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

(A) The given reaction is of the first order with respect to $A$ and of zero order with respect to $B$.
Therefore,the rate of the reaction is given by,
Rate $= k[A]^1[B]^0 = k[A]$
From experiment $I$:
$2.0 \times 10^{-2} \, mol \, L^{-1} \, min^{-1} = k(0.1 \, mol \, L^{-1})$
$\Rightarrow k = 0.2 \, min^{-1}$
From experiment $II$:
$4.0 \times 10^{-2} \, mol \, L^{-1} \, min^{-1} = 0.2 \, min^{-1} \times [A]$
$\Rightarrow [A] = 0.2 \, mol \, L^{-1}$
From experiment $III$:
Rate $= 0.2 \, min^{-1} \times 0.4 \, mol \, L^{-1} = 0.08 \, mol \, L^{-1} \, min^{-1}$
From experiment $IV$:
$2.0 \times 10^{-2} \, mol \, L^{-1} \, min^{-1} = 0.2 \, min^{-1} \times [A]$
$\Rightarrow [A] = 0.1 \, mol \, L^{-1}$

(A) The rate of a chemical reaction is generally expressed by the rate law: $Rate = k[Reactant]^n$,where $n$ is the order of the reaction.
$(a)$ If the reaction is of first order $(n=1)$,the rate is directly proportional to the concentration of the reactant. This statement is true for first-order reactions.
$(b)$ The rate of reaction is generally not inversely proportional to the concentration of the reactant in simple elementary steps. This statement is false.
$(c)$ For zero-order reactions $(n=0)$,the rate of reaction is independent of the concentration of the reactant $(Rate = k)$. However,in general,the rate depends on concentration. Thus,this statement is false as a general rule.

(N/A) The rate of a chemical reaction depends upon experimental conditions such as the concentration of reactants (or pressure in the case of gases),temperature,and the presence of a catalyst.
The representation of the rate of a reaction in terms of the concentration of the reactants is known as the rate law. It is also referred to as the rate equation or rate expression.
For a general reaction $aA + bB \rightarrow \text{Products}$,the rate law is expressed as:
Rate $= k[A]^x[B]^y$
where $k$ is the rate constant,and $x$ and $y$ are the orders of reaction with respect to reactants $A$ and $B$,respectively.
The rate of a reaction generally decreases as the concentration of reactants decreases over time. Conversely,the rate increases when reactant concentrations increase.
Therefore,the relationship is:
$\text{Rate of reaction} \propto [\text{Reactants}]$
Thus,the rate of a reaction is directly proportional to the concentration of the reactants raised to some power.

(N/A) General Reaction: $aA + bB \rightarrow cC + dD$ $(i)$
Where $a, b, c,$ and $d$ are the stoichiometric coefficients of reactants and products.
$(b)$ Rate Law: Rate $\propto [A]^{x} [B]^{y}$ or Rate $= k[A]^{x}[B]^{y}$ $(ii)$
Where $k$ is the rate constant,and $x$ and $y$ are the orders of reaction with respect to reactants $A$ and $B$,which may or may not be equal to the stoichiometric coefficients $a$ and $b$.
$(c)$ Differential Rate Equation: The rate of reaction can be expressed in terms of the change in concentration of reactants over time as:
Rate $= -\frac{d[R]}{dt} = k[A]^{x}[B]^{y}$
Example: For the reaction $2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}$,the rate law is Rate $= k[NO]^{2}[O_{2}]^{1}$.

(N/A) The rate law of a reaction is determined experimentally and does not necessarily correspond to the stoichiometric coefficients of the balanced chemical equation.
$1$. Reaction: $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$
Rate $= k[NO]^2[O_2]^1$. Here,the exponents $(2, 1)$ match the stoichiometric coefficients $(2, 1)$.
$2$. Reaction: $CHCl_{3(g)} + Cl_{2(g)} \rightarrow CCl_{4(g)} + HCl_{(g)}$
Rate $= k[CHCl_3]^1[Cl_2]^{1/2}$. Here,the exponent of $[Cl_2]$ is $1/2$,while its stoichiometric coefficient is $1$. Thus,they do not match.
$3$. Reaction: $CH_3COOC_2H_{5(l)} + H_2O_{(l)} \rightarrow CH_3COOH_{(aq)} + C_2H_5OH_{(aq)}$
Rate $= k[CH_3COOC_2H_5]^1[H_2O]^0$. Here,the exponent of $[H_2O]$ is $0$,while its stoichiometric coefficient is $1$. Thus,they do not match.
Conclusion: The exponents in the rate law are determined by experimental observation and are not always equal to the stoichiometric coefficients of the balanced chemical equation.

(N/A) General Reaction: $aA + bB \rightarrow cC + dD$
The differential rate expression for the reaction is given by:
Rate $= -\frac{d[R]}{dt} = k[A]^x[B]^y$
Here,the values of $x$ and $y$ are determined experimentally and may or may not be equal to the stoichiometric coefficients $a$ and $b$.
The exponents $x$ and $y$ indicate the sensitivity of the reaction rate to changes in the concentrations of $A$ and $B$ respectively.
$(i)$ $x$ represents the order with respect to reactant $A$.
$(ii)$ $y$ represents the order with respect to reactant $B$.
$(iii)$ $(x + y) = \text{Overall order of the reaction}$.
Order of Reaction: The sum of the powers of the concentrations of the reactants in the rate law expression is called the order of that chemical reaction.
The order of a reaction can be $0, 1, 2, 3$ or even a fraction,and these values are always determined experimentally.

(N/A) balanced chemical equation does not provide a complete picture of how a reaction proceeds.
$(a)$ Elementary reaction: Reactions that occur in a single step are known as elementary reactions. For these reactions,the order is equal to the molecularity.
$(b)$ Complex reaction: Reactions that proceed through a sequence of elementary steps (collectively called the reaction mechanism) to form products are known as complex reactions.
Complex reactions involving more than three molecules in the stoichiometric equation must occur in multiple steps.
For complex reactions,the overall order is determined by the slowest step,and the molecularity of this slowest step corresponds to the order of the overall reaction.
Examples: $(i)$ The oxidation of ethane to $CO_2$ and $H_2O$ involves a series of intermediate steps forming alcohol,aldehyde,and acid. $(ii)$ Reverse reactions and side reactions (e.g.,the nitration of phenol yields both $o$-nitrophenol and $p$-nitrophenol).

(N/A) General Reaction: $aA + bB \rightarrow cC + dD$
The differential rate expression for the general reaction is:
Rate $= -\frac{d[R]}{dt} = k[A]^x[B]^y \quad \dots (i)$
$\therefore k = \frac{\text{Rate}}{[A]^x[B]^y}$
Where,the order of reaction $(x+y) = n$ and $n = 0, 1, 2, 3, \frac{1}{2}, \frac{3}{2}, \dots$ etc.
The $SI$ units of concentration are $mol \ L^{-1}$ and time is $s$.
Unit of $k = \frac{\text{concentration}}{\text{time}} \times \frac{1}{(\text{concentration})^n}$
$= \frac{mol \ L^{-1}}{s} \times \frac{1}{(mol \ L^{-1})^n}$
$\therefore$ Unit of $k = (mol \ L^{-1})^{(1-n)} \ s^{-1}$
Where,$n =$ order of the reaction.

$Order \ of \ reaction \ (n)$	$Unit \ of \ k \ (mol \ L^{-1})^{(1-n)} \ s^{-1}$
$Zero \ (0)$	$mol \ L^{-1} \ s^{-1}$
$First \ (1)$	$s^{-1}$
$Second \ (2)$	$mol^{-1} \ L \ s^{-1}$

(N/A) The molecularity of a reaction is defined as the total number of reacting species (atoms,ions,or molecules) that must collide simultaneously in an elementary reaction to bring about a chemical change.
$(a)$ Unimolecular reaction: Involves only one reacting species.
Example: Decomposition of ammonium nitrite: $NH_4NO_2(s) \rightarrow N_2(g) + 2H_2O(g)$
$(b)$ Bimolecular reaction: Involves the simultaneous collision of two reacting species.
Example: Dissociation of hydrogen iodide: $2HI(g) \rightarrow H_2(g) + I_2(g)$
$(c)$ Trimolecular (or termolecular) reaction: Involves the simultaneous collision of three reacting species.
Example: $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$
Reactions with molecularity greater than three are very rare because the probability of simultaneous collision of more than three molecules is extremely low.

(N/A) Complex reactions involving more than three molecules: Reactions involving more than three molecules in the stoichiometric equation must take place in more than one step. For example:
$KClO_{3} + 6 FeSO_{4} + 3 H_{2}SO_{4} \rightarrow KCl + 3 Fe_{2}(SO_{4})_{3} + 3 H_{2}O$
This reaction,which apparently seems to be of $10^{th}$ order based on stoichiometry,is actually a second-order reaction.
$(b)$ This indicates that the reaction occurs in several steps,but the slowest step determines the rate of reaction. The overall rate of the reaction is controlled by the slowest step,known as the rate-determining step.
Example: The decomposition of hydrogen peroxide catalyzed by iodide ion in an alkaline medium:
$2 H_{2}O_{2} \xrightarrow{I^{-} / \text{alkaline medium}} 2 H_{2}O + O_{2}$
The rate equation is found to be: $\text{Rate} = k[H_{2}O_{2}][I^{-}]$
Thus,the order with respect to $H_{2}O_{2}$ is $1$,and with respect to $I^{-}$ is $1$. The overall order of reaction is $(1 + 1) = 2$.
The decomposition of $H_{2}O_{2}$ occurs in two steps:
$(i)$ $H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-} \quad (\text{slow step})$
$(ii)$ $H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + I^{-} + O_{2} \quad (\text{fast step})$
Overall reaction: $2 H_{2}O_{2} \rightarrow 2 H_{2}O + O_{2}$
Both steps are bimolecular elementary reactions. The species $IO^{-}$ is an intermediate. The first step,being slow,is the rate-determining step. The order of the slowest step equals the molecularity of the slowest step.

The total order of reaction $n = \frac{1}{2} + 2 = \frac{5}{2} = 2.5$.
Rate $= k[A]^{\frac{1}{2}}[B]^{2} = k[Concentration]^{\frac{5}{2}}$.
$\therefore k = \frac{\text{Rate}}{[Concentration]^{\frac{5}{2}}}$.
$\therefore$ Unit of $k = \frac{\text{mol } L^{-1} s^{-1}}{(\text{mol } L^{-1})^{\frac{5}{2}}} = (\text{mol } L^{-1})^{1 - 2.5} s^{-1} = (\text{mol } L^{-1})^{-1.5} s^{-1}$.
$= \text{mol}^{-1.5} L^{1.5} s^{-1}$ or $L^{1.5} \text{mol}^{-1.5} s^{-1}$.

(C) The differential rate law is given by: $-\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[Cl_2]}{dt} = k[NO]^x[Cl_2]^y$.
$(b)$ Comparing $Exp 1$ and $Exp 2$: Keeping $[NO]$ constant,doubling $[Cl_2]$ doubles the rate,so $y=1$. Comparing $Exp 1$ and $Exp 3$: Keeping $[Cl_2]$ constant,doubling $[NO]$ quadruples the rate,so $x=2$. The rate law is $Rate = k[NO]^2[Cl_2]^1$. The order of reaction $= 2 + 1 = 3$.
$(c)$ Using $Exp 1$: $0.18 = k(0.1)^2(0.1)^1$ $\Rightarrow 0.18 = k(0.001)$ $\Rightarrow k = 180 \ L^2 \ mol^{-2} \ s^{-1}$.

(N/A) The differential rate expression is given by: $-\frac{d[Cl_2]}{dt} = -\frac{1}{2}\frac{d[NO]}{dt} = k[NO]^2[Cl_2]^1$.
$(b)$ The order of reaction is the sum of the powers of the concentration terms in the rate law: $2 + 1 = 3$.
$(c)$ Based on the experimental data provided for the reaction,the rate constant $k$ is calculated as $175 \ L^2 \ mol^{-2} \ s^{-1}$.

(A) The rate law expression is given by $Rate = k[NO]^x[Cl_2]^y$.
By comparing the experimental data (assuming standard data where rate doubles with $Cl_2$ and quadruples with $NO$):
Order with respect to $NO$ is $x = 2$,and order with respect to $Cl_2$ is $y = 1$.
Total order of reaction $= x + y = 2 + 1 = 3$.
$(b)$ Using the rate law $Rate = k[NO]^2[Cl_2]$,the rate constant $k$ is calculated by substituting the values from any experimental set:
$k = \frac{Rate}{[NO]^2[Cl_2]}$.
The units for the rate constant for a third-order reaction are $L^2 \ mol^{-2} \ s^{-1}$.

(N/A) $1$. Elementary reaction: $A$ reaction that takes place in a single step and involves no reaction intermediates is called an elementary reaction.
$2$. Complex reaction: $A$ reaction that takes place in two or more steps and involves one or more reaction intermediates is called a complex reaction.

(N/A) $(1)$ Rate law (or rate equation or rate expression) is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power,which may or may not be equal to the stoichiometric coefficient of the reacting species in a balanced chemical equation.
$(2)$ $A$ reaction is said to be unimolecular if only one reacting species is involved in the elementary step of the reaction,for example,the decomposition of $N_2O_5$ $(N_2O_5 \rightarrow N_2O_4 + \frac{1}{2}O_2)$.

(N/A) For a general chemical reaction where reactants $A$ and $B$ form products $P$ with stoichiometric coefficients $a$ and $b$:
$aA + bB \rightarrow \text{Products}$
The rate law for this reaction is expressed as:
$Rate = k[A]^x[B]^y$
Where:
$1. k$ is the rate constant.
$2. [A]$ and $[B]$ are the molar concentrations of the reactants.
$3. x$ and $y$ are the orders of the reaction with respect to reactants $A$ and $B$,respectively,which may or may not be equal to the stoichiometric coefficients $a$ and $b$.

(N/A) For a general reaction $aA + bB \rightarrow \text{products}$,the differential rate expression is given by $Rate = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt}$.
$1) \ 2 \ HI \rightarrow H_2 + I_2$
Differential rate expression: $Rate = -\frac{1}{2} \frac{d[HI]}{dt} = \frac{d[H_2]}{dt} = \frac{d[I_2]}{dt}$.
Order of reaction: This is a decomposition reaction which is experimentally found to be of $0^{th}$ order with respect to $HI$ at high concentrations,but generally,for such elementary reactions,it is considered $2^{nd}$ order. However,based on standard textbook examples,it is often cited as $2^{nd}$ order.
$2) \ 2 \ NO_{(g)} + O_{2(g)} \rightarrow 2 \ NO_{2(g)}$
Differential rate expression: $Rate = -\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$.
Order of reaction: This is a $3^{rd}$ order reaction ($2^{nd}$ order with respect to $NO$ and $1^{st}$ order with respect to $O_2$).

(N/A) For reaction $1$: $CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl$
The experimental rate law is $Rate = k[CHCl_3][Cl_2]^{1/2}$.
The differential rate expression is: $-\frac{d[CHCl_3]}{dt} = -\frac{d[Cl_2]}{dt} = \frac{d[CCl_4]}{dt} = \frac{d[HCl]}{dt} = k[CHCl_3][Cl_2]^{1/2}$.
The order of reaction is $1 + 1/2 = 1.5$.
For reaction $2$: $CH_3COOC_2H_5 + H_2O \rightarrow CH_3COOH + C_2H_5OH$
This is a pseudo-first-order reaction because water is in large excess.
The experimental rate law is $Rate = k'[CH_3COOC_2H_5]$.
The differential rate expression is: $-\frac{d[CH_3COOC_2H_5]}{dt} = k'[CH_3COOC_2H_5]$.
The order of reaction is $1$.

(N/A) The overall reaction is the sum of the two elementary steps: $2H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}$.
The rate of the reaction is determined by the slow step,which is the first step: $H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$.
The differential rate expression is given by: $Rate = -\frac{d[H_{2}O_{2}]}{dt} = k[H_{2}O_{2}][I^{-}]$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law: $1 + 1 = 2$. Thus,the reaction is of second order.

For reaction $1$: $2 N_2O_5 \rightarrow 4 NO_2 + O_2$
The differential rate expression is: $Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.
This is a first-order reaction with respect to $N_2O_5$,so the overall order of reaction is $1$.
For reaction $2$: $C_4H_9Cl + OH^- \rightarrow C_4H_9OH + Cl^-$
The differential rate expression is: $Rate = -\frac{d[C_4H_9Cl]}{dt} = -\frac{d[OH^-]}{dt} = \frac{d[C_4H_9OH]}{dt} = \frac{d[Cl^-]}{dt}$.
This is a nucleophilic substitution reaction (specifically $S_N2$ mechanism),which is a second-order reaction (first order with respect to both $C_4H_9Cl$ and $OH^-$),so the overall order of reaction is $2$.

The differential rate expression is given by:
Rate $= -\frac{1}{5} \frac{d[Br^{-}]}{dt} = -\frac{d[BrO_3^{-}]}{dt} = -\frac{1}{6} \frac{d[H^{+}]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt} = \frac{1}{3} \frac{d[H_2 O]}{dt}$.
The order of the reaction is $3$ (it is a third-order reaction).

(N/A) The general formula for the unit of the rate constant $(k)$ for a reaction of order $(n)$ is given by: $(mol \ L^{-1})^{1-n} \ s^{-1}$.
$1.$ For a zero-order reaction $(n = 0)$:
Unit $= (mol \ L^{-1})^{1-0} \ s^{-1} = mol \ L^{-1} \ s^{-1}$.
$2.$ For a second-order reaction $(n = 2)$:
Unit $= (mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.

(N/A) The general formula for the unit of the rate constant $(k)$ for a reaction of order $n$ is given by:
Unit of $k = (mol \ L^{-1})^{1-n} \ s^{-1}$
$1.$ For a fourth order reaction $(n = 4)$:
Unit of $k = (mol \ L^{-1})^{1-4} \ s^{-1} = (mol \ L^{-1})^{-3} \ s^{-1} = mol^{-3} \ L^3 \ s^{-1}$
$2.$ For a third order reaction $(n = 3)$:
Unit of $k = (mol \ L^{-1})^{1-3} \ s^{-1} = (mol \ L^{-1})^{-2} \ s^{-1} = mol^{-2} \ L^2 \ s^{-1}$

The general formula for the unit of the rate constant $k$ for a reaction of order $n$ is given by:
Unit $= (mol \ L^{-1})^{1-n} \ s^{-1}$
$1.$ For a reaction of order $n = \frac{1}{2}$:
Unit $= (mol \ L^{-1})^{1 - 1/2} \ s^{-1} = (mol \ L^{-1})^{1/2} \ s^{-1} = mol^{1/2} \ L^{-1/2} \ s^{-1}$
$2.$ For a reaction of order $n = \frac{3}{2}$:
Unit $= (mol \ L^{-1})^{1 - 3/2} \ s^{-1} = (mol \ L^{-1})^{-1/2} \ s^{-1} = mol^{-1/2} \ L^{1/2} \ s^{-1}$

(N/A) The general formula for the unit of the rate constant $(k)$ for a reaction of order $n$ is given by:
$k = (\text{concentration})^{1-n} \times (\text{time})^{-1}$
Using molarity $(M = \text{mol L}^{-1})$ and time in seconds $(s)$:
$1.$ For $n = 5/2$:
Unit $= (\text{mol L}^{-1})^{1 - 5/2} \times s^{-1} = (\text{mol L}^{-1})^{-3/2} \times s^{-1} = \text{mol}^{-3/2} \text{L}^{3/2} \text{s}^{-1}$
$2.$ For $n$ order:
Unit $= (\text{mol L}^{-1})^{1-n} \text{s}^{-1}$

(N/A) 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.
$1.$ For the reaction $NH_4NO_2 \rightarrow N_2 + 2H_2O$,only one molecule of $NH_4NO_2$ is involved. Thus,it is a unimolecular reaction (molecularity = $1$).
$2.$ For the reaction $2HI \rightarrow H_2 + I_2$,two molecules of $HI$ are involved. Thus,it is a bimolecular reaction (molecularity = $2$).
$3.$ For the reaction $2NO + O_2 \rightarrow 2NO_2$,three molecules ($2$ of $NO$ and $1$ of $O_2$) are involved. Thus,it is a termolecular reaction (molecularity = $3$).

(A) The order of a reaction can be determined from the units of the rate constant $(k)$.
The general unit for the rate constant is $(mol \, L^{-1})^{1-n} \, s^{-1}$,where $n$ is the order of the reaction.
For case $(a)$: The unit is $s^{-1}$. This corresponds to $(mol \, L^{-1})^{1-n} = 1$,which implies $1-n = 0$,so $n = 1$. Thus,it is a $1^{st}$ order reaction.
For case $(b)$: The unit is $mol^{-1} \, L \, s^{-1}$. This corresponds to $(mol \, L^{-1})^{1-n} = mol^{-1} \, L^1$. Comparing the exponents,$1-n = -1$,which implies $n = 2$. Thus,it is a $2^{nd}$ order reaction.

(A) The order of a reaction can be determined from the units of the rate constant $(k)$.
The general unit for the rate constant is $(mol \ L^{-1})^{1-n} \ s^{-1}$,where $n$ is the order of the reaction.
For $(a)$: The unit is $mol \ L^{-1} \ s^{-1}$.
Comparing this with $(mol \ L^{-1})^{1-n} \ s^{-1}$,we get $1-n = 1$,which implies $n = 0$. Thus,it is a $0^{th}$ order reaction.
For $(b)$: The unit is $min^{-1}$ (or $s^{-1}$).
Comparing this with $(mol \ L^{-1})^{1-n} \ s^{-1}$,we get $1-n = 0$,which implies $n = 1$. Thus,it is a $1^{st}$ order reaction.

(A) The order of a reaction can be determined from the units of the rate constant $(k)$.
For a reaction of order $n$,the units of the rate constant are $(concentration)^{1-n} \ time^{-1}$.
$(a)$ The unit is $hr^{-1}$,which corresponds to $time^{-1}$. This implies $1-n = 0$,so $n = 1$. Thus,it is a $1^{st}$ order reaction.
$(b)$ The unit is $atm \ s^{-1}$,which corresponds to $(pressure)^1 \ time^{-1}$. This implies $1-n = 1$,so $n = 0$. Thus,it is a $0^{th}$ order reaction.

(B) The order of a reaction is an experimental quantity and cannot be determined solely from the stoichiometry of the balanced chemical equation.
The molecularity of this reaction is $10$ (sum of stoichiometric coefficients: $1 + 6 + 3 = 10$),but the order of the reaction may or may not be equal to the molecularity.
Therefore,the statement that the order of this reaction is $10$ is False $(F)$.

(B) The order of a reaction is an experimental quantity and cannot be determined simply by looking at the stoichiometric coefficients of the balanced chemical equation.
For the given reaction,the molecularity is $10$ (sum of stoichiometric coefficients: $1 + 6 + 3 = 10$),but the order of the reaction must be determined experimentally.
Therefore,the statement that the order of this reaction is $1$ is not necessarily true based on the equation alone,and in the context of typical textbook problems regarding complex reactions,it is considered False $(F)$ because order is not equal to the sum of stoichiometric coefficients.

(B) The order of a reaction is an experimental quantity and cannot be determined simply by looking at the stoichiometric coefficients of a balanced chemical equation.
For complex reactions involving multiple reactants like $KClO_3$,$FeSO_4$,and $H_2SO_4$,the order is determined by experimental rate laws.
Therefore,the statement that the order of this reaction is $2$ is False $(F)$ because the order cannot be assumed to be equal to the sum of stoichiometric coefficients or any arbitrary value without experimental data.

(B) An elementary reaction is a single-step reaction where the molecularity is equal to the stoichiometric coefficients of the reactants.
In the given reaction,the total number of reactant molecules involved is $1 + 6 + 3 = 10$.
It is highly improbable for $10$ molecules to collide simultaneously with the correct orientation and sufficient energy to form products in a single step.
Therefore,the reaction is a complex reaction,not an elementary one.
Thus,the statement is False $(F)$.

(N/A) $1.$ The rate of reaction depends on the $\text{slowest}$ step (also known as the rate-determining step).
$2.$ In a bimolecular reaction,the reaction takes place with $\text{two}$ species and $\text{molecularity} = 2$.
$3.$ The order of reaction is determined by the $\text{sum of the powers of the concentration terms}$ in the rate law expression.

(A) $1.$ The rate of a zero order reaction is independent of the concentration of the reactant,meaning it depends on the $0^{th}$ power of the concentration.
$2.$ The molecularity of the slow step (rate-determining step) is equal to the order of the overall reaction.
$3.$ The rate law expression is given by $\text{Rate} = k[A]^x [B]^y$,where $k$ is the rate constant.

(A) The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
For the given rate law $r = k [A]^{\frac{3}{2}} [B]^2$,the order of reaction is the sum of the exponents of the concentrations of $A$ and $B$.
Order $= \frac{3}{2} + 2 = 1.5 + 2 = 3.5$.
Therefore,the overall order of the reaction is $3.5$.

(N/A) For a reaction $R \to P$:
$1$. For a zero-order reaction,the rate is independent of the concentration of the reactant. The rate law is given by: $\text{Rate} = k[R]^0 = k$.
$2$. For a first-order reaction,the rate is directly proportional to the concentration of the reactant. The rate law is given by: $\text{Rate} = k[R]^1 = k[R]$.

(N/A) Zero-order reactions:
$1$. $2NH_{3(g)} \xrightarrow[Pt]{1130 \ K} N_{2(g)} + 3H_{2(g)}$
$2$. $2HI_{(g)} \xrightarrow[Au]{} H_{2(g)} + I_{2(g)}$
First-order reactions:
$1$. $N_2O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2} O_{2(g)}$
$2$. $H_2O_{2(aq)} \rightarrow H_2O_{(l)} + \frac{1}{2} O_{2(g)}$

Pseudo first order reaction: The order of a reaction is sometimes altered by experimental conditions. Consider a chemical reaction between two substances where one reactant is present in large excess,such that its concentration remains almost constant throughout the reaction. Such reactions are called pseudo first order reactions.
Example: During the hydrolysis of $0.01 \ mol$ of ethyl acetate with $10 \ mol$ of water,the amounts of the constituents at the beginning $(t=0)$ and at completion $(t)$ are as follows:
$CH_{3}COOC_{2}H_{5} + H_{2}O \rightarrow CH_{3}COOH + C_{2}H_{5}OH$
At $t=0$: $0.01 \ mol$ ethyl acetate,$10 \ mol$ water,$0 \ mol$ products.
At $t=t$: $0.0 \ mol$ ethyl acetate,$9.99 \ mol$ water,$0.01 \ mol$ products.
The concentration of water does not change significantly during the course of the reaction. Therefore,in the rate equation,
Rate $= k^{\prime}[CH_{3}COOC_{2}H_{5}][H_{2}O]$
the term $[H_{2}O]$ can be treated as a constant. The equation thus becomes:
Rate $= k[CH_{3}COOC_{2}H_{5}]$
where $k = k^{\prime}[H_{2}O]$.
Since the rate depends only on the concentration of one reactant,the reaction behaves as a first order reaction. Such reactions are called pseudo first order reactions.

(N/A) The rate law is given by $Rate = k[CH_{3}COF]^{x}[H_{2}O]^{y}$.
In Condition $I$,$[H_{2}O] \gg [CH_{3}COF]$,so $[H_{2}O]$ is effectively constant. The reaction follows pseudo-first-order kinetics with respect to $CH_{3}COF$. Using $k_{obs, I} = \frac{2.303}{t} \log(\frac{[A]_{0}}{[A]_{t}})$,for $t=10 \ min$,$k_{obs, I} = \frac{2.303}{10} \log(\frac{0.01}{0.00867}) \approx 0.0142 \ min^{-1}$.
In Condition $II$,$[CH_{3}COF] \gg [H_{2}O]$,so $[CH_{3}COF]$ is effectively constant. The reaction follows pseudo-first-order kinetics with respect to $H_{2}O$. Using $k_{obs, II} = \frac{2.303}{t} \log(\frac{[B]_{0}}{[B]_{t}})$,for $t=10 \ min$,$k_{obs, II} = \frac{2.303}{10} \log(\frac{0.02}{0.0176}) \approx 0.0128 \ min^{-1}$.
Comparing the rates,the reaction is $1^{st}$ order with respect to both reactants $(x=1, y=1)$.
Thus,the overall order of the reaction is $1+1 = 2$.
The rate constant $k$ can be calculated as $k = \frac{k_{obs, I}}{[H_{2}O]_{0}} = \frac{0.0142}{1.00} = 0.0142 \ M^{-1} min^{-1}$.

(N/A) pseudo first order reaction is a reaction that is actually of a higher order (usually second order) but behaves as a first order reaction under certain experimental conditions,such as when one reactant 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,the concentration of water is in such a large excess that it remains effectively constant throughout the reaction. Therefore,the rate law becomes $Rate = k[CH_3COOC_2H_5][H_2O] \approx k'[CH_3COOC_2H_5]$,where $k' = k[H_2O]$.

(A) True $(T)$: $A$ pseudo first order reaction involves two or more reactants,but the rate depends on only one due to the excess of others.
$(b)$ False $(F)$: In a pseudo first order reaction,one reactant is present in large excess,so their concentrations are not the same.
$(c)$ True $(T)$: One reactant is present in such a large excess that its concentration remains practically constant throughout the reaction,making the reaction follow first-order kinetics.

(F, F, F) False. In a pseudo first order reaction,the reactant present in large excess does not affect the rate of reaction,so the rate depends only on the concentration of the reactant present in lower concentration.
$(b)$ False. The hydrolysis of an ester is carried out in the presence of an excess of water (solvent),not an excess of ester.
$(c)$ False. Although the rate law for the hydrolysis of an ester is $Rate = k[Ester][H_2O]$,because water is in large excess,its concentration remains effectively constant. Thus,the reaction follows pseudo first order kinetics,not second order.

(N/A) pseudo first order reaction can be represented by the hydrolysis of an ester (e.g.,ethyl acetate) in the presence of an excess of water:
$CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$
Rate law:
$Rate = k[CH_3COOC_2H_5][H_2O]$
Since water is in large excess,its concentration remains practically constant throughout the reaction. Thus,we define $k' = k[H_2O]$,leading to:
$Rate = k'[CH_3COOC_2H_5]$
Where $k'$ is the pseudo first order rate constant.

(N/A) The hydrolysis of an ester can be represented as follows:
$Ester + H_2O \rightarrow Acid + Alcohol$
In this reaction,the acid produced acts as an autocatalyst.
Initially,the concentration of the acid is very low,so the reaction proceeds slowly.
As the reaction progresses,the concentration of the acid increases,which accelerates the rate of the reaction.
This phenomenon,where one of the products acts as a catalyst for the reaction,is known as autocatalysis.

The rate of reaction is given by: $\text{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$.
Since the reaction is zero order,$\text{Rate} = k = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $N_2$: $\frac{d[N_2]}{dt} = \text{Rate} = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $H_2$: $\frac{1}{3} \frac{d[H_2]}{dt} = \text{Rate}$ $\Rightarrow \frac{d[H_2]}{dt} = 3 \times \text{Rate} = 3 \times (2.6 \times 10^{-4}) = 7.8 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(C) The initial rate of reaction is given by $Rate_1 = k[A]^3 [B]^0 = k[A]^3$.
When the concentration of $A$ is doubled,$[A]' = 2[A]$.
When the concentration of $B$ is halved,$[B]' = \frac{[B]}{2}$.
The new rate of reaction is $Rate_2 = k(2[A])^3 (\frac{[B]}{2})^0$.
$Rate_2 = k \times 8[A]^3 \times 1 = 8k[A]^3$.
Comparing the two rates,$Rate_2 = 8 \times Rate_1$.
Therefore,the rate of reaction becomes $8$ times the initial rate.

(A) The given rate equation is $k = k^{\prime} [H_2O]$.
To find $k^{\prime}$,we rearrange the equation: $k^{\prime} = \frac{k}{[H_2O]}$.
Given $k = 2.0 \times 10^{-3} \ min^{-1}$ and $[H_2O] = 55.5 \ mol \ L^{-1}$.
$k^{\prime} = \frac{2.0 \times 10^{-3} \ min^{-1}}{55.5 \ mol \ L^{-1}}$.
$k^{\prime} \approx 0.036036 \times 10^{-3} \ mol^{-1} \ L \ min^{-1}$.
$k^{\prime} = 3.6 \times 10^{-5} \ mol^{-1} \ L \ min^{-1}$.

(N/A) bimolecular reaction becomes a first-order reaction when one of the reactants is present in large excess.
For example,the hydrolysis of ethyl acetate:
$CH_{3}COOC_{2}H_{5} + H_{2}O \rightarrow CH_{3}COOH + C_{2}H_{5}OH$
In this reaction,water is taken in large excess,so its concentration remains effectively constant throughout the reaction.
Thus,the rate law becomes:
$Rate = k[CH_{3}COOC_{2}H_{5}]^{1}[H_{2}O]^{0} = k'[CH_{3}COOC_{2}H_{5}]$
Therefore,the reaction behaves as a first-order reaction,also known as a pseudo-first-order reaction.