First Order reaction Questions in English

(D) For a first-order reaction,the relationship between the decay constant $(k)$ and the half-life $(t_{1/2})$ is given by the formula: $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 138.6 \ minutes$.
Substituting the value: $k = \frac{0.693}{138.6 \ min} = 0.005 \ min^{-1}$.
Therefore,the correct option is $D$.

(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 that $75\%$ of the reaction is completed in $32$ minutes,the remaining concentration $[A]_t = 100\% - 75\% = 25\% = 0.25 [A]_0$.
Substituting these values: $k = \frac{2.303}{32} \log \frac{[A]_0}{0.25 [A]_0} = \frac{2.303}{32} \log(4) = \frac{2.303 \times 0.602}{32}$.
The half-life $t_{1/2}$ is given by $t_{1/2} = \frac{0.693}{k}$.
Substituting $k$: $t_{1/2} = \frac{0.693 \times 32}{2.303 \times 0.602} \approx 16 \text{ minutes}$.
Alternatively,$75\%$ completion corresponds to $2$ half-lives $(2 \times t_{1/2} = 32 \text{ min})$,so $t_{1/2} = 16 \text{ minutes}$.

(B) Radioactive decay follows first-order kinetics because the rate of decay is directly proportional to the number of radioactive nuclei present at that time. Therefore,it is a $First \ order \ reaction$.

(A) The reaction is of first order because the unit of the rate constant is $s^{-1}$.
For a first-order reaction,the half-life period is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Substituting the given value of $k = 2.34 \ s^{-1}$:
$t_{1/2} = \frac{0.693}{2.34} \approx 0.296 \ s$.
Rounding to two decimal places,we get $0.30 \ s$.

(B) For a first order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
In the first case,$10\%$ is completed,so $[A]_t = 100 - 10 = 90$ and $t = 20 \text{ min}$.
$K = \frac{2.303}{20} \log \frac{100}{90} = \frac{2.303}{20} \log \frac{10}{9}$ ..... $(i)$
In the second case,$19\%$ is completed,so $[A]_t = 100 - 19 = 81$ and $t = ?$.
$K = \frac{2.303}{t} \log \frac{100}{81}$ ..... $(ii)$
Equating $(i)$ and $(ii)$:
$\frac{2.303}{20} \log \frac{10}{9} = \frac{2.303}{t} \log \frac{100}{81}$
$\frac{1}{20} \log \frac{10}{9} = \frac{1}{t} \log (\frac{10}{9})^2$
$\frac{1}{20} \log \frac{10}{9} = \frac{2}{t} \log \frac{10}{9}$
$\frac{1}{20} = \frac{2}{t}$
$t = 40 \text{ min}$.

(A) For a first-order reaction,the rate constant $K$ is given by the formula: $K = \frac{2.303}{t} \log_{10} \frac{p_0}{p_t}$.
Given: $p_0 = 500 \ atm$,$K = 3.38 \times 10^{-5} \ s^{-1}$,and $t = 10 \ minutes = 600 \ s$.
Substituting the values: $3.38 \times 10^{-5} = \frac{2.303}{600} \log_{10} \frac{500}{p_t}$.
$0.00880 = \log_{10} \frac{500}{p_t}$.
Taking the antilog: $10^{0.00880} = \frac{500}{p_t} \approx 1.0204$.
$p_t = \frac{500}{1.0204} \approx 490 \ atm$.

(D) For a first order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.
Thus,the unit of $K$ is $Time^{-1}$ (e.g.,$s^{-1}$ or $min^{-1}$).
Option $(d)$ states the unit is $mole^{-1} \ min^{-1}$,which is the unit for a second order reaction.
Therefore,statement $(d)$ is incorrect.

(C) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Given: $t = 2 \times 10^2 \ s$,$[A]_0 = 800 \ mol/dm^3$,$[A]_t = 50 \ mol/dm^3$.
Substituting the values: $k = \frac{2.303}{2 \times 10^2} \log_{10} \frac{800}{50}$.
$k = \frac{2.303}{200} \log_{10} 16$.
Since $\log_{10} 16 = \log_{10} 2^4 = 4 \times 0.3010 = 1.204$.
$k = \frac{2.303 \times 1.204}{200} = \frac{2.7728}{200} = 1.3864 \times 10^{-2} \ s^{-1}$.

(A) For a first-order reaction,the integrated rate equation is given by:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given:
$k = 6 \ min^{-1}$
$[A]_0 = 0.5 \ mol \ L^{-1}$
$[A]_t = 0.05 \ mol \ L^{-1}$
Substituting the values:
$6 = \frac{2.303}{t} \log \frac{0.5}{0.05}$
$6 = \frac{2.303}{t} \log(10)$
Since $\log(10) = 1$,we have:
$6 = \frac{2.303}{t}$
$t = \frac{2.303}{6} \approx 0.384 \ min$

(C) The rate law for a reaction of order $n$ is given by: $Rate = k[Concentration]^n$.
For a first order reaction,$n = 1$.
Therefore,$Rate = k[Concentration]^1$.
The unit of rate is $mole \ litre^{-1} \ sec^{-1}$ and the unit of concentration is $mole \ litre^{-1}$.
Substituting these units: $mole \ litre^{-1} \ sec^{-1} = k \times (mole \ litre^{-1})^1$.
Solving for $k$,we get: $k = sec^{-1}$.

(A) The decomposition of ammonium nitrite $(NH_4NO_2)$ is a first order reaction.
In this reaction,the rate is given by $Rate = k[NH_4NO_2]^1$.
Since the rate depends on the concentration of only one reactant raised to the power of $1$,it is a first order reaction.
Therefore,the correct option is $A$.

(C) The rate constant expression for a $1^{st}$ order reaction is given by:
$K = \frac{2.303}{t} \log_{10} \frac{a}{(a - x)}$
Where $a$ is the initial concentration and $(a - x)$ is the concentration at time $t$.

(A) For a first-order reaction,the half-life period $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given the rate constant $k = 6.2 \times 10^{-4} \ s^{-1}$.
Substituting the value of $k$ in the formula:
$t_{1/2} = \frac{0.693}{6.2 \times 10^{-4}} \ s$
$t_{1/2} = 1117.7 \ s$
Therefore,the correct option is $A$.

(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 $t = 30 \ min$ and the reaction is $30\%$ complete,so $[A]_t = 100 - 30 = 70$.
$k = \frac{2.303}{30} \log \frac{100}{70} = \frac{2.303}{30} \log(1.428) \approx \frac{2.303 \times 0.1547}{30} \approx 0.01188 \ min^{-1}$.
The half-life period $T_{1/2}$ is given by $T_{1/2} = \frac{0.693}{k}$.
$T_{1/2} = \frac{0.693}{0.01188} \approx 58.3 \ min$.
Rounding to the nearest option,the correct answer is $58.2 \ min$.

(A) The catalytic decomposition of hydrogen peroxide $(H_2O_2)$ is a first-order reaction.
This is evident from the rate law expression: $r = k[H_2O_2]^1$.
Therefore,the order of the reaction is $1$.

(A) For a $1^{st}$ order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Thus,the half-life $t_{1/2} = \frac{0.693}{k}$.
Since $k$ is a constant,$t_{1/2}$ is independent of the initial concentration of the reactant $[R]_0$.

(D) The rate law for a first-order reaction is given by $Rate = k[A]^1$.
Since the unit of $Rate$ is $mol \ L^{-1} \ s^{-1}$ and the unit of concentration $[A]$ is $mol \ L^{-1}$,we have:
$mol \ L^{-1} \ s^{-1} = k \times (mol \ L^{-1})^1$.
Therefore,the unit of $k$ is $s^{-1}$.

(C) For a first order reaction,the half-life $(t_{1/2})$ is given as $30 \ min$.
Since $50\%$ completion corresponds to one half-life,$t_{1/2} = 30 \ min$.
For $75\%$ completion,the reaction proceeds for two half-lives $(50\% + 25\%)$.
Total time $t = 2 \times t_{1/2} = 2 \times 30 \ min = 60 \ min$.
Alternatively,using the formula $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right)$,where $k = \frac{0.693}{30} \ min^{-1}$,we get $t = \frac{2.303}{k} \log \left( \frac{100}{100 - 75} \right) = \frac{2.303}{(0.693/30)} \log(4) = 60 \ min$.

(B) For a first-order reaction,the rate constant $K$ is given by the equation: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
At half-life $(t = t_{1/2})$,the concentration of the reactant becomes half of its initial concentration,i.e.,$[A]_t = \frac{[A]_0}{2}$.
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,$t_{1/2} = \frac{0.693}{K}$.

(B) For a first-order reaction,the half-life $(t_{1/2})$ is related to the decay constant $(k)$ by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 1.1 \times 10^{-9} \ s^{-1}$.
Substituting the value:
$t_{1/2} = \frac{0.693}{1.1 \times 10^{-9}} \ s$
$t_{1/2} = 0.63 \times 10^9 \ s = 6.3 \times 10^8 \ s$.

(C) For a $1^{st}$ order reaction,the rate law is given by $Rate = k[A]^1$.
Since $Rate$ has units of $\text{concentration} \times \text{time}^{-1}$ and $[A]$ has units of $\text{concentration}$,the unit of the rate constant $k$ is:
$k = \frac{\text{Rate}}{[A]} = \frac{\text{concentration} \times \text{time}^{-1}}{\text{concentration}} = \text{time}^{-1}$.
Therefore,the unit is per unit time.

(C) For a $1$st order reaction,the half-life $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Since this expression does not contain the initial concentration $[A]_0$,the half-life is independent of the initial concentration.

(C) 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 $t_{1/2} = 138.6 \ min$.
Substituting the value:
$K = \frac{0.693}{138.6 \ min} = 0.005 \ min^{-1}$
Therefore,the correct option is $(C)$.

(D) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k}$.
$(A)$ It is independent of initial concentration,which is true.
$(B)$ It depends on the rate constant $k$,which is temperature-dependent according to the Arrhenius equation $k = Ae^{-E_a/RT}$. Thus,it is not independent of temperature.
$(C)$ $A$ catalyst provides an alternative pathway with lower activation energy,increasing $k$ and decreasing $t_{1/2}$,which is true.
$(D)$ As temperature increases,$k$ increases,so $t_{1/2}$ decreases. Therefore,the statement that it increases with an increase in temperature is false.

(D) Since the given reaction is a first-order reaction,its half-life $(t_{1/2})$ remains constant.
For a first-order reaction,$75\%$ decomposition occurs in two half-lives $(2 \times t_{1/2})$.
Given that $75\%$ decomposition takes $24 \ minutes$,we have $2 \times t_{1/2} = 24 \ minutes$,so $t_{1/2} = 12 \ minutes$.
An hour is $60 \ minutes$,which corresponds to $60 / 12 = 5$ half-lives.
The amount of reactant remaining after $n$ half-lives is given by $(1/2)^n \times 100\%$.
For $n = 5$,the remaining amount is $(1/2)^5 \times 100\% = (1/32) \times 100\% = 3.125\%$.

Time $(minutes)$	Remaining Amount $(\%)$
$0$	$100$
$12$	$50$
$24$	$25$
$36$	$12.5$
$48$	$6.25$
$60$	$3.125$

Thus,the amount of oxide left is approximately $3\%$.
Hence,Option $D$ is the correct answer.

(C) For a first-order reaction,the rate of reaction is proportional to the first power of the concentration of the reactant.
The differential rate equation is: $-\frac{d[A]}{dt} = k[A]$.
Integrating this expression between the limits $t = 0$ at $[A] = [A]_0$ and $t = t$ at $[A] = [A]$,we get:
$\ln \frac{[A]_0}{[A]} = kt$.
Converting the natural logarithm to base $10$,we multiply by $2.303$:
$2.303 \, \log \frac{[A]_0}{[A]} = kt$.
Thus,the correct integrated rate equation is $kt = 2.303 \, \log \frac{[A]_0}{[A]}$.

(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,$t = t_{1/2}$ and $[A]_t = \frac{[A]_0}{2}$.
Substituting these values: $K = \frac{2.303}{t_{1/2}} \log \frac{[A]_0}{[A]_0/2} = \frac{2.303}{t_{1/2}} \log 2$.
Therefore,$t_{1/2} = \frac{2.303 \log 2}{K} = \frac{0.693}{K}$.

(B) For a $1$st order reaction,the integrated rate equation is given by:
$k = \frac{2.303}{t} \log \frac{a}{a - x}$
Rearranging this equation:
$\log (a - x) = \log a - \frac{kt}{2.303}$
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log (a - x)$,$x = t$,$m = -\frac{k}{2.303}$ (slope),and $c = \log a$ (intercept).
Since the plot of $\log (a - x)$ versus $t$ gives a straight line,the reaction is of the $1$st order.

(C) 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 $k = 3.46 \times 10^{-3} \ min^{-1}$.
Substituting the value of $k$:
$t_{1/2} = \frac{0.693}{3.46 \times 10^{-3}} \approx 200 \ min$
Therefore,the correct option is $(C)$.

(C) 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 $t_{1/2} = 69.35 \, \text{sec}$.
Substituting the value:
$k = \frac{0.693}{69.35} \approx 0.00999 \, \sec^{-1} \approx 0.01 \, \sec^{-1}$.
Therefore,the correct option is $(C)$.

(A) The number of half-lives $(n)$ is calculated as: $n = \frac{t}{t_{1/2}} = \frac{96 \ hr}{24 \ hr} = 4$.
Using the formula for the remaining amount of a substance: $N_t = N_0 \times (\frac{1}{2})^n$.
Substituting the values: $N_t = 10 \ g \times (\frac{1}{2})^4$.
$N_t = 10 \times \frac{1}{16} = 0.625 \ g$.
Thus,$0.625 \ g$ of $N_2O_5$ will remain.

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

(B) The decomposition of hydrogen peroxide $(H_2O_2)$ follows first-order kinetics.
The rate law for this reaction is given by $r = k[H_2O_2]^1$.
Therefore,it is a $1^{st}$ order reaction.

(D) For a first-order reaction,the half-life is given as $t_{1/2} = 120 \, min$.
The rate constant $k$ is calculated as $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{120} = 5.775 \times 10^{-3} \, min^{-1}$.
For $90\%$ decomposition,the amount remaining is $(a - x) = 100 - 90 = 10$.
Using the first-order integrated rate equation: $t = \frac{2.303}{k} \log_{10} \frac{a}{a - x}$.
Substituting the values: $t = \frac{2.303}{5.775 \times 10^{-3}} \log_{10} \frac{100}{10} = \frac{2.303}{5.775 \times 10^{-3}} \times 1 = 398.78 \, min$.
Rounding to the nearest value,we get $t \approx 400 \, minutes$.

(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 $t_{1/2} = 100 \, \text{sec}$.
Substituting the value:
$k = \frac{0.693}{100 \, \text{sec}} = 6.93 \times 10^{-3} \, \text{sec}^{-1}$
Therefore,the correct option is $(A)$.

(C) For a first order reaction,the integrated rate equation is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.
At half-life,$t = t_{1/2}$ and $x = \frac{a}{2}$,so $a-x = \frac{a}{2}$.
Substituting these values: $k = \frac{2.303}{t_{1/2}} \log \frac{a}{a/2} = \frac{2.303}{t_{1/2}} \log 2$.
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,$t_{1/2} = \frac{0.693}{k}$.

(B) For a first order reaction,the relationship between the rate constant $k$ and the half-life $t_{1/2}$ is given by:
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 480 \ s$.
Substituting the value:
$k = \frac{0.693}{480} \ s^{-1} = 1.44 \times 10^{-3} \ s^{-1}$.

(A) For a first order reaction,the half-life period is given by the formula $t_{1/2} = \frac{0.693}{k}$,where $k$ is the rate constant.
Since this expression does not contain the initial concentration $[A]_0$,the half-life period is independent of the initial concentration of the reactant.

(B) For a first-order reaction,the half-life $(t_{1/2})$ is given by the formula $t_{1/2} = \frac{0.693}{k}$.
This expression shows that the half-life is independent of the initial concentration of the reactant.
Therefore,the statement that the half-life depends on the initial concentration is incorrect.

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

(B) The unit of the rate constant $(min^{-1})$ indicates that the reaction is of the first order.
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 $k = 0.69 \times 10^{-1} \ min^{-1}$.
$t_{1/2} = \frac{0.693}{0.69 \times 10^{-1}} \ min \approx 10 \ min$.
To convert this into seconds: $10 \ min \times 60 \ \sec/min = 600 \ \sec$.

(D) For a first order reaction,the rate law is given by: $Rate = K[A]$.
Given that the rate constant $(K) = 3 \times 10^{-6} \ s^{-1}$ and the initial concentration $[A] = 0.10 \ M$.
Substituting these values into the rate equation:
$Rate = (3 \times 10^{-6} \ s^{-1}) \times (0.10 \ M) = 3 \times 10^{-7} \ M \ s^{-1}$.
Therefore,the initial rate of reaction is $3 \times 10^{-7} \ M \ s^{-1}$.

(D) For a first order reaction,the rate law is given by $r = k[A]$.
Given $r = 1.0 \times 10^{-2} \, mol \, L^{-1} \, min^{-1}$ and $[A] = 0.2 \, mol \, L^{-1}$.
Substituting these values,we get $1.0 \times 10^{-2} = k \times 0.2$.
Therefore,$k = \frac{1.0 \times 10^{-2}}{0.2} = 0.05 \, min^{-1}$.
The half-life period $t_{1/2}$ for a first order reaction is calculated as $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{0.05} = 13.86 \, min$.

(D) For a first order reaction,the half-life period $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given the decay constant $k = 1.155 \times 10^{-3} \, \sec^{-1}$.
Substituting the value of $k$:
$t_{1/2} = \frac{0.693}{1.155 \times 10^{-3}} = 600 \, \sec$
Therefore,the concentration of the reactants will be halved after $600 \, \sec$.

(D) For a first-order reaction,the half-life period $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Here,$k$ is the rate constant.
Since the expression for $t_{1/2}$ does not contain the initial concentration of the reactant $(CO)$,it implies that $t_{1/2} \propto (CO)^{0}$.
Therefore,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 law is given by $r = k[A]$.
Given the rate $r = 0.6932 \times 10^{-2} \, mol \, L^{-1} \, min^{-1}$ and initial concentration $[A] = 1 \, M$.
Calculating the rate constant $k$:
$k = \frac{r}{[A]} = \frac{0.6932 \times 10^{-2} \, mol \, L^{-1} \, min^{-1}}{1 \, mol \, L^{-1}} = 0.6932 \times 10^{-2} \, min^{-1}$.
For a first order reaction,the half-life $T_{1/2}$ is calculated as:
$T_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.6932 \times 10^{-2}} \approx 100 \, min$.

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \left( \frac{a}{a - x} \right)$.
For $75\%$ completion,$x = 0.75a$,so $a - x = 0.25a$ and $t = 30 \ min$:
$k = \frac{2.303}{30} \log \left( \frac{a}{0.25a} \right) = \frac{2.303}{30} \log(4) = \frac{2.303}{30} \times 0.6020$.
For $93.75\%$ completion,$x = 0.9375a$,so $a - x = 0.0625a$:
$k = \frac{2.303}{t'} \log \left( \frac{a}{0.0625a} \right) = \frac{2.303}{t'} \log(16) = \frac{2.303}{t'} \times 1.2040$.
Equating the two expressions for $k$:
$\frac{2.303}{30} \times 0.6020 = \frac{2.303}{t'} \times 1.2040$.
$t' = 30 \times \frac{1.2040}{0.6020} = 30 \times 2 = 60 \ min$.

(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 45 \, \text{min}$,so $k = \frac{0.693}{45} \, \text{min}^{-1}$.
For $99.9 \%$ completion,the remaining concentration is $a - 0.999a = 0.001a$.
The time taken is $t = \frac{2.303}{k} \log \left( \frac{a}{0.001a} \right) = \frac{2.303}{k} \log(10^3) = \frac{2.303 \times 3}{k}$.
Substituting $k = \frac{0.693}{45}$,we get $t = \frac{2.303 \times 3 \times 45}{0.693} \approx 448.5 \, \text{min}$.
Converting to hours: $t = \frac{448.5}{60} \approx 7.475 \, \text{hr} \approx 7.5 \, \text{hr}$.

(A) For a first order reaction,the rate constant $k$ is given by the formula:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given: $[A]_0 = 2.00 \, M$,$[A]_t = 0.15 \, M$,and $t = 200 \, \min$.
Substituting the values:
$k = \frac{2.303}{200} \log \left( \frac{2.00}{0.15} \right)$
$k = \frac{2.303}{200} \log(13.333)$
$k = \frac{2.303}{200} \times 1.1249$
$k \approx 1.29 \times 10^{-2} \, \min^{-1}$
Therefore,the correct option is $A$.

(C) For a first order reaction,the half-life period $(t_{1/2})$ is related to the rate constant $(k)$ by the formula: $t_{1/2} = \frac{0.693}{k}$.
Given $t_{1/2} = 693 \ sec$.
Substituting the value in the formula: $693 = \frac{0.693}{k}$.
Rearranging for $k$: $k = \frac{0.693}{693} = \frac{693 \times 10^{-3}}{693} = 10^{-3} \ sec^{-1}$.
Therefore,$k = 0.001 \ sec^{-1}$.