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

(N/A) The rate law is determined experimentally by measuring the initial rate of the reaction at different initial concentrations of the reactants.
$1$. Keep the concentration of one reactant constant and vary the concentration of the other reactant to observe the effect on the reaction rate.
$2$. For the reaction $2 \, NO \, (g) + O_2 \, (g) \to 2 \, NO_2 \, (g)$,if the concentration of $NO$ is doubled while keeping $[O_2]$ constant,the rate increases by a factor of $4$,indicating the order with respect to $NO$ is $2$.
$3$. If the concentration of $O_2$ is doubled while keeping $[NO]$ constant,the rate doubles,indicating the order with respect to $O_2$ is $1$.
$4$. Thus,the rate law is expressed as: $\text{Rate} = k[NO]^2[O_2]^1$.

(B) For an elementary reaction,the order and molecularity have the same value because the reaction occurs in a single step.

(C) Let the rate of reaction be $r_1 = k[A]^n$,where $n$ is the order of the reaction.
When the concentration of reactant $A$ is tripled,the new concentration is $[A'] = 3[A]$.
The new rate of reaction is $r_2 = k[3A]^n$.
Given that $r_2 = 27r_1$,we have:
$27r_1 = k[3A]^n$
Substituting $r_1 = k[A]^n$ into the equation:
$27(k[A]^n) = k(3^n)[A]^n$
$27 = 3^n$
Since $27 = 3^3$,we get $3^3 = 3^n$.
Therefore,$n = 3$.

(N/A) An elementary reaction is a single-step process where the order of the reaction is equal to the sum of the stoichiometric coefficients of the reactants.
For an elementary reaction,the order must be an integer (usually $1, 2, \text{or } 3$).
In the given rate law,$\text{Rate} = k[A]^1[B]^{3/2}$,the overall order of the reaction is $1 + 3/2 = 5/2 = 2.5$.
Since the order of the reaction is fractional,it cannot be an elementary reaction.

(N/A) No,the molecularity of a reaction can never be zero or a fractional number. Molecularity is defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary reaction,which must collide simultaneously to bring about a chemical reaction. Since at least one molecule must be present for a collision to occur,the molecularity must be at least $1$.

(B) 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.
For a reaction to occur,these species must collide effectively.
The probability of more than $3$ molecules colliding simultaneously is extremely low because it requires a precise orientation and energy alignment for all particles at the same instant.
Therefore,reactions with molecularity greater than $3$ are very rare.

(N/A) Molecularity is defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary reaction,which must collide in order to bring about a chemical reaction. Since a chemical reaction cannot occur without the participation of at least one reactant molecule,the minimum value for molecularity is $1$. Therefore,molecularity cannot be zero.

(N/A) An elementary reaction is a single-step process where the molecularity is defined as the number of reacting species taking part in the collision.
In contrast,a complex reaction occurs through a series of elementary steps.
Since each step in a complex reaction may have a different number of molecules involved,the molecularity of the overall complex reaction is not defined or is considered meaningless.
However,the order of a reaction is an experimentally determined quantity that depends on the rate-determining step (the slowest step) of the mechanism,making it applicable to both elementary and complex reactions.

(N/A) The balanced chemical equation represents the overall stoichiometry of a reaction but does not necessarily reflect the mechanism or the rate-determining step.
$1.$ The order of a reaction is an experimental quantity,whereas the stoichiometry is derived from the balanced equation.
$2.$ Many reactions are complex and occur in multiple steps. The rate of the overall reaction is determined by the slowest step in the mechanism.
$3.$ For example,consider the reaction: $CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl$. Based on the stoichiometry,one might incorrectly predict a second-order reaction,but the experimental rate law is $Rate = k[CHCl_3][Cl_2]^{1/2}$,which is of order $1.5$.
$4.$ Thus,the order of a reaction cannot be determined simply by looking at the balanced chemical equation.

(N/A) reaction that is actually of higher order but follows first order kinetics under certain conditions is called a pseudo first order reaction.
This usually happens when one of the reactants is present in a large excess.
Example: The acid-catalyzed hydrolysis of ethyl acetate $(CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH)$.
In this reaction,water is present in a large excess,so its concentration remains practically constant throughout the reaction.
The rate law is given by: $Rate = k[CH_3COOC_2H_5][H_2O]$.
Since $[H_2O]$ is constant,the rate becomes $Rate = k'[CH_3COOC_2H_5]$,where $k' = k[H_2O]$.
Thus,the reaction behaves as a first order reaction.

(A) From the rate law expression: $Rate = k [A]^x [B]^y$
Using the data from experiments $I, II,$ and $III$:
$6.00 \times 10^{-3} = k (0.1)^x (0.1)^y \dots(1)$
$2.40 \times 10^{-2} = k (0.1)^x (0.2)^y \dots(2)$
$1.20 \times 10^{-2} = k (0.2)^x (0.1)^y \dots(3)$
Dividing $(3)$ by $(1)$: $2^x = 2 \implies x = 1$
Dividing $(2)$ by $(1)$: $2^y = 4 \implies y = 2$
Thus,the rate law is $Rate = k [A]^1 [B]^2$.
For experiment $IV$:
$7.20 \times 10^{-2} = k (X)^1 (0.2)^2$
Using $k$ from experiment $I$: $k = \frac{6.00 \times 10^{-3}}{(0.1)(0.1)^2} = 6 \ L^2 \ mol^{-2} \ min^{-1}$
$7.20 \times 10^{-2} = 6 \times X \times 0.04 \implies X = \frac{7.20 \times 10^{-2}}{0.24} = 0.3 \ M$
For experiment $V$:
$2.88 \times 10^{-1} = 6 \times (0.3) \times Y^2$
$Y^2 = \frac{2.88 \times 10^{-1}}{1.8} = 0.16 \implies Y = 0.4 \ M$
Therefore,$X = 0.3$ and $Y = 0.4$.

(C) For a reaction,the rate law is given by: $\text{Rate} = k[\text{conc.}]^n$,where $n$ is the order of the reaction.
Taking logarithm on both sides:
$\log(\text{rate}) = \log(k) + n \log[\text{conc.}]$
This equation is of the form $y = mx + c$,where the slope $m$ is equal to the order of the reaction $n$.
Comparing the slopes of the lines in the graph:
The slope of line $D$ is the steepest,followed by $B$,then $A$,and finally $C$ is the least steep.
Therefore,the order of the reactions follows the sequence: $d > b > a > c$.

(C) The rate law is given by $r = k [NO]^{m} [Cl_{2}]^{n}$.
From the data:
$0.18 = k (0.1)^{m} (0.1)^{n} \quad \dots (1)$
$0.35 = k (0.1)^{m} (0.2)^{n} \quad \dots (2)$
$1.40 = k (0.2)^{m} (0.2)^{n} \quad \dots (3)$
Dividing equation $(2)$ by $(1)$:
$\frac{0.35}{0.18} \approx 2 = (\frac{0.2}{0.1})^{n} = 2^{n} \implies n = 1$.
Dividing equation $(3)$ by $(2)$:
$\frac{1.40}{0.35} = 4 = (\frac{0.2}{0.1})^{m} = 2^{m} \implies m = 2$.
The overall order of the reaction is $m + n = 2 + 1 = 3$.

(A) The given reaction is $2A + B_{2} \rightarrow 2AB$.
Since it is an elementary reaction,the rate law is given by $r_{1} = k[A]^{2}[B_{2}]$.
When the volume of the reaction vessel is reduced by a factor of $3$,the concentration of each reactant increases by a factor of $3$ (since $C = n/V$).
Let the new concentrations be $[A]' = 3[A]$ and $[B_{2}]' = 3[B_{2}]$.
The new rate $r_{2}$ is $r_{2} = k(3[A])^{2}(3[B_{2}])$.
$r_{2} = k \cdot 9[A]^{2} \cdot 3[B_{2}] = 27 \cdot k[A]^{2}[B_{2}] = 27 \cdot r_{1}$.
Thus,the rate of reaction increases by a factor of $27$.

(A) The rate law for the reaction is given by: $\text{Rate} = k[NO]^x[H_2]^y$.
Using data from Experiment $1$ and $2$:
$7 \times 10^{-9} = k(8 \times 10^{-5})^x(8 \times 10^{-5})^y$ ... $(i)$
$2.1 \times 10^{-8} = k(24 \times 10^{-5})^x(8 \times 10^{-5})^y$ ... $(ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{2.1 \times 10^{-8}}{7 \times 10^{-9}} = \left(\frac{24 \times 10^{-5}}{8 \times 10^{-5}}\right)^x$
$3 = (3)^x$
Therefore,$x = 1$.
The order of the reaction with respect to $NO$ is $1$.

(B) For a zero order reaction:
$Rate = k[Reactant]^0 = k$. Thus,graph $(a)$ (Rate vs Time) is constant,representing zero order.
$t_{1/2} = [A]_0 / (2k)$. Thus,graph $(b)$ ($t_{1/2}$ vs Initial concentration) is a straight line passing through the origin,representing zero order.
For a first order reaction:
$Rate = k[Concentration]$. Thus,graph $(e)$ (Rate vs Concentration) is a straight line passing through the origin,representing first order.
$[A]_t = [A]_0 e^{-kt}$. Thus,graph $(c)$ (Concentration vs Time) is an exponential decay curve,representing first order.
Graph $(d)$ shows concentration independent of time,which is not standard for these kinetics.
Therefore,$(a)$ and $(b)$ are zero order,while $(c)$ and $(e)$ are first order.

(B) The rate law for a reaction of order $n$ is given by: $\text{Rate} = k[A]^n$.
The unit of rate is $\text{mol } L^{-1} s^{-1}$.
The unit of concentration $[A]$ is $\text{mol } L^{-1}$.
Substituting these into the rate law: $(\text{mol } L^{-1}) s^{-1} = k(\text{mol } L^{-1})^n$.
Solving for $k$: $k = \frac{(\text{mol } L^{-1}) s^{-1}}{(\text{mol } L^{-1})^n} = (\text{mol } L^{-1})^{1-n} s^{-1} = \text{mol}^{1-n} L^{n-1} s^{-1}$.

(C) Let the rate law be $Rate = k[Cl_2]^x[NO]^y$.
Given that doubling $[Cl_2]$ increases the rate by a factor of $2$,we have $2^x = 2$,which implies $x = 1$.
When both concentrations are doubled,the rate increases by a factor of $8$,so $2^x \times 2^y = 8$.
Substituting $x = 1$,we get $2^1 \times 2^y = 8$,which simplifies to $2^y = 4$.
Therefore,$y = 2$.
The order of the reaction with respect to $NO$ is $2$.

(B) For a gaseous reaction,the half-life $t_{1/2}$ is related to the initial pressure $P_0$ by the relation: $t_{1/2} \propto \frac{1}{(P_0)^{n-1}}$,where $n$ is the order of the reaction.
Given:
$P_1 = 55.5 \ kPa$,$t_1 = 340 \ s$
$P_2 = 27.8 \ kPa$,$t_2 = 170 \ s$
Using the ratio formula:
$\frac{t_1}{t_2} = \left(\frac{P_2}{P_1}\right)^{n-1}$
Substituting the values:
$\frac{340}{170} = \left(\frac{27.8}{55.5}\right)^{n-1}$
$2 = \left(\frac{1}{2}\right)^{n-1}$
$2^1 = (2^{-1})^{n-1}$
$2^1 = 2^{-(n-1)}$
$1 = -(n-1)$
$1 = -n + 1$
$n = 0$
Thus,the order of the reaction is $0$.

(A) The half-life of a reaction of order $n$ is related to the initial concentration $[A_0]$ as $t_{\frac{1}{2}} \propto \frac{1}{[A_0]^{n-1}}$.
Given:
$t_{\frac{1}{2}, 1} = 100 \ s$ at $[A_0]_1 = 0.5 \ mol \ L^{-1}$
$t_{\frac{1}{2}, 2} = 50 \ s$ at $[A_0]_2 = 1.0 \ mol \ L^{-1}$
Taking the ratio:
$\frac{t_{\frac{1}{2}, 1}}{t_{\frac{1}{2}, 2}} = \left( \frac{[A_0]_2}{[A_0]_1} \right)^{n-1}$
$\frac{100}{50} = \left( \frac{1.0}{0.5} \right)^{n-1}$
$2 = (2)^{n-1}$
$2^1 = 2^{n-1}$
Comparing the exponents:
$n - 1 = 1$
$n = 2$
The order of the reaction is $2$.

(C) Let the rate law be $Rate = k[P_{NO}]^x [P_{H_2}]^y$.
Comparing experiments $1$ and $2$ where $[P_{H_2}]$ is constant:
$\frac{Rate_1}{Rate_2} = (\frac{P_{NO,1}}{P_{NO,2}})^x$
$\frac{0.135}{0.033} \approx 4.09 \approx 4$
$(\frac{40.0}{20.1})^x \approx 2^x$
$4 = 2^x \implies x = 2$.
Thus,the order of the reaction with respect to $NO$ is $2$.

(C) The rate law for the reaction is given by $r = k[X]^1[Y]^0 = k[X]$.
From experiment $I$,$2 \times 10^{-3} = k(0.1) \Rightarrow k = 2 \times 10^{-2} \ min^{-1}$.
For experiment $III$,the concentration of $X$ is $0.4 \ mol \ L^{-1}$.
Therefore,the initial rate is $r = k[X] = (2 \times 10^{-2})(0.4) = 0.8 \times 10^{-2} = 8 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.
Comparing this with $M \times 10^{-3}$,we get $M = 8$.
The ratio of the numerical value of $M$ to $0.2$ is $\frac{8}{0.2} = 40$.

(D) The rate law is given by $R = k[X]^x[Y]^y$.
From experiment $1$ and $2$,keeping $[Y]$ constant:
$\frac{R_2}{R_1} = \frac{k[0.50]^x[0.25]^y}{k[0.25]^x[0.25]^y} = \frac{4.0 \times 10^{-6}}{1.0 \times 10^{-6}}$
$2^x = 4$ $\Rightarrow 2^x = 2^2$ $\Rightarrow x = 2$.
From experiment $1$ and $3$,keeping $[X]$ constant:
$\frac{R_3}{R_1} = \frac{k[0.25]^x[0.50]^y}{k[0.25]^x[0.25]^y} = \frac{8.0 \times 10^{-6}}{1.0 \times 10^{-6}}$
$2^y = 8$ $\Rightarrow 2^y = 2^3$ $\Rightarrow y = 3$.
The overall order of the reaction is $x + y = 2 + 3 = 5$.

(D) The rate law for a second-order reaction is given by $r = k[A]^2$,where $k$ is the rate constant and $[A]$ is the concentration of the reactant.
If the concentration of the reactant is reduced to half,the new concentration becomes $[A]' = \frac{[A]}{2}$.
The new rate $r'$ will be:
$r' = k(\frac{[A]}{2})^2$
$r' = k \times \frac{[A]^2}{4}$
$r' = \frac{1}{4} \times k[A]^2$
$r' = \frac{1}{4} r$
Therefore,the rate decreases to one-fourth of its original value.

(B) For a reaction of order $n$,the half-life $t_{1/2}$ is related to the initial pressure $P_0$ as $t_{1/2} \propto (P_0)^{1-n}$.
Using the data from the table:
$\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left(\frac{(P_0)_1}{(P_0)_2}\right)^{1-n}$
Substituting the values for the first two entries:
$\frac{4}{2} = \left(\frac{50}{100}\right)^{1-n}$
$2 = \left(\frac{1}{2}\right)^{1-n}$
$2 = (2)^{-(1-n)}$
$2^1 = 2^{n-1}$
Equating the exponents:
$1 = n - 1$
$n = 2$
Therefore,the order of the reaction is $2$.

(A) The graph shows the rate of formation of the product versus time.
In region-$I$,the rate increases linearly with time,which does not correspond to standard integer-order kinetics.
In region-$II$,the rate approaches a constant value asymptotically.
In region-$III$,the rate becomes constant,which is characteristic of a zero-order reaction (where rate is independent of concentration).
Since the rate of reaction depends on the concentration of reactants,and the graph only shows rate versus time,the order of the reaction cannot be determined solely from this plot without knowing the concentration-time profile.
Therefore,only statement $B$ is correct.
The number of correct statements is $1$.

(C) The unit of the rate constant for a reaction of order $n$ is given by $(\text{mol} \ L^{-1})^{1-n} \ s^{-1}$.
Given the unit is $L \ mol^{-1} \ s^{-1}$,which is equivalent to $(\text{mol} \ L^{-1})^{-1} \ s^{-1}$.
Comparing the exponents: $1 - n = -1$.
Therefore,$n = 2$.
The reaction is of second order.

(C) . For a zero-order reaction,$t_{1/2} = \frac{[A]_0}{2K}$. As the concentration decreases,the half-life decreases. (Correct statement)
$B$. If the order with respect to a reactant is zero,it will not affect the rate of reaction. (Correct statement)
$C$. Order can be fractional,but molecularity is always a whole number and cannot be fractional. (Incorrect statement)
$D$. For a zero-order reaction,the unit is $mol \ L^{-1} s^{-1}$,and for a second-order reaction,the unit is $mol^{-1} L s^{-1}$. (Correct statement)
Thus,only statement $C$ is incorrect. The number of incorrect statements is $1$.

(A) The rate law for the reaction is given by $r = k[A]^1[B]^1$.
For the first experiment: $0.10 = k(20)(0.5)$ $\Rightarrow 0.10 = 10k$ $\Rightarrow k = 0.01 \ L \ mol^{-1} \ s^{-1}$.
For the second experiment: $0.40 = k(x)(0.5)$. Substituting $k = 0.01$: $0.40 = 0.01(x)(0.5)$ $\Rightarrow 0.40 = 0.005x$ $\Rightarrow x = 80 \ mol \ L^{-1}$.
For the third experiment: $0.80 = k(40)(y)$. Substituting $k = 0.01$: $0.80 = 0.01(40)(y)$ $\Rightarrow 0.80 = 0.4y$ $\Rightarrow y = 2 \ mol \ L^{-1}$.
Therefore,the values are $x = 80$ and $y = 2$.

(B) The rate-determining step $(RDS)$ is the slow step: $NOBr_2 + NO \rightarrow 2 \ NOBr$.
The rate law is given by: $r = k [NOBr_2] [NO]$ ---- $(i)$
From the fast equilibrium step: $K_{eq} = \frac{[NOBr_2]}{[NO] [Br_2]}$,which implies $[NOBr_2] = K_{eq} [NO] [Br_2]$ ---- $(ii)$
Substituting $(ii)$ into $(i)$:
$r = k \cdot K_{eq} [NO] [Br_2] [NO]$
$r = k' [NO]^2 [Br_2]^1$
The overall order of the reaction is the sum of the powers of the concentration terms in the rate law: $2 + 1 = 3$.

(D) The initial rate is given by $r = k[A]^2[B]$.
When the concentration of $A$ is tripled,the new concentration $[A'] = 3[A]$.
The new rate $r'$ is calculated as:
$r' = k[A']^2[B] = k(3[A])^2[B] = k(9[A]^2)[B] = 9k[A]^2[B]$.
Comparing the new rate with the initial rate,$r' = 9r$.
Therefore,the rate increases by a factor of $9$.

(B) The rate law for the reaction is given by $R = k[HI]^n$.
Using data from Experiment $1$ and Experiment $2$:
$\frac{R_2}{R_1} = \frac{k[HI]_2^n}{k[HI]_1^n}$
$\frac{3.0 \times 10^{-3}}{7.5 \times 10^{-4}} = \left(\frac{0.01}{0.005}\right)^n$
$4 = (2)^n$
Since $2^2 = 4$,we get $n = 2$.
Thus,the order of the reaction is $2$.

(B) The rate law is given by $r = K[A]^{x}[B]^{y}$.
Using experiment $(i)$ and $(iv)$ where $[B]$ is constant:
$6.0 \times 10^{-3} = K(0.1)^{x}(0.1)^{y}$
$2.40 \times 10^{-2} = K(0.4)^{x}(0.1)^{y}$
Dividing $(iv)$ by $(i)$ gives: $4 = (4)^{x}$,so $x = 1$.
Using experiment $(ii)$ and $(iii)$ where $[A]$ is constant:
$7.2 \times 10^{-2} = K(0.3)^{x}(0.2)^{y}$
$2.88 \times 10^{-1} = K(0.3)^{x}(0.4)^{y}$
Dividing $(iii)$ by $(ii)$ gives: $4 = (2)^{y}$,so $y = 2$.
Overall order $= x + y = 1 + 2 = 3$.

(C) The reaction is $2 \ A_{(g)} + B_{(g)} \longrightarrow C_{(g)}$.
At $t = 0$,$P_A = 1.5 \ atm$ and $P_B = 0.7 \ atm$. The initial rate is $r_1 = K(P_A)^2(P_B) = K(1.5)^2(0.7)$.
When $P_C = 0.5 \ atm$,the pressure of $A$ consumed is $2 \times 0.5 = 1.0 \ atm$ and $B$ consumed is $0.5 \ atm$.
Remaining pressures are $P_A = 1.5 - 1.0 = 0.5 \ atm$ and $P_B = 0.7 - 0.5 = 0.2 \ atm$.
The rate at this time is $r_2 = K(0.5)^2(0.2)$.
Taking the ratio: $\frac{r_1}{r_2} = \frac{K(1.5)^2(0.7)}{K(0.5)^2(0.2)} = \frac{2.25 \times 0.7}{0.25 \times 0.2} = \frac{1.575}{0.05} = 31.5$.
Since $31.5 = 315 \times 10^{-1}$,the value is $315$.

(D) For a reaction,if the time taken for the concentration to become $1/4$ of its initial value $(t_{75\%})$ is twice the time taken to become $1/2$ of its initial value $(t_{50\%})$,it indicates a first-order reaction with respect to $A$.
Mathematically,for first-order: $t_{75\%} = 2 \times t_{50\%}$.
Regarding $B$,the graph of concentration $[B]$ versus time $t$ is a straight line with a negative slope and a positive intercept. This represents a zero-order reaction with respect to $B$,as $[B]_t = [B]_0 - kt$.
Therefore,the overall order of the reaction $= 1 + 0 = 1$.

(D) The rate of formation of $B$ is given by the expression:
$\frac{d[B]}{dt} = K_1[A] - K_2[B]$
According to the steady-state approximation,the rate of formation of $B$ is set to zero:
$0 = K_1[A] - K_2[B]$
Rearranging the equation to solve for $[B]$:
$K_2[B] = K_1[A]$
$[B] = \frac{K_1}{K_2}[A]$

(D) The rate law for the reaction $aG + bH \rightarrow$ Products is given by: $\text{Rate} = k[G]^x[H]^y$.
From the given information:
$1)$ When both concentrations are doubled: $(2)^x(2)^y = 8$,which implies $2^{(x+y)} = 2^3$,so $x + y = 3$.
$2)$ When only $[G]$ is doubled: $(2)^x(1)^y = 2$,which implies $2^x = 2^1$,so $x = 1$.
Substituting $x = 1$ into the equation $x + y = 3$,we get $1 + y = 3$,which gives $y = 2$.
The overall order of the reaction is $x + y = 1 + 2 = 3$.
Therefore,the correct option is $(D)$.

(A) The rate law is given by $r = K [A]^{n_1} [B]^{n_2} [C]^{n_3}$.
Comparing experiments $1$ and $2$,$[A]$ and $[C]$ are constant,but $[B]$ changes and the rate remains constant,so $n_2 = 0$.
Comparing experiments $1$ and $3$,$[A]$ and $[B]$ are constant,$[C]$ doubles,and the rate doubles,so $n_3 = 1$.
Comparing experiments $1$ and $4$,$[B]$ and $[C]$ are constant,$[A]$ increases by $1.5$ times,and the rate increases by $1.5$ times,so $n_1 = 1$.
Thus,the rate law is $r = K [A] [C]$.
Using experiment $1$: $6.0 \times 10^{-5} = K \times 0.2 \times 0.1$,which gives $K = 3.0 \times 10^{-3} \ L \ mol^{-1} \ s^{-1}$.
Now,for $[A] = 0.15 \ mol \ dm^{-3}, [B] = 0.25 \ mol \ dm^{-3}, [C] = 0.15 \ mol \ dm^{-3}$:
$r = (3.0 \times 10^{-3}) \times 0.15 \times 0.15 = 6.75 \times 10^{-5} \ mol \ dm^{-3} \ s^{-1}$.
Comparing this with $Y \times 10^{-5}$,we get $Y = 6.75$.

(D) For reactant $P$,the time taken for $75 \%$ completion $(t_{75 \%})$ is twice the time taken for $50 \%$ completion $(t_{50 \%})$,i.e.,$t_{75 \%} = 2 \times t_{50 \%}$.
This relationship is characteristic of a first-order reaction.
Therefore,the order of the reaction with respect to $P$ is $1$.
For reactant $Q$,the graph shows that the concentration $[Q]$ decreases linearly with time. $A$ linear decrease in concentration with time indicates that the rate of reaction is independent of the concentration of $Q$.
Therefore,the order of the reaction with respect to $Q$ is $0$.
The overall order of the reaction is the sum of the individual orders: $1 + 0 = 1$.

(B) For the reaction $M \rightarrow N$,the rate law is given by $r = k[M]^x$,where $x$ is the order of the reaction with respect to $M$.
When the concentration of $M$ is doubled,the new rate $r'$ becomes $8r$.
So,$r' = k[2M]^x = 8r$.
Substituting $r = k[M]^x$ into the equation:
$k[2M]^x = 8 \times k[M]^x$
$(2)^x = 8$
$(2)^x = (2)^3$
Therefore,$x = 3$.

(C) The rate of the reaction is determined by the slowest step,which is the rate-determining step $(RDS)$:
$Rate = k_2[N_2O_2][H_2]$
From the fast equilibrium step,we have:
$K_{eq} = \frac{k_1}{k_{-1}} = \frac{[N_2O_2]}{[NO]^2}$
Therefore,$[N_2O_2] = \frac{k_1}{k_{-1}}[NO]^2$
Substituting this into the rate expression:
$Rate = k_2 \times \frac{k_1}{k_{-1}}[NO]^2[H_2]$
$Rate = k'[NO]^2[H_2]$
The order of the reaction is the sum of the exponents of the concentration terms in the rate law:
$Order = 2 + 1 = 3$

(C) For a reaction of order $n$,$t_{1/2} \propto [A]_0^{1-n}$.
Using the data: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{([A]_0)_1}{([A]_0)_2} \right)^{1-n}$.
$\frac{200}{100} = \left( \frac{0.100}{0.025} \right)^{1-n} \implies 2 = (4)^{1-n}$.
$2 = (2^2)^{1-n} = 2^{2-2n}$.
$1 = 2-2n \implies 2n = 1 \implies n = 1/2$.
Thus,statement $A$ is correct.
Since $n = 1/2$,$t_{1/2} \propto [A]_0^{1-1/2} = [A]_0^{1/2}$.
For $[A]_0 = 1 \text{ M}$: $\frac{200}{t_{1/2}} = \left( \frac{0.1}{1} \right)^{1/2} = \sqrt{0.1} = \frac{1}{\sqrt{10}}$.
$t_{1/2} = 200 \sqrt{10} \text{ min}$. Thus,statement $B$ is correct.
Statement $C$ is incorrect because the order of a reaction is a constant property for a given reaction under specific conditions.
For $[A]_0 = 1.6 \text{ M}$: $\frac{200}{t_{1/2}} = \left( \frac{0.1}{1.6} \right)^{1/2} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
$t_{1/2} = 200 \times 4 = 800 \text{ min}$. Thus,statement $D$ is correct.
Therefore,$A, B$ and $D$ are correct.

(A) The rate of the reaction is determined by the slow step: $\text{Rate} = k_2[A][B_2] \dots (1)$
From the fast equilibrium step,$A_2 \rightleftharpoons 2A$,the equilibrium constant is $K_{eq} = \frac{[A]^2}{[A_2]} = \frac{k_1}{k_{-1}}$.
Thus,$[A] = \sqrt{\frac{k_1}{k_{-1}}} [A_2]^{1/2}$.
Substituting this into equation $(1)$,we get $\text{Rate} = k_2 \sqrt{\frac{k_1}{k_{-1}}} [A_2]^{1/2} [B_2]^1$.
The overall order of the reaction is the sum of the exponents of the concentration terms: $0.5 + 1 = 1.5$.

(A) The initial rate of reaction is $r_1 = k[A]^{n}[B]^{m}$.
When the concentration of $A$ is doubled $(2[A])$ and the concentration of $B$ is halved $(\frac{[B]}{2})$,the new rate $r_2$ is given by:
$r_2 = k(2[A])^{n} \cdot \left(\frac{[B]}{2}\right)^{m}$
$r_2 = k \cdot 2^{n} \cdot [A]^{n} \cdot \frac{[B]^{m}}{2^{m}}$
$r_2 = 2^{(n-m)} \cdot k[A]^{n}[B]^{m}$
Therefore,the ratio $\frac{r_2}{r_1} = \frac{2^{(n-m)} k[A]^{n}[B]^{m}}{k[A]^{n}[B]^{m}} = 2^{(n-m)}$.

(B) The rate law for the reaction is given by $r = k[A]^2[B]^{1/2}$.
The order of a reaction is the sum of the powers of the concentration terms in the rate law expression.
Order of reaction $= 2 + 1/2 = 2.5$.

(C) The hydrolysis of cane sugar (sucrose) is represented by the equation: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 + C_6H_{12}O_6$.
Since water is present in large excess,its concentration remains practically constant during the reaction.
Therefore,the rate of reaction depends only on the concentration of sucrose,making it a pseudo first order reaction.

(C) The units of the rate constant $k = 2 \times 10^{-4} \ L \ mol^{-1} \ min^{-1}$ indicate that the reaction is of second order.
The rate law is $Rate = k[A]^2$.
First,convert the rate constant to $sec^{-1}$ units: $k = (2 \times 10^{-4} \ L \ mol^{-1} \ min^{-1}) / 60 = (1 / 30) \times 10^{-4} \ L \ mol^{-1} \ sec^{-1}$.
Given $Rate = (1 / 12) \times 10^{-5} \ M \ sec^{-1}$,we have $(1 / 12) \times 10^{-5} = (1 / 30) \times 10^{-4} \times [A]^2$.
Simplifying: $[A]^2 = [(1 / 12) \times 10^{-5}] / [(1 / 30) \times 10^{-4}] = (30 / 12) \times 10^{-1} = 2.5 \times 0.1 = 0.25$.
Thus,$[A] = \sqrt{0.25} = 0.5 \ M$.

(B) For a reaction of order $n$,the half-life is given by $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$.
Given that the plot of $[A]_0$ $Vs$ $\frac{1}{t_{1/2}}$ is a straight line with a positive slope,we have $\frac{1}{t_{1/2}} = k[A]_0$,which implies $t_{1/2} \propto \frac{1}{[A]_0}$.
This corresponds to a reaction of order $n = 2$.
Since $t_{1/2} \times [A]_0 = \text{constant}$,we can write: $(t_{1/2})_1 \times [A]_1 = (t_{1/2})_2 \times [A]_2$.
Substituting the given values: $20 \ min \times 1 \times 10^{-2} \ M = (t_{1/2})_2 \times 2 \times 10^{-2} \ M$.
Solving for $(t_{1/2})_2$,we get $(t_{1/2})_2 = \frac{20 \times 1 \times 10^{-2}}{2 \times 10^{-2}} \ min = 10 \ min$.

(A) The order of a reaction can be determined from the relationship between half-life $(t_{1/2})$ and initial pressure $(p_0)$: $t_{1/2} \propto (p_0)^{1-n}$,where $n$ is the order of the reaction.
Given that the half-life $(t_{1/2} = 84 \text{ min})$ remains constant despite changes in the initial pressure $(707, 79, 37.5 \text{ torr})$,it implies that $t_{1/2}$ is independent of the initial pressure.
For a first-order reaction,$t_{1/2} = \frac{0.693}{k}$,which is independent of the initial concentration or pressure.
Therefore,the reaction is of the first order $(n = 1)$.

(A) The rate of reaction is given by: $\text{Rate} = k[A]^x$.
The unit of rate is $\text{mol} \ L^{-1} \ sec^{-1}$.
The unit of concentration $[A]$ is $\text{mol} \ L^{-1}$.
Substituting these into the rate law equation:
$\text{mol} \ L^{-1} \ sec^{-1} = k \times (\text{mol} \ L^{-1})^x$.
Solving for $k$:
$k = \frac{\text{mol} \ L^{-1} \ sec^{-1}}{(\text{mol} \ L^{-1})^x} = (\text{mol} \ L^{-1})^{1-x} \ sec^{-1}$.
Expanding this,we get:
$k = \text{mol}^{1-x} \ L^{-(1-x)} \ sec^{-1} = \text{mol}^{1-x} \ L^{x-1} \ sec^{-1}$.