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

(B) According to the rate law,the rate of a chemical reaction is generally proportional to the concentration of the reactants raised to some power.
For a reaction $A \rightarrow \text{Products}$,the rate is given by $\text{rate} = k[A]^n$.
As the concentration $[A]$ increases,the rate of the reaction increases for reactions with order $n > 0$.
Note: For a zero-order reaction $(n = 0)$,the rate is independent of the concentration,i.e.,$\text{rate} = k[A]^0 = k$.

(B) The given rate law is $\text{Rate} = K_1[RCl]$.
This indicates that the reaction is of first order with respect to the alkyl halide $(RCl)$ and zero order with respect to sodium hydroxide $(NaOH)$.
If the concentration of the alkyl halide is reduced to one half,the new rate will be $\text{Rate}' = K_1 \times \frac{1}{2}[RCl] = \frac{1}{2} \times \text{Rate}$.
Therefore,the rate is halved on reducing the concentration of the alkyl halide to one half.

(B) Let the rate law be $Rate = k[A]^n$.
When concentration is doubled,$Rate' = k[2A]^n = 4 \times Rate$.
So,$2^n = 4$,which implies $n = 2$.
When concentration is tripled,$Rate'' = k[3A]^n = 9 \times Rate$.
So,$3^n = 9$,which implies $n = 2$.
Thus,the rate is proportional to the square of the concentration of $A$ $([A]^2)$.

(B) According to the rate law,the rate of a reaction is directly proportional to the concentration of the reactants raised to the power of their respective orders.
Therefore,if the concentration of the reactants is increased,the rate of reaction increases.

(D) The rate of reaction is given by $Rate = K [A] [B]$.
Since concentration is defined as $C = n/V$,if the volume $V$ is reduced to $1/4$ of its initial value,the concentration of each reactant increases by a factor of $4$ (i.e.,$[A]' = 4[A]$ and $[B]' = 4[B]$).
The new rate $Rate'$ is given by $Rate' = K [A]' [B]' = K (4[A]) (4[B]) = 16 K [A] [B]$.
Thus,the new rate is $16$ times the original rate.

(C) The rate law for the reaction is given by $Rate = k[A]^2[B]^1$.
Let the initial rate be $R_1 = k[A]^2[B]$.
When the concentration of $A$ is doubled $(2[A])$ and the concentration of $B$ is halved $(0.5[B])$,the new rate $R_2$ is:
$R_2 = k(2[A])^2(0.5[B])$
$R_2 = k(4[A]^2)(0.5[B])$
$R_2 = 2 \times k[A]^2[B]$
$R_2 = 2 \times R_1$.
Therefore,the rate of the reaction increases by two times.

(D) The rate of a reaction depends upon the concentration of the reactant. According to the rate law,the rate of a reaction is directly proportional to the concentration of the reactants raised to some power.

(A) The rate of a reaction is given by the rate law expression: $Rate = k[A]^x[B]^y$,where $x$ and $y$ are the orders of reaction with respect to $A$ and $B$ respectively.
Assuming the reaction is elementary,the rate law is $Rate = k[A]^1[B]^1$.
If the concentration of $A$ is doubled,the new rate becomes $Rate' = k[2A]^1[B]^1 = 2 \times k[A][B] = 2 \times Rate$.
Therefore,the rate of reaction is doubled.

(D) The rate of a chemical reaction is influenced by factors such as the nature of the reactants,their concentration,and the temperature of the reaction.
Molecularity is a theoretical concept representing the number of reacting species taking part in an elementary step. It does not directly determine the rate of the reaction,as the rate is determined by the rate law expression derived from experimental data.
Therefore,the correct option is $(d)$.

(C) The rate law for an elementary reaction is determined by the stoichiometry of the reactants.
For the reaction $A + 2B \rightarrow C + 2D$,the rate law is expressed as:
Rate $= K[A]^1[B]^2 = K[A][B]^2$.
Thus,the correct option is $(C)$.

(B) The given reaction is an example of a pseudo-first-order reaction.
In this reaction,water is present in large excess,so its concentration remains effectively constant throughout the reaction.
Therefore,the rate of the reaction depends only on the concentration of sucrose.
The rate law is expressed as $\text{Rate} = K [\text{sucrose}]^1 [\text{water}]^0$,which simplifies to $\text{Rate} = K [\text{sucrose}]$.
Thus,option $B$ is correct.

(A) Let the rate law be $Rate = k[A]^x[B]^y$.
From experiment $(1)$ and $(3)$:
$[A]$ changes from $0.012$ to $0.024$ (doubled),while $[B]$ is constant $(0.035)$. The rate remains $0.10$. Thus,$x = 0$.
From experiment $(1)$ and $(4)$:
$[A]$ is constant $(0.012)$,while $[B]$ changes from $0.035$ to $0.070$ (doubled). The rate changes from $0.10$ to $0.80$ (increased by $8$ times).
$2^y = 8 \implies y = 3$.
Therefore,the rate law is $Rate = k[B]^3$.

(A) Let the rate law be $Rate = k [A]^x [B_2]^y$.
From experiment $(1)$ and $(2)$,$[A]_0$ is constant $(0.50 \ M)$ and $[B_2]_0$ is doubled ($0.50 \ M$ to $1.00 \ M$). The rate increases from $1.6 \times 10^{-4}$ to $3.2 \times 10^{-4}$ (doubled).
Therefore,$2^y = 2$,which implies $y = 1$.
From experiment $(2)$ and $(3)$,$[B_2]_0$ is constant $(1.00 \ M)$ and $[A]_0$ is doubled ($0.50 \ M$ to $1.00 \ M$). The rate remains $3.2 \times 10^{-4}$ (no change).
Therefore,$2^x = 1$,which implies $x = 0$.
Substituting the values of $x$ and $y$ into the rate law,we get $Rate = k [A]^0 [B_2]^1 = k [B_2]$.

(C) For a first-order reaction,the rate of reaction is directly proportional to the concentration of the reactant: $Rate = k[Reactant]^1$.
When the volume of the reaction vessel is reduced to $1/3$ of its original volume $(V' = V/3)$,the concentration of the gaseous reactants increases by a factor of $3$ $(C' = n/V' = n/(V/3) = 3C)$.
Since the reaction is of first order with respect to the concentration,the new rate of reaction $(Rate')$ will be $k(3[Reactant])^1 = 3 \times k[Reactant] = 3 \times Rate$.
Therefore,the rate of reaction increases by $3$ times.

(B) Let the rate law be $r = k[A]^x[B]^y$.
According to the problem,when the concentration of $B$ is doubled,the rate doubles: $2r = k[A]^x(2[B])^y$.
This implies $2^y = 2$,so $y = 1$.
When the concentration of both $A$ and $B$ is doubled,the rate increases by a factor of $8$: $8r = k(2[A])^x(2[B])^y$.
Substituting $y = 1$,we get $8r = k(2^x[A]^x)(2[B]) = 2^{x+1}k[A]^x[B]$.
Since $r = k[A]^x[B]$,we have $8 = 2^{x+1}$,which means $2^3 = 2^{x+1}$.
Therefore,$x + 1 = 3$,so $x = 2$.
The rate law is $r = k[A]^2[B]$.

(B) Let the rate law be $rate = k[A]^x[B]^y$.
According to the problem,when $[A]$ is doubled and $[B]$ is constant,the rate increases by $4$ times:
$4 \times rate = k[2A]^x[B]^y$
$4 = 2^x \implies x = 2$.
When $[B]$ is doubled and $[A]$ is constant,the rate remains unaffected:
$rate = k[A]^x[2B]^y$
$1 = 2^y \implies y = 0$.
Substituting the values of $x$ and $y$ into the rate law expression:
$rate = k[A]^2[B]^0 = k[A]^2$.

(B) The velocity constant $K$ (also known as the rate constant) is a characteristic constant for a specific reaction at a given temperature.
It is independent of the concentration of reactants or products.
It depends primarily on the temperature of the reaction and the presence of a catalyst.
Therefore,the correct option is $(B)$.

(D) The rate law is given by $r = K[A]^n$.
When the concentration of $A$ is doubled,the new rate $r'$ becomes $4r$.
So,$4r = K[2A]^n$.
Dividing the new rate by the original rate: $\frac{4r}{r} = \frac{K[2A]^n}{K[A]^n}$.
$4 = 2^n$.
Since $4 = 2^2$,we get $n = 2$.

(D) The velocity constant $(K)$ is specific to a particular reaction at a given temperature.
It depends on the nature of the reactants and the temperature of the reaction.
It is defined as the rate of reaction when the concentration of each reactant is unity.
Therefore,statement $(D)$ is incorrect because $K$ is not a universal constant for all reactions; it varies from one reaction to another.

(C) The rate law for the given elementary reaction is: $\text{Rate} = K[A][B]$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law,which is $1 + 1 = 2$ (Second Order).
The general unit for the rate constant $K$ for a reaction of order $n$ is $(mol \ L^{-1})^{1-n} \ sec^{-1}$.
For a second-order reaction $(n = 2)$:
$K = (mol \ L^{-1})^{1-2} \ sec^{-1} = (mol \ L^{-1})^{-1} \ sec^{-1} = L \ mol^{-1} \ sec^{-1}$.
Therefore,option $C$ is the correct answer.

(C) The slowest step in a multi-step reaction mechanism is known as the $Rate \ determining \ step$.
Since the overall rate of the reaction cannot proceed faster than this slowest step,the rate of the reaction is determined by this step.

(C) The rate law is given as $Rate = K [CCl_3CHO]^1 [NO]^1$.
The overall order of the reaction is the sum of the powers of the concentration terms,which is $1 + 1 = 2$.
For a reaction of order $n$,the units of the rate constant $K$ are $(mol \ L^{-1})^{1-n} \ s^{-1}$.
Substituting $n = 2$,we get $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.

(A) For a first-order reaction,the rate law is given by: $\text{Rate} = k[N_2O_5]$.
Given,$k = 6.2 \times 10^{-1} \, s^{-1}$ and $[N_2O_5] = 1.25 \, mol \, L^{-1}$.
Substituting the values: $\text{Rate} = (6.2 \times 10^{-1} \, s^{-1}) \times (1.25 \, mol \, L^{-1})$.
$\text{Rate} = 7.75 \times 10^{-1} \, mol \, L^{-1} \, s^{-1}$.
Therefore,the correct option is $A$.

(D) For the given reaction,the rate law expression is given by: $\text{Rate} = k [N_2O_5]$.
Given that the rate constant $k = 3 \times 10^{-5} \, s^{-1}$ and the rate of reaction is $2.40 \times 10^{-5} \, mol \, L^{-1} \, s^{-1}$.
Substituting these values into the rate law equation:
$2.40 \times 10^{-5} = (3 \times 10^{-5}) \times [N_2O_5]$
Solving for $[N_2O_5]$:
$[N_2O_5] = \frac{2.40 \times 10^{-5}}{3 \times 10^{-5}} = 0.8 \, mol \, L^{-1}$.

(B) The rate law for the decomposition of $N_2O_5$ is given by $Rate = k[N_2O_5]$.
Given:
$Rate = 1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$
$k = 3.4 \times 10^{-5} \ s^{-1}$
Substituting the values in the rate equation:
$1.02 \times 10^{-4} = (3.4 \times 10^{-5}) \times [N_2O_5]$
$[N_2O_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}$
$[N_2O_5] = 3 \ mol \ L^{-1}$
Therefore,the correct option is $B$.

(A) The decomposition of $N_2O_5$ is a well-known first-order reaction.
For the reaction $2N_2O_5 \to 4NO_2 + O_2$,the rate law is experimentally determined to be $r = k[N_2O_5]$.
Therefore,the correct option is $A$.

(D) For the reaction $2A + B \to C$,the rate law is given as $\text{rate} = k[A][B]$.
According to the Arrhenius equation,$k = Ae^{-E_a/RT}$.
This shows that the rate constant $k$ depends only on temperature and the activation energy $(E_a)$,not on the initial concentrations of the reactants $A$ and $B$.
Therefore,the value of $k$ is independent of the initial concentrations of $A$ and $B$.

(C) For a reaction of order $n$,the unit of the rate constant $k$ is given by $(mol \, L^{-1})^{1-n} \, s^{-1}$.
For a second-order reaction,$n = 2$.
Substituting $n = 2$ into the formula: $k = (mol \, L^{-1})^{1-2} \, s^{-1} = (mol \, L^{-1})^{-1} \, s^{-1}$.
Since concentration is expressed in $mol \, L^{-1}$,the unit of the rate constant is $(\text{concentration})^{-1} \, s^{-1}$.
Thus,the dimension involves both time $(s^{-1})$ and concentration $(mol^{-1} \, L)$.
Therefore,the correct option is $C$.

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

(C) The overall order of a reaction is the sum of the exponents of the concentration terms in the rate law expression.
For option $C$,the rate law is $\text{Rate} = K[x]^{1.5}[y]^{-1}[z]^0$.
The overall order $= 1.5 + (-1) + 0 = 0.5$.
Therefore,option $C$ is correct.

(C) The rate law for the reaction is given by $r = k[A]^m$,where $m$ is the order of the reaction with respect to $A$.
According to the problem,when the concentration of $A$ is increased four times,the rate becomes $2r$.
So,$2r = k[4A]^m$.
Dividing the second equation by the first: $\frac{2r}{r} = \frac{k[4A]^m}{k[A]^m}$.
$2 = 4^m$.
Taking the logarithm or expressing $4$ as $2^2$: $2^1 = (2^2)^m = 2^{2m}$.
Equating the exponents: $1 = 2m$,which gives $m = 0.5$.

(B) For the decomposition of ammonium nitrate $(NH_4NO_2 \rightarrow N_2 + 2H_2O)$, the volume of $N_2$ gas evolved at time $t$ $(V_t)$ is proportional to the amount of reactant consumed.
The rate constant $K$ for a first-order reaction is given by the formula: $K = \frac{2.303}{t} \log \frac{V_{\infty}}{V_{\infty} - V_t}$.
Substituting the given values ($V_{\infty} = 35.05$ $cc$) into the equation for different time intervals yields a constant value for $K$.
Since the data fits the first-order integrated rate equation, the order of the reaction is $1$.

(A) The rate law for the acid-catalyzed hydrolysis of ethyl acetate is given by:
$\text{Rate} = k[CH_3COOEt][H_2O]$
In this reaction,water $(H_2O)$ is present in large excess,so its concentration remains effectively constant throughout the reaction.
Therefore,the rate expression simplifies to:
$\text{Rate} = k'[CH_3COOEt]$
where $k' = k[H_2O]$.
Since the rate depends on the concentration of only one reactant,this is a pseudo-first-order reaction.
Thus,the order of the reaction is $1$.

(D) Let the rate law be $r = k[A]^n$.
When the concentration of $A$ is increased $10$ folds,the new rate $r'$ becomes $100r$.
So,$100r = k[10A]^n$.
Dividing the two equations: $\frac{100r}{r} = \frac{k[10A]^n}{k[A]^n}$.
$100 = (10)^n$.
Since $100 = 10^2$,we get $n = 2$.

(C) The reaction $C_{12}H_{22}O_{11} + H_2O \to C_6H_{12}O_6 + C_6H_{12}O_6$ involves water in large excess.
Since the concentration of water remains practically constant during the reaction,the rate of reaction depends only on the concentration of cane sugar.
Therefore,it follows first-order kinetics and is classified as a pseudo-unimolecular reaction.

(B) The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
Given the rate expression: $\frac{dc}{dt} = K[E]^{3/2}[D]^{3/2}$.
Order of reaction = $\frac{3}{2} + \frac{3}{2} = \frac{6}{2} = 3$.
Therefore,the correct option is $(B)$.

(C) The molecularity of a reaction is defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary reaction step.
For the decomposition of $N_2O_5$,the elementary step involves two molecules of $N_2O_5$ colliding to form the products.
Therefore,the reaction is bimolecular.

(B) The molecularity of a reaction is defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary reaction,which must collide simultaneously in order to bring about a chemical reaction.
Since the reaction involves two different reactants,at least two molecules must be involved in the elementary step.
Therefore,the reaction cannot be unimolecular (molecularity = $1$).
Thus,option $(B)$ is correct.

(B) The overall order of a reaction is defined as the sum of the powers to which the concentration terms are raised in the rate law expression.

(D) The reaction for the inversion of cane sugar is: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 + C_6H_{12}O_6$.
Since water is present in large excess,its concentration remains practically constant during the reaction.
Therefore,the rate law is given by: $\text{Rate} = K [C_{12}H_{22}O_{11}] [H_2O]^0$.
As the rate depends only on the concentration of one reactant (cane sugar),it is a pseudo-unimolecular reaction.

(B) The order of a reaction with respect to a reactant is defined by the exponent of its concentration term in the rate law expression.
For a reaction $r = k[X]^n$,if the order with respect to $X$ is $2$,then $n = 2$.
Therefore,the rate of the reaction is proportional to $[X]^2$,i.e.,$r \propto [X]^2$.

(C) For a reaction of order $n$,the half-life period $(t_{1/2})$ is related to the initial concentration $(a)$ as: $t_{1/2} \propto \frac{1}{a^{n-1}}$.
Given that $t_{1/2} \propto \frac{1}{a^1}$,we have $n-1 = 1$,which implies $n = 2$.
Therefore,the reaction is of the second order.

(NONE) The correct answer is not explicitly listed as a 'wrong' statement among the options provided,as all statements $A, B, C,$ and $D$ are scientifically correct.
$1$. Molecularity is the number of reacting species taking part in an elementary reaction,which must be a whole number.
$2$. Order is an experimental quantity,while molecularity is a theoretical concept; they are often different.
$3$. Zero-order reactions are common (e.g.,photochemical reactions).
$4$. Order is determined by the rate-determining step in a reaction mechanism.
However,if the question implies identifying a false statement,all provided options are actually true in chemical kinetics.

(B) The rate law is given by $Rate = k[A]^1[B]^0$.
Since the sum of the exponents of the concentration terms in the rate law is $1 + 0 = 1$,the order of the reaction is $1$.
Molecularity is defined as the number of reacting species taking part in an elementary reaction. Since $A$ and $B$ are involved in the reaction $A + B \to \text{products}$,the molecularity is $2$.

(D) For a reaction of order $n$,the rate law is given by $r = k[A]^n$.
Integrating the rate equation for an initial concentration $a$,the half-life period $T$ is related to the initial concentration by the expression $T \propto \frac{1}{a^{n - 1}}$.
This relationship holds for all orders $n \neq 1$.
Therefore,the correct option is $D$.

(B) The concentration of sucrose decreases from $0.4 \ M$ to $0.2 \ M$ in $1 \ hour$. This means the half-life $(t_{1/2})$ is $1 \ hour$.
When the concentration is further reduced from $0.2 \ M$ to $0.1 \ M$,it takes another $1 \ hour$ (total $2 \ hours$).
Since the half-life is independent of the initial concentration,the reaction follows first-order kinetics.
Therefore,the order of the reaction is $1$.

(B) The correct answer is $(B)$.
The hydrolysis of methyl acetate in a dilute aqueous solution is represented by the equation: $CH_3COOCH_3 + H_2O \xrightarrow{H^{+}} CH_3COOH + CH_3OH$.
In this reaction,water is present in large excess,so its concentration remains practically constant during the reaction.
Therefore,the rate of the reaction depends only on the concentration of methyl acetate,making it a pseudo-$unimolecular$ reaction.

(B) The reaction involves two reactant molecules in the rate-determining step,so it is bimolecular.
From the rate law $\frac{dx}{dt} = k[CH_3COOC_2H_5][NaOH]$,the sum of the powers of the concentration terms is $1 + 1 = 2$.
Therefore,the reaction is of second order.
Thus,the correct option is $B$.

(D) The rate law for the reaction is given by $Rate = k[HI]^2$.
Since the sum of the powers of the concentration terms in the rate law expression is $2$,the order of the reaction is $2$.
Therefore,the reaction is of second order.

(B) The hydrolysis of sucrose (inversion of 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.
Thus,the rate law is: $\text{Rate} = K[C_{12}H_{22}O_{11}]^1[H_2O]^0$.
This is a pseudo-first order reaction,which is a type of first order reaction.