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

(B) unimolecular reaction is one in which only one reactant molecule is involved in the elementary step of the reaction.
In the reaction $N_2O_5 \to N_2O_4 + \frac{1}{2}O_2$,only one molecule of $N_2O_5$ undergoes decomposition.
Therefore,it is a unimolecular reaction.
Thus,the correct option is $(B)$.

(C) The order of a reaction is an experimental quantity that depends on the mechanism of the reaction and the concentration of the reactants.
It does not depend on the surface area of the reactants.
Therefore,the order of the reaction remains constant.
Hence,Option $C$ is the correct answer.

(B) The inversion of sugar is represented by the reaction: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 + C_6H_{12}O_6$.
Since two molecules ($C_{12}H_{22}O_{11}$ and $H_2O$) are involved in the elementary step of the reaction,the molecularity is $2$.
Although it is a pseudo-first-order reaction because the concentration of water is in large excess and remains constant,the molecularity remains $2$.

(A) 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.
For the given reaction $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$,the stoichiometric coefficients indicate that $1$ molecule of $H_2$ and $1$ molecule of $Br_2$ are involved,so the molecularity is $1 + 1 = 2$.
The order of reaction is the sum of the powers of the concentration terms in the rate law expression.
Given $\text{rate} = K[H_2]^1[Br_2]^{1/2}$,the order is $1 + \frac{1}{2} = \frac{3}{2}$.
Therefore,the molecularity is $2$ and the order is $\frac{3}{2}$.

(C) The rate of reaction is defined by the rate law expression,where the sum of the exponents of the concentration terms is the order of reaction.
For the reaction $A + B \xrightarrow{K} C$,the rate expression is $\text{Rate} = K[A]^x[B]^y$.
In option $C$,the expression is $\frac{-d[A]}{dt} = K[A][B]^0$. The order of reaction is the sum of exponents,which is $1 + 0 = 1$.
However,the option indicates the order as $2$,which is incorrect.

(C) 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.
$1$. Molecularity is always a whole number $(1, 2, 3, \dots)$. It can never be zero,fractional,or negative.
$2$. It is a theoretical concept derived from the reaction mechanism.
$3$. Therefore,the statement that molecularity may be fractional is incorrect.
Thus,the correct option is $C$.

(A) The rate law for the reaction is given by $Rate = k[C_6H_5N_2^+Cl^-]^n$,where $n$ is the order of the reaction.
Given that when the initial concentration is doubled,the rate of evolution of $N_2$ (which is the rate of reaction) becomes two times faster.
Let the initial rate be $R_1 = k[C]^n$ and the new rate be $R_2 = k[2C]^n$.
According to the problem,$R_2 = 2R_1$.
Therefore,$k[2C]^n = 2 \times k[C]^n$.
$2^n = 2^1$.
Thus,$n = 1$.
Hence,it is a first order reaction.

(C) In the reaction $A + B \to \text{Products}$,the rate law is given by $Rate = k[A]^x[B]^y$.
When $B$ is taken in excess,its concentration remains effectively constant throughout the reaction.
Thus,the rate equation becomes $Rate = k'[A]^x$,where $k' = k[B]^y$.
This type of reaction,where a higher-order reaction appears to be first-order due to the excess of one reactant,is known as a pseudo-unimolecular reaction.

(C) For a $second$ order reaction,the rate law is $Rate = k[A]^2$.
The half-life period $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{1}{k[A]_0}$.
Therefore,$t_{1/2}$ is inversely proportional to the initial concentration of reactants $([A]_0)$.

(B) The reaction is $RCl + H_2O \to ROH + HCl$.
Since water is present in a large excess,its concentration remains effectively constant throughout the reaction.
Therefore,the rate of the reaction depends only on the concentration of $RCl$,making it a pseudo-first-order reaction,so the order is $1$.
Molecularity is defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary step. Here,both $RCl$ and $H_2O$ are involved in the collision,so the molecularity is $2$.

(B) The rate law for the reaction is given by $r = k[C_A]^{3/2} [C_B]^{-1/2}$.
The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
Order $= \frac{3}{2} + (-\frac{1}{2}) = \frac{3-1}{2} = \frac{2}{2} = 1$.
Therefore,the order of the reaction is $1$.

(C) For a chemical reaction with the rate expression $\text{Rate} = k[A]^m[B]^n$,the order of the reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
Therefore,the order of the reaction $= m + n$.

(D) For a second order reaction,the rate law is given by: $r = k[A]^2$.
When the concentration of $A$ is doubled,the new concentration becomes $[A]' = 2[A]$.
The new rate $r'$ is: $r' = k[2A]^2 = 4k[A]^2$.
Therefore,$r' = 4r$.
The rate of formation of $B$ increases by a factor of $4$.

(C) The rate law for the given reaction is determined experimentally as $r = k[FeCl_3]^2 [SnCl_2]^1$.
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,it is a third-order reaction.

(C) The rate law expression is given by: $\text{Rate} = k [A]^x [B]^y$.
Given that the reaction is first-order with respect to $A$ $(x = 1)$ and second-order with respect to $B$ $(y = 2)$.
Substituting these values into the rate law expression,we get: $\text{Rate} = k [A]^1 [B]^2$ or $\text{Rate} = k [A] [B]^2$.

(D) The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
It is an experimental quantity and can be zero,a whole number,a fraction,or even negative.

(A) The rate of the reaction is determined by the slowest step,which is step $(ii)$.
Rate $r = k[A][B_2]$.
Since $A$ is an intermediate,we use the equilibrium from the fast step $(i)$:
$K_{eq} = \frac{[A]^2}{[A_2]} \implies [A] = K_{eq}^{1/2} [A_2]^{1/2}$.
Substituting this into the rate expression:
$r = k \times K_{eq}^{1/2} [A_2]^{1/2} [B_2] = k'[A_2]^{1/2} [B_2]^1$.
The overall order of the reaction is the sum of the exponents of the concentration terms:
Order $= 0.5 + 1 = 1.5$.

(B) The rate expression for the reaction is given as: Rate $= k[A]^{1/2}[B]^{3/2}$.
For a rate law of the form Rate $= k[A]^x[B]^y$,the overall order of the reaction is the sum of the exponents of the concentration terms,which is $x + y$.
Here,$x = 1/2$ and $y = 3/2$.
Therefore,the order of the reaction $= 1/2 + 3/2 = 4/2 = 2$.

(C) The rate law for the reaction is given by: $\text{Rate} = k[A]^n$,where $n$ is the order of the reaction.
Let the initial concentration of $A$ be $[A]_1 = P$ and the initial rate be $r_1 = kP^n$.
When the concentration is increased by a factor of $1.5$,the new concentration is $[A]_2 = 1.5P$.
The new rate becomes $r_2 = 2.25 r_1 = k(1.5P)^n$.
Dividing the expression for $r_2$ by $r_1$ gives: $\frac{r_2}{r_1} = \frac{k(1.5P)^n}{kP^n} = (1.5)^n$.
Since $\frac{r_2}{r_1} = 2.25$,we have $(1.5)^n = 2.25$.
Since $(1.5)^2 = 2.25$,it follows that $n = 2$.
Therefore,the order of the reaction is $2$.

(D) The reactions that are bimolecular but follow first-order kinetics are known as $Pseudo-unimolecular$ reactions.
This occurs when one of the reactants is present in large excess,making its concentration effectively constant throughout the reaction.

(B) The rate law for the reaction is given as $\text{Rate} = k[B]^2$.
Since the concentration of $A$ is taken in excess,its concentration remains effectively constant throughout the reaction.
The rate of the reaction depends only on the concentration of $B$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law expression.
Here,the power of $[B]$ is $2$.
Therefore,the order of the reaction is $2$.

(A) The rate law for the reaction can be written as: $Rate = k[A]^x[B]^y$.
Given that doubling the concentration of $A$ doubles the rate,we have $2 = (2)^x$,which implies $x = 1$.
Given that doubling the concentration of $B$ does not change the rate,we have $1 = (2)^y$,which implies $y = 0$.
The overall order of the reaction is $x + y = 1 + 0 = 1$.

(B) The correct answer is $(B)$.
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.
Since it represents a count of particles,molecularity can never be zero or fractional; it must always be a whole number $(1, 2, 3, \dots)$.

(B) The order of a reaction is the sum of the powers of the concentration terms in the rate law expression.
Given rate expression: $\text{rate} = K[A]^{3/2}[B]^{-1}$.
Order of reaction $= \frac{3}{2} + (-1) = \frac{3}{2} - \frac{2}{2} = \frac{1}{2}$.
Therefore,the correct option is $B$.

(C) The rate law for the reaction can be expressed as: $\text{Rate} = k[NO]^x[Cl_2]^y$.
$1$. When $[Cl_2]$ is doubled and $[NO]$ is constant,the rate becomes $2$ times: $2 = (2)^y$,which implies $y = 1$.
$2$. When $[NO]$ is doubled and $[Cl_2]$ is constant,the rate becomes $4$ times: $4 = (2)^x$,which implies $x = 2$.
$3$. The overall order of the reaction is the sum of the exponents: $\text{Order} = x + y = 2 + 1 = 3$.

(D) For a second order reaction,the integrated rate equation is given by: $k = \frac{1}{t} \left( \frac{1}{[A]_t} - \frac{1}{[A]_0} \right)$.
Given: $k = 8 \times 10^{-5} \ M^{-1} \ min^{-1}$,$[A]_0 = 1 \ M$,and $[A]_t = 0.5 \ M$.
Substituting the values into the formula:
$8 \times 10^{-5} = \frac{1}{t} \left( \frac{1}{0.5} - \frac{1}{1} \right)$.
$8 \times 10^{-5} = \frac{1}{t} (2 - 1)$.
$8 \times 10^{-5} = \frac{1}{t}$.
$t = \frac{1}{8 \times 10^{-5}} = 0.125 \times 10^{5} \ min = 1.25 \times 10^{4} \ min$.

(D) The half-life period $(t_{1/2})$ for a reaction of order $n$ is given by the relation $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$.
For a second-order reaction $(n=2)$,$t_{1/2} = \frac{1}{k[A]_0}$.
Given the expression $t_{1/2} = \frac{1}{Ka}$,where $a$ represents the initial concentration $[A]_0$,the reaction is of second order.
Therefore,the correct option is $(D)$.

(C) The reaction $2NO + O_2 \to 2NO_2$ follows the rate law expression: $Rate = k[NO]^2[O_2]^1$.
The overall order of the reaction is the sum of the powers of the concentration terms in the rate law expression.
$\text{Order} = 2 + 1 = 3$.
Therefore,it is a third-order reaction.

(D) . 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. Therefore,it is determined by the relative concentration of reactants.

(C) The general unit for the rate constant $K$ of a reaction of order $n$ is $(mol \, L^{-1})^{1-n} \, s^{-1}$.
For a second-order reaction,$n = 2$.
Substituting $n = 2$ into the formula: $(mol \, L^{-1})^{1-2} \, s^{-1} = (mol \, L^{-1})^{-1} \, s^{-1} = mol^{-1} \, L \, s^{-1}$.
Comparing this with the given options,option $C$ has the unit $L \, mol^{-1} \, s^{-1}$,which is equivalent to $mol^{-1} \, L \, s^{-1}$.

(A) The rate law expression is given as $R = k[A]^1[B]^2$.
The order of a reaction is the sum of the powers of the concentration terms in the rate law expression.
Therefore,the order of the reaction $= 1 + 2 = 3$.

(A) The general unit for the rate constant of a reaction of order $n$ is $(mol \, L^{-1})^{1-n} \, sec^{-1}$.
For a first order reaction $(n = 1)$,the unit is $(mol \, L^{-1})^{1-1} \, sec^{-1} = sec^{-1}$.
For a zero order reaction $(n = 0)$,the unit is $(mol \, L^{-1})^{1-0} \, sec^{-1} = mol \, L^{-1} \, sec^{-1}$,which is equivalent to $M \, sec^{-1}$ since $M = mol \, L^{-1}$.
Therefore,the units are $sec^{-1}$ and $M \, sec^{-1}$ respectively.

(C) The decomposition of $N_2O_5$ follows first-order kinetics,where the rate law is given by $Rate = k[N_2O_5]$.
Since two molecules of $N_2O_5$ are involved in the stoichiometric equation,the reaction is bimolecular.
Therefore,the reaction is bimolecular and first order.

(D) For an elementary reaction,the molecularity is defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary step,which must collide simultaneously in order to bring about a chemical reaction.
Given the elementary reaction: $2A + B \to C + D$.
The number of reacting species is $2$ molecules of $A$ and $1$ molecule of $B$.
Therefore,the total molecularity $= 2 + 1 = 3$.

(C) For a zero-order reaction,the rate of reaction is independent of the concentration of the reacting species. Since the given reaction is photochemical $(hv)$,it follows zero-order kinetics where the rate depends on the intensity of light rather than the concentration of reactants.

(C) The rate law for the reaction is given by $R = K[A]^n$,where $n$ is the order of reaction.
Case $1$: When the concentration is increased $2$ times,the rate increases $4$ times.
$4R = K[2A]^n$
$4 = 2^n$
$2^2 = 2^n \Rightarrow n = 2$
Case $2$: When the concentration is increased $3$ times,the rate increases $9$ times.
$9R = K[3A]^n$
$9 = 3^n$
$3^2 = 3^n \Rightarrow n = 2$
Thus,the order of reaction is $2$.

(C) 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 in order to bring about a chemical reaction. Since it represents a count of particles,it must always be a whole number $(1, 2, 3, \dots)$. It can never be zero,negative,or a fraction.

(C) The rate law for the reaction is given by: $R = k[NO]^2[O_2]$.
When the volume is reduced to half,the concentration of each reactant doubles because $C = \frac{n}{V}$.
Let the new concentrations be $[NO]' = 2[NO]$ and $[O_2]' = 2[O_2]$.
The new rate $R'$ is: $R' = k(2[NO])^2(2[O_2])$.
$R' = k \times 4[NO]^2 \times 2[O_2] = 8 \times k[NO]^2[O_2]$.
Therefore,$R' = 8R$. The rate of reaction increases to eight times its initial value.

(D) The reaction $H_2 + Br_2 \rightarrow 2HBr$ follows a rate law of $Rate = k[H_2][Br_2]^{1/2}$.
Thus,the order of the reaction is $1 + 0.5 = 1.5$.
Therefore,the statement in option $D$ is wrongly matched.

(A) In a chemical reaction,if one of the reactants is present in large excess,the reaction follows first-order kinetics even if the molecularity is higher. This is known as a pseudo unimolecular reaction.
The acid-catalyzed hydrolysis of methyl acetate is a classic example:
$CH_3COOCH_3 + H_2O \xrightarrow{H^{+}} CH_3COOH + CH_3OH$
Here,water is present in large excess,so its concentration remains effectively constant during the reaction. The rate law is given by $Rate = k[CH_3COOCH_3]$. Thus,the reaction is of the first order,making it a pseudo unimolecular reaction.

(B) The relationship between half-life $(t_{1/2})$ and initial concentration $(a)$ for a reaction of order $n$ is given by:
$t_{1/2} \propto a^{1-n}$
Taking the ratio for two different concentrations:
$\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left(\frac{a_1}{a_2}\right)^{1-n}$
Substituting the given values:
$\frac{0.1}{0.4} = \left(\frac{200}{50}\right)^{1-n}$
$\frac{1}{4} = (4)^{1-n}$
$(4)^{-1} = (4)^{1-n}$
Comparing the powers: $-1 = 1 - n \Rightarrow n = 2$.
Thus,the reaction is of the second order.

(B) The rate law for the reaction is given by: $\text{Rate} = k[A]^n$,where $n$ is the order of reaction.
For the first condition: $2.4 = k(2.2)^n$ ... $(i)$
For the second condition,$[A] = 2.2/2 = 1.1 \, mM$ and $\text{Rate} = 0.6 \, mM \, s^{-1}$:
$0.6 = k(1.1)^n$ ... $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{2.4}{0.6} = \frac{k(2.2)^n}{k(1.1)^n}$
$4 = (2)^n$
$2^2 = 2^n$
Therefore,$n = 2$.
The order of reaction with respect to $A$ is $2$.

(D) The correct answer is $(D)$.
Order of a reaction is an experimental quantity and it can be zero,an integer,or even a fractional value.
Therefore,the statement that order cannot be fractional is incorrect.

(B) The rate law for the reaction with respect to reactant $B$ is given by $R = k[B]^n$,where $n$ is the order of reaction with respect to $B$.
According to the problem,if the concentration of $B$ is doubled $([B]' = 2[B])$,the rate becomes one-fourth of the original rate $(R' = \frac{1}{4}R)$.
Substituting these into the rate law:
$\frac{1}{4}R = k(2[B])^n$
Dividing the new rate equation by the original rate equation:
$\frac{\frac{1}{4}R}{R} = \frac{k(2[B])^n}{k[B]^n}$
$\frac{1}{4} = 2^n$
$2^{-2} = 2^n$
Therefore,$n = -2$.

(C) The rate law for the reaction $A + B \to C$ can be expressed as: $\text{Rate} = k[A]^x[B]^y$.
According to the problem,doubling the concentration of $A$ increases the rate by $4$ times $(2^x = 4)$,which implies $x = 2$.
Doubling the concentration of $B$ doubles the reaction rate $(2^y = 2)$,which implies $y = 1$.
The overall order of the reaction is the sum of the exponents: $x + y = 2 + 1 = 3$.

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

(D) For a reaction of order $n$,the half-life $(t_{1/2})$ is related to the initial concentration $([A]_0)$ by the expression: $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$.
Given that $t_{1/2} \propto \frac{1}{[A]_0^3}$,we can equate the exponents: $n - 1 = 3$.
Therefore,$n = 4$.
The order of the reaction is $4$.

(D) For parallel first-order reactions,the percentage distribution of products is determined by the ratio of their respective rate constants to the total rate constant.
$\% \text{ distribution of } B = \frac{k_1}{k_1 + k_2} \times 100$
Given $k_1 = 1.26 \times 10^{-4} \ s^{-1}$ and $k_2 = 0.38 \times 10^{-4} \ s^{-1}$:
$\% \text{ of } B = \frac{1.26 \times 10^{-4}}{1.26 \times 10^{-4} + 0.38 \times 10^{-4}} \times 100 = \frac{1.26}{1.64} \times 100 \approx 76.83\%$
$\% \text{ of } C = 100 - 76.83\% = 23.17\%$
Thus,the percentage distribution is $76.83\% \ B$ and $23.17\% \ C$.

(C) The rate law is $\frac{dx}{dt} = K[A]^n$.
Taking the logarithm on both sides: $\log \left( \frac{dx}{dt} \right) = n \log [A] + \log K$.
This equation is in the form of a straight line $y = mx + c$,where $y = \log \left( \frac{dx}{dt} \right)$,$x = \log [A]$,slope $m = n$,and intercept $c = \log K$.
Thus,by plotting $\log \left( \frac{dx}{dt} \right)$ versus $\log [A]$,the slope gives the order of reaction $(n)$ and the intercept gives the logarithm of the rate constant $(\log K)$.
Therefore,both the order of reaction and the rate constant can be determined.

(C) For a second-order reaction,the integrated rate equation is given by: $k = \frac{1}{t} \left( \frac{1}{[A]_t} - \frac{1}{[A]_0} \right)$.
Given: $k = 8 \times 10^{-5} \, M^{-1} \, \text{min}^{-1}$,$[A]_0 = 1 \, M$,and $[A]_t = 0.5 \, M$.
Substituting the values: $8 \times 10^{-5} = \frac{1}{t} \left( \frac{1}{0.5} - \frac{1}{1} \right)$.
$8 \times 10^{-5} = \frac{1}{t} (2 - 1)$.
$8 \times 10^{-5} = \frac{1}{t} \times 1$.
$t = \frac{1}{8 \times 10^{-5}} = \frac{10^5}{8} = 0.125 \times 10^5 = 1.25 \times 10^4 \, \text{min}$.