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

(C) The rate law for the reaction is given by: $\text{Rate} = k[A]^x$ ...$(1)$
When the concentration of $A$ is doubled,the rate increases by $1.59$ times:
$1.59 \times \text{Rate} = k[2A]^x$ ...$(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{1.59 \times \text{Rate}}{\text{Rate}} = \frac{k[2A]^x}{k[A]^x}$
$1.59 = (2)^x$
Taking $\log$ on both sides:
$\log(1.59) = x \log(2)$
$0.2014 = x \times 0.3010$
$x = \frac{0.2014}{0.3010} \approx 0.669$
Since $0.669 \approx \frac{2}{3}$,the order of the reaction is $\frac{2}{3}$.

(C) To determine the reaction mechanism,we analyze the rate law based on the experimental data.
Let the rate law be $\text{Rate} = k[S]^x[Nu]^y$.
From experiment $1$ and $2$,when $[S]$ is doubled and $[Nu]$ is constant,the rate doubles ($2.2 \times 10^{-3}$ to $4.4 \times 10^{-3}$),so $x = 1$.
From experiment $1$ and $3$,when $[Nu]$ is doubled and $[S]$ is constant,the rate doubles ($2.2 \times 10^{-3}$ to $4.4 \times 10^{-3}$),so $y = 1$.
The overall rate law is $\text{Rate} = k[S]^1[Nu]^1$.
Since the reaction is first order with respect to both the substrate and the nucleophile,the total order is $2$.
This indicates a bimolecular nucleophilic substitution reaction,which is the $S_{N}2$ mechanism.

(C) Higher order reactions $(>3)$ are rare because the probability of simultaneous collision of more than three reacting species is extremely low,making the frequency of such effective collisions negligible.

(C) Esterification is a second-order reaction.
$CH_3COOH + C_2H_5OH \rightarrow CH_3COOC_2H_5 + H_2O$
The rate law for this reaction is $r = k[CH_3COOH][C_2H_5OH]$.
When each solution is diluted with an equal volume of water,the total volume doubles,so the concentration of each reactant is reduced to half of its initial value.
Let the initial concentrations be $[A]_0$ and $[B]_0$. The initial rate is $r = k[A]_0[B]_0$.
After dilution,the new concentrations are $[A]' = \frac{[A]_0}{2}$ and $[B]' = \frac{[B]_0}{2}$.
The new rate is $r' = k \times (\frac{[A]_0}{2}) \times (\frac{[B]_0}{2}) = \frac{1}{4} \times k[A]_0[B]_0 = \frac{1}{4}r$.
Thus,the rate of reaction becomes $0.25$ times the initial rate.

(B) Given,
The reaction is $PCl_5 \longrightarrow PCl_3 + Cl_2$.
Rate $= 1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Rate constant $(k) = 3.4 \times 10^{-5} \ s^{-1}$.
The rate law for this first-order reaction is:
$\text{Rate} = k [PCl_5]$
Substituting the given values:
$1.02 \times 10^{-4} = 3.4 \times 10^{-5} \times [PCl_5]$
$[PCl_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}$
$[PCl_5] = 3 \ mol \ L^{-1}$.

(B) The given rate law is $\text{rate} = k[CH_3COOC_2H_5][NaOH]$.
Since the sum of the powers of the concentration terms is $1 + 1 = 2$,the reaction is of second order.
The unit of the rate constant $k$ for a reaction of order $n$ is given by $(mol \ L^{-1})^{1-n} \ s^{-1}$.
For $n = 2$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.

(A) For the reaction,$A + 2B \rightarrow$ Products.
Since the half-life $(t_{1/2})$ remains constant when the concentration of $B$ is increased,the reaction is of $1^{st}$ order with respect to $B$.
Since the rate remains constant when the concentration of $A$ is doubled,the reaction is of $0^{th}$ order with respect to $A$.
The rate law expression is: $\text{Rate} = k[A]^0[B]^1$.
The overall order of the reaction is $0 + 1 = 1$.
The unit of the rate constant $(k)$ for a $1^{st}$ order reaction is $s^{-1}$.

(C) The rate law is given by $r = k[A][B]^{2}$.
When the volume of the vessel is reduced to half $(V_{2} = V_{1}/2)$,the concentration of the gaseous reactants doubles because concentration is inversely proportional to volume $(C = n/V)$.
Thus,the new concentrations are $[A]' = 2[A]$ and $[B]' = 2[B]$.
The new rate $r'$ is calculated as:
$r' = k[A]'([B]')^{2} = k(2[A])(2[B])^{2}$.
$r' = k \times 2[A] \times 4[B]^{2} = 8 \times k[A][B]^{2}$.
Since $r = k[A][B]^{2}$,we have $r' = 8r$.
Therefore,the reaction rate increases by $8$ times.

(C) pseudo first order reaction is a reaction that appears to be of first order but is actually of a higher order.
In such reactions,one of the reactants is present in a large excess,so its concentration remains effectively constant throughout the reaction.
The rate constant $(k')$ of a pseudo first order reaction is defined as $k' = k[B]$,where $[B]$ is the concentration of the reactant present in excess.
Therefore,the rate constant depends on the concentration of the reactant present in excess.

(B) For the reaction $m A \rightarrow x B$,the rate law is given by $r = k[A]^{2}$.
If the concentration of $A$ is doubled,the new concentration becomes $[A]' = 2[A]$.
The new rate $r'$ is given by $r' = k[A]'{}^{2} = k(2[A])^{2}$.
$r' = 4k[A]^{2}$.
Since $r = k[A]^{2}$,we have $r' = 4r$.
Therefore,the reaction rate will be quadrupled.

(D) The rate law for the esterification reaction is $Rate = k[CH_3COOH][CH_3OH]$.
In the first case,the total volume is $100 \text{ cm}^3 + 100 \text{ cm}^3 = 200 \text{ cm}^3$.
In the second case,each solution is diluted with an equal volume of water before mixing. Thus,$100 \text{ cm}^3$ of acid is diluted to $200 \text{ cm}^3$ and $100 \text{ cm}^3$ of alcohol is diluted to $200 \text{ cm}^3$. When these are mixed,the total volume becomes $400 \text{ cm}^3$.
Since the total volume of the mixture is doubled,the concentration of each reactant ($CH_3COOH$ and $CH_3OH$) becomes half of its initial concentration in the mixture.
Let the initial concentrations be $[C_1]$ and $[C_2]$. Then $Rate_1 = k[C_1][C_2]$.
In the second case,the new concentrations are $[C_1/2]$ and $[C_2/2]$.
$Rate_2 = k[C_1/2][C_2/2] = \frac{1}{4} k[C_1][C_2] = 0.25 \times Rate_1$.
Thus,the rate becomes $0.25$ times the initial rate.

(C) The rate law for the reaction $A \rightarrow B$ is given by: $\text{Rate} = k[A]^{n}$,where $n$ is the order of the reaction.
For the first condition: $2 \times 10^{-3} = k(0.05)^{n} \quad \dots(i)$
For the second condition: $1.6 \times 10^{-2} = k(0.1)^{n} \quad \dots(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{1.6 \times 10^{-2}}{2 \times 10^{-3}} = \frac{k(0.1)^{n}}{k(0.05)^{n}}$
$8 = (\frac{0.1}{0.05})^{n}$
$8 = (2)^{n}$
Since $8 = 2^{3}$,we have $2^{3} = 2^{n}$.
Therefore,$n = 3$.

(D) reaction is said to be of second order if its rate depends on the concentration of two reactants or the square of the concentration of one reactant.
The saponification of an ester,such as the reaction between methyl acetate and sodium hydroxide,follows second-order kinetics.
The rate law for this reaction is $Rate = k[CH_{3}COOCH_{3}][NaOH]$.
Thus,$CH_{3}COOCH_{3} + NaOH \longrightarrow CH_{3}COONa + CH_{3}OH$ is a second-order reaction.

(C) For a reaction of $n$th order,the integrated rate law is given by $k = \frac{1}{(n-1)t} [\frac{1}{(A_t)^{n-1}} - \frac{1}{(A_0)^{n-1}}]$.
At half-life,$t = t_{1/2}$ and $A_t = A_0 / 2$.
Substituting these values,we get $t_{1/2} \propto \frac{1}{(A_0)^{n-1}}$,which can also be written as $t_{1/2} \propto (A_0)^{1-n}$.
Therefore,the half-life period is directly proportional to $a^{1-n}$ where $a$ is the initial concentration.

(C) The half-life period $(t_{1/2})$ of an $n$th order reaction is related to the initial concentration $(a)$ as:
$t_{1/2} \propto \frac{1}{a^{n-1}}$
Given that $t_{1/2} \propto \frac{1}{a^5}$,we can compare the exponents:
$n - 1 = 5$
$n = 6$
Therefore,the order of the reaction is $6$.

(D) Let the order of reaction with respect to $A$ and $B$ be $m$ and $n$ respectively.
Then,$rate = k[A]^{m}[B]^{n}$
From the table:
$2 = k[0.2]^{m}[0.2]^{n}$ $(i)$
$4 = k[0.2]^{m}[0.4]^{n}$ $(ii)$
$36 = k[0.6]^{m}[0.4]^{n}$ $(iii)$
On comparing Eqs. $(i)$ and $(ii)$:
$\frac{4}{2} = \frac{k[0.2]^{m}[0.4]^{n}}{k[0.2]^{m}[0.2]^{n}}$
$2 = (\frac{0.4}{0.2})^{n} = 2^{n}$
$n = 1$
On comparing Eqs. $(ii)$ and $(iii)$:
$\frac{36}{4} = \frac{k[0.6]^{m}[0.4]^{n}}{k[0.2]^{m}[0.4]^{n}}$
$9 = (\frac{0.6}{0.2})^{m} = 3^{m}$
$3^{2} = 3^{m}$
$m = 2$
Therefore,the rate law is $r = k[A]^{2}[B]$.

(D) The rate law for the reaction is given by: $Rate = k[AB]^n$.
Using the data from the table:
$2.75 \times 10^{-8} = k(0.20)^n$ --- $(i)$
$11.0 \times 10^{-8} = k(0.40)^n$ --- $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{11.0 \times 10^{-8}}{2.75 \times 10^{-8}} = (\frac{0.40}{0.20})^n$
$4 = 2^n$
Since $4 = 2^2$,we have $2^2 = 2^n$.
Therefore,$n = 2$. The reaction is of second order.

(A) The rate law is determined experimentally by measuring the rate of reaction at different concentrations of reactants. Therefore,the statement that the rate law cannot be determined experimentally is incorrect. Complex reactions often exhibit fractional orders,bimolecular reactions involve the collision of two species,and molecularity is a concept defined specifically for elementary reactions.

(C) The rate law for the reaction is given by $\text{Rate} = k[A][B]^2$.
Given values are:
$k = 5.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$
$[A] = 0.05 \ mol \ L^{-1} = 5 \times 10^{-2} \ mol \ L^{-1}$
$[B] = 0.1 \ mol \ L^{-1} = 1 \times 10^{-1} \ mol \ L^{-1}$
Substituting these values into the rate equation:
$\text{Rate} = (5.0 \times 10^{-6}) \times (0.05) \times (0.1)^2$
$\text{Rate} = 5.0 \times 10^{-6} \times 5 \times 10^{-2} \times 1 \times 10^{-2}$
$\text{Rate} = 25 \times 10^{-10} \ mol \ L^{-1} \ s^{-1}$
$\text{Rate} = 2.50 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$

(D) The correct answer is $D$.
Molecularity is a theoretical concept 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.
It is determined by examining the balanced chemical equation of an elementary step and cannot be determined experimentally,unlike the order of a reaction which is an experimental quantity.

(C) The rate law is given by: $\text{Rate} = k[A][B]^2$.
Given values are: $k = 5.0 \times 10^{-6} \, mol^{-2} \, L^2 \, s^{-1}$,$[A] = 0.05 \, mol \, L^{-1}$,and $[B] = 0.1 \, mol \, L^{-1}$.
Substituting these values into the rate equation:
$\text{Rate} = (5.0 \times 10^{-6}) \times (0.05) \times (0.1)^2$
$= (5.0 \times 10^{-6}) \times (5.0 \times 10^{-2}) \times (1.0 \times 10^{-2})$
$= 25.0 \times 10^{-10} \, mol \, L^{-1} \, s^{-1}$
$= 2.50 \times 10^{-9} \, mol \, L^{-1} \, s^{-1}$.

(A) The given reaction is $2 \ FeCl_3 + SnCl_2 \rightarrow 2 \ FeCl_2 + SnCl_4$.
Experimentally,the rate of this reaction is found to be $Rate = k[FeCl_3]^2[SnCl_2]^1$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law expression.
Order $= 2 + 1 = 3$.
Therefore,this is a third-order reaction.

(C) Let the rate law be $r = k[I_2]^x [H^{+}]^y [CH_3COCH_3]^z$.
Comparing experiments $2$ and $3$: When $[H^{+}]$ and $[CH_3COCH_3]$ are constant,doubling $[I_2]$ ($0.01$ to $0.02$) does not change the rate $(0.192)$. Thus,$2^x = 1 \Rightarrow x = 0$.
Comparing experiments $1$ and $2$: When $[I_2]$ and $[CH_3COCH_3]$ are constant,doubling $[H^{+}]$ ($0.1$ to $0.2$) doubles the rate ($0.096$ to $0.192$). Thus,$2^y = 2 \Rightarrow y = 1$.
Comparing experiments $2$ and $4$: When $[I_2]$ and $[H^{+}]$ are constant,doubling $[CH_3COCH_3]$ ($0.1$ to $0.2$) doubles the rate ($0.192$ to $0.384$). Thus,$2^z = 2 \Rightarrow z = 1$.
Total order $= x + y + z = 0 + 1 + 1 = 2$.
The orders are $0, 1, 1, 2$.

(A) Let $R_1, R_2$,and $R_3$ be the rates of three reactions of first,second,and third-order,respectively,and $k$ be the rate constant for all three reactions.
The rate laws are:
$R_1 = k[A]^1$
$R_2 = k[A]^2$
$R_3 = k[A]^3$
where $[A]$ is the concentration of reactant $A$ in moles per litre.
Given that $[A] = 1$,we have:
$R_1 = k(1)^1 = k$
$R_2 = k(1)^2 = k$
$R_3 = k(1)^3 = k$
Therefore,$R_1 = R_2 = R_3$.

(B) For the reaction $N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate of reaction is given by:
Rate $= -\frac{d[N_2O_5]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt} = 2 \frac{d[O_2]}{dt}$
Substituting the given rate expressions:
$K_1[N_2O_5] = \frac{1}{2} K_2[N_2O_5] = 2K_3[N_2O_5]$
Dividing by $[N_2O_5]$ gives $K_1 = \frac{1}{2} K_2 = 2K_3$.
Multiplying the entire relation by $2$,we get $2K_1 = K_2 = 4K_3$.

(A) The rate of a reaction is determined by the slowest step in the mechanism.
The rate law for the slow step is $R = k_3[A][B]$.
Since $B$ is an intermediate formed in the fast equilibrium step $A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} B$,we have the equilibrium constant $K_{eq} = \frac{[B]}{[A]} = \frac{k_1}{k_2}$.
This gives $[B] = \frac{k_1}{k_2}[A]$.
Substituting this into the rate expression: $R = k_3[A](\frac{k_1}{k_2}[A]) = \frac{k_3 k_1}{k_2}[A]^2$.
Thus,the rate is proportional to $[A]^2$.

(A) For the reaction $A_2 + B_2 \rightarrow 2 AB$,the rate of reaction $R$ is defined as $R = k[A_2]^x [B_2]^y$.
The rate of formation of $AB$ is given as $\frac{d[AB]}{dt} = 2R$.
From the table:
$1.25 \times 10^{-4} = k(0.1)^x(0.1)^y \dots (i)$
$2.5 \times 10^{-4} = k(0.2)^x(0.1)^y \dots (ii)$
$5.0 \times 10^{-4} = k(0.2)^x(0.2)^y \dots (iii)$
Dividing $(ii)$ by $(i)$,we get $2^x = 2$,so $x = 1$.
Dividing $(iii)$ by $(ii)$,we get $2^y = 2$,so $y = 1$.
Thus,the rate law is $R = k[A_2][B_2]$.
Using values from $(i)$: $1.25 \times 10^{-4} = k(0.1)(0.1) = k(0.01)$.
$k = \frac{1.25 \times 10^{-4}}{0.01} = 1.25 \times 10^{-2} \ L \ mol^{-1} s^{-1}$.

(A) The rate law is given by $-\frac{d[HI]}{dt} = k[HI]^2$.
Here,the rate of reaction has units of concentration per unit time,i.e.,$mol \ L^{-1} \ s^{-1}$.
The concentration $[HI]$ has units of $mol \ L^{-1}$.
Substituting these into the rate law:
$mol \ L^{-1} \ s^{-1} = k \times (mol \ L^{-1})^2$.
Therefore,$k = \frac{mol \ L^{-1} \ s^{-1}}{(mol \ L^{-1})^2} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.

(C) The unit of the rate constant indicates the order of the reaction.
For a reaction $A \rightarrow P$,the rate $r = k[A]^n$.
Reaction $1$: $k_1 = 1 \ s^{-1}$,which corresponds to a $1^{st}$ order reaction. Thus,$r_1 = k_1[A]^1 = 1 \times 1 = 1 \ M \ s^{-1}$.
Reaction $2$: $k_2 = 0.1 \ L \ mol^{-1} \ s^{-1}$,which corresponds to a $2^{nd}$ order reaction. Thus,$r_2 = k_2[A]^2 = 0.1 \times 1^2 = 0.1 \ M \ s^{-1}$.
Reaction $3$: $k_3 = 0.01 \ L^2 \ mol^{-2} \ s^{-1}$,which corresponds to a $3^{rd}$ order reaction. Thus,$r_3 = k_3[A]^3 = 0.01 \times 1^3 = 0.01 \ M \ s^{-1}$.
Comparing the rates:
$r_1 = 1, r_2 = 0.1, r_3 = 0.01$.
$r_1 = 10 \ r_2 \implies r_2 = \frac{r_1}{10}$.
$r_1 = 100 \ r_3 \implies r_3 = \frac{r_1}{100}$.
$r_2 = 10 \ r_3 \implies r_3 = \frac{r_2}{10}$.
From the options,$r_1 = 100 \ r_3$ and $r_2 = 10 \ r_3$ (or $r_3 = r_2 / 10$) is consistent with option $C$.

(A) The order of a reaction is defined as the sum of the powers of the concentration terms of the reactants in the rate law expression.
For a rate law given by $Rate = K[A]^x[B]^y$,the overall order of the reaction is $x + y$.
Given the rate expression: $Rate = K[A]^{\frac{1}{2}}[B]^{\frac{3}{2}}$.
Here,the exponents are $x = \frac{1}{2}$ and $y = \frac{3}{2}$.
Overall order of reaction $= \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$.
Therefore,the reaction is of $second$ order.

(B) The given sequence of reaction is $A$ $\xrightarrow{k_1} X$ $\xrightarrow{k_2} Y$ $\xrightarrow{k_3} Z$.
In a sequence of consecutive reactions,the slowest step is the rate-determining step.
The rate of a step is directly proportional to its rate constant $k$.
Given $k_3 > k_2 > k_1$,the step with the smallest rate constant $k_1$ is the slowest step.
Therefore,the conversion $A \longrightarrow X$ is the rate-determining step.

(D) The rate law expression is given by $\text{Rate} = k[A]^\alpha[B]^\beta$.
From entry $1$ and $2$,$[A]$ is constant. Taking the ratio:
$\frac{4 \times 10^{-2}}{2 \times 10^{-2}} = \frac{k[0.02]^\alpha[0.04]^\beta}{k[0.02]^\alpha[0.02]^\beta}$
$2 = [2]^\beta \Rightarrow \beta = 1$.
From entry $2$ and $3$,$[B]$ is constant. Taking the ratio:
$\frac{8 \times 10^{-2}}{4 \times 10^{-2}} = \frac{k[0.04]^\alpha[0.04]^\beta}{k[0.02]^\alpha[0.04]^\beta}$
$2 = [2]^\alpha \Rightarrow \alpha = 1$.
Substituting $\alpha = 1$ and $\beta = 1$ into entry $1$:
$2 \times 10^{-2} = k[0.02]^1[0.02]^1$
$k = \frac{2 \times 10^{-2}}{4 \times 10^{-4}} = \frac{200}{4} = 50 \text{ } M^{-1}s^{-1}$.

(D) For the given reaction: $CH_3COOC_2H_5(aq) + NaOH(aq) \longrightarrow CH_3COONa(aq) + C_2H_5OH(aq)$
$1$. 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. Here,two molecules ($CH_3COOC_2H_5$ and $NaOH$) are involved,so the molecularity is $2$.
$2$. The rate law for this saponification reaction is experimentally determined as: $Rate = k[CH_3COOC_2H_5]^1[NaOH]^1$.
$3$. The order of the reaction is the sum of the powers of the concentration terms in the rate law expression,which is $1 + 1 = 2$.
Therefore,the reaction is of second order and has a molecularity of $2$.

(B) The rate constant $(k)$ of a chemical reaction is a characteristic property that depends only on temperature and the nature of the reactants,not on the concentration of the reactants or the time elapsed.
For any order of reaction,the rate constant remains constant at a given temperature.
Therefore,the rate constant at $t_2 = 20 \ min$ will be the same as the rate constant at $t_1 = 10 \ min$.
Thus,the rate constant is $10^{-2} \ min^{-1}$.

(B) The half-life $(t_{1/2})$ of an $n^{th}$ order reaction is related to the initial concentration $(a)$ of the reactant as $t_{1/2} \propto a^{1-n}$.
For two different initial concentrations,we have the relation: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left(\frac{a_2}{a_1}\right)^{n-1}$.
Given: $(t_{1/2})_1 = 5 \ \text{min}$,$a_1 = 0.1 \ M$ and $(t_{1/2})_2 = 50 \ \text{min}$,$a_2 = 0.01 \ M$.
Substituting the values: $\frac{5}{50} = \left(\frac{0.01}{0.1}\right)^{n-1}$.
$\frac{1}{10} = (0.1)^{n-1}$.
$0.1^1 = (0.1)^{n-1}$.
Comparing the exponents: $1 = n - 1$,which gives $n = 2$.

(B) The overall reaction is $2 A + B \longrightarrow D + E$.
The proposed mechanism is:
$1. A + B \longrightarrow C + D$ (slow step)
$2. A + C \longrightarrow E$ (fast step)
The rate of a reaction is determined by the slowest step,known as the Rate Determining Step ($R$.$D$.$S$).
Since the first step $(A + B \longrightarrow C + D)$ is the slow step,the rate law is derived from the reactants involved in this step.
Therefore,the rate law is $r = K[A][B]$.

(B) The temperature coefficient is defined as the ratio of rate constants at temperatures differing by $10^{\circ} C$: $\text{Temperature coefficient} = \frac{k_{(t+10)}}{k_t}$.
Given,$\text{Temperature coefficient} = 2$,$k_{(27^{\circ} C)} = 10^{-3} \ min^{-1}$,and we need to find $k_{(17^{\circ} C)}$.
Substituting the values: $2 = \frac{k_{(27^{\circ} C)}}{k_{(17^{\circ} C)}}$.
$2 = \frac{10^{-3}}{k_{(17^{\circ} C)}}$.
$k_{(17^{\circ} C)} = \frac{10^{-3}}{2} = 0.5 \times 10^{-3} = 5 \times 10^{-4} \ min^{-1}$.

(A) Let the rate law be $\frac{-d[NO]}{dt} = k[NO]^x[H_2]^y$.
From experiment $1$ and $2$,$[H_2]$ is constant,so $\frac{43.2 \times 10^{-5}}{4.8 \times 10^{-5}} = (\frac{3 \times 10^{-2}}{1 \times 10^{-2}})^x \implies 9 = 3^x \implies x = 2$.
From experiment $2$ and $3$,$[NO]$ is constant,so $\frac{86.4 \times 10^{-5}}{43.2 \times 10^{-5}} = (\frac{2 \times 10^{-3}}{1 \times 10^{-3}})^y \implies 2 = 2^y \implies y = 1$.
Thus,the rate law is $\frac{-d[NO]}{dt} = k[NO]^2[H_2]$.

(A) 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 in order to bring about a chemical reaction.
Molecularity is always a whole number $(1, 2, 3, ...)$ and can never be zero,fractional,or negative.
Therefore,the statement that it may be a fractional number is incorrect.

(C) Let the rate law be $Rate = k[A]^x[B]^y$.
From experiments $2$ and $3$ (where $[B]$ is constant):
$\frac{10}{1} = \frac{k(10)^x(1)^y}{k(1)^x(1)^y}$ $\Rightarrow 10 = 10^x$ $\Rightarrow x = 1$.
From experiments $1$ and $2$ (where $[A]$ is constant):
$\frac{100}{1} = \frac{k(1)^x(10)^y}{k(1)^x(1)^y}$ $\Rightarrow 100 = 10^y$ $\Rightarrow 10^2 = 10^y$ $\Rightarrow y = 2$.
Thus,the order with respect to $A$ is $1$ and with respect to $B$ is $2$.

(B) The rate law is given by $r = k[A]^{\alpha}[B]^{\beta}$.
When the concentration of $B$ is doubled,the rate does not change,which implies that the reaction is of zero order with respect to $B$. Therefore,$\beta = 0$.
When the concentrations of both $A$ and $B$ are doubled,the rate increases by a factor of $4$:
$\frac{r_2}{r_1} = \frac{k[2A]^{\alpha}[2B]^{\beta}}{k[A]^{\alpha}[B]^{\beta}} = 4$
$2^{\alpha} \cdot 2^{\beta} = 4$
Since $\beta = 0$,we have $2^{\alpha} = 4$,which gives $\alpha = 2$.
The overall order of the reaction is $\alpha + \beta = 2 + 0 = 2$.
The unit of the rate constant for a second-order reaction is $L \ mol^{-1} \ s^{-1}$.

(D) The rate law for the reaction is given by $r = k[A]^n$.
Taking the square root on both sides,we get $\sqrt{r} = \sqrt{k} \times [A]^{n/2}$.
The given graph is a plot of $\sqrt{r_0}$ versus $[A]_0$,which is a straight line passing through the origin.
Comparing this with the equation $y = mx$,we have $y = \sqrt{r_0}$,$x = [A]_0$,and slope $m = \sqrt{k}$.
From the graph,the slope is $4.0$.
Therefore,$\sqrt{k} = 4.0$,which implies $k = (4.0)^2 = 16.0$.
Also,comparing the powers of $[A]$,we have $n/2 = 1$,which gives $n = 2$.
Thus,the order of the reaction is $n = 2$ and the rate constant is $k = 16.0 \ dm^3 \ mol^{-1} \ min^{-1}$.
Hence,option $(d)$ is the correct answer.

(C) Given the rate law: $Rate = k[H^{+}]^n$.
At $pH = 3$,the concentration of hydrogen ions is $[H^{+}]_1 = 10^{-3} \ M$.
At $pH = 1$,the concentration of hydrogen ions is $[H^{+}]_2 = 10^{-1} \ M$.
It is given that the rate increases $100$ times,so $Rate_2 = 100 \times Rate_1$.
Using the ratio: $\frac{Rate_2}{Rate_1} = \left(\frac{[H^{+}]_2}{[H^{+}]_1}\right)^n$.
Substituting the values: $100 = \left(\frac{10^{-1}}{10^{-3}}\right)^n$.
$100 = (10^2)^n$.
$10^2 = 10^{2n}$.
Comparing the exponents,$2 = 2n$,which gives $n = 1$.

(B) The reaction is $CH_{3}COOC_{2}H_{5} + H_{2}O \xrightarrow{H^+} CH_{3}COOH + C_{2}H_{5}OH$.
In a standard pseudo-first order reaction,water is in excess,making the order with respect to ester $1$.
However,if the reaction is carried out with a large excess of ester,the concentration of the ester remains effectively constant throughout the reaction.
Therefore,the rate of reaction becomes independent of the concentration of the ester,resulting in a zero-order reaction with respect to the ester.