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

(B) The rate law for the reaction is given by $r = k[x]^1[y]^2$.
Substituting the given values: $5.4 \times 10^{-2} = k(0.2)(0.1)^2$.
$5.4 \times 10^{-2} = k(0.2)(0.01)$.
$5.4 \times 10^{-2} = k(0.002)$.
$k = \frac{5.4 \times 10^{-2}}{2 \times 10^{-3}} = 27 \ mol^{-2} \ dm^6 \ sec^{-1}$.

(B) The rate law for the reaction is given by $r = k[A]^1[B]^1$.
Given: $r = 15 \times 10^{-2} \ mol \ dm^{-3} \ sec^{-1}$,$[A] = 0.3 \ mol \ dm^{-3}$,and $[B] = 0.05 \ mol \ dm^{-3}$.
Substituting the values into the rate law: $15 \times 10^{-2} = k \times (0.3) \times (0.05)$.
$15 \times 10^{-2} = k \times (0.015)$.
$k = \frac{15 \times 10^{-2}}{15 \times 10^{-3}} = 10 \ mol^{-1} \ dm^3 \ sec^{-1}$.

(A) The rate law for the reaction is given by $Rate = k[X]^1[Y]^0 = k[X]$.
Given that $Rate = 1.2 \times 10^{-4} \ mol \ dm^{-3} \ sec^{-1}$ and $[X] = 0.6 \ mol \ dm^{-3}$.
Substituting these values into the rate law: $1.2 \times 10^{-4} = k(0.6)$.
Solving for $k$: $k = \frac{1.2 \times 10^{-4}}{0.6} = 2 \times 10^{-4} \ sec^{-1}$.

(C) The initial rate law is given by $r = K[A][B][C]^2$.
If the concentration of $A$ is halved,the new concentration becomes $[A]' = \frac{[A]}{2}$.
The new rate of reaction $r'$ is given by $r' = K[A]'][B][C]^2 = K(\frac{[A]}{2})[B][C]^2$.
Substituting the initial rate expression,we get $r' = \frac{1}{2} K[A][B][C]^2 = \frac{1}{2} r$.
Therefore,the rate of reaction decreases $\frac{1}{2}$ times.

(D) The general unit of the rate constant $k$ for a reaction of order $n$ is given by the formula: $k = (mol \ L^{-1})^{1-n} s^{-1}$.
For the unit to be $s^{-1}$,the exponent of the concentration term $(mol \ L^{-1})$ must be $0$.
Therefore,$1 - n = 0$,which implies $n = 1$.
Thus,the reaction is a first-order reaction.

(C) The rate law is given by $r = k[A]^2[B]^0$.
Since the order of the reaction with respect to $B$ is $0$,the rate of the reaction does not depend on the concentration of $B$.
Therefore,the rate of reaction is independent of the concentration of $B$.

(D) The unit of the rate constant $(k)$ is given by the formula $M^{(1-n)} \cdot sec^{-1}$,where $n$ is the order of the reaction.
Given that the rate constant $k = x \ sec^{-1}$,we can compare the units.
$M^{(1-n)} \cdot sec^{-1} = M^0 \cdot sec^{-1}$.
Equating the exponents of $M$,we get $1 - n = 0$.
Therefore,$n = 1$.

(B) The given rate law is $r=k[H_2][I_2]$.
$1$. The order of reaction with respect to $H_2$ is $1$.
$2$. The order of reaction with respect to $I_2$ is $1$.
$3$. The overall order of reaction is the sum of the powers of the concentration terms in the rate law,which is $1 + 1 = 2$.
Therefore,the statement that the overall order of reaction is $1$ is $NOT$ true.

(B) The rate law for the reaction is given by:
$r = k[A]^2 [B]^2$
Substituting the given values:
$3.6 \times 10^{-2} = k(0.2)^2 (0.1)^2$
$3.6 \times 10^{-2} = k(0.04)(0.01)$
$3.6 \times 10^{-2} = k(4 \times 10^{-4})$
$k = \frac{3.6 \times 10^{-2}}{4 \times 10^{-4}} = 0.9 \times 10^2 = 90 \ mol^{-3} \ dm^9 \ sec^{-1}$

(B) The decomposition reaction of hydrogen peroxide is given by: $H_2O_2 \longrightarrow H_2O + \frac{1}{2} O_2$.
Since the reaction is of the first order,the rate of reaction depends on the first power of the concentration of the reactant.
Therefore,the rate law equation is expressed as: $r = k[H_2O_2]^1$ or $r = k[H_2O_2]$.

(B) The initial rate law is given by $r = k[A][B][C]^2$.
When the concentrations of $A$ and $B$ are doubled,the new concentrations become $[A]' = 2[A]$ and $[B]' = 2[B]$.
The new rate $r_{\text{new}}$ is given by $r_{\text{new}} = k[2A][2B][C]^2$.
Simplifying this,we get $r_{\text{new}} = 4 \times k[A][B][C]^2$.
Since $r = k[A][B][C]^2$,we have $r_{\text{new}} = 4r$.

(C) The given rate law is $\text{rate} = k[NO]^2[O_2]$.
In this expression,the exponent of $[NO]$ is $2$,which means the reaction is second order with respect to $NO$.
The exponent of $[O_2]$ is $1$,which means the reaction is first order with respect to $O_2$.
The overall order of the reaction is the sum of the exponents of the concentration terms in the rate law: $2 + 1 = 3$.
Therefore,the reaction is second order in $NO$,first order in $O_2$,and third order overall.

(A) The overall order of a reaction is defined as the sum of the exponents of the concentration terms in the rate law expression.
Given the rate law: $\text{Rate} = k[A]^{2}[B]^{1}[C]^{0}$.
The exponents are $2, 1, \text{ and } 0$.
Overall order $= 2 + 1 + 0 = 3$.

(C) The decomposition of gaseous acetaldehyde is represented by the equation: $CH_{3}CHO_{(g)} \longrightarrow CH_{4_{(g)}} + CO_{(g)}$
According to the experimental rate law for this reaction,the rate is given by: $\text{Rate} = k[CH_{3}CHO]^{3/2}$
The order of reaction is the sum of the powers of the concentration terms in the rate law expression.
Therefore,the order of the reaction is $1.5$.

(D)
Given rate law: $\text{rate}_{1} = k[A]^{x}$
When concentration is increased by $10$ times,the new rate is $\text{rate}_{2} = 100 \times \text{rate}_{1}$
So,$k[10A]^{x} = 100 \times k[A]^{x}$
Dividing both sides by $k[A]^{x}$:
$10^{x} = 100$
$10^{x} = 10^{2}$
Therefore,$x = 2$
The order of reaction is $2$.

(A) The given reaction is: $O_{3(g)} + O_{(g)} \longrightarrow 2O_{2(g)}$
$i$. The rate law is given as: $\text{rate} = k[O_3]^1[O]^1$.
$ii$. The order of the reaction is the sum of the powers of the concentration terms in the rate law: $1 + 1 = 2$.
$iii$. The molecularity is the number of reactant species taking part in the elementary step: $1 \text{ molecule of } O_3 + 1 \text{ atom of } O = 2$.
Therefore,the molecularity is $2$ and the order is $2$.

(D) Let the rate law be $r = k[A]^{x}[B]^{y}$.
Given that when $[A]$ is doubled at constant $[B]$,the rate increases by a factor of $4$:
$4r = k[2A]^{x}[B]^{y}$ $\Rightarrow 4 = 2^{x}$ $\Rightarrow x = 2$.
Given that when $[B]$ is doubled at constant $[A]$,the rate is doubled:
$2r = k[A]^{x}[2B]^{y}$ $\Rightarrow 2 = 2^{y}$ $\Rightarrow y = 1$.
Substituting the values of $x$ and $y$ into the rate law expression,we get $r = k[A]^{2}[B]^{1}$.

(B) The rate law expression is given as: $\text{Rate} = \frac{+d[C]}{dt} = k[A]^1[B]^1$.
The order of reaction with respect to $A$ is $1$.
The order of reaction with respect to $B$ is $1$.
The total order of the reaction is the sum of the exponents of the concentration terms in the rate law: $1 + 1 = 2$.

(B) The rate law is given as $\text{rate} = k[A][B]$.
Rearranging the formula to solve for $[A]$,we get $[A] = \frac{\text{rate}}{k[B]}$.
Substituting the given values: $[A] = \frac{0.25 \ mol \ dm^{-3} \ s^{-1}}{6.25 \ mol^{-1} \ dm^3 \ s^{-1} \times 0.25 \ mol \ dm^{-3}}$.
$[A] = \frac{0.25}{6.25 \times 0.25} \ mol \ dm^{-3} = \frac{1}{6.25} \ mol \ dm^{-3} = 0.16 \ mol \ dm^{-3}$.

(D) The decomposition of hydrogen peroxide $(H_2 O_2)$ is a well-studied chemical reaction.
Experimentally,the rate of decomposition of $H_2 O_2$ is found to be proportional to the square of the concentration of $H_2 O_2$.
The rate law is given by: $\text{Rate} = k [H_2 O_2]^2$.
Since the exponent of the concentration term in the rate law is $2$,the order of the reaction is $2$.

(C) Let the rate law be $Rate = k[A]^x[B]^y$.
Given that when $[A]$ is doubled keeping $[B]$ constant,the rate doubles.
$R_1 = k[A]^x[B]^y$ ... $(i)$
$2R_1 = k[2A]^x[B]^y$ ... $(ii)$
Dividing $(ii)$ by $(i)$:
$\frac{2R_1}{R_1} = \frac{k[2A]^x[B]^y}{k[A]^x[B]^y}$
$2 = 2^x$
$x = 1$
Therefore,the order of reaction with respect to $A$ is $1$.

(B) The rate law for the reaction is given by: $r = k[A]^1[B]^2$.
If the concentration of $B$ is increased $3$ times,the new concentration becomes $[B'] = 3[B]$.
The new rate is $r_{new} = k[A][3B]^2$.
$r_{new} = k[A] \times 9[B]^2 = 9 \times (k[A][B]^2)$.
$r_{new} = 9r$.
Therefore,the reaction rate increases $9$ times.

(D) reaction intermediate is a substance that is produced in one step of a reaction mechanism and consumed in a subsequent step.
In the given mechanism:
Step $i$: $2 ClO^{-} \rightarrow ClO_2^{-} + Cl^{-}$
Step $ii$: $ClO_2^{-} + ClO^{-} \rightarrow ClO_3^{-} + Cl^{-}$
The species $ClO_2^{-}$ is produced in step $i$ and consumed in step $ii$.
Therefore,$ClO_2^{-}$ is the reaction intermediate.

(A) Reaction intermediates are substances that are produced in one step of a reaction mechanism and consumed in a subsequent step.
In the given mechanism,$F_{(g)}$ is produced in step $(I)$ and consumed in step $(II)$.
Therefore,$F_{(g)}$ is the reaction intermediate.

(B) reaction intermediate is a substance that is produced in one step of a reaction mechanism and consumed in a subsequent step.
In the given mechanism:
Step $(i)$: $2 SO_{2(g)} + 2 NO_{2(g)} \rightarrow 2 SO_{3(g)} + 2 NO_{(g)}$
Step $(ii)$: $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$
Here,$NO_{(g)}$ is produced in step $(i)$ and consumed in step $(ii)$.
$NO_{2(g)}$ acts as a catalyst because it is consumed in step $(i)$ and regenerated in step $(ii)$.
Therefore,$NO_{(g)}$ is the reaction intermediate.

(D) The unit of the rate constant $K$ for a reaction of order $n$ is given by the formula: $(mol \ L^{-1})^{1-n} \ S^{-1}$.
Given the unit is $mol^{-3/2} \ L^{3/2} \ S^{-1}$,we can equate the powers:
$(mol \ L^{-1})^{1-n} = mol^{-3/2} \ L^{3/2}$.
This implies $1 - n = -3/2$.
Solving for $n$: $n = 1 + 3/2 = 5/2 = 2.5$.
Therefore,the order of the reaction is $2.5$.

(A) The rate law for the reaction is given by: $Rate = k[A]^1[B]^2$.
If the concentration of $B$ is increased two times,the new concentration becomes $[B'] = 2[B]$.
The new rate $Rate'$ will be: $Rate' = k[A]^1[2B]^2 = 4 \times k[A]^1[B]^2$.
Therefore,the rate of reaction increases by $4-$Times.
Hence,the correct option is $A$.

(C) The overall order of the reaction is $n = \frac{1}{2} + \frac{3}{2} = 2$.
The unit of the rate constant for a reaction of order $n$ is given by the formula $(Mol \cdot L^{-1})^{1-n} \cdot Sec^{-1}$.
Substituting $n = 2$ into the formula,we get $(Mol \cdot L^{-1})^{1-2} \cdot Sec^{-1} = (Mol \cdot L^{-1})^{-1} \cdot Sec^{-1} = Mol^{-1} \cdot L \cdot Sec^{-1}$.

(A) The rate law for the reaction is given by $r_1 = k[A]^1[B]^2$.
When the concentrations of both $A$ and $B$ are doubled,the new concentrations become $[A]' = 2[A]$ and $[B]' = 2[B]$.
The new rate $r_2$ is calculated as:
$r_2 = k[2A]^1[2B]^2$
$r_2 = k \times 2[A] \times 4[B]^2$
$r_2 = 8 \times k[A][B]^2$
$r_2 = 8r_1$.
Therefore,the rate increases by $8$ times.

(B) 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 second-order reaction $(n = 2)$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.
Since the given unit is $L \ mol^{-1} \ s^{-1}$,the reaction is of the second order.

(B) For a reaction $A \rightarrow B$,the rate law is given by: $\text{Rate} = k[A]^n$,where $n$ is the order of reaction.
Given that when the concentration $[A]$ is increased by $9$ times,the rate increases by $3$ times.
So,$3 \times \text{Rate} = k(9[A])^n$.
Dividing this by the original rate equation: $\frac{3 \times \text{Rate}}{\text{Rate}} = \frac{k(9[A])^n}{k[A]^n}$.
$3 = 9^n$.
Since $9 = 3^2$,we have $3 = (3^2)^n = 3^{2n}$.
Comparing the exponents: $1 = 2n$,which gives $n = 1/2$.
Therefore,the order of reaction is $1/2$.

(B)
Unit of rate constant $(k) = (\text{mol L}^{-1})^{1-n} \text{s}^{-1}$
Unit of rate of reaction $= \text{mol L}^{-1} \text{s}^{-1}$
Given that the units are the same,we equate them:
$(\text{mol L}^{-1})^{1-n} \text{s}^{-1} = \text{mol L}^{-1} \text{s}^{-1}$
This implies $1-n = 1$,which gives $n = 0$.
Therefore,the order of the reaction is $0$.

(B) The unit of the rate constant is $L^2 \ mol^{-2} \ sec^{-1}$,which corresponds to a $3^{rd}$ order reaction $(n = 3)$.
For a reaction of order $n$,the half-life period is given by $t_{1/2} \propto [R_0]^{1-n}$.
Substituting $n = 3$,we get $t_{1/2} \propto [R_0]^{1-3} = [R_0]^{-2}$.

(A) In a multi-step reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.
Given the mechanism:
$(i)$ $ClO^{-} + ClO^{-} \xrightarrow{K_{1}} ClO_{2}^{-} + Cl^{-}$ (Slow step)
$(ii)$ $ClO_{2}^{-} + ClO^{-} \xrightarrow{K_{2}} ClO_{3}^{-} + Cl^{-}$ (Fast step)
The rate of the reaction is determined by the slow step $(i)$.
Therefore,the rate law expression is given by: $\text{Rate} = K_{1}[ClO^{-}][ClO^{-}] = K_{1}[ClO^{-}]^{2}$.

(C) The rate law for the reaction is given by $Rate = K[X]^m[Y]^n$.
Given that the total order of the reaction is $3$,so $m + n = 3$.
The order of reaction with respect to $X$ is $m = 2$.
Substituting $m = 2$ into the equation $2 + n = 3$,we get $n = 1$.
Therefore,the rate equation is $Rate = K[X]^2[Y]^1$.
Since the rate of reaction can be expressed as $-\frac{d[X]}{dt}$,the differential rate equation is $-\frac{d[X]}{dt} = K[X]^2[Y]$.

(A) The rate law is given by $v_1 = K[A]^2[B]^0 = K[A]^2$.
When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is doubled $([B]' = 2[B])$,the new rate $v_2$ is:
$v_2 = K(2[A])^2(2[B])^0$
$v_2 = K(4[A]^2)(1)$
$v_2 = 4 \times K[A]^2$
$v_2 = 4 \times v_1$.
Therefore,the rate of reaction becomes $4$ times the initial rate.

(B) For an $n^{th}$ order reaction,the relationship between half-life $(t_{1/2})$ and initial concentration $([R]_0)$ is given by:
$t_{1/2} \propto [R]_0^{1-n}$
For an $(n-1)^{th}$ order reaction,substitute $n$ with $(n-1)$:
$t_{1/2} \propto [R]_0^{1-(n-1)}$
$t_{1/2} \propto [R]_0^{1-n+1}$
$t_{1/2} \propto [R]_0^{2-n}$

(D) The general formula for the rate constant $K$ of an $n^{th}$ order reaction is given by:
$K = (\text{mole} / \text{litre})^{1-n} \cdot s^{-1}$
For a fourth-order reaction,$n = 4$.
Substituting $n = 4$ into the formula:
$K = (\text{mole} / \text{litre})^{1-4} \cdot s^{-1}$
$K = (\text{mole} / \text{litre})^{-3} \cdot s^{-1}$
Therefore,the correct option is $D$.

(C) The esterification reaction involves the reaction between an ester and water in the presence of an acid catalyst: $RCOOR' + H_2O \xrightarrow{H^+} RCOOH + R'OH$.
Since water is present in large excess,its concentration remains practically constant during the reaction.
Therefore,the rate of the reaction depends only on the concentration of the ester.
Such reactions,which are theoretically second-order but behave as first-order,are known as pseudo-first-order reactions.

(A) The general formula for the unit of the rate constant $K$ for a reaction of order $n$ is given by:
$Unit = \left(\frac{mol}{L}\right)^{1-n} \cdot s^{-1}$
For a third-order reaction,$n = 3$.
Substituting $n = 3$ into the formula:
$Unit = \left(\frac{mol}{L}\right)^{1-3} \cdot s^{-1} = \left(\frac{mol}{L}\right)^{-2} \cdot s^{-1}$
This can also be written as:
$\left(\frac{L}{mol}\right)^2 \cdot s^{-1}$
Comparing this with the given options,option $A$ is $\left(\frac{L}{mol}\right)^2 \cdot s^{-1}$,which matches our derived result.
Therefore,the correct option is $A$.

(D) The rate law for the reaction is given by: $Rate = k[A]^2[B]^0 = k[A]^2$.
When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is halved $([B]' = 0.5[B])$,the new rate $(Rate')$ is:
$Rate' = k(2[A])^2(0.5[B])^0 = k(4[A]^2)(1) = 4k[A]^2$.
Comparing the new rate with the initial rate,we get $Rate' = 4 \times Rate$.
Therefore,the rate becomes $4$ times the original rate.

(A) The unit of the rate constant $K$ for a reaction of order $n$ is given by the formula: $(mol \ L^{-1})^{1-n} \ s^{-1}$.
Given the unit is $L^2 \ mol^{-2} \ s^{-1}$,which can be rewritten as $mol^{-2} \ L^2 \ s^{-1}$.
Comparing this with $(mol \ L^{-1})^{1-n} \ s^{-1}$,we have:
$(mol \ L^{-1})^{1-n} = mol^{-2} \ L^2$.
This implies $1-n = -2$.
Solving for $n$,we get $n = 1 + 2 = 3$.
Therefore,the order of the reaction is $3$.

(B) The rate law for the reaction is given by $Rate = k[A]^1[B]^2$.
If the concentration of $B$ is increased three times,the new concentration becomes $[B]' = 3[B]$.
The new rate of reaction $Rate'$ will be $Rate' = k[A]^1(3[B])^2$.
$Rate' = k[A]^1(9[B]^2) = 9 \times k[A]^1[B]^2$.
$Rate' = 9 \times Rate$.
Therefore,the rate of reaction increases by $9$ times.
Note: The rate constant $k$ remains unaffected by changes in concentration.

(A) 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 second-order reaction $(n = 2)$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.
Since the given unit is $L \ mol^{-1} \ s^{-1}$,the reaction is of the second order.

(C) The given reaction is $C_2H_{4(g)} + H_{2(g)} \rightarrow C_2H_{6(g)}$.
This is a hydrogenation reaction which typically follows second-order kinetics.
The general formula for the unit of the rate constant $k$ for an $n^{th}$ order reaction is $(mol \cdot L^{-1})^{1-n} \cdot s^{-1}$.
For a second-order reaction,$n = 2$.
Substituting $n = 2$ into the formula: $(mol \cdot L^{-1})^{1-2} \cdot s^{-1} = (mol \cdot L^{-1})^{-1} \cdot s^{-1} = mol^{-1} \cdot L \cdot s^{-1}$.
Therefore,the correct unit is $mol^{-1} \cdot L \cdot s^{-1}$,which corresponds to option $C$.

(A) The general formula for the unit of the rate constant $(k)$ for a reaction of order $(n)$ is given by: $(Concentration)^{1-n} \times Time^{-1}$.
For a second-order reaction,$n = 2$.
Substituting $n = 2$ into the formula: $(Mol \ L^{-1})^{1-2} \times S^{-1} = (Mol \ L^{-1})^{-1} \times S^{-1}$.
This simplifies to: $Mol^{-1} \ L^1 \ S^{-1}$ or $Mol^{-1} \ L \ S^{-1}$.
Therefore,the correct option is $A$.

(D) bimolecular reaction is one in which two reactant species collide simultaneously to form products.
In the reaction $2NO + O_2 \rightarrow 2NO_2$,the rate-determining step involves the collision of $2$ molecules of $NO$ and $1$ molecule of $O_2$,but specifically,the reaction $2NO + O_2 \rightarrow 2NO_2$ is often cited as a termolecular reaction in elementary steps.
However,looking at the options provided:
Option $A$ involves $4$ molecules ($1$ $N_2$ + $3$ $H_2$).
Option $B$ is a unimolecular decomposition.
Option $C$ is a unimolecular decomposition.
Option $D$ involves $3$ molecules ($2$ $NO$ + $1$ $O_2$).
Actually,in many contexts,the reaction $2NO + O_2 \rightarrow 2NO_2$ is considered termolecular. If we re-evaluate the options for a bimolecular process,none strictly fit perfectly as elementary bimolecular reactions except if we consider the mechanism of $2NO + O_2 \rightarrow 2NO_2$ as $2NO + O_2 \rightarrow 2NO_2$ being a complex reaction where the elementary step $NO + O_2 \rightarrow NO_3$ is bimolecular.
Given the standard textbook examples,$2NO + O_2 \rightarrow 2NO_2$ is the most common example used to discuss molecularity higher than $2$ or complex kinetics. If the question implies which reaction involves two species in its rate-determining step,$D$ is the most appropriate choice.

(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 second-order reaction $(n = 2)$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.
Since the given unit is $L \ mol^{-1} \ s^{-1}$,the reaction is of the second order.

(D) For a reaction of order $n$,the half-life time $(t_{1/2})$ is related to the initial concentration $[R]_0$ by the expression:
$t_{1/2} \propto [R]_0^{1-n}$.
Comparing this with the given relation $t_{1/2} \propto \frac{1}{[R]_0^{n-1}}$,we can rewrite the given relation as:
$t_{1/2} \propto [R]_0^{-(n-1)} = [R]_0^{1-n}$.
Since the given expression matches the standard formula for a reaction of order $n$,the order of the reaction is $n$.

(A) An elementary reaction is a single-step reaction where the rate of reaction is directly proportional to the product of the concentrations of the reactants raised to the power of their stoichiometric coefficients.
Therefore,for an elementary reaction,the order of the reaction is equal to its molecularity.