times 10^{-3}} = \frac{1}{2} \times 10^3 = 500 \ s$

(A) For a first order reaction,the integrated rate law is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
At half-life,$t = t_{1/2}$ and $[A]_t = \frac{[A]_0}{2}$.
Substituting these values into the equation:
$k = \frac{2.303}{t_{1/2}} \log \frac{[A]_0}{[A]_0 / 2} = \frac{2.303}{t_{1/2}} \log 2$.
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,$t_{1/2} = \frac{0.693}{k}$.
This expression shows that the half-life of a first order reaction is independent of the initial concentration of the reactant.

(D) For a $1^{st}$ order reaction,the rate equation is given by:
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
Given that the reaction is $\frac{2}{3}$ complete,the amount reacted $x = \frac{2}{3} a$,where $a$ is the initial concentration.
Therefore,the remaining concentration $[A]_t = a - \frac{2}{3} a = \frac{a}{3}$.
Substituting the values into the equation:
$t = \frac{2.303}{4.3 \times 10^{-4}} \log \frac{a}{a/3}$
$t = \frac{2.303}{4.3 \times 10^{-4}} \log 3$
Using $\log 3 \approx 0.4771$:
$t = \frac{2.303 \times 0.4771}{4.3 \times 10^{-4}} \approx \frac{1.0988}{4.3 \times 10^{-4}} \approx 2555 \, s$
Rounding to the given options,$t = 2.5 \times 10^{3} \, s$.

(B) The reaction is $A_{(g)} \rightarrow 2 B_{(g)} + C_{(g)}$.
Let $P$ be the decrease in pressure of $A$ at time $t$.

Initial pressure	$P_0$	$0$	$0$
Pressure at time $t$	$P_0 - P$	$2P$	$P$

Total pressure $P_t = (P_0 - P) + 2P + P = P_0 + 2P$.
Therefore,$2P = P_t - P_0$,which gives $P = \frac{P_t - P_0}{2}$.
The pressure of $A$ at time $t$ is $P_A = P_0 - P = P_0 - \frac{P_t - P_0}{2} = \frac{2P_0 - P_t + P_0}{2} = \frac{3P_0 - P_t}{2}$.
The integrated rate equation for a first order reaction is $k = \frac{2.303}{t} \log \left( \frac{P_0}{P_A} \right)$.
Substituting $P_A$:
$k = \frac{2.303}{t} \log \left( \frac{P_0}{\frac{3P_0 - P_t}{2}} \right) = \frac{2.303}{t} \log \left( \frac{2P_0}{3P_0 - P_t} \right)$.