First Order reaction Questions in English

(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,$[A]_t = [A]_0 - 0.99[A]_0 = 0.01[A]_0$. Thus,$64 = \frac{2.303}{k} \log \frac{100}{1} = \frac{2.303}{k} \times 2$.
For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$. Thus,$t = \frac{2.303}{k} \log \frac{1000}{1} = \frac{2.303}{k} \times 3$.
Dividing the two equations: $\frac{t}{64} = \frac{3}{2}$.
Therefore,$t = 64 \times 1.5 = 96 \text{ min}$.

(A) For the reaction $A_{(g)} \rightarrow 2B_{(g)}$,let the initial concentration of $A$ be $[A]_0 = \frac{2 \ mol}{5 \ L} = 0.4 \ M$.
Let $x$ be the amount of $A$ reacted at time $t = 20 \ min$.
At $t = 20 \ min$,$[A]_t = 0.4 - x$ and $[B]_t = 2x$.
Given that $[A]_t = \frac{[B]_t}{2}$,we have $0.4 - x = \frac{2x}{2} = x$.
Thus,$2x = 0.4$,which means $x = 0.2 \ M$.
At $t = 20 \ min$,$[A]_t = 0.4 - 0.2 = 0.2 \ M$.
Since $[A]_t = \frac{[A]_0}{2}$,the time $t = 20 \ min$ is exactly the half-life $t_{1/2}$ of the reaction.
Therefore,the half-life time is $20 \ min$.

(A) Let the initial mass of $A_2$ be $x \ g$ and $B_2$ be $y \ g$.
Given: $x + y = 10 \ g$ (Equation $1$).
For first-order kinetics,the remaining amount is given by $N = N_0 \times (1/2)^n$,where $n$ is the number of half-lives.
For $A_2$: Half-life $t_{1/2} = 2 \ hours$. In $6 \ hours$,$n = 6/2 = 3$ half-lives. Remaining $A_2 = x \times (1/2)^3 = x/8$.
For $B_2$: Half-life $t_{1/2} = 3 \ hours$. In $6 \ hours$,$n = 6/3 = 2$ half-lives. Remaining $B_2 = y \times (1/2)^2 = y/4$.
Total remaining mass after $6 \ hours = x/8 + y/4 = 2 \ g$ (Equation $2$).
From Equation $1$,$y = 10 - x$.
Substitute into Equation $2$: $x/8 + (10 - x)/4 = 2$.
Multiply by $8$: $x + 2(10 - x) = 16$.
$x + 20 - 2x = 16$.
$-x = -4$,so $x = 4 \ g$.

(A) The rate law for the reaction is given by: $\text{Rate} = k[A]^1[B]^0 = k[A]$.
Since the reaction is first order with respect to $A$,the integrated rate equation for $A$ is: $\ln(a_t) = \ln(a_0) - kt$.
Taking the exponential on both sides,we get: $a_t = a_0 e^{-kt}$.

(D) For a first-order reaction,the half-life is given by $t_{50\%} = \frac{\ln 2}{k}$.
For $t_{87.5\%}$,the amount remaining is $100 - 87.5 = 12.5\%$,which is $(1/8)$ of the initial amount. Thus,$t_{87.5\%} = \frac{\ln 8}{k} = \frac{3 \ln 2}{k} = 3 \cdot t_{50\%}$.
For $t_{90\%}$,the amount remaining is $100 - 90 = 10\%$,which is $(1/10)$ of the initial amount. Thus,$t_{90\%} = \frac{\ln 10}{k} \approx \frac{2.303}{k} \approx 3.32 \cdot t_{50\%}$.
Therefore,the statement $t_{90\%} = 4 \cdot t_{50\%}$ is $INCORRECT$.

(B) For the reaction $2A \rightarrow B$,the rate of reaction is given by $Rate = -\frac{1}{2} \frac{d[A]}{dt} = \frac{d[B]}{dt} = K[A]$.
Given $K = 10^{-2} \ min^{-1}$,this is a first-order reaction with respect to $A$.
$(1)$ The rate of formation of $B$ is $\frac{d[B]}{dt} = K[A]$. Since $K$ is the rate constant of the reaction,$K_B$ (rate of formation of $B$) depends on the concentration of $A$,so option $A$ is incorrect.
$(2)$ Half-life $t_{0.5} = \frac{\ln 2}{K} = \frac{0.7}{10^{-2}} = 70 \ min$. Thus,option $B$ is correct.
$(3)$ For a first-order reaction,$t_{0.75} = 2 \times t_{0.5} = 2 \times 70 = 140 \ min$. Thus,option $C$ is incorrect.

(C) For the reaction $C_2H_4O_{(g)} \to CH_{4(g)} + CO_{(g)}$:
At $t=0$,pressure of $C_2H_4O = P_0 = 80 \ mm$,$P_{CH_4} = 0$,$P_{CO} = 0$. Total pressure $P_t = 80 \ mm$.
At $t=20 \ min$,pressure of $C_2H_4O = 80-x$,$P_{CH_4} = x$,$P_{CO} = x$. Total pressure $P_t = (80-x) + x + x = 80+x = 120 \ mm$.
Thus,$x = 40 \ mm$.
The pressure of reactant remaining is $P_0 - x = 80 - 40 = 40 \ mm$.
Using the first-order rate constant formula $k = \frac{2.303}{t} \log \frac{P_0}{P_0-x} = \frac{2.303}{20} \log \frac{80}{40} = \frac{2.303}{20} \log 2$.
Since $\log 2 \approx 0.3010$,$k = \frac{2.303 \times 0.3010}{20} \approx \frac{0.693}{20} \ min^{-1}$.
The half-life is $t_{1/2} = \frac{0.693}{k} = \frac{0.693 \times 20}{0.693} = 20 \ min$.

(A) For a first$-order$ reaction,the half$-life$ is $T_{50} = \frac{0.693}{k}$.
Average life is $T_{av} = \frac{1}{k} \approx 1.44 \times T_{50}$.
For $75\%$ completion,the time taken is $T_{75} = \frac{2.303}{k} \log(\frac{100}{100-75}) = \frac{2.303}{k} \log(4) = \frac{2.303 \times 0.602}{k} \approx 2 \times T_{50}$.
Comparing the values: $T_{50} \approx 1.00 \times T_{50}$,$T_{av} \approx 1.44 \times T_{50}$,and $T_{75} \approx 2.00 \times T_{50}$.
Therefore,the increasing order is $T_{50} < T_{av} < T_{75}$.

(A) Let the initial pressure of $N_2O$ be $P_0$. The reaction is $N_2O_{(g)} \to N_{2(g)} + \frac{1}{2}O_{2(g)}$.
At $t=0$,pressure is $P_0, 0, 0$. Total pressure $P_0 = P_{\infty} / 1.5 = \frac{2}{3}P_{\infty}$.
At time $t$,pressure is $(P_0 - x), x, 0.5x$. Total pressure $P_t = P_0 - x + x + 0.5x = P_0 + 0.5x$.
So,$0.5x = P_t - P_0$,which means $x = 2(P_t - P_0)$.
The pressure of $N_2O$ at time $t$ is $P_{N_2O} = P_0 - x = P_0 - 2(P_t - P_0) = 3P_0 - 2P_t$.
Substituting $P_0 = \frac{2}{3}P_{\infty}$,we get $P_{N_2O} = 3(\frac{2}{3}P_{\infty}) - 2P_t = 2P_{\infty} - 2P_t$.
For a first-order reaction,$K = \frac{1}{t} \ln \left( \frac{P_0}{P_{N_2O}} \right) = \frac{1}{t} \ln \left( \frac{2P_{\infty}/3}{2P_{\infty} - 2P_t} \right) = \frac{1}{t} \ln \left( \frac{2P_{\infty}}{3(2P_{\infty} - 2P_t)} \right) = \frac{1}{t} \ln \left( \frac{P_{\infty}}{3P_{\infty} - 3P_t} \right)$.

(A) For the first order reaction $A \xrightarrow{K_1} B$,the rate constant $K_1$ is given by: $K_1 = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right)$.
Given $90\%$ of $A$ has reacted,$[A]_t = 10\% \text{ of } [A]_0$,so $K_1 = \frac{2.303}{t} \log \left( \frac{100}{10} \right) = \frac{2.303}{t} \times 1 = \frac{2.303}{t}$.
For the reaction $B \xrightarrow{K_2} \text{Product}$,the rate constant $K_2$ is given by: $K_2 = \frac{2.303}{2t} \log \left( \frac{[B]_0}{[B]_{2t}} \right)$.
Given $99\%$ of $B$ has reacted,$[B]_{2t} = 1\% \text{ of } [B]_0$,so $K_2 = \frac{2.303}{2t} \log \left( \frac{100}{1} \right) = \frac{2.303}{2t} \times 2 = \frac{2.303}{t}$.
Therefore,the ratio $\frac{K_1}{K_2} = \frac{2.303/t}{2.303/t} = 1$.

(B) The reaction is $C_4H_8 \rightarrow 2C_2H_4$.
Let the initial moles of $C_4H_8$ be $a$.
At time $t$,let $x$ moles of $C_4H_8$ react.
Remaining moles of $C_4H_8 = a - x$.
Moles of $C_2H_4$ formed = $2x$.
Given that the molar ratio $\frac{[C_2H_4]}{[C_4H_8]} = 1$,so $\frac{2x}{a - x} = 1$.
$2x = a - x$ $\Rightarrow 3x = a$ $\Rightarrow x = \frac{a}{3}$.
Remaining concentration of $C_4H_8$ at time $t$ is $[A]_t = a - \frac{a}{3} = \frac{2a}{3}$.
Using the first-order rate equation $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$:
$t = \frac{2.303}{2.3 \times 10^{-4}} \log \frac{a}{2a/3} = 10013 \times \log(1.5)$.
$t = 10013 \times 0.1761 \approx 1763.3 \text{ s}$.
Converting to minutes: $t = \frac{1763.3}{60} \approx 29.39 \text{ min} \approx 30 \text{ min}$.

(C) For a first order reaction,the amount remaining after $n$ half-lives is given by $[A] = [A]_0 \times (1/2)^n$.
Given $[A]_0 = 2 \ M$ and $[A] = 0.125 \ M$.
$0.125 = 2 \times (1/2)^n$
$(1/2)^n = 0.125 / 2 = 0.0625 = 1/16 = (1/2)^4$.
Thus,$n = 4$ half-lives.
Total time taken is $1 \text{ hour} = 60 \text{ min}$.
Since $n \times t_{1/2} = 60 \text{ min}$,we have $4 \times t_{1/2} = 60 \text{ min}$.
Therefore,$t_{1/2} = 60 / 4 = 15 \text{ min}$.

(C) Initial moles of $N_2O_5 = \frac{10.8 \text{ g}}{108 \text{ g/mol}} = 0.1 \text{ mol}$.
Number of half-lives $n = \frac{t}{t_{1/2}} = \frac{9.6 \text{ h}}{2.4 \text{ h}} = 4$.
Moles of $N_2O_5$ remaining after $4$ half-lives $= 0.1 \times (\frac{1}{2})^4 = \frac{0.1}{16} = 0.00625 \text{ mol}$.
Moles of $N_2O_5$ reacted $= 0.1 - 0.00625 = 0.09375 \text{ mol}$.
From the reaction $N_2O_5 \to 2NO_2 + \frac{1}{2} O_2$,$1 \text{ mole}$ of $N_2O_5$ produces $0.5 \text{ mole}$ of $O_2$.
Moles of $O_2$ produced $= 0.5 \times 0.09375 = 0.046875 \text{ mol}$.
Volume of $O_2$ at $STP = 0.046875 \text{ mol} \times 22.4 \text{ L/mol} = 1.05 \text{ L}$.

(A) For a first order reaction,$k = \frac{0.693}{t_{1/2}}$. Given $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{3}{2}$,so $\frac{k_2}{k_1} = \frac{3}{2}$.
For the first reaction,$t_1 = \frac{2.303}{k_1} \log \left( \frac{100}{100 - 25} \right) = \frac{2.303}{k_1} \log \left( \frac{4}{3} \right)$.
For the second reaction,$t_2 = \frac{2.303}{k_2} \log \left( \frac{100}{100 - 75} \right) = \frac{2.303}{k_2} \log (4)$.
Taking the ratio: $\frac{t_1}{t_2} = \frac{k_2}{k_1} \times \frac{\log(4/3)}{\log(4)} = \frac{3}{2} \times \frac{0.602 - 0.477}{0.602} = \frac{3}{2} \times \frac{0.125}{0.602} \approx 1.5 \times 0.2076 \approx 0.311$.
Thus,the ratio is approximately $0.3 : 1$.

(D) For a $1^{st}$ order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{K}$.
Since $t_{1/2}$ is independent of the initial concentration $(a)$,the graph of $t_{1/2}$ versus $a$ is a straight line parallel to the x-axis.
Therefore,the slope of this line is Zero.

(B) For a first order reaction,the number of half-lives $(n)$ in $60 \ \text{min}$ is $n = \frac{t}{t_{1/2}} = \frac{60}{10} = 6$.
The remaining fraction of reactant is given by $\frac{[A]_t}{[A]_0} = (\frac{1}{2})^n = (\frac{1}{2})^6 = \frac{1}{64}$.
The fraction of reactant degraded $(x)$ is $1 - \text{remaining fraction} = 1 - \frac{1}{64} = \frac{63}{64}$.

(A) The reaction is a first-order reaction because the half-life is independent of the initial concentration.
For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{V_{\infty}}{V_{\infty} - V_t}$,where $V_{\infty}$ is the total volume of $N_2$ at completion and $V_t$ is the volume at time $t$.
Given: $V_{\infty} = 100 \, \text{L}$,$V_t = 20 \, \text{L}$,and $t = 10 \, \text{min}$.
Substituting the values:
$k = \frac{2.303}{10} \log \frac{100}{100 - 20}$
$k = \frac{2.303}{10} \log \frac{100}{80}$
$k = \frac{2.303}{10} \log \frac{5}{4} \, \text{min}^{-1}$

(A) For a first order reaction,the time required for $75\%$ completion is twice the half-life $(t_{1/2})$.
$t_{75\%} = 2 \times t_{1/2}$
Given $t_{75\%} = 32 \, \text{min}$.
$32 = 2 \times t_{1/2}$
$t_{1/2} = \frac{32}{2} = 16 \, \text{min}$.
Since $50\%$ completion is equivalent to the half-life $(t_{1/2})$,the reaction is $50\%$ complete in $16$ minutes.

(C) For a first-order reaction $A \to B$,the rate of reaction is given by $Rate = k[A]$.
Since $[A] = [A]_0 e^{-kt}$,the rate of reaction is $Rate = k[A]_0 e^{-kt}$.
The rate of formation of $B$ is equal to the rate of reaction,so $R = k[A]_0 e^{-kt}$.
This equation represents an exponential decay curve,where the rate $R$ decreases exponentially with time $t$.

(A) For a first-order reaction,the integrated rate law is given by: $[A]_t = [A]_0 e^{-kt}$
Given that $t = \frac{1}{k}$,we substitute this value into the equation:
$[A]_t = [A]_0 e^{-k \times (1/k)}$
$[A]_t = [A]_0 e^{-1}$
$[A]_t = \frac{[A]_0}{e}$

(A) For a first order reaction,the rate constant $k$ is given by the formula:
$k = \frac{2.303}{t} \log_{10} \frac{a}{a-x}$
Given $t = 10 \, \text{min}$ and $\frac{a}{a-x} = 8$.
Substituting these values into the equation:
$k = \frac{2.303}{10} \log_{10} 8$
Since $8 = 2^3$,we can write $\log_{10} 8 = \log_{10} 2^3 = 3 \log_{10} 2$.
Therefore,$k = \frac{2.303 \times 3 \log 2}{10}$.

(B) For a first-order reaction,the half-life $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Here,$k$ is the rate constant.
Since $t_{1/2}$ is independent of the initial concentration $(a)$,the graph of $t_{1/2}$ versus $a$ will be a straight line parallel to the concentration axis ($a$-axis).
Therefore,the correct curve is the one showing $t_{1/2}$ as constant with respect to $a$.

(C) The given rate law $\frac{-d[A]}{dt} = k[A]$ indicates a first-order reaction with respect to $A$.
For a first-order reaction,the integrated rate equation is $[A] = [A]_0 e^{-kt}$.
The half-life $t_{1/2}$ for a first-order reaction is given by $t_{1/2} = \frac{\ln 2}{k}$,which implies $k = \frac{\ln 2}{t_{1/2}}$.
Substituting the given time $t = \frac{t_{1/2}}{\ln 2}$ into the rate equation:
$[A] = [A]_0 e^{-k \times (\frac{t_{1/2}}{\ln 2})}$
$[A] = [A]_0 e^{-(\frac{\ln 2}{t_{1/2}}) \times (\frac{t_{1/2}}{\ln 2})}$
$[A] = [A]_0 e^{-1} = \frac{[A]_0}{e}$.

(C) For a first order reaction,the half-life $t_{1/2} = 50\, min$.
The rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{50\, min}$.
For $80\%$ completion,the remaining concentration is $100\% - 80\% = 20\%$.
The time $t$ is given by the formula: $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right)$.
Substituting the values: $t = \frac{2.303}{(0.693/50)} \log \left( \frac{100}{20} \right)$.
$t = \frac{2.303 \times 50}{0.693} \times \log(5)$.
Since $\log(5) \approx 0.699$ and $\frac{2.303}{0.693} \approx 3.32$,we get $t \approx 3.32 \times 50 \times 0.699 \approx 116.1\, min$.
Rounding to the nearest integer,$t \approx 117\, min$.

(D) The given reaction is a first-order reaction as the rate law is $r = K[A]$.
For a first-order reaction,the integrated rate equation is: $\ln \frac{[A]_0}{[A]_t} = Kt$ or $\ln \frac{n_0}{n_t} = Kt$.
Given: $K = 0.023 \ s^{-1}$,$t = 50 \ s$,$n_0 = 2.5 \ moles$.
Substituting the values: $\ln \frac{2.5}{n_t} = 0.023 \times 50$.
$\ln \frac{2.5}{n_t} = 1.15$.
Taking the exponential of both sides: $\frac{2.5}{n_t} = e^{1.15} \approx 3.158$.
$n_t = \frac{2.5}{3.158} \approx 0.79 \ moles$.

(C) For a first-order reaction: $k = \frac{2.303}{t} \log \frac{P_i}{P_{SO_2Cl_2}}$
Given reaction: $SO_2Cl_{2(g)} \to SO_{2(g)} + Cl_{2(g)}$
At $t = 0$,$P_i = 0.5 \text{ atm}$.
At $t = 100 \text{ s}$,$P_{total} = 0.6 \text{ atm}$.
Let $x$ be the decrease in pressure of $SO_2Cl_2$.
$P_{total} = (P_i - x) + x + x = P_i + x$.
$0.6 = 0.5 + x \Rightarrow x = 0.1 \text{ atm}$.
$P_{SO_2Cl_2} = P_i - x = 0.5 - 0.1 = 0.4 \text{ atm}$.
$k = \frac{2.303}{100} \log \frac{0.5}{0.4} = \frac{2.303}{100} \log 1.25$.
$k = \frac{2.303 \times 0.0969}{100} = 2.23 \times 10^{-3} \text{ s}^{-1}$.

(A) The given rate law is $\frac{-d[A]}{dt} = K[A]$.
This represents a first-order reaction with respect to reactant $A$.
The integrated rate equation for a first-order reaction is $[A]_t = [A]_0 e^{-Kt}$.
Given $[A]_0 = Co$ and $t = \frac{1}{K}$.
Substituting these values into the equation:
$[A]_t = Co \, e^{-K \times (\frac{1}{K})}$
$[A]_t = Co \, e^{-1}$
$[A]_t = \frac{Co}{e}$.

(B) The decomposition reaction is: $N_2O_5(g) \to 2NO_2(g) + \frac{1}{2}O_2(g)$
At $t = 0$: $P_{N_2O_5} = 50 \, mm \, Hg$,$P_{NO_2} = 0$,$P_{O_2} = 0$. Total pressure $P_0 = 50 \, mm \, Hg$.
At $t = 50 \, min$: Let $p_1$ be the decrease in pressure of $N_2O_5$. The partial pressures are: $P_{N_2O_5} = 50 - p_1$,$P_{NO_2} = 2p_1$,$P_{O_2} = 0.5p_1$.
Total pressure $P_t = (50 - p_1) + 2p_1 + 0.5p_1 = 50 + 1.5p_1 = 87.5 \, mm \, Hg$.
$1.5p_1 = 37.5 \implies p_1 = 25 \, mm \, Hg$.
Since $p_1 = 25$ is half of the initial pressure $50$,the half-life $t_{1/2} = 50 \, min$.
At $t = 100 \, min$ $(2 \times t_{1/2})$,the remaining pressure of $N_2O_5$ is $50 \times (1/2)^2 = 12.5 \, mm \, Hg$.
Thus,$50 - p_2 = 12.5 \implies p_2 = 37.5 \, mm \, Hg$.
Total pressure at $100 \, min = 50 + 1.5p_2 = 50 + 1.5(37.5) = 50 + 56.25 = 106.25 \, mm \, Hg$.

(C) The half-life $t_{1/2} = 10 \ days$.
The rate constant $k$ is given by $k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{10} = 0.0693 \ days^{-1}$.
For a first-order reaction,the time $t$ required for a fraction $x$ of the reactant to be converted is given by $t = \frac{1}{k} \ln \left( \frac{a}{a - x} \right)$.
Here,$x = \frac{1}{4}a$,so the remaining amount is $a - x = a - \frac{1}{4}a = \frac{3}{4}a$.
Thus,$t = \frac{1}{0.0693} \ln \left( \frac{a}{3a/4} \right) = \frac{1}{0.0693} \ln \left( \frac{4}{3} \right)$.
$t = \frac{\ln 4 - \ln 3}{0.0693} = \frac{2 \ln 2 - \ln 3}{0.0693} = \frac{2(0.693) - 1.1}{0.0693} = \frac{1.386 - 1.1}{0.0693} = \frac{0.286}{0.0693} \approx 4.13 \ days$.
The closest option is $4.1 \ days$.

(D) For a first-order reaction,the half-life $(t_{1/2})$ is constant.
Given that $50\%$ of the reaction is complete in $100 \ s$,this is the half-life,$t_{1/2} = 100 \ s$.
After another $100 \ s$ (total $200 \ s$),the remaining concentration becomes $A_0/4$,which means $75\%$ of the reaction is complete.
Since the time taken for the second half-life is also $100 \ s$,the reaction follows first-order kinetics.

(A) For the reaction $2N_2O_5 \,(g) \to 4NO_2 \,(g) + O_2 \,(g)$,let the initial pressure of $N_2O_5$ be $P_0 = 50 \, mmHg$.
At $t = 30 \, min$,let the pressure of $N_2O_5$ decrease by $2p$. The total pressure is $P_t = (50 - 2p) + 4p + p = 50 + 3p = 87.5 \, mmHg$.
Thus,$3p = 37.5 \, mmHg$,so $p = 12.5 \, mmHg$.
The pressure of $N_2O_5$ remaining at $t = 30 \, min$ is $50 - 2(12.5) = 25 \, mmHg$.
Since the pressure of $N_2O_5$ halves in $30 \, min$,the half-life $t_{1/2} = 30 \, min$.
At $t = 60 \, min$ (which is $2 \times t_{1/2}$),the pressure of $N_2O_5$ remaining is $50 / 4 = 12.5 \, mmHg$.
Let $50 - 2p' = 12.5$,so $2p' = 37.5$,which means $p' = 18.75 \, mmHg$.
The total pressure at $t = 60 \, min$ is $50 + 3p' = 50 + 3(18.75) = 50 + 56.25 = 106.25 \, mmHg$.

(C) Given half-life $t_{1/2} = 15 \ minutes$.
Total time $T = 1 \ hr = 60 \ minutes$.
The number of half-lives $n$ is calculated as $n = \frac{T}{t_{1/2}} = \frac{60}{15} = 4$.
The amount of substance remaining is given by the formula $\frac{N}{N_0} = (\frac{1}{2})^n$.
Substituting the value of $n$,we get $\frac{N}{N_0} = (\frac{1}{2})^4 = \frac{1}{16}$.
Therefore,the amount of substance left after $1 \ hour$ is $\frac{1}{16}$ of the original amount.

(B) The rate law is given by $R = k[A]^x [B]^y$.
From experiments $I$ and $II$,$[A]$ is constant and $[B]$ changes,but the rate remains constant. Thus,$y = 0$ (zero order with respect to $B$).
From experiments $I$ and $III$,when $[A]$ doubles ($0.10$ to $0.20$),the rate doubles ($6.93 \times 10^{-3}$ to $1.386 \times 10^{-2}$). Thus,$x = 1$ (first order with respect to $A$).
The rate equation is $R = k[A]$.
Using experiment $I$: $6.93 \times 10^{-3} = k(0.10) \Rightarrow k = 6.93 \times 10^{-2} \ min^{-1}$.
For a first-order reaction,the half-life is $t_{1/2} = \frac{\ln(2)}{k} = \frac{0.693}{6.93 \times 10^{-2}} = 10 \ min$.

(C) For a first order reaction,the rate is given by $\text{Rate} = K[A]_0$.
Given $\text{Rate} = 0.6932 \times 10^{-2} \ mol \ L^{-1} \ min^{-1}$ and $[A]_0 = 0.1 \ M$.
Calculating the rate constant $K$: $K = \frac{\text{Rate}}{[A]_0} = \frac{0.6932 \times 10^{-2}}{0.1} = 0.6932 \times 10^{-1} \ min^{-1}$.
The half-life $t_{1/2}$ for a first order reaction is given by $t_{1/2} = \frac{0.693}{K}$.
Substituting the value of $K$: $t_{1/2} = \frac{0.693}{0.6932 \times 10^{-1}} \approx \frac{0.693}{0.06932} \approx 10 \ min$.
Hence,option $C$ is correct.

(B) For the reaction: $A_{(g)} \longrightarrow B_{(g)} + C_{(g)}$
At $t = 0$: Initial pressure of $A = P_i$,$P_B = 0$,$P_C = 0$.
At time $t$: Let the decrease in pressure of $A$ be $x$. Then,$P_A = P_i - x$,$P_B = x$,and $P_C = x$.
The total pressure at time $t$ is $P_t = P_A + P_B + P_C = (P_i - x) + x + x = P_i + x$.
From this,$x = P_t - P_i$.
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,$K = \frac{2.303}{t} \log \frac{P_{\text{initial}}}{P_{\text{final}}}$.
Substituting the values,$K = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}$.

(C) For a first order reaction,the integrated rate equation is:
$K = \frac{2.303}{t} \log \frac{a}{a-x}$
Given: $K = 5.78 \times 10^{-5} \ s^{-1}$,$t = 10 \ hours = 10 \times 3600 \ s = 36000 \ s$.
Substituting the values:
$5.78 \times 10^{-5} = \frac{2.303}{36000} \log \frac{a}{a-x}$
$\log \frac{a}{a-x} = \frac{5.78 \times 10^{-5} \times 36000}{2.303} \approx \frac{2.0808}{2.303} \approx 0.903$
Since $\log 8 \approx 0.903$,we have $\log \frac{a}{a-x} = \log 8$.
Therefore,$\frac{a}{a-x} = 8$,which implies the remaining fraction $\frac{a-x}{a} = \frac{1}{8}$.

(B) For a first-order reaction,the concentration of the reactant at time $t$ is given by the integrated rate law: $[A]_t = [A]_0 e^{-kt}$.
Here,$[A]_0$ is the initial concentration and $[A]_t$ is the concentration at time $t$.
The amount of substance dissociated is $[A]_0 - [A]_t = [A]_0 - [A]_0 e^{-kt} = [A]_0(1 - e^{-kt})$.
The degree of dissociation $(\alpha)$ is defined as the ratio of the amount dissociated to the initial amount:
$\alpha = \frac{[A]_0(1 - e^{-kt})}{[A]_0} = 1 - e^{-kt}$.

(C) For a first-order reaction,the time $t$ required for a fraction of reaction to complete is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $t_{0.5}$ (half-life),$[A]_t = 0.5[A]_0$,so $t_{0.5} = \frac{2.303}{k} \log 2$.
For $t_{0.75}$,$[A]_t = 0.25[A]_0$,so $t_{0.75} = \frac{2.303}{k} \log \frac{[A]_0}{0.25[A]_0} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} \log 2^2 = 2 \times \frac{2.303}{k} \log 2$.
Therefore,$t_{0.75} = 2 \times t_{0.5}$.
The ratio $\frac{t_{0.75}}{t_{0.5}} = \frac{2}{1}$ or $2 : 1$.

(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 the half-life is independent of the initial concentration $[A]$,changing the concentration will not affect the half-life.
Therefore,the half-life remains $5 \ minutes$.

(C) 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: $[A]_0 = 0.4 \, M$,$[A]_t = 0.1 \, M$,and $t = 20 \, \min$.
Substituting the values: $K = \frac{2.303}{20} \log \frac{0.4}{0.1} = \frac{2.303}{20} \log 4$
Since $\log 4 = 2 \log 2 \approx 2 \times 0.3010 = 0.6020$:
$K = \frac{2.303 \times 0.6020}{20} \approx \frac{1.386}{20} = 0.0693 \, \min^{-1}$.

(C) The reaction is $2H_2O_{2(aq)} \to 2H_2O_{(l)} + O_{2(g)}$.
Given the rate constant $k = 3 \times 10^{-3} \ min^{-1}$.
Convert $k$ to $s^{-1}$: $k = \frac{3 \times 10^{-3}}{60} \ s^{-1} = 5 \times 10^{-5} \ s^{-1}$.
The rate law for the first-order reaction is $\text{Rate} = k[H_2O_2]$.
Given $\text{Rate} = 2 \times 10^{-4} \ M \ s^{-1}$.
$[H_2O_2] = \frac{\text{Rate}}{k} = \frac{2 \times 10^{-4} \ M \ s^{-1}}{5 \times 10^{-5} \ s^{-1}} = 4 \ M$.

(B) For a first-order reaction,the amount of reactant remaining after $n$ half-lives is given by the formula: $\text{Fraction remaining} = \frac{1}{2^n}$.
Here,$n = 3$ half-lives.
Therefore,$\text{Fraction remaining} = \frac{1}{2^3} = \frac{1}{8} = 0.125$.
To find the percentage,multiply by $100$: $0.125 \times 100 = 12.5\%$.

(D) For a first-order reaction,the integrated rate equation is given by: $\ln \frac{a}{a-x} = kt$
Rearranging the equation: $\frac{a-x}{a} = e^{-kt}$
This can be written as: $1 - \frac{x}{a} = e^{-kt}$
The degree of dissociation,denoted by $\alpha$,is defined as the fraction of the initial concentration that has reacted,i.e.,$\alpha = \frac{x}{a}$.
Substituting this into the equation: $1 - \alpha = e^{-kt}$
Therefore,$\alpha = 1 - e^{-kt}$

(C) For a first-order reaction,the half-life is given by $t_{1/2} = 120 \ min$.
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 $10\%$ of the initial concentration $([A]_0 = 100, [A]_t = 10)$.
Using the first-order rate equation: $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]_t}$.
$t_{90\%} = \frac{2.303}{5.775 \times 10^{-3}} \log \frac{100}{10} = \frac{2.303}{5.775 \times 10^{-3}} \times 1 = 398.8 \ min$.

(C) For a first order reaction,the rate law is given by $r = k[A]$.
Given: $r = 2.0 \times 10^{-5} \ M \ sec^{-1}$ and $[A] = 0.01 \ M$.
Substituting the values: $2.0 \times 10^{-5} = k \times 0.01$.
$k = \frac{2.0 \times 10^{-5}}{10^{-2}} = 2.0 \times 10^{-3} \ sec^{-1}$.
The half-life period $t_{1/2}$ for a first order reaction is $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{2.0 \times 10^{-3}} = 0.3465 \times 10^3 = 346.5 \ sec$.

(A) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{K}$.
Given $t_{1/2} = 6.93 \times 10^{-3} \, \text{min}$,we have $6.93 \times 10^{-3} = \frac{0.693}{K}$.
Thus,$K = \frac{0.693}{6.93 \times 10^{-3}} = 10^2 \, \text{min}^{-1}$.
Substituting $K$ into the given equation: $\log_{10} (10^2) = 12 - \frac{6 \times 10^3}{T}$.
$2 \log_{10} 10 = 12 - \frac{6 \times 10^3}{T}$.
Since $\log_{10} 10 = 1$,we get $2 = 12 - \frac{6 \times 10^3}{T}$.
$\frac{6 \times 10^3}{T} = 12 - 2 = 10$.
$T = \frac{6 \times 10^3}{10} = 600 \, K$.

(C) For a first-order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
At $t_{1/4}$,the amount of reactant consumed is $\frac{1}{4}[A]_0$,so the remaining concentration is $[A]_t = [A]_0 - \frac{1}{4}[A]_0 = \frac{3}{4}[A]_0$.
Substituting this into the equation:
$t_{1/4} = \frac{2.303}{k} \log \frac{[A]_0}{\frac{3}{4}[A]_0} = \frac{2.303}{k} \log \frac{4}{3}$.

(B) For a first order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
For $75\%$ completion,$[A]_t = 100 - 75 = 25$,so $t_{75\%} = \frac{2.303}{K} \log \frac{100}{25} = \frac{2.303}{K} \log 4 = 100 \ min$ ...... $(i)$
For $87.5\%$ completion,$[A]_t = 100 - 87.5 = 12.5$,so $t_{87.5\%} = \frac{2.303}{K} \log \frac{100}{12.5} = \frac{2.303}{K} \log 8$ ...... $(ii)$
Dividing $(ii)$ by $(i)$:
$\frac{t_{87.5\%}}{100} = \frac{\log 8}{\log 4} = \frac{\log 2^3}{\log 2^2} = \frac{3 \log 2}{2 \log 2} = 1.5$
$t_{87.5\%} = 1.5 \times 100 = 150 \ min$.

(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.
For $10 \%$ completion,$x = 10$ and $t = 20 \, \text{min}$.
$k = \frac{2.303}{20} \log \frac{100}{90} = \frac{2.303}{20} \times (\log 10 - \log 9) = \frac{2.303}{20} \times (1 - 0.9542) = \frac{2.303 \times 0.0458}{20} \approx 0.00527 \, \text{min}^{-1}$.
Now,for $19 \%$ completion,$x = 19$ and $a = 100$.
$t = \frac{2.303}{k} \log \frac{100}{100-19} = \frac{2.303}{0.00527} \log \frac{100}{81}$.
$t = \frac{2.303}{0.00527} \times (2 - \log 81) = \frac{2.303}{0.00527} \times (2 - 1.9085) = \frac{2.303 \times 0.0915}{0.00527} \approx 40 \, \text{min}$.

(A) For the reaction $A_{(g)} \to 2B_{(g)} + C_{(g)}$:
At $t=0$,pressure is $P_0, 0, 0$ for $A, B, C$ respectively.
At time $t$,let the pressure of $A$ decrease by $x$. Then $P_A = P_0 - x$,$P_B = 2x$,and $P_C = x$.
The total pressure $P_t = (P_0 - x) + 2x + x = P_0 + 2x$.
From this,$2x = P_t - P_0$,so $x = \frac{P_t - P_0}{2}$.
The pressure of $A$ at time $t$ is $P_A = P_0 - x = P_0 - \frac{P_t - P_0}{2} = \frac{2P_0 - P_t + P_0}{2} = \frac{3P_0 - P_t}{2}$.
For a first-order reaction,$K = \frac{1}{t} \ln \left( \frac{P_0}{P_A} \right) = \frac{1}{t} \ln \left( \frac{P_0}{(3P_0 - P_t)/2} \right) = \frac{1}{t} \ln \left( \frac{2P_0}{3P_0 - P_t} \right)$.