First Order reaction Questions in English

(B) For a $1^{st}$ order reaction,the rate constant $k$ is given by:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given: $t = 1.388 \, h = 1.388 \times 3600 \, s = 4996.8 \, s$.
If $75\%$ is consumed,then $[A]_t = [A]_0 - 0.75[A]_0 = 0.25[A]_0$.
$k = \frac{2.303}{4996.8} \log \frac{[A]_0}{0.25[A]_0} = \frac{2.303}{4996.8} \log(4) = \frac{2.303 \times 0.602}{4996.8} \approx 2.775 \times 10^{-4} \, s^{-1} \approx 2.8 \times 10^{-4} \, s^{-1}$.

(A) For the reaction: $((CH_3)_2CHN = NCH(CH_3)_2)_{(g)} \rightarrow (N_2)_{(g)} + (C_6H_{14})_{(g)}$
At $t=0$: Initial pressure = $P_o$,$N_2$ = $0$,$C_6H_{14}$ = $0$
At time $t$: Pressure = $(P_o - x)$,$N_2$ = $x$,$C_6H_{14}$ = $x$
The total pressure $P_t$ at time $t$ is given by:
$P_t = (P_o - x) + x + x = P_o + x$
Therefore,$x = P_t - P_o$
The partial pressure of the reactant at time $t$ is $(P_o - x) = P_o - (P_t - P_o) = 2P_o - P_t$
For a first-order reaction,the rate constant $K$ is:
$K = \frac{2.303}{t} \log \frac{P_o}{P_o - x}$
Substituting $(P_o - x) = 2P_o - P_t$:
$K = \frac{2.303}{t} \log \frac{P_o}{2P_o - P_t}$

(A) The rate constant $K$ for a first-order reaction is given by the formula: $K = \frac{2.303}{t} \log \frac{a}{a-x}$.
Given that $t = 10 \text{ minutes}$ and $\frac{a}{a-x} = 8$.
Substituting the values: $K = \frac{2.303}{10} \log 8$.
Since $8 = 2^3$,we have $\log 8 = \log(2^3) = 3 \log 2$.
Therefore,$K = \frac{2.303 \times 3 \log 2}{10}$.

(C) The rate law for the given reaction is $Rate = k[N_2O_5]^1$ because the unit of the rate constant $(s^{-1})$ indicates a first-order reaction.
Given:
$Rate = 2.40 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$
$k = 3.0 \times 10^{-5} \ s^{-1}$
Substituting the values into the rate equation:
$2.40 \times 10^{-5} = (3.0 \times 10^{-5}) \times [N_2O_5]$
$[N_2O_5] = \frac{2.40 \times 10^{-5}}{3.0 \times 10^{-5}}$
$[N_2O_5] = 0.8 \ mol \ L^{-1}$

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 69.3 \ min$,so $k = \frac{0.693}{69.3} = 0.01 \ min^{-1}$.
For $80\%$ completion,the remaining concentration is $100 - 80 = 20\%$.
The time $t$ is calculated using the formula: $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right)$.
Substituting the values: $t = \frac{2.303}{0.01} \log \left( \frac{100}{20} \right)$.
$t = 230.3 \times \log(5) = 230.3 \times 0.69897 \approx 160.97 \ min$.

(A) For a first-order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$.
For reaction $A$,$94\%$ has reacted,so the remaining amount is $100 - 94 = 6\%$. Thus,$K_1 = \frac{2.303}{t} \log \frac{100}{6}$.
For reaction $B$,$50\%$ has reacted,so the remaining amount is $100 - 50 = 50\%$. Thus,$K_2 = \frac{2.303}{t} \log \frac{100}{50}$.
Taking the ratio: $\frac{K_1}{K_2} = \frac{\log(100/6)}{\log(100/50)} = \frac{\log(16.67)}{\log(2)} = \frac{1.2219}{0.3010} \approx 4.06$.

(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: $[A]_0 = 1.0 \, M$,$[A]_t = 0.25 \, M$,$t = 20 \, \text{min}$.
Substituting the values:
$k = \frac{2.303}{20} \log \frac{1.0}{0.25}$
$k = \frac{2.303}{20} \log(4)$
Since $\log(4) \approx 0.6021$:
$k = \frac{2.303 \times 0.6021}{20} \approx 0.06931 \, \text{min}^{-1}$.

(A) The decomposition of ammonium nitrite $(NH_4NO_2)$ is a well-known example of a first-order reaction.
The rate law for this reaction is given by $Rate = k[NH_4NO_2]^1$.
Other options like the decomposition of $HI$ or $NO_2$ at high temperatures are typically second-order reactions,and the reaction of $NO$ with $O_2$ is a third-order reaction.

(C) For a first-order reaction,the half-life $(t_{1/2})$ is the time taken for the concentration to reduce to half of its initial value.
Given that the concentration decreases from $0.8 \, M$ to $0.4 \, M$ in $15 \, \text{minutes}$,the half-life is $t_{1/2} = 15 \, \text{minutes}$.
To find the time taken for the concentration to change from $0.1 \, M$ to $0.025 \, M$,we note that $0.025 \, M$ is $\frac{1}{4}$ of $0.1 \, M$ (i.e.,$0.1$ $\rightarrow 0.05$ $\rightarrow 0.025$).
This corresponds to two half-lives.
Therefore,the total time required is $2 \times t_{1/2} = 2 \times 15 \, \text{minutes} = 30 \, \text{minutes}$.

(A) The unit of the rate constant is $sec^{-1}$,which indicates that the reaction is a $1^{st}$ order reaction.
The rate law for a $1^{st}$ order reaction is given by: $\text{Rate} = k[A]^1$.
If the initial concentration of $A$ is tripled (i.e.,$[A]_{new} = 3[A]_{old}$),the new rate becomes: $\text{Rate}_{new} = k(3[A]_{old}) = 3 \times \text{Rate}_{old}$.
Therefore,the rate increases by a factor of $3$.

(C) For a first-order reaction,the rate law is given by $r = k[A]$.
Given: $r = 1.5 \times 10^{-2} \, mol \, L^{-1} \, min^{-1}$ and $[A] = 0.5 \, M$.
Substituting the values: $1.5 \times 10^{-2} = k \times 0.5$.
Therefore,the rate constant $k = \frac{1.5 \times 10^{-2}}{0.5} = 3 \times 10^{-2} \, min^{-1}$.
The half-life $(t_{1/2})$ for a first-order reaction is calculated as $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{3 \times 10^{-2}} = 23.1 \, min$.

(D) For a first-order reaction,the rate equation is $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]_t}$.
Given that the reaction is $1/4^{th}$ complete,$[A]_t = [A]_0 - \frac{1}{4}[A]_0 = \frac{3}{4}[A]_0$.
Substituting the values: $t = \frac{2.303}{K} \log \frac{[A]_0}{\frac{3}{4}[A]_0}$.
$t = \frac{2.303}{K} \log \frac{4}{3}$.

(B) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{a}{a-x}$.
For $75\%$ completion,$x = 0.75a$,so $a-x = 0.25a$ and $t = 32 \text{ min}$.
$K = \frac{2.303}{32} \log \frac{100}{25} = \frac{2.303}{32} \log 4 = \frac{2.303 \times 0.602}{32}$.
Now,for $50\%$ completion (half-life,$t_{1/2}$),the formula is $t_{1/2} = \frac{0.693}{K}$.
Substituting $K = \frac{2.303 \times 0.602}{32} \approx \frac{0.693}{32} \times 2 = \frac{0.693}{16}$.
Therefore,$t_{1/2} = \frac{0.693}{0.693 / 16} = 16 \text{ minutes}$.

(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Substituting the given value: $k = \frac{0.693}{69.3 \ s} = 0.01 \ s^{-1} = 10^{-2} \ s^{-1}$.
The rate of reaction $r$ is given by $r = k[A]$.
Substituting the values: $r = (10^{-2} \ s^{-1}) \times (0.10 \ mol \ L^{-1}) = 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

(D) For a first-order reaction,the concentration decreases as follows: $1$ $\xrightarrow{t_{1/2}} 1/2$ $\xrightarrow{t_{1/2}} 1/4$ $\xrightarrow{t_{1/2}} 1/8$ $\xrightarrow{t_{1/2}} 1/16$.
Time taken to reach $1/4$ of initial concentration is $2 \times t_{1/2} = 20 \ min$,so $t_{1/2} = 10 \ min$.
Time taken to reach $1/16$ of initial concentration is $4 \times t_{1/2} = 4 \times 10 \ min = 40 \ min$.

(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 flask $I$: $[A]_0 = 1 \ M$,$[A]_t = 0.25 \ M$,$t = 8 \ hours$.
$k = \frac{2.303}{8} \log(\frac{1}{0.25}) = \frac{2.303}{8} \log(4) = \frac{2.303 \times 0.602}{8} \approx 0.1733 \ h^{-1}$.
For flask $II$: $[A]_0 = 0.6 \ M$,$[A]_t = 0.3 \ M$.
Since $[A]_t = \frac{[A]_0}{2}$,the time taken is the half-life $t_{1/2}$.
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.1733} \approx 4.0 \ hours$.

(A) The rate law for a first-order reaction is given by: $Rate = k[N_2O_5]^1$
Given,$k = 6.2 \times 10^{-4} \, s^{-1}$ and $[N_2O_5] = 1.25 \, mol \, L^{-1}$.
Substituting the values: $Rate = (6.2 \times 10^{-4} \, s^{-1}) \times (1.25 \, mol \, L^{-1})$
$Rate = 7.75 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$.

(B) For a first-order reaction,the integrated rate equation is given by: $Kt = \ln(a) - \ln(a - x)$.
Rearranging this,we get: $\ln(a - x) = -Kt + \ln(a)$.
Converting to base $10$ logarithm: $\log(a - x) = -\frac{Kt}{2.303} + \log(a)$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(a - x)$ and $x = t$,the slope is $-\frac{K}{2.303}$.
Since the plot of $\log(a - x)$ versus $t$ is a straight line,the reaction follows first-order kinetics.

(D) For a first-order reaction,the half-life period is given by $t_{1/2} = \frac{0.693}{K}$.
Since $0.693$ is a constant,$t_{1/2} \times K = 0.693$,which is a constant at a given temperature.
Option $A$ is incorrect because a first-order reaction never goes to completion in finite time.
Option $B$ is incorrect because for a first-order reaction,$t_{1/2}$ is independent of the initial concentration $a$.
Option $C$ is incorrect because the unit of $K$ for a first-order reaction is $\text{s}^{-1}$.

(B) For a reaction of order $n$,the unit of the rate constant is $(mol \ L^{-1})^{1-n} \ s^{-1}$.
For a first-order reaction,$n = 1$.
Substituting $n = 1$ into the formula: $(mol \ L^{-1})^{1-1} \ s^{-1} = (mol \ L^{-1})^0 \ s^{-1} = s^{-1}$.
Therefore,the unit of the rate constant for a first-order reaction is $s^{-1}$.

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
The half-life $t_{1/2} = \frac{0.693}{k}$ is independent of the initial concentration of the reactant.
Therefore,statement $C$ is incorrect because the half-life of a first-order reaction is independent of the initial concentration.
For $99.6\%$ completion,the remaining concentration is $0.4\%$ of the initial concentration $([A]_t = 0.004[A]_0)$.
Using $t = \frac{2.303}{k} \log(\frac{[A]_0}{[A]_t}) = \frac{2.303}{k} \log(\frac{1}{0.004}) = \frac{2.303}{k} \log(250) \approx \frac{2.303}{k} \times 2.398 \approx \frac{5.52}{k}$.
Since $t_{1/2} = \frac{0.693}{k}$,then $8 \times t_{1/2} = 8 \times \frac{0.693}{k} \approx \frac{5.54}{k}$,which is approximately equal to the time for $99.6\%$ completion.

(D) For a first-order reaction,the rate law is given by $r = k[A]$.
Given $r = 1 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1}$ and $[A] = 0.2 \, \text{M}$.
Substituting the values: $1 \times 10^{-2} = k \times 0.2$.
$k = \frac{1 \times 10^{-2}}{0.2} = 5 \times 10^{-2} \, \text{min}^{-1}$.
The half-life $t_{1/2}$ is calculated as $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{5 \times 10^{-2}} = \frac{69.3}{5} = 13.86 \, \text{min} \approx 14 \, \text{min}$.

(B) For a first-order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $93.75\%$ completion,$[A]_t = 100 - 93.75 = 6.25$.
$t_{93.75\%} = \frac{2.303}{k} \log \frac{100}{6.25} = \frac{2.303}{k} \log 16 = \frac{2.303}{k} \log 2^4 = 4 \times \frac{2.303 \times 0.3010}{k}$.
Since $t_{0.5} = \frac{0.693}{k} = \frac{2.303 \times 0.3010}{k}$,we get $t_{93.75\%} = 4 \times t_{0.5}$.

(B) For a first-order reaction,the half-life period is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Given,$k = 0.6932 \ hr^{-1}$.
Substituting the value: $t_{1/2} = \frac{0.693}{0.6932} \approx 1 \ hr$.

(C) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{a}{a-x}$.
For the first part,$a = 100$,$x = 90$,and $t = 10 \ hours$:
$K = \frac{2.303}{10} \log \frac{100}{100-90} = \frac{2.303}{10} \log 10 = \frac{2.303}{10} \times 1 = 0.2303 \ h^{-1}$.
For the second part,$a = 100$,$x = 99.9$,and we need to find $t$:
$t = \frac{2.303}{K} \log \frac{100}{100-99.9} = \frac{2.303}{0.2303} \log \frac{100}{0.1} = \frac{2.303}{0.2303} \log 1000 = 10 \times 3 = 30 \ hours$.

(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 $99\%$ completion,$[A]_t = 100 - 99 = 1$,so $K = \frac{2.303}{32} \log \frac{100}{1} = \frac{2.303}{32} \times 2$ ........... $(1)$
For $99.9\%$ completion,$[A]_t = 100 - 99.9 = 0.1$,so $K = \frac{2.303}{t} \log \frac{100}{0.1} = \frac{2.303}{t} \times 3$ ............ $(2)$
Equating $(1)$ and $(2)$:
$\frac{2.303 \times 2}{32} = \frac{2.303 \times 3}{t}$
$\frac{2}{32} = \frac{3}{t}$
$t = \frac{3 \times 32}{2} = 48 \ \text{minutes}$.

(D) For a first-order reaction,the integrated rate equation is given by: $\log(a - x) = -\frac{kt}{2.303} + \log a$.
This equation is of the form $y = mx + c$,where the slope $m = -\frac{k}{2.303}$. Thus,a plot of $\log(a - x)$ versus $t$ is a straight line with a negative slope. This matches graph $A$.
Additionally,for a first-order reaction,the half-life period is given by: $t_{1/2} = \frac{0.693}{k}$.
This shows that $t_{1/2}$ is independent of the initial concentration $a$. Thus,a plot of $t_{1/2}$ versus $a$ is a horizontal line parallel to the $a$-axis. This matches graph $C$.
Therefore,both graphs $A$ and $C$ represent a first-order reaction.

(A) For a first-order reaction,the rate constant $k$ is related to the half-life $t_{1/2}$ by the formula:
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 20 \ min$.
Substituting the value:
$k = \frac{0.693}{20} = 0.03465 \ min^{-1}$
Therefore,the correct option is $A$.

(B) For a first-order reaction,the relationship between initial concentration $[A]_0$ and concentration at time $t$,$[A]_t$,is given by $[A]_t = \frac{[A]_0}{2^n}$,where $n$ is the number of half-lives.
Given: $[A]_0 = 0.08 \ mol/L$,$[A]_t = 0.01 \ mol/L$,and $t_{1/2} = 10 \ minutes$.
$2^n = \frac{[A]_0}{[A]_t} = \frac{0.08}{0.01} = 8$
Since $2^3 = 8$,we have $n = 3$.
The total time $t$ is given by $t = n \times t_{1/2}$.
$t = 3 \times 10 \ minutes = 30 \ minutes$.

(A) Let the initial pressure of $A$ be $P_0$.
Reaction: $A(g) \rightarrow 2B(g) + C(g)$
At $t=0$: $P_0, 0, 0$ (Total pressure $= P_0$)
At $t=10 \, min$: $P_0-x, 2x, x$ (Total pressure $= P_0 + 2x = 176 \, mm$)
At $t=\infty$: $0, 2P_0, P_0$ (Total pressure $= 3P_0 = 270 \, mm$)
$(1)$ Initial pressure $P_0 = 270 / 3 = 90 \, mm$.
$(2)$ At $t=10 \, min$,$P_0 + 2x = 176 \implies 90 + 2x = 176 \implies 2x = 86 \implies x = 43 \, mm$.
Pressure of $A$ remaining $= P_0 - x = 90 - 43 = 47 \, mm$.
$(3)$ Rate constant $k = \frac{2.303}{t} \log \frac{P_0}{P_0-x} = \frac{2.303}{10} \log \frac{90}{47} = \frac{2.303}{10} \times 0.2821 = 6.496 \times 10^{-2} \, min^{-1}$.

(A) For a first-order reaction,the formula is $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]_t}$.
Here,$[R]_0$ is the initial concentration and $[R]_t$ is the concentration at time $t$.
Given,$k = 60 \ s^{-1}$ and $[R]_t = \frac{[R]_0}{10}$.
Substituting the values: $t = \frac{2.303}{60} \log \frac{[R]_0}{[R]_0/10}$.
$t = \frac{2.303}{60} \log(10)$.
Since $\log(10) = 1$,$t = \frac{2.303}{60} \approx 3.838 \times 10^{-2} \ s$.

(A) For a first-order reaction,the integrated rate law is given by $[A]_t = [A]_0 e^{-kt}$.
The degree of dissociation (or fraction reacted) $\alpha$ is given by $\alpha = \frac{[A]_0 - [A]_t}{[A]_0} = 1 - \frac{[A]_t}{[A]_0} = 1 - e^{-kt}$.
Thus,option $A$ is correct.
In the Arrhenius equation $k = A e^{-E_a/RT}$,the pre-exponential factor $A$ has the same units as the rate constant $k$. For a first-order reaction,the unit of $k$ is $s^{-1}$,which is the reciprocal of time,not time itself. Therefore,option $B$ is incorrect.
For a first-order reaction,the plot of $\ln[A]$ versus time is a straight line,not the reciprocal of concentration. Therefore,option $C$ is incorrect.

(D) For a first-order reaction,the half-life period $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$.
Since the half-life of a first-order reaction is independent of the initial concentration of the reactant,the initial concentration of $0.5 \ M$ does not affect the value of $t_{1/2}$.
Therefore,the correct expression is $\frac{\ln 2}{k}$.

(A) For a first-order reaction,the rate constant $k$ is related to the half-life $t_{1/2}$ by the formula:
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 1386 \ s$.
Substituting the value:
$k = \frac{0.693}{1386} \ s^{-1}$
$k = 0.0005 \ s^{-1}$
$k = 0.5 \times 10^{-3} \ s^{-1}$

(B) For a first-order reaction,the integrated rate equation is given by:
$\ln \frac{C_t}{C_0} = -k_1 t$
Rearranging this,we get:
$\ln \frac{C_0}{C_t} = k_1 t$
Or,in exponential form:
$C_t = C_0 e^{-k_1 t}$
Which can be rewritten as:
$C_t e^{k_1 t} = C_0$
Thus,option $B$ is the correct representation.

(C) For a first order reaction,the rate constant is given by $K = \frac{2.303}{t} \log \frac{a}{a-x}$.
For $90\%$ completion,$x = 0.9a$,so $a-x = 0.1a$. Thus,$t_{90\%} = \frac{2.303}{K} \log \frac{a}{0.1a} = \frac{2.303}{K} \log 10 = \frac{2.303}{K}$.
We know that the half-life is $t_{1/2} = \frac{0.693}{K}$,which implies $K = \frac{0.693}{t_{1/2}}$.
Substituting $K$ in the expression for $t_{90\%}$:
$t_{90\%} = \frac{2.303}{0.693 / t_{1/2}} = \frac{2.303}{0.693} \times t_{1/2} \approx 3.32 \times t_{1/2}$.
Therefore,the time is approximately $3.3$ times the half-life.

(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:
$k = 10^{-2} \ sec^{-1}$
$[A]_0 = 20 \ g$
$[A]_t = 5 \ g$
Substituting the values:
$t = \frac{2.303}{10^{-2}} \log \frac{20}{5}$
$t = 230.3 \times \log(4)$
$t = 230.3 \times 0.6021$
$t \approx 138.6 \ sec$

(C) For a first-order reaction,the rate $R$ at time $t$ is given by $R = k[A]_t$,where $[A]_t = [A]_0 e^{-kt}$.
Thus,$R_t = k[A]_0 e^{-kt} = R_0 e^{-kt}$.
Given $R_{10} = 0.04 \ mol \ L^{-1} \ s^{-1}$ and $R_{20} = 0.03 \ mol \ L^{-1} \ s^{-1}$.
Taking the ratio: $\frac{R_{10}}{R_{20}} = \frac{e^{-k(10)}}{e^{-k(20)}} = e^{10k} = \frac{0.04}{0.03} = 1.333$.
Taking natural log on both sides: $10k = \ln(1.333) = 0.2877$.
$k = 0.02877 \ s^{-1}$.
The half-life period is $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02877} \approx 24.1 \ s$.

(A) For a first-order reaction,the half-life $t_{1/2}$ is constant.
After $2 \ hours$,the reaction is $50\%$ complete,meaning $t_{1/2} = 2 \ hours$.
After another $2 \ hours$ (total $4 \ hours$),the remaining $50\%$ of the reactant is halved again,making the reaction $75\%$ complete $(50\% + 25\% = 75\%)$.
Since the time taken for each successive half-life is constant $(2 \ hours)$,the reaction follows first-order kinetics.

(C) Enzyme-catalysed reactions follow first-order kinetics.
The concentration of the substance decreases from $1.28 \; mg \; L^{-1}$ to $0.04 \; mg \; L^{-1}$.
We can calculate the number of half-lives $(n)$ as follows:
$1.28$ $\xrightarrow{t_{1/2}} 0.64$ $\xrightarrow{t_{1/2}} 0.32$ $\xrightarrow{t_{1/2}} 0.16$ $\xrightarrow{t_{1/2}} 0.08$ $\xrightarrow{t_{1/2}} 0.04$
This sequence shows that the concentration reduces by half $5$ times,so $n = 5$.
The total time required is $t = n \times t_{1/2}$.
$t = 5 \times 138 \; s = 690 \; s$.

(B) Given: Half-life period $(t_{1/2})$ = $1386 \ s$.
For a first-order reaction,the relationship between half-life and rate constant $(k)$ is given by:
$t_{1/2} = \frac{0.693}{k}$
Rearranging for $k$:
$k = \frac{0.693}{t_{1/2}}$
Substituting the value:
$k = \frac{0.693}{1386} \ s^{-1}$
$k = 0.0005 \ s^{-1}$
$k = 5.0 \times 10^{-4} \ s^{-1} = 0.5 \times 10^{-3} \ s^{-1}$.

(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 that $60\%$ is completed in $60 \ min$,$[A]_t = 100 - 60 = 40$.
$k = \frac{2.303}{60} \log \frac{100}{40} = \frac{2.303}{60} \log 2.5$.
Using $\log 2.5 = \log(10/4) = \log 10 - \log 4 = 1 - 0.60 = 0.40$.
$k = \frac{2.303 \times 0.40}{60} = \frac{0.9212}{60} \ min^{-1}$.
For $50\%$ completion $(t_{1/2})$,$t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693 \times 60}{0.9212} \approx 45.12 \ min$.
Thus,the reaction is completed in approximately $45 \ min$.

(C) For a first-order reaction,the rate constant $k$ is given by the equation:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
At half-life $(t_{1/2})$,the concentration of the reactant $[A]_t = \frac{[A]_0}{2}$.
Substituting this into the equation:
$t_{1/2} = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0 / 2}$
$t_{1/2} = \frac{2.303}{k} \log 2$
Since $2.303 \log 2 = \ln 2$,we get:
$t_{1/2} = \frac{\ln 2}{k}$
The half-life of a first-order reaction is independent of the initial concentration of the reactant.

(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 6.93 \, \text{min}$,so $k = \frac{0.693}{6.93} = 0.1 \, \text{min}^{-1}$.
The integrated rate equation for a first-order reaction is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
For $99 \%$ completion,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.
Substituting the values: $0.1 = \frac{2.303}{t} \log \frac{100}{1}$.
$0.1 = \frac{2.303 \times 2}{t}$.
$t = \frac{4.606}{0.1} = 46.06 \, \text{min}$.

(B) For a first order reaction,the rate constant $k$ is given by:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Substituting the given values:
$k = \frac{2.303}{40} \log \frac{0.1}{0.025}$
$k = \frac{2.303}{40} \log 4$
$k = \frac{2.303 \times 0.6020}{40} \approx 3.47 \times 10^{-2} \ min^{-1}$
The rate of reaction $R$ is given by $R = k[A]$.
For $[A] = 0.01 \ M$:
$R = (3.47 \times 10^{-2}) \times 0.01$
$R = 3.47 \times 10^{-4} \ M/min$.

(D) The decomposition reaction is: $H_2O_{2(aq)} \rightarrow H_2O_{(l)} + \frac{1}{2} O_{2(g)}$
For a first order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given $[A]_0 = 0.5 \ M$,$[A]_t = 0.125 \ M$,and $t = 50 \ min$:
$k = \frac{2.303}{50} \log \frac{0.5}{0.125} = \frac{2.303}{50} \log(4) = \frac{2.303 \times 0.602}{50} \approx 0.0277 \ min^{-1}$
When $[H_2O_2] = 0.05 \ M$,the rate of disappearance of $H_2O_2$ is:
$Rate_{H_2O_2} = k[H_2O_2] = 0.0277 \times 0.05 = 1.385 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$
From the stoichiometry,$-\frac{d[H_2O_2]}{dt} = 2 \frac{d[O_2]}{dt}$,so the rate of formation of $O_2$ is:
$\frac{d[O_2]}{dt} = \frac{1}{2} \times Rate_{H_2O_2} = \frac{1.385 \times 10^{-3}}{2} = 6.925 \times 10^{-4} \ mol \ L^{-1} \ min^{-1} \approx 6.93 \times 10^{-4} \ mol \ min^{-1}$

(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 $20\%$ completion,$[A]_t = 100 - 20 = 80$,so $K = \frac{2.303}{t_{20}} \log \frac{100}{80} = \frac{2.303}{t_{20}} \log 1.25$.
For $60\%$ completion,$[A]_t = 100 - 60 = 40$,so $K = \frac{2.303}{t_{60}} \log \frac{100}{40} = \frac{2.303}{t_{60}} \log 2.5$.
Equating the two expressions for $K$: $\frac{1}{t_{20}} \log 1.25 = \frac{1}{t_{60}} \log 2.5$.
Therefore,$\frac{t_{60}}{t_{20}} = \frac{\log 2.5}{\log 1.25} = \frac{\log(10/4)}{\log(10/8)} = \frac{\log 10 - \log 4}{\log 10 - \log 8} = \frac{1 - 2 \log 2}{1 - 3 \log 2}$.
Given $\log 2 = 0.3$,we have $\frac{t_{60}}{t_{20}} = \frac{1 - 2(0.3)}{1 - 3(0.3)} = \frac{1 - 0.6}{1 - 0.9} = \frac{0.4}{0.1} = 4$.

(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 100 \ min$,so $k = \frac{0.693}{100} = 6.93 \times 10^{-3} \ min^{-1}$.
The integrated rate equation for a first order reaction is $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
Here,$[A]_t = 10 \% \text{ of } [A]_0$,so $\frac{[A]_0}{[A]_t} = \frac{100}{10} = 10$.
Substituting the values: $t = \frac{2.303}{6.93 \times 10^{-3}} \log(10)$.
Since $\log(10) = 1$,$t = \frac{2.303}{0.00693} \approx 332.3 \ min$.
Thus,the time required is approximately $332 \ min$.