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

(C) The rate law for the reaction $3X \to 4Y$ is given by $-\frac{1}{3} \frac{d[X]}{dt} = k[X]^n$.
For an $n$-th order reaction where $n \neq 1$,the integrated rate law is $\frac{1}{[X]^{n-1}} - \frac{1}{[X]_0^{n-1}} = 3(n-1)kt$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \frac{1}{[X]^3}$,we identify $n-1 = 3$,so $n = 4$.
The slope of the graph is $3(n-1)k = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
Substituting $n-1 = 3$,we get $3(3)k = \frac{1}{\sqrt{3}}$,which implies $9k = \frac{1}{\sqrt{3}}$,or $k = \frac{1}{9\sqrt{3}}$.
The rate of reaction $r$ is given by $r = k[X]^n = \frac{1}{9\sqrt{3}} \times (0.1)^4$.
$r = \frac{1}{9\sqrt{3}} \times 10^{-4} \ M \ \min^{-1}$.

(D) For a reaction of order $n \neq 1$,the integrated rate law is $t = \frac{1}{k(n-1)} [\frac{1}{C_t^{n-1}} - \frac{1}{C_0^{n-1}}]$.
For $t_{1/2}$,$C_t = C_0 / 2$:
$t_{1/2} = \frac{1}{k(n-1)} [\frac{2^{n-1}}{C_0^{n-1}} - \frac{1}{C_0^{n-1}}] = \frac{2^{n-1} - 1}{k(n-1)C_0^{n-1}}$.
For $t_{7/8}$,$C_t = C_0 / 8 = C_0 / 2^3$:
$t_{7/8} = \frac{1}{k(n-1)} [\frac{8^{n-1}}{C_0^{n-1}} - \frac{1}{C_0^{n-1}}] = \frac{(2^3)^{n-1} - 1}{k(n-1)C_0^{n-1}} = \frac{2^{3(n-1)} - 1}{k(n-1)C_0^{n-1}}$.
Dividing $t_{7/8}$ by $t_{1/2}$:
$\frac{t_{7/8}}{t_{1/2}} = \frac{2^{3(n-1)} - 1}{2^{n-1} - 1}$.
Therefore,$t_{7/8} = t_{1/2} \ [\frac{2^{3(n-1)} - 1}{2^{n-1} - 1}]$.

(A) The rate of a reaction is determined by the slowest step in the mechanism.
From Step $1$ (slow step),the rate law is given by: Rate $= k [NO]^2 [H_2]$.
Since the rate is directly proportional to the concentration of $H_2$ (first order with respect to $H_2$),doubling the concentration of $H_2$ while keeping $[NO]$ constant will double the rate of the reaction.
Therefore,both statement $A$ and statement $C$ are correct.

(C) The rate law is given by $r = k[A]^x[B]^y$.
From the data:
$1) \ 5.0 \times 10^{-5} = k(0.2)^x(0.3)^y$
$2) \ 5.0 \times 10^{-5} = k(0.2)^x(0.1)^y$
$3) \ 1.4 \times 10^{-4} = k(0.4)^x(0.05)^y$
Dividing $(1)$ by $(2)$:
$1 = (0.3/0.1)^y = 3^y \implies y = 0$.
Dividing $(3)$ by $(1)$:
$\frac{1.4 \times 10^{-4}}{5.0 \times 10^{-5}} = \frac{k(0.4)^x(0.05)^0}{k(0.2)^x(0.3)^0}$
$2.8 = (0.4/0.2)^x = 2^x$
Taking $\log$ on both sides: $x \log 2 = \log 2.8 \implies x = \frac{\log 2.8}{\log 2} \approx 1.48 \approx \frac{3}{2}$.
Thus,the order with respect to $A$ is $\frac{3}{2}$ and with respect to $B$ is $0$.

(NONE) For a reaction of order $n$,the half-life is related to the initial concentration by the relation: $t_{1/2} \propto [A_0]^{1-n}$.
Given data:
$1$) $[A_0]_1 = 0.1 \ M$,$(t_{1/2})_1 = 100 \ s$
$2$) $[A_0]_2 = 0.025 \ M$,$(t_{1/2})_2 = 50 \ s$
Taking the ratio: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{[A_0]_1}{[A_0]_2} \right)^{1-n}$
$\frac{100}{50} = \left( \frac{0.1}{0.025} \right)^{1-n}$
$2 = (4)^{1-n}$
$2 = (2^2)^{1-n} = 2^{2-2n}$
Equating exponents: $1 = 2 - 2n \implies 2n = 1 \implies n = 0.5$.
Since $n = 0.5$,the reaction is of order $0.5$. None of the provided options $A$ or $B$ are correct. For $n=0.5$,the unit of $k$ is $M^{1-0.5} \ s^{-1} = M^{0.5} \ s^{-1}$. Thus,$D$ is incorrect. Option $C$ is also incorrect based on the calculated order.

(C) The rate of the reaction is determined by the slowest step,which is Step-$2$: $r = k_2[AE][A]$.
Since $AE$ is an intermediate,we express its concentration using the equilibrium constant $K_{eq}$ from the fast Step-$1$: $K_{eq} = \frac{[AE]}{[A][E]}$,which gives $[AE] = K_{eq}[A][E]$.
Substituting this into the rate expression: $r = k_2(K_{eq}[A][E])[A] = k_2 K_{eq}[A]^2[E]$.
Letting $k = k_2 K_{eq}$,the rate law becomes $r = k[A]^2[E]$.

(D) We know that the relationship between half-life $(t_{1/2})$ and initial pressure $(P_{A_0})$ for a reaction of order $n$ is given by:
$t_{1/2} \propto (P_{A_0})^{1-n}$
Therefore,$\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{(P_{A_0})_1}{(P_{A_0})_2} \right)^{1-n}$
Given: $(t_{1/2})_1 = 100 \text{ s}$,$(P_{A_0})_1 = 0.1 \text{ atm}$
$(t_{1/2})_2 = 50 \text{ s}$,$(P_{A_0})_2 = 0.025 \text{ atm}$
Substituting the values:
$\frac{100}{50} = \left( \frac{0.1}{0.025} \right)^{1-n}$
$2 = (4)^{1-n}$
$2^1 = (2^2)^{1-n}$
$2^1 = 2^{2(1-n)}$
Comparing the powers: $1 = 2(1-n)$
$0.5 = 1 - n$
$n = 1 - 0.5 = 0.5$
Thus,the order of the reaction is $0.5$.

(A) The rate law of a chemical reaction is determined experimentally and depends on the mechanism of the reaction.
Species appearing in the balanced chemical equation do not necessarily appear in the rate law,especially if they are intermediates or if the reaction is complex.
Therefore,option $A$ is incorrect.
Bimolecular elementary reactions involve the collision of two particles,making them second order,so $B$ is correct.
Alkaline hydrolysis of an ester (saponification) follows second-order kinetics,so $C$ is correct.
For elementary reactions,the order and molecularity are indeed the same,so $D$ is correct.

(C) For an $n$-th order reaction,the half-life $t_{1/2}$ is related to the initial pressure $P_0$ as $t_{1/2} \propto \frac{1}{P_0^{n-1}}$.
Therefore,$\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{(P_0)_2}{(P_0)_1} \right)^{n-1}$.
Substituting the given values:
$\frac{135}{112.5} = \left( \frac{300}{250} \right)^{n-1}$.
$1.2 = (1.2)^{n-1}$.
Comparing the exponents,$n - 1 = 1$,which gives $n = 2$.

(C) The rate law for the reaction is given by $r = k[A]^x$,where $x$ is the order of the reaction.
Initially,$r = k[A]^x$ $(1)$.
When the concentration is doubled,the new rate becomes $7r$,so $7r = k[2A]^x$ $(2)$.
Dividing equation $(2)$ by equation $(1)$: $\frac{7r}{r} = \frac{k[2A]^x}{k[A]^x}$.
$7 = 2^x$.
Taking the logarithm on both sides: $\log(7) = x \log(2)$.
$x = \frac{\log(7)}{\log(2)} \approx \frac{0.845}{0.301} \approx 2.81$.
Since $2 < 2.81 < 3$,the order of the reaction is between $2$ and $3$.

(B) The rate law is expressed as $r = k[NO]^x[Cl_2]^y$.
From experiment $I$ and $II$,keeping $[NO]$ constant:
$\frac{3 \times 10^{-3}}{1 \times 10^{-3}} = \left(\frac{0.15}{0.05}\right)^y \implies 3 = 3^y \implies y = 1$.
From experiment $I$ and $III$,keeping $[Cl_2]$ constant:
$\frac{9 \times 10^{-3}}{1 \times 10^{-3}} = \left(\frac{0.15}{0.05}\right)^x \implies 9 = 3^x \implies x = 2$.
Thus,the rate law is $r = k[NO]^2[Cl_2]^1$ or $r = k[Cl_2][NO]^2$.

(C) The rate of a reaction is determined by its rate-determining step.
Given the rate-determining step is $Y + \frac{1}{2}Z \to Q$,the rate law is expressed as: $Rate = k[Y][Z]^{1/2}$.
If the concentration of $Z$ is doubled,the new rate $(r')$ becomes:
$r' = k[Y][2Z]^{1/2} = k[Y] \times \sqrt{2} \times [Z]^{1/2}$.
$r' = \sqrt{2} \times Rate = 1.414 \times Rate$.
Therefore,the rate of reaction will become $1.414$ times the original rate.

(A) The rate law is given by $r = k[A]^x[B]^y$.
Comparing experiments $I$ and $II$:
$[A]$ is doubled,$[B]$ is constant,and the rate $r$ is doubled.
Therefore,$x = 1$.
Comparing experiments $II$ and $III$:
$[A]$ is constant,$[B]$ is doubled,and the rate $r$ is doubled.
Therefore,$y = 1$.
The rate law is $r = k[A][B]$. Statement $(b)$ is correct.
Using experiment $I$ to find $k$:
$2 \times 10^{-4} = k \times (1 \times 10^{-2}) \times (2 \times 10^{-2})$
$2 \times 10^{-4} = k \times (2 \times 10^{-4})$
$k = 1 \ mol^{-1} \ L \ sec^{-1}$. Statement $(a)$ is correct.
If both $[A]$ and $[B]$ are doubled:
$r' = k(2[A])(2[B]) = 4k[A][B] = 4r$.
Thus,the rate increases four times. Statement $(c)$ is correct.
Therefore,all statements $(a), (b),$ and $(c)$ are correct.

(D) The rate of the reaction is determined by the slow step $(II)$:
$r = k_2 [N_2O_2] [H_2]$
Since $N_2O_2$ is an intermediate,we express its concentration using the equilibrium constant $(K_{eq})$ from the fast step $(I)$:
$K_{eq} = \frac{[N_2O_2]}{[NO]^2} \implies [N_2O_2] = K_{eq} [NO]^2$
Substituting this into the rate expression:
$r = k_2 (K_{eq} [NO]^2) [H_2]$
Let $k = k_2 K_{eq}$,then the rate law is:
$r = k [NO]^2 [H_2]$

(D) The rate constant $(k)$ is a characteristic property of a reaction at a given temperature.
It is independent of the initial concentration of the reactants for all orders of reactions.
It depends only on the temperature and the presence of a catalyst.

(C) For a reaction of order $n$,the half-life period $(t_{1/2})$ is related to the initial pressure $(P)$ as $(t_{1/2}) \propto P^{1-n}$.
Using the data: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = (\frac{P_1}{P_2})^{1-n}$.
Taking the first two values: $\frac{3.52}{1.76} = (\frac{50}{100})^{1-n} \implies 2 = (\frac{1}{2})^{1-n} = 2^{n-1}$.
Comparing exponents: $n-1 = 1$,so $n = 2$.
Now,for $t_{1/2} = 2.5 \ hrs$,we use the relation $(t_{1/2}) \propto P^{1-2} \implies (t_{1/2}) \propto P^{-1} \implies (t_{1/2}) \times P = \text{constant}$.
Using the first data point: $3.52 \times 50 = 2.5 \times P_2$.
$P_2 = \frac{3.52 \times 50}{2.5} = \frac{176}{2.5} = 70.4 \ mm \ Hg$.
Rounding to the nearest integer,the pressure is $70 \ mm \ Hg$.

(A) The rate of reaction increases with temperature according to the formula: $r_{new} = r_{initial} \times \mu^{(\Delta T / 10)}$.
Here,$\mu = 2$ (temperature coefficient),$T_1 = 273 \ K$,and $T_2 = 313 \ K$.
$\Delta T = T_2 - T_1 = 313 - 273 = 40 \ K$.
Substituting the values: $r_{new} = R_0 \times (2)^{(40 / 10)}$.
$r_{new} = R_0 \times (2)^4 = 16 \ R_0$.

(C) $(I)$ Since $t_{1/2}$ is independent of concentration of $Zn$ at constant $pH$,the order with respect to $[Zn]$ is $1$.
$(II)$ Let the rate law be $r = k[Zn]^1[H^{+}]^x$.
When $pH = 3$,$[H^{+}] = 10^{-3} \ M$. When $pH = 2$,$[H^{+}] = 10^{-2} \ M$.
Given that for constant $[Zn]$,the rate increases by $100$ times when $pH$ changes from $3$ to $2$:
$\frac{r_2}{r_1} = \frac{k[Zn][10^{-2}]^x}{k[Zn][10^{-3}]^x} = 100$
$10^x = 100 \Rightarrow x = 2$.
Thus,the rate law is $\frac{dx}{dt} = k[Zn][H^{+}]^2$. Statement $(B)$ is correct.
$(III)$ Checking statement $(C)$: If $[Zn]$ becomes $4[Zn]$ and $[H^{+}]$ becomes $\frac{[H^{+}]}{2}$,then $r' = k[4Zn][\frac{H^{+}}{2}]^2 = k[4Zn][\frac{[H^{+}]^2}{4}] = k[Zn][H^{+}]^2 = r$. Thus,$(C)$ is correct.
$(IV)$ Checking statement $(D)$: If $[H^{+}]$ is doubled at constant $[Zn]$,then $r' = k[Zn][2H^{+}]^2 = 4k[Zn][H^{+}]^2 = 4r$. Thus,$(D)$ is correct.
Therefore,statements $(B)$,$(C)$,and $(D)$ are correct.

(A) The half-life of a reaction of order $n$ is given by the relation $t_{1/2} \propto a^{1-n}$,where $a$ is the initial concentration.
For plot $(I)$,$t_{1/2}$ is independent of $[A]$,so $1-n = 0 \implies n = 1$.
For plot $(II)$,$t_{1/2} \propto [A]$,so $1-n = 1 \implies n = 0$.
For plot $(III)$,$t_{1/2} \propto [A]^1$,which is equivalent to $t_{1/2} \propto (1/[A])^{-1}$. Alternatively,for $n=2$,$t_{1/2} \propto 1/a$,so $t_{1/2} \propto [A]^{-1}$. The plot shows $t_{1/2}$ vs $1/[A]$ is linear,which corresponds to $n=2$ because $t_{1/2} = 1/(k[A])$.
Thus,the orders are $1, 0, 2$ respectively.

(A) For a multi-step reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.
In the given mechanism,step $(a)$ is the slow step:
$Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
The rate law for this reaction is given by the rate of the slow step:
$Rate = k[Br^{-}][H^{+}][H_2O_2]$
The order of the reaction is the sum of the powers of the concentration terms in the rate law expression:
$Order = 1 + 1 + 1 = 3$
Therefore,the order of the reaction is $3$.

(B) The rate law for the reaction is given by: $\text{rate} = k[A]^1[B]^1 = k[A][B]$.
Initial rate: $R_1 = k[A]_0[B]_0 = 4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
When $50\%$ of the reactants are converted into products,the concentration of each reactant becomes half of its initial concentration:
$[A] = \frac{[A]_0}{2}$ and $[B] = \frac{[B]_0}{2}$.
New rate: $R_2 = k \left( \frac{[A]_0}{2} \right) \left( \frac{[B]_0}{2} \right) = \frac{1}{4} k[A]_0[B]_0$.
Substituting the initial rate value:
$R_2 = \frac{1}{4} \times (4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}) = 1 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
Therefore,the correct option is $B$.

(A) For reaction $-I$: The rate law is $-\frac{d[A]}{dt} = k_1[A]^0$. Since the order is $0$,the unit of $k_1$ is $\text{concentration} \times \text{time}^{-1} = M \ sec^{-1}$.
For reaction $-II$: The rate law is $-\frac{d[B]}{dt} = k_2[B]^1$. Since the order is $1$,the unit of $k_2$ is $\text{time}^{-1} = sec^{-1}$.

(C) For the reaction $AB_5 \to AB + 4B$,the rate of reaction is given by:
$Rate = -\frac{d[AB_5]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$
Given that $-\frac{d[AB_5]}{dt} = K[AB_5]$ and $\frac{d[B]}{dt} = K_1[AB_5]$,we substitute these into the rate expression:
$K[AB_5] = \frac{1}{4} (K_1[AB_5])$
Dividing both sides by $[AB_5]$,we get:
$K = \frac{1}{4} K_1 \Rightarrow K_1 = 4K$

(D) The rate law for the given reaction is expressed as $Rate = k[I^{-}][OCl^{-}]$.
The order of reaction with respect to $I^{-}$ is $1$ and with respect to $OCl^{-}$ is $1$.
The overall order of reaction is the sum of the exponents of the concentration terms in the rate law expression.
Overall order $= 1 + 1 = 2$.

(A) Let the rate law be $r = k[A_2]^a[B_2]^b$.
From the table:
$1) 0.04 = k(0.2)^a(0.2)^b$
$2) 0.04 = k(0.1)^a(0.4)^b$
$3) 0.08 = k(0.2)^a(0.4)^b$
Dividing $(3)$ by $(1)$:
$\frac{0.08}{0.04} = \frac{k(0.2)^a(0.4)^b}{k(0.2)^a(0.2)^b} \implies 2 = (2)^b \implies b = 1$.
Dividing $(3)$ by $(2)$:
$\frac{0.08}{0.04} = \frac{k(0.2)^a(0.4)^b}{k(0.1)^a(0.4)^b} \implies 2 = (2)^a \implies a = 1$.
Thus,$a = 1$ and $b = 1$.

(C) The rate-determining step $(RDS)$ is the slow step of the mechanism.
For option $C$:
Step $1$: $X + X \rightleftharpoons X_2$ (fast equilibrium)
Step $2$: $X_2 \to Q + T$ (slow,$RDS$)
Step $3$: $T + G \to 2M$ (fast)
The rate of the reaction is determined by the slow step: $r = k_2[X_2]$.
From the fast equilibrium in step $1$,the equilibrium constant $K_C = \frac{[X_2]}{[X]^2}$,which gives $[X_2] = K_C[X]^2$.
Substituting this into the rate expression: $r = k_2 K_C [X]^2 = k[X]^2$.
This matches the given rate law.

(B) The rate law for a reaction is an experimental quantity and does not necessarily depend on the stoichiometric coefficients of the reactants in the balanced chemical equation.
Therefore,the statement that the rate law depends on the concentrations of all reactants appearing in the stoichiometric equation is false.

(D) The order of a reaction is an experimental quantity and can be zero,an integer,or a fraction.
Since the order of a reaction can indeed be fractional,the statement that 'Order cannot be fractional' is incorrect.
Therefore,option $D$ is the correct answer.

(A) The reaction is $X + Y \to Z$.
Molecularity is defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary reaction,which must collide simultaneously to result in a chemical reaction.
Since the mechanism of the reaction is not provided,the molecularity of the overall reaction cannot be defined.
The order of a reaction is the sum of the powers of the concentration terms in the rate law expression.
Given $\text{rate} \propto [X]$,the rate law is $\text{rate} = k[X]^1[Y]^0$.
Thus,the order of the reaction is $1$.

(D) For an elementary reaction,the order of the reaction with respect to each reactant is equal to its stoichiometric coefficient in the balanced chemical equation.
Since the reaction is between $A$ and $B$ and is of second order,the sum of the stoichiometric coefficients must be $2$.
For an elementary reaction $aA + bB \rightarrow \text{Products}$,the rate law is $r = k[A]^a[B]^b$.
Given that the reaction is elementary and second order,the sum of the powers $a + b = 2$.
Option $D$ represents $r = k[A]^1[B]^1$,where $a=1$ and $b=1$,so $a+b=2$.
Therefore,the rate equation $r = k[A][B]$ is consistent with an elementary second-order reaction.

(A) For a reaction of $n^{th}$ order,the half-life period $(t_{1/2})$ is given by the relation:
$t_{1/2} \propto \frac{1}{a^{n-1}}$
This implies that $t_{1/2}$ is inversely proportional to $a^{n-1}$.

(C) For a second-order reaction,the integrated rate equation is given by: $\frac{1}{(a-x)} = \frac{1}{a} + Kt$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = K$ and the intercept $c = \frac{1}{a}$.
Given $\theta = \tan^{-1}(1/2)$,the slope is $K = \tan \theta = 1/2 = 0.5 \ L \ mol^{-1} \ min^{-1}$.
The intercept $OA = \frac{1}{a} = 2 \ L \ mol^{-1}$,which gives the initial concentration $a = 1/2 = 0.5 \ mol \ L^{-1}$.
The rate of reaction at the start $(t=0)$ is given by $r = K[A]_0^2$.
Substituting the values: $r = 0.5 \times (0.5)^2 = 0.5 \times 0.25 = 0.125 \ mol \ L^{-1} \ min^{-1}$.

(A) In a multi-step reaction,the rate-determining step is the slowest step of the reaction mechanism.
Since the rate constant $K$ is directly proportional to the rate of the reaction,the step with the smallest rate constant will be the slowest step.
Given the condition $K_3 > K_2 > K_1$,the rate constant $K_1$ is the smallest.
Therefore,step $1$ is the slowest step and acts as the rate-determining step.

(C) The half-life period $(t_{1/2})$ for a reaction of order $n$ is related to the initial concentration $(a)$ as $t_{1/2} \propto a^{1-n}$.
Given the expression $t_{1/2} = \frac{1}{k \cdot a}$,we can write this as $t_{1/2} \propto a^{-1}$.
Comparing the exponents of $a$,we get $1 - n = -1$.
Solving for $n$,we find $n = 2$.
Therefore,the reaction is of the second order.

(B) The rate law for the reaction is given as $Rate = k[H_{2}][ICl]$.
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 first step is the rate-determining step (slow step). The rate of the reaction is determined by this step: $Rate = k[H_{2}][ICl]$.
This matches the given information that the reaction is first order with respect to both $H_{2}$ and $ICl$. Therefore,only Mechanism $B$ is consistent.

(C) For mechanism $A$:
The rate is determined by the slow step: $r = k_2[Cl_2][HS^-]$.
From the fast equilibrium step,$K_{eq} = \frac{[H^+][HS^-]}{[H_2S]}$,so $[HS^-] = K_{eq} \frac{[H_2S]}{[H^+]}$.
Substituting this into the rate expression: $r = k_2 K_{eq} [Cl_2] \frac{[H_2S]}{[H^+]} = K' \frac{[Cl_2][H_2S]}{[H^+]}$. This does not match the given rate law.
For mechanism $B$:
The rate is determined by the slow step: $r = k_1[Cl_2][H_2S]$.
This matches the given rate law $r = K[Cl_2][H_2S]$.
Therefore,only mechanism $B$ is consistent.

(D) The given rate law is $\text{rate} = k[P]^1[Q]^{0.5}[R]^{0.5}$.
$1$. The order with respect to $P$ is $1$,which is correct.
$2$. The total order of the reaction is the sum of the exponents of the concentration terms: $1 + 0.5 + 0.5 = 2$,which is correct.
$3$. The order with respect to $Q$ is $0.5$ and with respect to $R$ is $0.5$,which is correct.
$4$. The unit of the rate constant $k$ for a reaction of order $n$ is $(mol\ L^{-1})^{1-n} s^{-1}$. For $n = 2$,the unit is $(mol\ L^{-1})^{1-2} s^{-1} = mol^{-1}\ L\ s^{-1}$.
Therefore,the statement in option $D$ is incorrect because $mol\ L^{-1}\ s^{-1}$ is the unit for a zero-order reaction.

(A) In a multi-step reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.
The slow step is: $2B \xrightarrow{k} B_2$.
The rate law for this step is: $Rate = k[B]^2$.
Comparing this to the general rate law expression $Rate = k[A]^x[B]^y$:
Order with respect to $A$ $(x)$ = $0$.
Order with respect to $B$ $(y)$ = $2$.
Overall order of the reaction = $x + y = 0 + 2 = 2$.
Therefore,the correct option is $A$.

(D) The rate law is given by $Rate = k[P]^1[Q]^{0.5}[R]^{0.5}$.
$1$. The order with respect to $P$ is $1$,which is correct.
$2$. The total order of the reaction is the sum of the exponents: $1 + 0.5 + 0.5 = 2$,which is correct.
$3$. The order with respect to $Q$ and $R$ is $0.5$ each,which is correct.
$4$. The unit of the rate constant $k$ for a reaction of order $n$ is $(mol \, L^{-1})^{1-n} s^{-1}$. For $n = 2$,the unit is $(mol \, L^{-1})^{1-2} s^{-1} = (mol \, L^{-1})^{-1} s^{-1} = L \, mol^{-1} s^{-1}$. Therefore,the statement in option $D$ is wrong.

(B) The rate laws for the reactions are given by:
$r_1 = k[A]^1$
$r_2 = k[A]^2$
$r_3 = k[A]^3$
Given that the rate constants are equal $(k_1 = k_2 = k_3 = k)$ and the concentration $[A] < 1 \ M$.
Let $[A] = x$,where $0 < x < 1$.
Then $r_1 = kx$,$r_2 = kx^2$,and $r_3 = kx^3$.
Since $x < 1$,it follows that $x > x^2 > x^3$.
Therefore,$kx > kx^2 > kx^3$,which implies $r_1 > r_2 > r_3$.

(D) For a reaction of $n^{th}$ order,the half-life $t_{1/2}$ is related to the initial concentration $a$ as $t_{1/2} \propto a^{1-n}$.
This implies $t_{1/2} \times a^{n-1} =$ constant.
Given the condition $a^2 t_{1/2} =$ constant,we compare the exponents of $a$.
Here,$n-1 = 2$.
Therefore,$n = 3$.
Thus,it is a third-order reaction.

(B) Let the order of reaction with respect to $A$ and $B$ be $\alpha$ and $\beta$ respectively.
The rate law is given by: $\frac{d[C]}{dt} = k[A]^{\alpha}[B]^{\beta}$
From the given data:
$1.2 \times 10^{-3} = k[0.1]^{\alpha}[0.1]^{\beta} \quad (i)$
$1.2 \times 10^{-3} = k[0.1]^{\alpha}[0.2]^{\beta} \quad (ii)$
$2.4 \times 10^{-3} = k[0.2]^{\alpha}[0.1]^{\beta} \quad (iii)$
Dividing $(ii)$ by $(i)$:
$\frac{1.2 \times 10^{-3}}{1.2 \times 10^{-3}} = \frac{k[0.1]^{\alpha}[0.2]^{\beta}}{k[0.1]^{\alpha}[0.1]^{\beta}}$
$1 = 2^{\beta} \implies \beta = 0$
Dividing $(iii)$ by $(i)$:
$\frac{2.4 \times 10^{-3}}{1.2 \times 10^{-3}} = \frac{k[0.2]^{\alpha}[0.1]^{\beta}}{k[0.1]^{\alpha}[0.1]^{\beta}}$
$2 = 2^{\alpha} \implies \alpha = 1$
Thus,the rate law is: $\frac{d[C]}{dt} = k[A]^1[B]^0 = k[A]$.

(C) The rate of the reaction is determined by the slow step:
$r = k_2 [NOBr_2][NO]$
Since $NOBr_2$ is an intermediate,we express it in terms of reactants using the fast equilibrium step:
$K_{eq} = \frac{[NOBr_2]}{[NO][Br_2]}$
Therefore,$[NOBr_2] = K_{eq} [NO][Br_2]$
Substituting this into the rate expression:
$r = k_2 (K_{eq} [NO][Br_2]) [NO]$
$r = (k_2 K_{eq}) [NO]^2 [Br_2]$
Letting $K = k_2 K_{eq}$,the rate law is:
$r = K [NO]^2 [Br_2]$

(D) The rate constant $(k)$ is a characteristic property of a reaction at a given temperature.
It does not depend on the concentration of the reactants.
Therefore,changing the concentration of $B$ will change the reaction rate,but the rate constant $(k)$ remains unchanged.

(C) The initial rate is given by $R = K[A][B]$.
When the concentration of $A$ is kept constant and the concentration of $B$ is doubled,the new concentration of $B$ becomes $[B]' = 2[B]$.
The new rate $R'$ is given by $R' = K[A][B]' = K[A](2[B])$.
Substituting the initial rate expression into the new rate expression,we get $R' = 2 \times (K[A][B]) = 2R$.
Therefore,the rate of the reaction is doubled.

(B) The rate of reaction is defined as the rate of disappearance of reactant divided by its stoichiometric coefficient,which is equal to the rate of appearance of product divided by its stoichiometric coefficient.
For the reaction $2N_2O_5 \to 4NO_2 + O_2$,the rate expression is:
$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt}$
Given that $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$ and $\frac{d[NO_2]}{dt} = k'[N_2O_5]$,substitute these into the rate expression:
$\frac{1}{2} (k[N_2O_5]) = \frac{1}{4} (k'[N_2O_5])$
$\frac{k}{2} = \frac{k'}{4}$
$k' = 2k$ or $2k = k'$.

(D) The initial rate is given by: $Rate_1 = k[A]^n[B]^m$
When the concentration of $A$ is doubled $(2[A])$ and the concentration of $B$ is halved $([B]/2)$,the new rate is: $Rate_2 = k[2A]^n[\frac{1}{2}B]^m$
Taking the ratio of the new rate to the initial rate:
$\frac{Rate_2}{Rate_1} = \frac{k[2A]^n[\frac{1}{2}B]^m}{k[A]^n[B]^m}$
$= (2)^n \times (\frac{1}{2})^m$
$= 2^n \times 2^{-m} = 2^{(n-m)}$

(D) The rate law for the reaction $A \to B$ is given by $r = k[A]^x$,where $x$ is the order of reaction.
For the first condition: $2.40 \times 10^{-4} = k(2 \times 10^{-3})^x$ $(i)$
For the second condition: $0.60 \times 10^{-4} = k(1 \times 10^{-3})^x$ $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{2.40 \times 10^{-4}}{0.60 \times 10^{-4}} = \frac{k(2 \times 10^{-3})^x}{k(1 \times 10^{-3})^x}$
$4 = (2)^x$
Since $4 = 2^2$,we have $x = 2$.
Therefore,the order of reaction with respect to $A$ is $2$.

(B) The rate law for the reaction is given by: $Rate = K [A]^x [B]^y$
Initial rate: $0.3 = K [A]^x [B]^y$ $(1)$
When both concentrations are doubled: $2.4 = K [2A]^x [2B]^y = K [A]^x [B]^y \times 2^{x+y} = 0.3 \times 2^{x+y}$
$2^{x+y} = 2.4 / 0.3 = 8 = 2^3$,so $x + y = 3$ $(2)$
When only $A$ is doubled: $0.6 = K [2A]^x [B]^y = K [A]^x [B]^y \times 2^x = 0.3 \times 2^x$
$2^x = 0.6 / 0.3 = 2 = 2^1$,so $x = 1$
Substituting $x = 1$ into $x + y = 3$,we get $y = 2$
Thus,the order with respect to $A$ is $1$ and with respect to $B$ is $2$.

(D) For the elementary reaction $A_2 \underset{k_{-1}}{\overset{k_1}{\longleftrightarrow}} 2A$,the rate of reaction can be expressed in terms of the change in concentration of reactants and products.
The rate of disappearance of $A_2$ is $-\frac{d[A_2]}{dt} = k_1[A_2] - k_{-1}[A]^2$.
The rate of appearance of $A$ is $\frac{1}{2} \frac{d[A]}{dt} = k_1[A_2] - k_{-1}[A]^2$.
Multiplying both sides by $2$,we get $\frac{d[A]}{dt} = 2k_1[A_2] - 2k_{-1}[A]^2$.