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

(A) The rate law for the reaction can be expressed as: $\text{Rate} = k[A]^x[B]^y$.
$1$. When the concentration of $A$ is doubled,the rate doubles: $2 \times \text{Rate} = k[2A]^x[B]^y$,which implies $2^x = 2$,so $x = 1$.
$2$. When the concentration of $B$ is doubled,the rate remains unchanged: $\text{Rate} = k[A]^x[2B]^y$,which implies $2^y = 1$,so $y = 0$.
$3$. The overall order of the reaction is $x + y = 1 + 0 = 1$.

(B) The given rate law is $\text{Rate} = K[A]^2[B]$.
When a reactant is taken in large excess,its concentration remains effectively constant throughout the reaction.
Since $[A]$ is in large excess,$[A] \approx \text{constant}$.
Therefore,the rate law becomes $\text{Rate} = K' [B]$,where $K' = K[A]^2$.
This is a pseudo-first-order reaction.
Thus,the order of the reaction with respect to $[B]$ is $1$.

(D) The rate law is given by $Rate = k [A] [B]$.
Since concentration $[C] = n/V$,if the volume $V$ is reduced to $V/4$,the concentration of each reactant becomes $4$ times the initial concentration.
New rate $Rate' = k [4A] [4B] = 16 \times k [A] [B]$.
Therefore,the rate of the reaction becomes $16$ times the initial rate.

(C) The rate law for the reaction is given by $r = k[A]^2[B]^3$.
When the concentrations of both $A$ and $B$ are doubled,the new rate $r'$ becomes:
$r' = k[2A]^2[2B]^3$
$r' = k \times 4[A]^2 \times 8[B]^3$
$r' = 32 \times k[A]^2[B]^3$
$r' = 32r$
Therefore,the rate will increase by a factor of $32$.

(C) The relationship between half-life $(t_{1/2})$ and initial concentration $(a)$ for a reaction of order $n$ is given by: $t_{1/2} \propto a^{1-n}$.
Given:
For $a_1 = 50 \, \text{mm}$, $t_{1/2,1} = 4 \, \text{hours}$.
For $a_2 = 100 \, \text{mm}$, $t_{1/2,2} = 2 \, \text{hours}$.
Using the ratio: $\frac{t_{1/2,1}}{t_{1/2,2}} = \left( \frac{a_1}{a_2} \right)^{1-n}$.
Substituting the values: $\frac{4}{2} = \left( \frac{50}{100} \right)^{1-n}$.
$2 = \left( \frac{1}{2} \right)^{1-n} = (2^{-1})^{1-n} = 2^{n-1}$.
Comparing the exponents: $1 = n - 1$, which gives $n = 2$.

(B) For a reaction of order $n$ (where $n \neq 1$),the integrated rate law is given by:
$k = \frac{1}{(n-1)t} \left( \frac{1}{[R]^{n-1}} - \frac{1}{[R]_0^{n-1}} \right)$
At half-life,$t = t_{1/2}$ and $[R] = \frac{[R]_0}{2}$.
Substituting these values:
$k = \frac{1}{(n-1)t_{1/2}} \left( \frac{1}{([R]_0/2)^{n-1}} - \frac{1}{[R]_0^{n-1}} \right)$
$k = \frac{1}{(n-1)t_{1/2}} \left( \frac{2^{n-1} - 1}{[R]_0^{n-1}} \right)$
Rearranging for $t_{1/2}$:
$t_{1/2} = \frac{2^{n-1} - 1}{k(n-1)} \times \frac{1}{[R]_0^{n-1}}$
Therefore,$t_{1/2} \propto \frac{1}{[R]_0^{n-1}}$ or $t_{1/2} \propto [R]_0^{1-n}$.

(B) The general rate law is given by: $\text{Rate} = K[A]^x[B]^y$.
When the concentration of $A$ is doubled,the rate increases by $4$ times $(2^2 = 4)$,which implies the order with respect to $A$ is $x = 2$.
When the concentration of $B$ is doubled,the rate remains unchanged,which implies the order with respect to $B$ is $y = 0$.
Substituting these values into the rate law: $\text{Rate} = K[A]^2[B]^0 = K[A]^2$.

(D) The rate constant for a second-order reaction with equal initial concentrations is given by $k = \frac{1}{t} \cdot \frac{x}{a(a-x)}$.
Let the initial concentration $a = 1$.
For $60\%$ completion,$x = 0.6$ and $t = 3000 \ s$.
$k = \frac{1}{3000} \cdot \frac{0.6}{1(1-0.6)} = \frac{1}{3000} \cdot \frac{0.6}{0.4} = \frac{1}{3000} \cdot 1.5 = \frac{1}{2000} \ s^{-1}$.
For $20\%$ completion,$x = 0.2$ and $a = 1$.
$t = \frac{1}{k} \cdot \frac{x}{a(a-x)} = 2000 \cdot \frac{0.2}{1(1-0.2)} = 2000 \cdot \frac{0.2}{0.8} = 2000 \cdot 0.25 = 500 \ s$.

(C) The rate law for the reaction $2A + B \rightarrow A_2B$ is given by $Rate = k[A]^x[B]^y$.
Assuming the reaction is elementary,the rate law is $Rate = k[A]^2[B]^1$.
Let the initial rate be $R_1 = k[A]^2[B]$.
When the concentration of $A$ is doubled $(2[A])$ and $B$ is halved $([B]/2)$,the new rate $R_2$ is:
$R_2 = k(2[A])^2([B]/2) = k(4[A]^2)([B]/2) = 2k[A]^2[B]$.
Comparing $R_2$ with $R_1$,we get $R_2 = 2R_1$.
Therefore,the rate of the reaction increases by $2$ times.

(D) The rate of the reaction is determined by the slowest step,which is Step-$II$: $\text{Rate} = K_2[X][B]$.
From the fast equilibrium step (Step-$I$): $K_{eq} = \frac{[X]}{[A]^2}$,so $[X] = K_{eq}[A]^2$.
Substituting the value of $[X]$ into the rate expression for the slow step:
$\text{Rate} = K_2(K_{eq}[A]^2)[B] = K[A]^2[B]$,where $K = K_2 \times K_{eq}$.

(B) The molecularity of a reaction 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.
For a reaction involving two different reactants,at least two molecules must collide to form products.
Therefore,such a reaction must be at least bimolecular (molecularity $\ge 2$).
It cannot be a unimolecular reaction because a unimolecular reaction involves the decomposition or rearrangement of a single reactant molecule.

(C) The initial rate is $R_1 = k[A]^n[B]^m$.
When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is halved $([B]' = \frac{1}{2}[B])$,the new rate $R_2$ is:
$R_2 = k(2[A])^n(\frac{1}{2}[B])^m$
$R_2 = k \times 2^n \times [A]^n \times (\frac{1}{2})^m \times [B]^m$
$R_2 = 2^n \times 2^{-m} \times k[A]^n[B]^m$
$R_2 = 2^{(n-m)} \times R_1$
Therefore,the ratio of the new rate to the initial rate is $\frac{R_2}{R_1} = 2^{(n-m)}$.

(D) The rate-determining step is the second reaction: $NOBr_{2(g)} + NO_{(g)} \rightarrow 2 NOBr_{(g)}$.
The rate law for this step is: $Rate = k[NOBr_2][NO]$.
Since $NOBr_2$ is an intermediate,we express its concentration using the equilibrium constant $(K_c)$ of the first step: $NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$.
$K_c = \frac{[NOBr_2]}{[NO][Br_2]} \implies [NOBr_2] = K_c[NO][Br_2]$.
Substituting this into the rate law: $Rate = k \times K_c[NO][Br_2] \times [NO] = k'[NO]^2[Br_2]$.
Thus,the order of the reaction with respect to $NO_{(g)}$ is $2$.

(C) The rate of the reaction is determined by the slow step:
$Rate = k[NOBr_2][NO]$
Since $NOBr_2$ is an intermediate,we express its concentration using the equilibrium constant $K_{eq}$ from the first step:
$K_{eq} = \frac{[NOBr_2]}{[NO][Br_2]}$
$[NOBr_2] = K_{eq}[NO][Br_2]$
Substituting this into the rate equation:
$Rate = k \cdot K_{eq}[NO][Br_2] \cdot [NO]$
$Rate = k' [NO]^2 [Br_2]^1$
The order of reaction with respect to $NO_{(g)}$ is the exponent of $[NO]$,which is $2$.

(B) For a first-order reaction,the rate of reaction is given by $Rate = k[A]^1$. As the concentration of the reactant $[A]$ decreases over time,the rate of the reaction also decreases. Therefore,the statement that the rate remains constant is incorrect. The half-life of a first-order reaction is $t_{1/2} = 0.693/k$,which is independent of the initial concentration. The unit of the rate constant $k$ for a reaction of order $n$ is $(mol \text{ } L^{-1})^{1-n} \text{ } s^{-1}$. For $n=2$,the unit is $(mol \text{ } L^{-1})^{-1} \text{ } s^{-1} = mol^{-1} \text{ } L \text{ } s^{-1}$.

(A) The rate law is given as $Rate = K [A]^{3/2} [B]^{-1/2}$.
The order of reaction is the sum of the powers of the concentration terms in the rate law expression.
Order $= \frac{3}{2} + (-\frac{1}{2}) = \frac{3-1}{2} = \frac{2}{2} = 1$.

(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. Since it represents a count of particles,it must always be a whole number (integer) and cannot be zero or a fraction.

(C) The initial rate is given by: $V_1 = K[A]^n[B]^m$ $(I)$
The new rate $V_2$ is calculated by substituting the new concentrations: $V_2 = K(2[A])^n([B]/2)^m$ $(II)$
Taking the ratio of equation $(II)$ to equation $(I)$:
$\frac{V_2}{V_1} = \frac{K(2[A])^n([B]/2)^m}{K[A]^n[B]^m} = \frac{2^n[A]^n \times [B]^m \times 2^{-m}}{[A]^n[B]^m} = 2^n \times 2^{-m} = 2^{n-m}$

(C) For a reaction of order $n$,the time required for a definite fraction of the reaction to complete is given by the relation: $t \propto \frac{1}{[A]_0^{n-1}}$,where $[A]_0$ is the initial concentration of the reactant.
Given that the time is inversely proportional to the initial concentration,we have $t \propto \frac{1}{[A]_0^1}$.
Comparing the two expressions,we get $n - 1 = 1$,which implies $n = 2$.
Therefore,the reaction is of the second order.

(D) The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
Given rate law: $r = K[A]^{1/2}[B]^{1/3}[C]^{1/4}$.
Order of reaction = $1/2 + 1/3 + 1/4$.
To add these fractions,find the least common multiple of the denominators $(2, 3, 4)$,which is $12$.
Order = $(6/12) + (4/12) + (3/12) = 13/12$.

(B) The rate of the reaction is determined by the slow step,which is the rate-determining step $(RDS)$.
Rate $= k[A_2][B]$.
Since $A_2$ is an intermediate,we express its concentration using the equilibrium constant $K_{eq}$ from the fast step:
$K_{eq} = \frac{[A_2]}{[A]^2} \implies [A_2] = K_{eq}[A]^2$.
Substituting this into the rate equation:
Rate $= k \cdot K_{eq}[A]^2[B] = k'[A]^2[B]^1$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law:
Order $= 2 + 1 = 3$.

(B) The relationship between half-life $(t_{1/2})$ and initial concentration $(a)$ for a reaction of order $n$ is given by: $t_{1/2} \propto \frac{1}{a^{n-1}}$.
Given that when initial concentration $a_2 = 2a_1$,the half-life becomes $(t_{1/2})_2 = \frac{(t_{1/2})_1}{2}$.
Substituting these values into the ratio: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{a_2}{a_1} \right)^{n-1}$.
$\frac{(t_{1/2})_1}{(t_{1/2})_1 / 2} = \left( \frac{2a_1}{a_1} \right)^{n-1}$.
$2 = (2)^{n-1}$.
Comparing the exponents: $n - 1 = 1$,which gives $n = 2$.
Therefore,the order of the reaction is $2$.

(B) The general formula for the unit of the rate constant $(k)$ for a reaction of order $(n)$ is given by: $(mol \cdot L^{-1})^{1-n} \cdot s^{-1}$.
For a second-order reaction,$n = 2$.
Substituting $n = 2$ into the formula: $(mol \cdot L^{-1})^{1-2} \cdot s^{-1} = (mol \cdot L^{-1})^{-1} \cdot s^{-1} = mol^{-1} \cdot L \cdot s^{-1}$.

(D) The rate law for the reaction is given as $r = k[A]^x[B]^y[C]^z$.
Given the expression $r \propto [A]^1[B]^2$,the exponents of the concentrations are $1$ for $A$ and $2$ for $B$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law expression.
Order $= 1 + 2 = 3$.

(C) The rate law for the elementary reaction $A + B \rightarrow AB$ is given by $r = k[A][B]$.
If the concentrations of $A$ and $B$ are doubled,the new rate $r'$ is:
$r' = k[2A][2B] = 4k[A][B] = 4r$.
Therefore,the rate of the reaction becomes $4$ times the original rate.

(A) The rate law for the reaction is $r = k[A][B]^2$.
Let the initial concentrations be $[A] = x$ and $[B] = y$.
Initial rate $r_1 = k(x)(y)^2 = kxy^2$.
New concentrations are $[A]' = 4x$ and $[B]' = y/2$.
New rate $r_2 = k(4x)(y/2)^2 = k(4x)(y^2/4) = kxy^2$.
Since $r_2 = r_1$,the rate remains constant.

(B) The rate-determining step is the slow step of the reaction mechanism.
In this mechanism,the slow step is: $P + Q \to R + S$.
The rate of the reaction is determined by the slow step,so the rate law is given by the concentration of the reactants involved in the slow step.
Therefore,the rate law expression is $r = k[P][Q]$.

(C) The rate of the reaction is determined by the slow step $(ii)$:
$r = k[X][Y_2]$ $---$ $(1)$
From the fast equilibrium step $(i)$,we have:
$K_{eq} = \frac{[X]^2}{[X_2]}$
$[X]^2 = K_{eq}[X_2]$
$[X] = K_{eq}^{1/2}[X_2]^{1/2}$ $---$ $(2)$
Substituting equation $(2)$ into equation $(1)$:
$r = k \cdot K_{eq}^{1/2}[X_2]^{1/2}[Y_2]^1$
$r = k'[X_2]^{1/2}[Y_2]^1$
The overall order of the reaction is the sum of the exponents of the concentration terms in the rate law:
Order $= 0.5 + 1 = 1.5$

(D) The half-life period $(T_{1/2})$ of a reaction is related to the initial concentration $([A]_0)$ by the expression $T_{1/2} \propto [A]_0^{1-n}$,where $n$ is the order of the reaction.
For a first-order reaction $(n=1)$,$T_{1/2} = \frac{\ln 2}{k}$.
Since this expression does not contain the initial concentration term,the half-life of a first-order reaction is independent of the initial concentration of the reactant.
Therefore,if doubling the initial concentration does not affect the half-life,the reaction must be of the first order.

(C) The rate law for the reaction is given by: $\text{Rate} = k[A]^2[B]^3$.
Let the initial rate be $r_1 = k[A]^2[B]^3$.
When the concentrations of both $A$ and $B$ are doubled,the new concentrations are $[A'] = 2[A]$ and $[B'] = 2[B]$.
The new rate $r_2$ is: $r_2 = k[2A]^2[2B]^3$.
$r_2 = k \cdot 4[A]^2 \cdot 8[B]^3$.
$r_2 = 32 \cdot k[A]^2[B]^3$.
$r_2 = 32 \cdot r_1$.
Therefore,the rate increases by a factor of $32$.

(D) Let the order of reaction with respect to $A$ and $B$ be $x$ and $y$ respectively.
So,the rate law can be given as:
$R = k[A]^{x}[B]^{y} \dots (i)$
When the concentration of only $B$ is doubled,the rate is doubled:
$2R = k[A]^{x}[2B]^{y} \dots (ii)$
Dividing $(ii)$ by $(i)$:
$2 = 2^{y} \Rightarrow y = 1$
When concentrations of both $A$ and $B$ are doubled,the rate increases by a factor of $8$:
$8R = k[2A]^{x}[2B]^{y} \dots (iii)$
Dividing $(iii)$ by $(i)$:
$8 = 2^{x} \times 2^{y}$
Substituting $y = 1$:
$8 = 2^{x} \times 2^{1}$ $\Rightarrow 4 = 2^{x}$ $\Rightarrow x = 2$
Thus,the rate law is $R = k[A]^{2}[B]$.

(D) The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
It is an experimental quantity and is not necessarily related to the stoichiometric coefficients of the balanced chemical equation.
The order of a reaction can be zero,a whole number,or even a fraction.
Therefore,the statement that the order of reaction is always a whole number is incorrect.

(B) The rate of reaction is defined as:
$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Substituting the given rate expressions:
$\frac{1}{2} k[N_2O_5] = \frac{1}{4} k'[N_2O_5] = k''[N_2O_5]$
Dividing by $[N_2O_5]$:
$\frac{k}{2} = \frac{k'}{4} = k''$
From $\frac{k}{2} = \frac{k'}{4}$,we get $k' = 2k$.
From $\frac{k}{2} = k''$,we get $k'' = \frac{k}{2}$.

(D) Let the order of reaction with respect to $A$ be $x$ and with respect to $B$ be $y$.
The rate law is given by: $\text{Rate} = k[A]^x[B]^y$
Using the data from the table:
$I. \ 6.0 \times 10^{-3} = k(0.1)^x(0.1)^y$
$II. \ 7.2 \times 10^{-2} = k(0.3)^x(0.2)^y$
$III. \ 2.88 \times 10^{-1} = k(0.3)^x(0.4)^y$
$IV. \ 2.40 \times 10^{-2} = k(0.4)^x(0.1)^y$
Dividing Eq. $(I)$ by Eq. $(IV)$:
$\frac{6.0 \times 10^{-3}}{2.40 \times 10^{-2}} = \left(\frac{0.1}{0.4}\right)^x \left(\frac{0.1}{0.1}\right)^y$
$0.25 = (0.25)^x \implies x = 1$
Dividing Eq. $(II)$ by Eq. $(III)$:
$\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}} = \left(\frac{0.3}{0.3}\right)^x \left(\frac{0.2}{0.4}\right)^y$
$0.25 = (0.5)^y \implies (0.5)^2 = (0.5)^y \implies y = 2$
Therefore,the rate law is: $\text{Rate} = k[A]^1[B]^2 = k[A][B]^2$.

(A) For the reaction,$A + B \longrightarrow$ Products.
Let the rate law be: $\text{Rate} = k[A]^x[B]^y$.
From observation $(i)$,doubling $[A]$ doubles the rate: $2 \times \text{Rate} = k[2A]^x[B]^y \implies 2 = 2^x \implies x = 1$.
From observation $(ii)$,doubling both $[A]$ and $[B]$ increases the rate by a factor of $8$: $8 \times \text{Rate} = k[2A]^1[2B]^y$.
Dividing this by the original rate law: $8 = 2^1 \times 2^y \implies 8 = 2 \times 2^y \implies 4 = 2^y \implies y = 2$.
Thus,the rate law is: $\text{Rate} = k[A][B]^2$.

(C) Let the rate law be $\text{Rate} = k [CH_3COCH_3]^x [Br_2]^y [H^+]^z$.
Comparing experiments $(1)$ and $(2)$: $[CH_3COCH_3]$ and $[H^+]$ are constant,while $[Br_2]$ doubles. The rate remains $5.7 \times 10^{-5} M s^{-1}$. Thus,$y = 0$.
Comparing experiments $(2)$ and $(3)$: $[CH_3COCH_3]$ is constant,$[Br_2]$ is constant,and $[H^+]$ doubles. The rate increases from $5.7 \times 10^{-5}$ to $1.14 \times 10^{-4}$ (a factor of $2$). Thus,$z = 1$.
Comparing experiments $(1)$ and $(4)$: $[Br_2]$ is constant. $[CH_3COCH_3]$ increases by $4/3$ and $[H^+]$ increases by $4$. The rate increases from $5.7 \times 10^{-5}$ to $3.04 \times 10^{-4}$ (a factor of $\approx 5.33$). Since $5.33 = (4/3) \times 4$,we find $x = 1$.
Therefore,the rate law is $\text{Rate} = k [CH_3COCH_3][H^+]$.

(D) The rate law for the reaction is given as $Rate = k[H_2]^1[ICl]^1$.
For Mechanism $A$,which is a single-step elementary reaction,the rate law would be $Rate = k[H_2][ICl]^2$. This does not match the given rate law.
For Mechanism $B$,the rate-determining step is the slow step: $H_{2(g)} + ICl_{(g)} \rightarrow HCl_{(g)} + HI_{(g)}$.
The rate law for this elementary step is $Rate = k[H_2][ICl]$.
This matches the given rate law,which is first order with respect to both $H_2$ and $ICl$. Therefore,only Mechanism $B$ is consistent with the given information.

(A) Since the reaction is $2nd$ order with respect to $CO$,the rate law is given as:
$r = k[CO]^2$
Let the initial concentration of $CO$ be $a$,so $[CO] = a$.
$r_1 = k(a)^2 = ka^2$
When the concentration is doubled,$[CO] = 2a$.
Therefore,$r_2 = k(2a)^2 = 4ka^2$.
Comparing the two rates,$r_2 = 4r_1$.
Thus,the rate of reaction increases by a factor of $4$.

(B) Step $(i)$: $NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$ (Fast equilibrium)
Step $(ii)$: $NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g)}$ (Slow,rate-determining step)
The rate law is determined by the slow step: $\text{Rate} = k[NOBr_2][NO]$.
Since $NOBr_2$ is an intermediate,we express its concentration using the equilibrium constant $K_C$ from step $(i)$: $K_C = \frac{[NOBr_2]}{[NO][Br_2]}$.
Therefore,$[NOBr_2] = K_C[NO][Br_2]$.
Substituting this into the rate law: $\text{Rate} = k \cdot K_C[NO][Br_2][NO] = k'[NO]^2[Br_2]$.
The order of the reaction with respect to $NO_{(g)}$ is $2$.

(B) For a reaction,the rate law can be expressed as $Rate = k[A]^x[B]^y$.
$1$. When the concentration of $B$ is doubled,the half-life does not change. This indicates that the reaction is of first order with respect to $B$ $(y=1)$,as the half-life of a first-order reaction is independent of the initial concentration of the reactant.
$2$. When the concentration of $A$ is doubled,the rate increases by two times. This indicates that the reaction is of first order with respect to $A$ $(x=1)$.
$3$. The overall order of the reaction is $n = x + y = 1 + 1 = 2$.
$4$. The unit of the rate constant for a reaction of order $n$ is $(mol \ L^{-1})^{1-n} \ s^{-1}$. For $n=2$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.

(D) The rate-determining step $(RDS)$ is the slow step of the reaction mechanism.
For mechanism $A$:
The slow step is $Cl_2 + H_2S \rightarrow H^{+} + Cl^{-} + Cl^{+} + HS^{-}$.
The rate law derived from this step is $\text{rate} = k[Cl_2][H_2S]$,which matches the given rate equation.
For mechanism $B$:
The slow step is $Cl_2 + HS^{-} \rightarrow 2Cl^{-} + H^{+} + S$.
The rate law is $\text{rate} = k[Cl_2][HS^{-}]$.
From the fast equilibrium $H_2S \rightleftharpoons H^{+} + HS^{-}$,the equilibrium constant is $K = \frac{[H^{+}][HS^{-}]}{[H_2S]}$,so $[HS^{-}] = \frac{K[H_2S]}{[H^{+}]}$.
Substituting this into the rate law gives $\text{rate} = k' \frac{[Cl_2][H_2S]}{[H^{+}]}$,which does not match the given rate equation.
Therefore,only mechanism $A$ is consistent.

(D) Let the rate of reaction be $\frac{d[C]}{dt} = k[A]^{x}[B]^{y}.$
From the given data:
$1.2 \times 10^{-3} = k[0.1]^{x}[0.1]^{y} \quad (i)$
$1.2 \times 10^{-3} = k[0.1]^{x}[0.2]^{y} \quad (ii)$
$2.4 \times 10^{-3} = k[0.2]^{x}[0.1]^{y} \quad (iii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{1.2 \times 10^{-3}}{1.2 \times 10^{-3}} = \frac{k[0.1]^{x}[0.1]^{y}}{k[0.1]^{x}[0.2]^{y}}$
$1 = (0.5)^{y} \Rightarrow y = 0.$
Dividing equation $(i)$ by $(iii)$:
$\frac{1.2 \times 10^{-3}}{2.4 \times 10^{-3}} = \frac{k[0.1]^{x}[0.1]^{y}}{k[0.2]^{x}[0.1]^{y}}$
$0.5 = (0.5)^{x} \Rightarrow x = 1.$
Thus,the rate law is $\frac{d[C]}{dt} = k[A]^{1}[B]^{0} = k[A].$

(C) Higher order $(> 3)$ reactions are rare because the probability of simultaneous collision of all the reacting species is extremely low.
For a reaction to occur,molecules must collide with sufficient energy and proper orientation.
As the number of molecules involved in a single step increases,the likelihood of them colliding at the same time and place decreases significantly.

(A) The rate of reaction is defined as the rate of disappearance of reactant divided by its stoichiometric coefficient.
For reaction $(a)$: $2N_2O_5 \rightarrow 4NO_{2(g)} + O_{2(g)}$
Rate $= -\frac{1}{2} \frac{d[N_2O_5]}{dt} = k_{rate}[N_2O_5]$
Given $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$,so Rate $= \frac{k}{2} [N_2O_5]$.
For reaction $(b)$: $N_2O_5 \rightarrow 2NO_{2(g)} + 1/2 O_{2(g)}$
Rate $= -\frac{d[N_2O_5]}{dt} = k'_{rate}[N_2O_5]$
Given $-\frac{d[N_2O_5]}{dt} = k'[N_2O_5]$,so Rate $= k' [N_2O_5]$.
Since the rate of the reaction is the same regardless of how the stoichiometry is written,we equate the rates:
$\frac{k}{2} [N_2O_5] = k' [N_2O_5]$
$k = 2k'$

(C) The rate of the reaction is determined by the slow step,which is step $(II)$:
$Rate = k[NOBr_2][NO]$
Since $NOBr_2$ is an intermediate,we use the equilibrium from step $(I)$ to express its concentration:
$K_{eq} = \frac{[NOBr_2]}{[NO][Br_2]}$
$[NOBr_2] = K_{eq}[NO][Br_2]$
Substituting this into the rate equation:
$Rate = k \cdot K_{eq}[NO][Br_2] \cdot [NO]$
$Rate = k'[NO]^2[Br_2]$
The overall order of the reaction is the sum of the powers of the concentration terms in the rate law:
$Order = 2 + 1 = 3$

(B) The rate law is given by $r = k[A][B]^{0.5}$.
Concentration is defined as $C = n/V$,where $n$ is the number of moles and $V$ is the volume.
If the volume $V$ is reduced to $V/4$,the concentration becomes $C' = n/(V/4) = 4(n/V) = 4C$.
Substituting the new concentrations $[A]' = 4[A]$ and $[B]' = 4[B]$ into the rate law:
$r' = k[A]'([B]')^{0.5} = k(4[A])(4[B])^{0.5}$.
$r' = k \times 4[A] \times 2[B]^{0.5} = 8 \times k[A][B]^{0.5}$.
$r' = 8r$.
Therefore,the rate of reaction increases by $8$ times.

(C) Let the rate law be $r = k[A]^x[B]^y$.
From experiment $1$ and $2$,$[B]$ is constant. Comparing the rates:
$\frac{8.4 \times 10^{-6}}{4.2 \times 10^{-6}} = (\frac{2.0}{1.0})^x \implies 2 = 2^x \implies x = 1$.
From experiment $1$ and $3$,$[A]$ is constant. Comparing the rates:
$\frac{5.6 \times 10^{-6}}{4.2 \times 10^{-6}} = (\frac{0.20}{0.15})^y \implies \frac{4}{3} = (\frac{4}{3})^y \implies y = 1$.
Thus,the rate law is $r = k[A]^1[B]^1$ or $r = k[A][B]$.

(B) For a $2^{nd}$ order reaction,the integrated rate law is $\frac{1}{[R]} = kt + \frac{1}{[R]_0}$.
Comparing this with the equation of a straight line $y = mx + c$,we get the slope $k = \tan(\theta) = \tan(\tan^{-1}(0.5)) = 0.5 \ L \ mol^{-1} \ min^{-1}$ and the intercept $\frac{1}{[R]_0} = 2 \ L \ mol^{-1}$.
Thus,$[R]_0 = \frac{1}{2} = 0.5 \ M$. This makes option $B$ incorrect and option $C$ correct.
The half-life is $t_{1/2} = \frac{1}{k[R]_0} = \frac{1}{0.5 \times 0.5} = \frac{1}{0.25} = 4 \ min$. So,option $A$ is correct.
For $75\%$ completion,$[R] = [R]_0 - 0.75[R]_0 = 0.25[R]_0 = 0.125 \ M$.
Using the rate law: $\frac{1}{0.125} = 0.5 \times t + 2 \implies 8 = 0.5t + 2 \implies 0.5t = 6 \implies t = 12 \ min$.
Since $12 \ min = \frac{12}{60} \ hours = 0.2 \ hours$,option $D$ is correct.
Therefore,the incorrect statement is $B$.

(C) The rate of reaction is defined as $Rate = -\frac{d[X_{2}]}{dt} = -\frac{d[Y_{2}]}{dt} = \frac{1}{2} \frac{d[XY]}{dt}$.
Let the rate law be $Rate = k[X_{2}]^{a}[Y_{2}]^{b}$.
From the data:
$1$) $5 \times 10^{-6} = k(0.1)^{a}(0.1)^{b}$
$2$) $10^{-5} = k(0.2)^{a}(0.1)^{b}$
Dividing $(2)$ by $(1)$: $2 = 2^{a} \implies a = 1$.
$3$) $4 \times 10^{-5} = k(0.2)^{a}(0.2)^{b}$
Dividing $(3)$ by $(2)$: $4 = 2^{b} \implies b = 2$.
Thus,$Rate = k[X_{2}][Y_{2}]^{2}$.
The rate of appearance of $XY$ is $\frac{d[XY]}{dt} = 2 \times Rate = 2k[X_{2}][Y_{2}]^{2}$.
Using the first data point: $5 \times 10^{-6} = 2k(0.1)(0.1)^{2} = 2k(10^{-3})$.
$k = \frac{5 \times 10^{-6}}{2 \times 10^{-3}} = 2.5 \times 10^{-3} \ M^{-2} \ sec^{-1}$.

(B) The given expression is $\ln k_t = \ln k_0 + \left( \frac{\ln (2.5)}{10} \right) \times t$.
Taking the exponential of both sides,we get $k_t = k_0 \times (2.5)^{t/10}$.
The temperature coefficient is defined as the ratio of rate constants at temperatures differing by $10 \, ^{\circ}C$,i.e.,$\frac{k_{t+10}}{k_t}$.
Substituting the values: $\frac{k_{t+10}}{k_t} = \frac{k_0 \times (2.5)^{(t+10)/10}}{k_0 \times (2.5)^{t/10}} = (2.5)^{(t+10-t)/10} = (2.5)^{10/10} = 2.5$.
Thus,the temperature coefficient is $2.5$.