Rate law , Rate constant , Order of Reaction and Molecularity Questions in English

(B) The rate law is expressed as $r = k[A]^x[B]^y$.
From the data:
$0.045 = k(0.05)^x(0.05)^y$ ...... $(1)$
$0.090 = k(0.10)^x(0.05)^y$ ...... $(2)$
$0.72 = k(0.20)^x(0.10)^y$ ...... $(3)$
Dividing $(2)$ by $(1)$:
$\frac{0.090}{0.045} = \left( \frac{0.10}{0.05} \right)^x$ $\Rightarrow 2 = 2^x$ $\Rightarrow x = 1$.
Dividing $(3)$ by $(2)$:
$\frac{0.72}{0.090} = \left( \frac{0.20}{0.10} \right)^x \left( \frac{0.10}{0.05} \right)^y$
$8 = (2)^1 \times (2)^y$
$8 = 2 \times 2^y$ $\Rightarrow 4 = 2^y$ $\Rightarrow y = 2$.
Thus,the rate law is $r = k[A][B]^2$.

(A) Applying the steady state approximation to the intermediate $B$:
The rate of change of concentration of $B$ is given by: $\frac{d[B]}{dt} = k_1 [A] - k_2 [B]$
According to the steady state approximation,the rate of formation of $B$ is set to zero: $\frac{d[B]}{dt} = 0$
Therefore,$0 = k_1 [A] - k_2 [B]$
Rearranging the equation to solve for $[B]$:
$k_2 [B] = k_1 [A]$
$[B] = \left( \frac{k_1}{k_2} \right) [A]$

(A) For reaction $(i)$,the plot is $\ln [R]$ versus time,which is a straight line with a negative slope. The integrated rate equation for a first-order reaction is $\ln [R]_t = -Kt + \ln [R]_0$. This matches the form $y = mx + c$,where $y = \ln [R]_t$ and $x = t$. Thus,reaction $(i)$ is of first order.
For reaction $(ii)$,the plot is $[R]$ versus time,which is a straight line with a negative slope. The integrated rate equation for a zero-order reaction is $[R]_t = -Kt + [R]_0$. This matches the form $y = mx + c$,where $y = [R]_t$ and $x = t$. Thus,reaction $(ii)$ is of zero order.
Therefore,the orders of reactions $(i)$ and $(ii)$ are $1$ and $0$ respectively.

(A) For $t \le 1 \ hour$,the population grows as $N(t) = N_0 \exp(t)$. Thus,$\frac{N_0}{N} = \exp(-t)$. This is a decreasing exponential function.
For $t > 1 \ hour$,the population follows $\frac{dN}{dt} = -5N^2$. Integrating this from $t=1$ to $t$,we get $\int_{N_1}^{N} \frac{dN}{N^2} = \int_{1}^{t} -5 dt$,where $N_1 = N_0 \exp(1)$.
This gives $[-\frac{1}{N}]_{N_1}^{N} = -5(t-1)$,so $\frac{1}{N} - \frac{1}{N_1} = 5(t-1)$.
Multiplying by $N_0$,we get $\frac{N_0}{N} = \frac{N_0}{N_1} + 5N_0(t-1) = \exp(-1) + 5N_0(t-1)$.
This is a linear function with a positive slope for $t > 1$. Therefore,the plot of $\frac{N_0}{N}$ vs. $t$ decreases until $t=1$ and then increases linearly.

(C) The rate-determining step is the slow step,which is $(i)$ $NO_2 \to NO + O$.
The rate law is determined by this slow step: $Rate = k[NO_2]^1$.
Since the rate depends only on the concentration of $NO_2$,the order of the reaction is $1$.
For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k}$,which is independent of the initial concentration of the reactant.
Therefore,the reaction is zero order with respect to $F_2$ because the rate expression does not contain $[F_2]$.

(D) For an elementary reaction,the rate law is determined directly from the stoichiometry of the balanced chemical equation.
Since the reaction $2NO_{(g)} + Cl_{2(g)} \longrightarrow 2NOCl_{(g)}$ is given as an elementary step,the rate law is $Rate = k[NO]^2[Cl_2]^1$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law,which is $2 + 1 = 3$.
The molecularity of an elementary reaction is the number of reacting species taking part in the elementary step,which is $2 + 1 = 3$.
Therefore,for any elementary reaction,the order of the reaction is equal to its molecularity.

(B) The rate law for the reaction is given by: $\text{Rate} = k[HI]^n$.
Using the data from the table:
For experiment $1$: $8.80 \times 10^{-10} = k(0.0500)^n$ (Equation $1$)
For experiment $2$: $3.52 \times 10^{-9} = k(0.1000)^n$ (Equation $2$)
Dividing Equation $2$ by Equation $1$:
$\frac{3.52 \times 10^{-9}}{8.80 \times 10^{-10}} = \left(\frac{0.1000}{0.0500}\right)^n$
$4 = (2)^n$
Since $2^2 = 4$,we find that $n = 2$.
Thus,the order of the reaction with respect to $HI$ is $2.0$.

(D) The expression for the initial rate is $r_{0} = k(P_{A})_{0}(P_{B})_{0}^{2} = k(0.60)(0.80)^{2} \dots (1)$
After some time,the partial pressure of $C$ is $0.20 \ atm$.
According to the stoichiometry of the reaction $A_{(g)} + 2B_{(g)} \longrightarrow C_{(g)} + D_{(g)}$,the partial pressures of $A$ and $B$ are $P_{A} = 0.60 - 0.20 = 0.40 \ atm$ and $P_{B} = 0.80 - 2(0.20) = 0.40 \ atm$,respectively.
The expression for the rate at this instant is $r = k(P_{A})(P_{B})^{2} = k(0.40)(0.40)^{2} \dots (2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{r}{r_{0}} = \frac{k(0.40)(0.40)^{2}}{k(0.60)(0.80)^{2}} = \frac{0.40 \times 0.16}{0.60 \times 0.64} = \frac{0.064}{0.384} = \frac{1}{6}$

(A) The rate of a reaction is determined by the slowest step in the mechanism,which is called the rate-determining step.
In the given mechanism:
Step $1$: $2B \xrightarrow{k} B_2$ $[Slow]$
Step $2$: $B_2 + A \to D$ $[Fast]$
The rate law is determined by the slow step: $\text{Rate} = k[B]^2$.
Comparing this to the general rate law expression $\text{Rate} = k[A]^x[B]^y$:
Order with respect to $A$ $(x)$ = $0$.
Order with respect to $B$ $(y)$ = $2$.
Overall order of reaction = $x + y = 0 + 2 = 2$.
Thus,the rate law expression is $k[B]^2$,order with respect to $A$ is $0$,order with respect to $B$ is $2$,and overall order is $2$.

(B) The rate law for a chemical reaction is expressed as $\text{rate} = K [A]^p [B]^q$.
The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
Here,the powers are $p$ and $q$.
Therefore,the order of the reaction is $(p+q)$.

(C) For a reaction of order $n$,the half-life $t_{1/2}$ is related to the initial concentration $a$ by the expression $t_{1/2} \propto \frac{1}{a^{n-1}}$.
This is a standard relationship derived from the integrated rate law for an $n^{th}$ order reaction where $n \neq 1$.
Therefore,the order of the reaction is $n$.
Thus,Option $C$ is correct.

(C) Molecularity is defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary reaction,which must collide simultaneously in order to bring about a chemical reaction.
If two different reactants take part in a reaction,at least two molecules must collide for the reaction to occur.
Therefore,the minimum number of reacting species is $2$.
Since molecularity is the sum of the number of reacting species,the molecularity of such a reaction cannot be $1$.

(C) Let the rate law be $r = k[A]^n$,where $n$ is the order of the reaction.
Initial rate: $r_1 = k[A]^n$
New rate: $r_2 = 2r_1 = k[8A]^n$
Dividing the two equations: $\frac{2r_1}{r_1} = \frac{k[8A]^n}{k[A]^n}$
$2 = (8)^n$
Since $8 = 2^3$,we have $2^1 = (2^3)^n = 2^{3n}$
Equating the exponents: $1 = 3n$
Therefore,$n = \frac{1}{3}$.

(A) For a second-order reaction,the half-life $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{1}{k[A]_0}$.
Given: $t_{1/2} = 30 \ min$ and $[A]_0 = 0.1 \ M$.
Substituting the values: $30 = \frac{1}{k \times 0.1}$.
$k = \frac{1}{30 \times 0.1} = \frac{1}{3} = 0.333 \ M^{-1} \ min^{-1}$.

(A) For plot $(i)$,the rate $-\frac{d[A]}{dt}$ is directly proportional to $[A]$. This represents a $1^{st}$ order reaction,where $rate = k[A]$.
For plot $(ii)$,the concentration $[A]$ decreases linearly with time $t$. This represents a zero order reaction,where $[A]_t = -kt + [A]_0$.
For plot $(iii)$,the graph of $\frac{1}{[A]}$ versus $t$ is a straight line with a positive slope. This represents a $2^{nd}$ order reaction,where $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$.
Thus,the reaction orders are $1, 0, 2$ respectively.

(D) The rate of the reaction is determined by the slow step,which is step $(ii)$.
Rate $= k_2[M][Z]$ $....$ $(1)$
Since step $(i)$ is a very rapid equilibrium,we can write the equilibrium constant expression:
$K_{eq} = \frac{[M]}{[X][Y]}$
$[M] = K_{eq}[X][Y]$ $....$ $(2)$
Substituting the value of $[M]$ from equation $(2)$ into equation $(1)$:
Rate $= k_2 K_{eq} [X][Y][Z]$
Let $k = k_2 K_{eq}$,then:
Rate $= k[X][Y][Z]$

(A) For a reaction of order $n$,the half-life $(t_{1/2})$ is related to the initial pressure $(P_0)$ as: $(t_{1/2}) \propto (P_0)^{1-n}$.
Therefore,$\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left(\frac{P_1}{P_2}\right)^{1-n} = \left(\frac{P_2}{P_1}\right)^{n-1}$.
Given: $(t_{1/2})_1 = 440 \ sec$,$P_1 = 364 \ mm \ Hg$; $(t_{1/2})_2 = 880 \ sec$,$P_2 = 182 \ mm \ Hg$.
Substituting the values: $\frac{440}{880} = \left(\frac{182}{364}\right)^{n-1}$.
$\frac{1}{2} = \left(\frac{1}{2}\right)^{n-1}$.
Comparing the exponents: $n-1 = 1$,which gives $n = 2$.

(B) The rate law for the reaction is $r = k [NO]^2 [O_2]^1$.
When the volume of the vessel is doubled,the concentration of each reactant is halved because $[C] = n/V$.
New concentration $[NO]' = [NO]/2$ and $[O_2]' = [O_2]/2$.
New rate $r' = k ([NO]/2)^2 ([O_2]/2)^1$.
$r' = k ([NO]^2 / 4) ([O_2] / 2) = (1/8) k [NO]^2 [O_2] = (1/8) r$.
Thus,the rate reduces to one-eighth of its initial rate.

(D) The rate law is $R = K[A][B]^2$.
At $t = 0$: $[A]_0 = C_0, [B]_0 = 2C_0$.
At $t = 30 \min$: $[C] = C_0/4$.
From the stoichiometry of the reaction $2A + B \rightarrow C + D$:
$[A] = [A]_0 - 2[C] = C_0 - 2(C_0/4) = C_0/2$.
$[B] = [B]_0 - [C] = 2C_0 - C_0/4 = 7C_0/4$.
Substituting these into the rate law:
$R = K(C_0/2)(7C_0/4)^2 = K(C_0/2)(49C_0^2/16) = 49KC_0^3/32$.

(D) For an $n^{th}$ order reaction,the half-life is related to the initial concentration $a$ by the expression: $t_{1/2} \propto a^{1-n}$.
This can be written as $t_{1/2} \times a^{n-1} =$ constant.
Given the condition $t_{1/2} \times a^2 =$ constant.
Comparing the two expressions: $n-1 = 2$.
Therefore,$n = 3$.
Thus,it is a third-order reaction.

(A) In a reaction mechanism,the rate-determining step $(RDS)$ is the slowest step.
For the given reaction,the slow step is: $NO_2 + F_2 \to NO_2F + F$.
The rate of the overall reaction is determined by the rate of this slow step.
Therefore,the rate law is given by: $r = K [NO_2][F_2]$.

(A) The given rate equation is $\frac{-dc}{dt} = \frac{K_1 C}{1 + K_2 C}$.
When the concentration $(C)$ is very high,the term $K_2 C$ becomes much larger than $1$ (i.e.,$K_2 C \gg 1$).
Therefore,the denominator $1 + K_2 C$ can be approximated as $K_2 C$.
Substituting this into the rate equation: $\frac{-dc}{dt} \approx \frac{K_1 C}{K_2 C} = \frac{K_1}{K_2}$.
Since the rate is now independent of the concentration $(C)$,the reaction follows zero-order kinetics.
Thus,the order of the reaction is $0$.

(A) Let the rate law be $Rate = k[A]^x[B]^y$.
From the data:
$1$) $1.2 \times 10^{-3} = k(0.05)^x(0.05)^y$
$2$) $2.4 \times 10^{-3} = k(0.10)^x(0.05)^y$
$3$) $1.2 \times 10^{-3} = k(0.05)^x(0.10)^y$
Dividing equation $(2)$ by $(1)$:
$\frac{2.4 \times 10^{-3}}{1.2 \times 10^{-3}} = (\frac{0.10}{0.05})^x$ $\Rightarrow 2 = 2^x$ $\Rightarrow x = 1$.
Dividing equation $(3)$ by $(1)$:
$\frac{1.2 \times 10^{-3}}{1.2 \times 10^{-3}} = (\frac{0.10}{0.05})^y$ $\Rightarrow 1 = 2^y$ $\Rightarrow y = 0$.
Thus,the order with respect to $A$ is $1$ and with respect to $B$ is $0$.

(C) The rate law is given by $R = K[A]^{\alpha}[B_2]^{\beta}$.
From the experimental data:
$1.6 \times 10^{-4} = K[0.50]^{\alpha}[0.50]^{\beta}$ ...... $(i)$
$3.2 \times 10^{-4} = K[0.50]^{\alpha}[1.00]^{\beta}$ ...... $(ii)$
$3.2 \times 10^{-4} = K[1.00]^{\alpha}[1.00]^{\beta}$ ...... $(iii)$
Divide $(iii)$ by $(ii)$:
$\frac{3.2 \times 10^{-4}}{3.2 \times 10^{-4}} = \frac{K[1.00]^{\alpha}[1.00]^{\beta}}{K[0.50]^{\alpha}[1.00]^{\beta}}$
$1 = (2)^{\alpha} \implies \alpha = 0$.
Divide $(ii)$ by $(i)$:
$\frac{3.2 \times 10^{-4}}{1.6 \times 10^{-4}} = \frac{K[0.50]^{\alpha}[1.00]^{\beta}}{K[0.50]^{\alpha}[0.50]^{\beta}}$
$2 = (2)^{\beta} \implies \beta = 1$.
Thus,the rate law is $R = K[A]^0[B_2]^1$,which simplifies to $R = K[B_2]$.

(D) The order of reaction is the sum of the powers of the concentration terms in the rate law expression.
Order $= \frac{1}{2} + \frac{1}{2} + \frac{1}{4} = 1 + \frac{1}{4} = \frac{5}{4}$.

(A) The rate law is given by $R = K[A]^{\alpha}[B]^{\beta}$.
From the given data:
$0.10 = K[0.012]^{\alpha}[0.035]^{\beta}$ $(i)$
$0.80 = K[0.024]^{\alpha}[0.070]^{\beta}$ $(ii)$
$0.10 = K[0.024]^{\alpha}[0.035]^{\beta}$ $(iii)$
$0.80 = K[0.012]^{\alpha}[0.070]^{\beta}$ $(iv)$
Dividing $(iii)$ by $(i)$:
$\frac{0.10}{0.10} = \frac{K[0.024]^{\alpha}[0.035]^{\beta}}{K[0.012]^{\alpha}[0.035]^{\beta}}$
$1 = 2^{\alpha} \Rightarrow \alpha = 0$.
Dividing $(iv)$ by $(i)$:
$\frac{0.80}{0.10} = \frac{K[0.012]^{\alpha}[0.070]^{\beta}}{K[0.012]^{\alpha}[0.035]^{\beta}}$
$8 = 2^{\beta} \Rightarrow \beta = 3$.
Thus,the rate law is $R = K[A]^0[B]^3$ or $R = K[B]^3$.

(C) The rate law is given by $Rate = k[A]^x[B]^y$.
From observation $1$: $2 \times 10^{-3} = k(0.1)^x(0.1)^y$ $(i)$
From observation $2$: $3.2 \times 10^{-3} = k(0.4)^x(0.1)^y$ $(ii)$
Dividing $(ii)$ by $(i)$: $1.6 = (4)^x \implies 1.6 = (2^2)^x = 2^{2x}$. Since $1.6 = 2^{0.678}$,this suggests a typo in the provided data. Assuming standard values where $Rate_2 = 8 \times 10^{-3}$,then $4 = 4^x \implies x = 1$.
From observation $3$: $8 \times 10^{-3} = k(0.1)^x(0.2)^y$ $(iii)$
Dividing $(iii)$ by $(i)$: $4 = (2)^y \implies y = 2$.
Order of reaction $= x + y = 1 + 2 = 3$.

(B) The rate law is given by $r = k[A]^{\alpha}[B]^{\beta} \dots (i)$
When the concentration of $A$ is tripled and $B$ is doubled,the new rate $r'$ is $72r$:
$72r = k[3A]^{\alpha}[2B]^{\beta} \dots (ii)$
Dividing equation $(ii)$ by equation $(i)$:
$72 = 3^{\alpha} \times 2^{\beta}$
Testing the options:
For option $B$ $(\alpha = 2, \beta = 3)$:
$3^{2} \times 2^{3} = 9 \times 8 = 72$
Thus,the order with respect to $A$ is $2$ and with respect to $B$ is $3$.

(B) The given rate law is $-\frac{d[C]}{dt} = \frac{k_1 [C]}{1 + k_2 [C]}$.
When the concentration $[C]$ is very high,the term $k_2 [C]$ becomes much larger than $1$ (i.e.,$k_2 [C] \gg 1$).
Therefore,the denominator $1 + k_2 [C]$ can be approximated as $k_2 [C]$.
Substituting this into the rate law: Rate $\approx \frac{k_1 [C]}{k_2 [C]} = \frac{k_1}{k_2}$.
Since the rate is now independent of the concentration $[C]$,the rate can be written as Rate $= k'[C]^0$.
Thus,the order of the reaction is $0$.

(B) For an $n^{th}$ order reaction,the integrated rate law is given by:
$K = \frac{1}{t(n-1)} \left[ \frac{1}{[A]_t^{n-1}} - \frac{1}{[A]_0^{n-1}} \right]$
For $100\%$ completion,the concentration of the reactant $[A]_t$ becomes $0$.
Substituting $[A]_t = 0$ into the equation:
$K = \frac{1}{t_{100\%}(n-1)} \left[ \frac{1}{0^{n-1}} - \frac{1}{[A]_0^{n-1}} \right]$
Since $n < 1$,$n-1$ is negative. Let $m = 1-n$,where $m > 0$.
Then $n-1 = -m$.
$K = \frac{1}{t_{100\%}(-m)} \left[ \frac{1}{0^{-m}} - \frac{1}{[A]_0^{-m}} \right]$
$K = \frac{1}{t_{100\%}(-m)} \left[ 0 - [A]_0^m \right]$
$K = \frac{[A]_0^m}{t_{100\%} \times m}$
$t_{100\%} = \frac{[A]_0^m}{K \times m}$
Substituting $m = 1-n$ back:
$t_{100\%} = \frac{[A]_0^{1-n}}{K(1-n)}$

(C) The general unit of the rate constant $k$ for a reaction of order $n$ is given by the formula: $(mol \ L^{-1})^{1-n} \ s^{-1}$.
Given the unit is $mol^{-2} \ L^{2} \ s^{-1}$,we can equate the powers of concentration:
$(mol \ L^{-1})^{1-n} = mol^{-2} \ L^{2}$.
Comparing the powers of $mol$:
$1-n = -2$.
Solving for $n$:
$n = 1 + 2 = 3$.
Therefore,the order of the reaction is $3$.

(D) For an elementary reaction $A_{(g)} + 2B_{(g)} \to \text{product}$,the rate law is given by $Rate = K[A][B]^2$.
Since the order with respect to $A$ is $1$ and with respect to $B$ is $2$,the overall order is $3$.
If $[B]$ is in excess,$[B]$ remains approximately constant,so the rate becomes $Rate = K'[A]$,where $K' = K[B]^2$. This is a Pseudo first-order reaction.
For a first-order reaction,$t_{1/2} = \frac{0.693}{K'}$,which is independent of the initial concentration of $A$.
If $[A]$ is in excess,the rate becomes $Rate = K''[B]^2$,where $K'' = K[A]$. This is a Pseudo second-order reaction.
Therefore,all statements $A, B, C,$ and $D$ are technically correct descriptions of the kinetics under the given conditions. However,in the context of standard multiple-choice questions,if one must be chosen as 'incorrect' due to a potential typo in the question's premise,option $B$ is often cited in textbooks regarding the dependence of $t_{1/2}$ on the rate constant $K'$,but strictly speaking,none are incorrect. Given the options,$D$ is the most standard description of a pseudo-order reaction.

(B) Let the rate law be $r = k [A]^x [B]^y$.
Comparing $Exp$ $1$ and $Exp$ $2$ (where $[B]_0$ is constant):
$\frac{r_2}{r_1} = \frac{k [A]_2^x [B]_2^y}{k [A]_1^x [B]_1^y} \implies \frac{0.80}{0.10} = (\frac{0.024}{0.012})^x$
$8 = (2)^x \implies x = 3$.
Comparing $Exp$ $1$ and $Exp$ $3$ (where $[A]_0$ is constant):
$\frac{r_3}{r_1} = \frac{k [A]_3^x [B]_3^y}{k [A]_1^x [B]_1^y} \implies \frac{0.10}{0.10} = (\frac{0.070}{0.035})^y$
$1 = (2)^y \implies y = 0$.
Thus,the rate law is $r = k [A]^3$.

(D) The rate of the reaction is determined by the slow step,which is the rate-determining step.
From step $(ii)$,the rate expression is: Rate $= k_1[M][Z]$ ...... $(1)$
Since step $(i)$ is a very rapid equilibrium,we can write the equilibrium constant expression:
$K_{eq} = \frac{[M]}{[X][Y]}$
Therefore,$[M] = K_{eq}[X][Y]$ ...... $(2)$
Substituting the value of $[M]$ from equation $(2)$ into equation $(1)$:
Rate $= k_1 K_{eq} [X][Y][Z]$
Let $k = k_1 K_{eq}$,then the rate law becomes:
Rate $= k[X][Y][Z]$

(D) Let the rate law be $Rate = k[A]^x[B]^y$.
From experiment $1$: $2 \times 10^{-3} = k(0.1)^x(0.1)^y$ $(i)$
From experiment $2$: $8 \times 10^{-3} = k(0.4)^x(0.1)^y$ (ii)
Dividing (ii) by $(i)$: $4 = (4)^x \implies x = 1$.
From experiment $3$: $1.6 \times 10^{-2} = k(0.1)^x(0.2)^y$ (iii)
Dividing (iii) by $(i)$: $8 = (2)^y \implies y = 3$.
Total order of reaction = $x + y = 1 + 3 = 4$.

(D) The rate law for the reaction $3 \, A_{(g)} \rightarrow B_{(g)} + C_{(g)}$ is given by $-\frac{1}{3} \frac{d[A]}{dt} = K[A]^2$.
Given $K = 10^{-14} \, L/mol \cdot min$ and $[A] = 0.5 \, M$.
Substituting the values: $-\frac{d[A]}{dt} = 3 \times K \times [A]^2$.
$-\frac{d[A]}{dt} = 3 \times 10^{-14} \times (0.5)^2 = 3 \times 10^{-14} \times 0.25 = 7.5 \times 10^{-15} \, M/min$.
To convert the rate from $M/min$ to $M/sec$,divide by $60$:
$-\frac{d[A]}{dt} = \frac{7.5 \times 10^{-15}}{60} = 1.25 \times 10^{-16} \, M/sec$.
Since this value is not among the options,the correct choice is $D$.

(C) For an $n^{th}$ order reaction $(n \neq 1)$,the integrated rate law is given by:
$K = \frac{1}{t(n - 1)} \left[ \frac{1}{A_t^{n - 1}} - \frac{1}{A_0^{n - 1}} \right]$
For $t_{0.5}$ (half-life),$A_t = \frac{A_0}{2}$,so:
$K t_{0.5} = \frac{1}{n - 1} \left[ \frac{2^{n - 1} - 1}{A_0^{n - 1}} \right]$
For $t_{0.875}$,$A_t = A_0 - 0.875 A_0 = 0.125 A_0 = \frac{A_0}{8}$,so:
$K t_{0.875} = \frac{1}{n - 1} \left[ \frac{8^{n - 1} - 1}{A_0^{n - 1}} \right]$
Dividing the two expressions:
$\frac{t_{0.875}}{t_{0.5}} = \frac{8^{n - 1} - 1}{2^{n - 1} - 1}$

(B) The rate law for the reaction is given by: $\text{Rate} = k[N_2O_5]^n$.
Assuming a first-order reaction with respect to $N_2O_5$ (as the unit of rate constant $s^{-1}$ indicates a first-order reaction),the rate law is $\text{Rate} = k[N_2O_5]$.
Rearranging to find the concentration: $[N_2O_5] = \frac{\text{Rate}}{k}$.
Substituting the given values: $[N_2O_5] = \frac{1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}}{3.4 \times 10^{-5} \ s^{-1}} = 3 \ mol \ L^{-1}$.

(B) The rate law is given by: $\text{Rate} = K[A]^{x}[B]^{y}$
From experiment $(1)$: $5 \times 10^{-4} = K[2.5 \times 10^{-4}]^{x}[3 \times 10^{-5}]^{y} \quad \dots (i)$
From experiment $(2)$: $4 \times 10^{-3} = K[5 \times 10^{-4}]^{x}[6 \times 10^{-5}]^{y} \quad \dots (ii)$
From experiment $(3)$: $1.6 \times 10^{-2} = K[1 \times 10^{-3}]^{x}[6 \times 10^{-5}]^{y} \quad \dots (iii)$
Dividing $(iii)$ by $(ii)$:
$\frac{1.6 \times 10^{-2}}{4 \times 10^{-3}} = \left(\frac{1 \times 10^{-3}}{5 \times 10^{-4}}\right)^{x} \implies 4 = 2^{x} \implies x = 2$
Dividing $(ii)$ by $(i)$:
$\frac{4 \times 10^{-3}}{5 \times 10^{-4}} = \left(\frac{5 \times 10^{-4}}{2.5 \times 10^{-4}}\right)^{2} \cdot \left(\frac{6 \times 10^{-5}}{3 \times 10^{-5}}\right)^{y}$
$8 = 2^{2} \cdot 2^{y} \implies 8 = 4 \cdot 2^{y} \implies 2 = 2^{y} \implies y = 1$
Thus,the order with respect to $A$ is $2$ and with respect to $B$ is $1$.

(C) The rate law is given as $\text{Rate} = K \frac{[I^{-}]^1 [OCl^{-}]^1}{[OH^{-}]^1}$.
The overall order of a reaction is the sum of the powers of the concentration terms in the rate law expression.
$\text{Order} = 1 + 1 + (-1) = 1$.
Therefore,the overall order of the reaction is $1$.

(A) The rate of the reaction is determined by the slow step $(iii)$:
Rate $= k_5 [COCl][Cl_2]$
Assuming equilibrium for step $(ii)$:
$K_{eq2} = \frac{k_3}{k_4} = \frac{[COCl]}{[Cl][CO]} \Rightarrow [COCl] = \frac{k_3}{k_4} [Cl][CO]$
Assuming equilibrium for step $(i)$:
$K_{eq1} = \frac{k_1}{k_2} = \frac{[Cl]^2}{[Cl_2]} \Rightarrow [Cl] = \left( \frac{k_1}{k_2} \right)^{1/2} [Cl_2]^{1/2}$
Substituting $[Cl]$ into the expression for $[COCl]$:
$[COCl] = \frac{k_3}{k_4} \times \left( \frac{k_1}{k_2} \right)^{1/2} [Cl_2]^{1/2} [CO]$
Substituting $[COCl]$ into the rate expression:
Rate $= k_5 \times \left( \frac{k_3}{k_4} \times \left( \frac{k_1}{k_2} \right)^{1/2} [Cl_2]^{1/2} [CO] \right) \times [Cl_2]$
Rate $= k_5 \times \frac{k_3}{k_4} \times \left( \frac{k_1}{k_2} \right)^{1/2} [CO][Cl_2]^{3/2}$

(A) The reaction is $2A + B \to A_2B$.
Given the stoichiometry,the change in concentration is $\Delta[A] = [A]_0 - [A]_t = 0.2 - 0.12 = 0.08 \ mol/L$.
Since $2 \ mol$ of $A$ react with $1 \ mol$ of $B$,the amount of $B$ consumed is $\frac{1}{2} \times 0.08 = 0.04 \ mol/L$.
Therefore,the concentration of $B$ at this time is $[B]_t = [B]_0 - 0.04 = 0.4 - 0.04 = 0.36 \ mol/L$.
The rate law is given by $\text{Rate} = K[A][B]^2$.
Substituting the values: $\text{Rate} = (2.0 \times 10^{-6}) \times (0.12) \times (0.36)^2$.
$\text{Rate} = 2.0 \times 10^{-6} \times 0.12 \times 0.1296 = 3.1104 \times 10^{-8} \ mol \ L^{-1} \ s^{-1}$.

(C) For a reaction of order $n$,the half-life is given by $t_{1/2} \propto \frac{1}{a^{n-1}}$,where $a$ is the initial concentration.
For a second-order reaction $(n=2)$,$t_{1/2} \propto \frac{1}{a}$.
Let the initial concentration be $a_0$.
After the $1^{st}$ half-life,the concentration becomes $a_1 = \frac{a_0}{2}$.
After the $2^{nd}$ half-life,the concentration becomes $a_2 = \frac{a_1}{2} = \frac{a_0}{4}$.
After the $3^{rd}$ half-life,the concentration becomes $a_3 = \frac{a_2}{2} = \frac{a_0}{8}$.
Since $t_{1/2} = \frac{1}{k \cdot a}$,the $n^{th}$ half-life is $t_{n} = \frac{1}{k \cdot a_{n-1}}$.
Given $t_1 = 20 \ s = \frac{1}{k \cdot a_0}$,then $k = \frac{1}{20 \cdot a_0}$.
The $3^{rd}$ half-life is $t_3 = \frac{1}{k \cdot a_2} = \frac{1}{k \cdot (a_0/4)} = \frac{4}{k \cdot a_0} = 4 \times 20 \ s = 80 \ s$.

(C) The rate constant $(k)$ of a chemical reaction is a characteristic property that depends only on the temperature and the nature of the reactants.
It is independent of the concentration of the reactants.
Therefore,if the concentration of a first-order reaction is increased by $x$ times,the rate constant $(k)$ remains unchanged.

(B) The rate law for the reaction $A + 2B \to C + D$ can be written as $Rate = k[A]^x[B]^y$,where $x$ and $y$ are the orders with respect to $A$ and $B$ respectively.
According to the problem,when $[A]$ is increased $9$ times,the rate increases $3$ times:
$3 \times Rate = k(9[A])^x[B]^y$
$3 = 9^x \implies 3 = (3^2)^x \implies 3^1 = 3^{2x} \implies x = \frac{1}{2}$.
When $[B]$ is increased $2$ times,the rate increases $2$ times:
$2 \times Rate = k[A]^x(2[B])^y$
$2 = 2^y \implies y = 1$.
The total order of the reaction is $x + y = \frac{1}{2} + 1 = \frac{3}{2}$.

(C) The rate law is given by $Rate = k[A]^x[B]^y$.
From observation $1$ and $2$,$[B]$ is constant. When $[A]$ doubles,the rate increases from $2 \times 10^{-3}$ to $4 \times 10^{-3}$ (doubles). Thus,$2^x = 2$,which means $x = 1$.
From observation $1$ and $3$,$[A]$ is constant. When $[B]$ doubles,the rate increases from $2 \times 10^{-3}$ to $8 \times 10^{-3}$ (quadruples). Thus,$2^y = 4$,which means $y = 2$.
The overall order of reaction is $x + y = 1 + 2 = 3$.

(A) For a reaction mechanism to be consistent with the rate law,the rate-determining step (the slowest step) must involve the reactants in the same stoichiometric ratio as the rate law.
In Mechanism $I$,the slow step is $NO_{2(g)} + O_{3(g)} \to NO_{3(g)} + O_{2(g)}$. The rate of this step is $R = K[NO_2][O_3]$,which matches the given rate law.
In Mechanism $II$,the slow step is $NO_{2(g)} + [O] \to NO_3$. From the fast equilibrium $O_{3(g)} \rightleftharpoons O_{2(g)} + [O]$,we have $K_{eq} = \frac{[O_2][O]}{[O_3]}$,so $[O] = K_{eq} \frac{[O_3]}{[O_2]}$. Substituting this into the rate expression for the slow step,$R = K'[NO_2][O] = K'[NO_2] \times K_{eq} \frac{[O_3]}{[O_2]} = K'' \frac{[NO_2][O_3]}{[O_2]}$. This does not match the given rate law.
Therefore,only Mechanism $I$ is consistent.

(D) For a second order reaction,the rate law is given by: $-\frac{d[X]}{dt} = k[X]^2$
Integrating this expression: $\int_{[X]_0}^{[X]} -\frac{d[X]}{[X]^2} = \int_{0}^{t} k dt$
This results in: $\frac{1}{[X]} - \frac{1}{[X]_0} = kt$
Rearranging into the equation of a straight line $(y = mx + c)$: $\frac{1}{[X]} = kt + \frac{1}{[X]_0}$
Here,$y = \frac{1}{[X]}$,$m = k$,$x = t$,and $c = \frac{1}{[X]_0}$.
Thus,a plot of $\frac{1}{[X]}$ against time $t$ yields a straight line.

(D) The given reaction is $2N_2O_5 \to 4NO_2 + O_2$.
The unit of the rate constant $(sec^{-1})$ indicates that it is a first-order reaction.
The rate law is given by: $\text{Rate} = k[N_2O_5]$.
Given:
$k = 3.0 \times 10^{-5} \, sec^{-1}$
$\text{Rate} = 2.4 \times 10^{-5} \, M \, sec^{-1}$
Substituting the values into the rate law:
$2.4 \times 10^{-5} = (3.0 \times 10^{-5}) \times [N_2O_5]$
$[N_2O_5] = \frac{2.4 \times 10^{-5}}{3.0 \times 10^{-5}}$
$[N_2O_5] = 0.8 \, M$.

(B) The reaction is $RCl + H_2O \longrightarrow ROH + HCl$.
Since water is present in large excess,its concentration remains effectively constant during the reaction.
Therefore,the rate law is given by $Rate = k[RCl]^1[H_2O]^0 = k'[RCl]^1$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law,which is $1 + 0 = 1$.
The molecularity is the number of reacting species taking part in an elementary step,which is $2$ ($RCl$ and $H_2O$).
Thus,molecularity is $2$ and order is $1$.