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

(B) The rate law for the first-order reaction is $r = k[H_2O_2]$.
Given: $k = 3 \times 10^{-3} \ min^{-1} = \frac{3 \times 10^{-3}}{60} \ s^{-1} = 5 \times 10^{-5} \ s^{-1}$.
Given: $r = 2 \times 10^{-4} \ M \ s^{-1}$.
Substituting the values in the rate law: $2 \times 10^{-4} = (5 \times 10^{-5}) \times [H_2O_2]$.
$[H_2O_2] = \frac{2 \times 10^{-4}}{5 \times 10^{-5}} = \frac{20 \times 10^{-5}}{5 \times 10^{-5}} = 4 \ M$.

(B) The rate of the reaction is determined by the slow step $(ii)$:
$r = K_3 [N_2O_2] [Cl_2]$
Since $N_2O_2$ is an intermediate,we express it in terms of reactants using the fast equilibrium step $(i)$:
$K_{eq} = \frac{K_1}{K_2} = \frac{[N_2O_2]}{[NO]^2}$
$[N_2O_2] = \frac{K_1}{K_2} [NO]^2$
Substituting this into the rate expression:
$r = K_3 \left( \frac{K_1}{K_2} [NO]^2 \right) [Cl_2]$
$r = \frac{K_1 K_3}{K_2} [NO]^2 [Cl_2]$

(D) reaction intermediate is a species that is produced in one step of a reaction mechanism and consumed in a subsequent step.
In step $(i)$,$Cl_{(g)}$ is produced.
In step $(ii)$,$Cl_{(g)}$ is consumed.
Therefore,$Cl_{(g)}$ is the reaction intermediate.

(A) The rate law for the reaction is given by: $\text{Rate} = k[A]^1[B]^2[C]^0$.
When the concentration of all reactants is doubled,the new concentrations are $[A]' = 2[A]$,$[B]' = 2[B]$,and $[C]' = 2[C]$.
The new rate $(\text{Rate})_1$ is: $(\text{Rate})_1 = k[2A]^1[2B]^2[2C]^0$.
$(\text{Rate})_1 = k \times 2[A] \times 4[B]^2 \times 1$.
$(\text{Rate})_1 = 8 \times k[A][B]^2$.
Therefore,the rate of reaction increases $8$ times.

(A) An elementary reaction is a reaction that occurs in a single step.
Option $A$ represents the decomposition of ethyl iodide,which is a unimolecular elementary reaction.
Options $B$,$C$,and $D$ are complex reactions that occur through multiple steps.

(A) The rate law is given by $r = k[A]^2[B]$.
Given values are $k = 6.25 \ mol^{-2} \ dm^6 \ s^{-1}$,$[A] = 1 \ mol \ dm^{-3}$,and $[B] = 0.2 \ mol \ dm^{-3}$.
Substituting these values into the rate law equation:
$r = 6.25 \times (1)^2 \times (0.2)$
$r = 6.25 \times 0.2 = 1.250 \ mol \ dm^{-3} \ s^{-1}$.

(A) The initial rate of reaction is given by $r_1 = k[A][B]^2$.
When the concentration of $A$ is doubled,the new concentration is $[A]' = 2[A]$.
The new rate of reaction is $r_2 = k[2A][B]^2 = 2 \times k[A][B]^2$.
Therefore,$r_2 = 2 \times r_1$.
Thus,the rate of reaction increases by a factor of $2$.

(A) The rate law is given as $\text{rate} = k[A][B]^2$.
Given values are $k = 6.25 \ mol^{-2} \ dm^6 \ s^{-1}$,$[A] = 1 \ M$,and $[B] = 0.2 \ M$.
Substituting these values into the rate equation:
$\text{Rate} = 6.25 \times (1) \times (0.2)^2$
$\text{Rate} = 6.25 \times 1 \times 0.04$
$\text{Rate} = 0.25 \ mol \ dm^{-3} \ s^{-1}$.

(B) The given rate law is: $\text{Rate} = k[CH_3Br][OH^{-}]$
If the concentration of both reactants is doubled,the new concentrations become $2[CH_3Br]$ and $2[OH^{-}]$.
New rate $(\text{Rate})_{1} = k \times (2[CH_3Br]) \times (2[OH^{-}])$
$(\text{Rate})_{1} = 4 \times k[CH_3Br][OH^{-}]$
$(\text{Rate})_{1} = 4 \times \text{Rate}$
Therefore,the rate of reaction increases by a factor of $4$.

(C) For a second order reaction,the rate law is given by $R = k[A]^2$.
Given:
$k = 1.62 \ M^{-1} \ s^{-1}$
$[A] = 2 \times 10^{-3} \ M$
Substituting the values into the rate law:
$R = 1.62 \times (2 \times 10^{-3})^2$
$R = 1.62 \times 4 \times 10^{-6}$
$R = 6.48 \times 10^{-6} \ M \ s^{-1}$

(A) The given reaction is $2 \ NOBr_{(g)} \rightarrow 2 \ NO_{(g)} + Br_{2_{(g)}}$.
Given: $k = 1.62 \ M^{-1} \ s^{-1}$ and $[NOBr] = 2.00 \times 10^{-3} \ M$.
The rate law is $r = k[NOBr]^{2}$.
Substituting the values:
$r = 1.62 \ M^{-1} \ s^{-1} \times (2.00 \times 10^{-3} \ M)^{2}$
$r = 1.62 \times 4.00 \times 10^{-6} \ M \ s^{-1}$
$r = 6.48 \times 10^{-6} \ M \ s^{-1}$.

(B) The initial rate of the reaction is given by: $\text{Rate}_1 = k[A]^2[B]$.
When the concentration of $A$ is doubled,the new concentration is $[A]' = 2[A]$.
The new rate of the reaction is: $\text{Rate}_2 = k[2A]^2[B]$.
Simplifying the expression: $\text{Rate}_2 = k \times 4[A]^2[B] = 4 \times \text{Rate}_1$.
Therefore,the rate of reaction increases by a factor of $4$.

(A) The reaction for the inversion of cane sugar is: $C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Fructose})$.
In this reaction,$H_2O$ is present in large excess,so its concentration remains effectively constant during the reaction.
Therefore,the rate of the reaction depends only on the concentration of sucrose,making it a pseudo-first order reaction.

(A) The rate of a chemical reaction depends upon the slowest step in a multistep mechanism,which is known as the rate-determining step.
Therefore,the overall rate of a multistep reaction is equal to the rate of the slowest step.

(A) The rate law is given as $r = k[C_2H_5I]^1$. The exponent of the concentration term in the rate law is $1$,so the order of the reaction is $1$.
Since the reaction involves the decomposition of a single molecule of $C_2H_5I$ in the elementary step,the molecularity is $1$.
Therefore,both the order and molecularity of this reaction are $1$.

(B) The given reaction is $2NO_{2(g)} \longrightarrow 2NO_{(g)} + O_{2(g)}$.
Molecularity is the number of reacting species taking part in an elementary reaction,which must collide simultaneously in order to bring about a chemical reaction.
Here,$2$ molecules of $NO_2$ are involved in the elementary step,so the molecularity is $2$.
The rate law for this elementary reaction is $Rate = k[NO_2]^2$.
Since the power of the concentration term in the rate law is $2$,the order of the reaction is $2$.
Therefore,the order is $2$ and the molecularity is $2$.

(C) Given the rate law: $r_1 = k[A]^x$ and $r_2 = k[10A]^x$.
According to the problem,$r_2 = 100r_1$.
Substituting the expressions: $k[10A]^x = 100 \times k[A]^x$.
This simplifies to: $10^x = 100$.
Since $100 = 10^2$,we have $10^x = 10^2$.
Therefore,$x = 2$.
The order of the reaction is $2$.

(C) For an elementary reaction,the order is equal to the sum of the stoichiometric coefficients of the reactants in the rate law expression.
Given rate law: $r = k[O_3]^1[O]^1$.
Order = $1 + 1 = 2$.
Molecularity is the number of reacting species (atoms,ions,or molecules) taking part in an elementary reaction.
In the reaction $O_{3(g)} + O_{(g)} \rightarrow 2O_{2(g)}$,there are two reactant species ($O_3$ and $O$).
Therefore,molecularity = $2$.
Thus,the order is $2^{nd}$ and the molecularity is $2$.

(C) The rate law is given by $r = k[A][B]$.
For option $A$: $r' = k[2A][B] = 2r$ (Rate changes).
For option $B$: $r' = k[A][2B] = 2r$ (Rate changes).
For option $C$: $r' = k[A/2][2B] = k[A][B] = r$ (Rate remains unchanged).
For option $D$: $r' = k[A][B/2] = 0.5r$ (Rate changes).
Therefore,the condition that does not affect the rate of reaction is $C$.

(B) The given rate law is $r = k[A][B]$.
Let the initial rate be $r_1 = k[A][B]$.
For option $A$: $r_2 = k(2[A])(2[B]) = 4k[A][B] = 4r_1$.
For option $B$: $r_2 = k(2[A])[B] = 2k[A][B] = 2r_1$.
For option $C$: $r_2 = k(0.5[A])(2[B]) = k[A][B] = r_1$.
For option $D$: $r_2 = k[A](0.5[B]) = 0.5k[A][B] = 0.5r_1$.
Thus,the rate of reaction doubles when the concentration of $A$ is doubled and the concentration of $B$ is kept constant.

(D) 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 law: $r = k[A]^{1.5}[B]^{2.5}$.
Order of reaction = $1.5 + 2.5 = 4$.
Therefore,the correct option is $D$.

(A) The initial rate is given by $r_1 = k[A]^a[B]^b$.
When the concentration of $A$ is doubled $([A]' = 2[A])$ and $B$ is halved $([B]' = \frac{1}{2}[B])$,the new rate $r_2$ is:
$r_2 = k(2[A])^a(\frac{1}{2}[B])^b$
$r_2 = k \cdot 2^a \cdot [A]^a \cdot (2^{-1})^b \cdot [B]^b$
$r_2 = 2^a \cdot 2^{-b} \cdot k[A]^a[B]^b$
$r_2 = 2^{a-b} \cdot r_1$
Therefore,the ratio $\frac{r_2}{r_1} = 2^{a-b}$.

(A) The rate law is given by the expression $R = k[NO_2]^2[CO]^0$.
Comparing this with the general rate law $R = k[A]^x[B]^y$,we can see that the exponent of the concentration of $CO$ is $0$.
Therefore,the order of the reaction with respect to $CO$ is $0$.

(A) The given rate law is $r = k[A]^2[B]$.
Given values are $k = 6.25 \ M^{-2} \ s^{-1}$,$[A] = 1 \ M$,and $[B] = 0.2 \ M$.
Substituting these values into the rate equation:
$r = 6.25 \times (1)^2 \times (0.2)$
$r = 6.25 \times 1 \times 0.2$
$r = 1.25 \ M \ s^{-1}$.
Therefore,the correct option is $A$.

(B) The unit of the rate constant for a reaction of order $n$ is given by $(\text{concentration})^{1-n} \times (\text{time})^{-1}$.
For a first-order reaction $(n=1)$,the unit is $(\text{time})^{-1}$.
Given the unit of the rate constant is $hour^{-1}$,which matches the unit for a first-order reaction.
Therefore,the order of the reaction is $1$.

(B) The rate law for the reaction $H_{2(g)} + I_{2(g)} \longrightarrow 2 HI_{(g)}$ is given by $Rate = k[H_2][I_2]$.
Since the sum of the powers of the concentration terms in the rate law is $1 + 1 = 2$,this reaction is a second-order reaction.
Option $B$ is the correct answer.

(B) Given rate of reaction $r = 3.6 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.
Rate law is $r = k[A][B]^2$.
Given concentrations are $[A] = 0.2 \ M$ and $[B] = 0.1 \ M$.
Substituting the values in the rate law:
$3.6 \times 10^{-2} = k \times (0.2) \times (0.1)^2$
$3.6 \times 10^{-2} = k \times (0.2) \times (0.01)$
$3.6 \times 10^{-2} = k \times (0.002)$
$k = \frac{3.6 \times 10^{-2}}{2 \times 10^{-3}}$
$k = 1.8 \times 10^1 = 18 \ mol^{-2} \ dm^6 \ s^{-1}$.
Thus,the correct option is $B$.

(C) The given reaction is $2 NO_{2(g)} \longrightarrow 2 NO_{(g)} + O_{2(g)}$.
This reaction is a well-known example of a second-order reaction at temperatures below $500 \ K$.
The rate law for this reaction is expressed as: $\text{Rate} = k[NO_2]^2$.
Since the exponent of the concentration term $[NO_2]$ is $2$,the order of the reaction is $2$.

(A) For a complex reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.
In the given mechanism,the first step is the slow step:
$NO_2Cl_{(g)} \longrightarrow NO_{2(g)} + Cl_{(g)}$ (slow)
Therefore,the rate of the reaction depends only on the concentration of the reactant involved in this slow step.
The rate law is given by: $r = k[NO_2Cl]$.

(C) The unit of the rate constant $(k)$ for a reaction of order $n$ is given by $(mol \ L^{-1})^{1-n} \,s^{-1}$.
For a first-order reaction $(n=1)$, the unit is $(mol \ L^{-1})^{1-1} \,s^{-1} = s^{-1}$.
Since the given unit of the rate constant is $s^{-1}$, the reaction is a first-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.
It is an experimental quantity and is not necessarily related to the stoichiometric coefficients of the balanced chemical equation.
Crucially,the order of a reaction can be zero,a fraction,or a whole number.
Therefore,the statement that it is always a whole number is incorrect.

(B) The given reaction is $2 N_2O_{5(g)} \longrightarrow 4 NO_{2(g)} + O_{2(g)}$.
Since the unit of the rate constant is $s^{-1}$,it is a first-order reaction.
The rate law expression is given by: $\text{Rate} = k[N_2O_5]$.
Substituting the given values: $1.02 \times 10^{-4} = 3.4 \times 10^{-5} \times [N_2O_5]$.
Therefore,$[N_2O_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}} = 3.0 \ mol \ L^{-1}$.

(B) The rate law expression is defined as $r = k [A]^x [B]^y$.
Given that the rate is proportional to the square of $[NO_2]$,the order with respect to $NO_2$ is $2$ $(x = 2)$.
Given that the rate is independent of $[CO]$,the order with respect to $CO$ is $0$ $(y = 0)$.
Therefore,the rate law equation is $r = k [NO_2]^2 [CO]^0$.

(B) The decomposition of hydrogen peroxide $(H_2O_2)$ is a well-known example of a first-order reaction.
The rate law for this reaction is given by: $\text{Rate} = k[H_2O_2]^1$.
Therefore,the order of the reaction is $1$.

(C) 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 experimentally determined quantity and can be an integer,a fraction,or zero. It is $NOT$ a theoretical quantity,as it cannot be determined simply from the stoichiometric coefficients of a balanced chemical equation.

(A) The correct answer is $(A)$.
In a multistep reaction,the overall rate of the reaction is determined by the rate of the slowest step,which is known as the rate-determining step.
This step acts as a bottleneck,limiting the overall speed at which the entire reaction proceeds.
Even if other steps are faster,the overall rate cannot exceed the rate of the slowest step.

(B) The rate law is given by: $\text{rate} = k[A]^2[B]$.
Substituting the given values: $0.22 \ mol \ L^{-1} \ s^{-1} = k \times (1 \ mol \ L^{-1})^2 \times (0.25 \ mol \ L^{-1})$.
$0.22 = k \times 1 \times 0.25$.
$k = \frac{0.22}{0.25} = 0.88 \ mol^{-2} \ L^2 \ s^{-1}$.

(D) The rate of reaction is given by the expression: $Rate = k[A]^x[B]^y$.
Here,$k$ is the rate constant.
$1$. The rate constant $k$ is independent of the concentration of reactants.
$2$. It varies with temperature according to the Arrhenius equation.
$3$. When the concentration of each reactant is unity,the rate of reaction is equal to the rate constant $(Rate = k)$.
$4$. The rate of reaction is directly proportional to the product of concentrations of reacting species,not the rate constant itself.
Therefore,statement $D$ is incorrect.

(B) The given rate law is $\text{rate} = k[NO]^2[H_2]$.
From the rate law expression,the exponent of $[H_2]$ is $1$,so the order of reaction with respect to $H_2$ is $1$.
The overall order of reaction is the sum of the exponents of the concentration terms in the rate law.
$\text{Overall order} = 2 + 1 = 3$.
Therefore,the order with respect to $H_2$ is $1$ and the overall order is $3$.

(C) The rate law for the reaction is given by: $Rate = k[A]^1[B]^2$
Rearranging to solve for the rate constant $k$: $k = \frac{Rate}{[A][B]^2}$
Substituting the given values: $k = \frac{3.6 \times 10^{-2} \ mol \ dm^{-3} \ sec^{-1}}{(0.2 \ mol \ dm^{-3})(0.1 \ mol \ dm^{-3})^2}$
$k = \frac{3.6 \times 10^{-2}}{0.2 \times 0.01} = \frac{3.6 \times 10^{-2}}{0.002} = 18 \ mol^{-2} \ dm^6 \ sec^{-1}$

(C) The rate law for the reaction is given by: $Rate = k[A]^1[B]^2$
Given: $Rate = 7.2 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$,$[A] = 0.4 \ mol \ dm^{-3}$,and $[B] = 0.1 \ mol \ dm^{-3}$.
Rearranging the rate law to solve for the rate constant $k$: $k = \frac{Rate}{[A][B]^2}$
Substituting the values: $k = \frac{7.2 \times 10^{-2}}{(0.4)(0.1)^2} = \frac{7.2 \times 10^{-2}}{0.4 \times 0.01} = \frac{7.2 \times 10^{-2}}{4 \times 10^{-3}} = 1.8 \times 10^1 = 18 \ mol^{-2} \ dm^6 \ s^{-1}$.

(D) Molecularity is the number of reacting species taking part in an elementary reaction,which must collide simultaneously in order to bring about a chemical reaction. For the given reaction,there are $2$ reactant molecules ($1$ molecule of $H_2$ and $1$ molecule of $Br_2$),so the molecularity is $2$ (Bimolecular).
Order of reaction is the sum of the powers of the concentration terms in the rate law expression.
Given rate law: $r = k[H_2]^1[Br_2]^{\frac{1}{2}}$.
Order $= 1 + \frac{1}{2} = \frac{3}{2}$.

(C) The rate law for the reaction is given by: $\text{Rate} = k[A]^2[B]$
Rearranging to solve for the rate constant $k$: $k = \frac{\text{Rate}}{[A]^2[B]}$
Substituting the given values: $k = \frac{1.8 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}}{(0.2 \ mol \ dm^{-3})^2 \times (0.1 \ mol \ dm^{-3})}$
$k = \frac{1.8 \times 10^{-2}}{0.04 \times 0.1} = \frac{1.8 \times 10^{-2}}{4 \times 10^{-3}} = 4.5 \ mol^{-2} \ dm^6 \ s^{-1}$

(A) The rate law expression is determined by the order of the reaction with respect to each reactant.
Given that the reaction is second order in $NO$ and first order in $Cl_2$,the rate law is expressed as:
Rate $= k[NO]^2 [Cl_2]^1 = k[NO]^2 [Cl_2]$

(A) The given rate law is $\text{rate} = k[A]^2[B]$.
Rearranging for the rate constant $k$,we get $k = \frac{\text{rate}}{[A]^2[B]}$.
Substituting the given values: $k = \frac{0.24 \ mol \ dm^{-3} \ s^{-1}}{(0.5 \ mol \ dm^{-3})^2 \times (0.2 \ mol \ dm^{-3})}$.
$k = \frac{0.24}{0.25 \times 0.2} = \frac{0.24}{0.05} = 4.8 \ mol^{-2} \ dm^6 \ s^{-1}$.

(D) The rate constant $k$ is a characteristic constant for a given reaction at a specific temperature.
$(a)$ It is independent of the concentration of reactants.
$(b)$ It varies with temperature according to the Arrhenius equation.
$(c)$ If the concentration of all reactants is unity,the rate of reaction equals the rate constant.
$(d)$ The unit of the rate constant depends on the order of the reaction $(n)$ and is given by $(mol \ L^{-1})^{1-n} \ s^{-1}$. Therefore,statement $(d)$ is $NOT$ true.

(B) The rate law for the reaction is given by: $Rate = k[A]^1[B]^0 = k[A]$.
Given that $Rate = 6 \times 10^{-4} \ mol \ dm^{-3} \ s^{-1}$ and $[A] = 0.3 \ M$ (or $0.3 \ mol \ dm^{-3}$).
Substituting these values into the rate law equation:
$k = \frac{Rate}{[A]} = \frac{6 \times 10^{-4} \ mol \ dm^{-3} \ s^{-1}}{0.3 \ mol \ dm^{-3}} = 2 \times 10^{-3} \ s^{-1}$.
Thus,the correct option is $B$.

(A) The rate law expression for a reaction is given by: $\text{Rate} = k[A]^x[B]^y$,where $x$ and $y$ are the orders of reaction with respect to reactants $A$ and $B$ respectively.
Given the reaction: $CHCl_{3(g)} + Cl_{2(g)} \rightarrow CCl_{4(g)} + HCl_{(g)}$.
The order with respect to $CHCl_{3(g)}$ is $1$.
The order with respect to $Cl_{2(g)}$ is $1/2$.
Substituting these values into the rate law expression,we get: $\text{Rate} = k[CHCl_3]^1[Cl_2]^{1/2}$ or $\text{Rate} = k[CHCl_3][Cl_2]^{1/2}$.

(B) The initial rate is given by $\text{Rate} = k[A][B]$.
To increase the rate by a factor of $2$,the new rate $(\text{Rate})_1$ must be $2 \times \text{Rate}$.
If $[A]$ is doubled and $[B]$ is kept constant,the new rate is $(\text{Rate})_1 = k(2[A])[B] = 2k[A][B] = 2 \times \text{Rate}$.
Thus,the rate increases by a factor of $2$ when $[A]$ is doubled and $[B]$ is kept constant.

(C) The rate law is given by $R = k[x][y]$.
If the concentration of $x$ is doubled and the concentration of $y$ is kept constant,the new rate $R'$ becomes:
$R' = k[2x][y] = 2k[x][y] = 2R$.
Thus,the rate of reaction doubles.