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

(C) For a reaction of order $n$,the half-life $T_{1/2}$ is related to the initial concentration $a_0$ by the expression: $T_{1/2} \propto \frac{1}{a_0^{n-1}}$.
From the given graph,$T_{1/2}$ is directly proportional to $\frac{1}{a_0}$.
This means $T_{1/2} \propto (a_0)^{-1}$.
Comparing the exponents of $a_0$,we get $n - 1 = 1$,which implies $n = 2$.
Therefore,the order of the reaction is $2$.

(C) The rate law can be expressed as $Rate = k[A]^x[B]^y$.
When the concentration of $A$ is doubled,the rate doubles,which implies $2^x = 2$,so $x = 1$.
When both concentrations are increased by four times,the rate increases by four times: $4^x \times 4^y = 4$.
Substituting $x = 1$,we get $4^1 \times 4^y = 4$,which means $4^y = 1$,so $y = 0$.
The total order of the reaction is $x + y = 1 + 0 = 1$.

(B) The reaction is $RCl + H_2O \rightarrow ROH + HCl$.
Since water is present in large excess,its concentration remains practically constant during the reaction.
This makes the reaction a pseudo-first-order reaction,where the order of reaction is $1$.
However,the molecularity is determined by the number of reacting species colliding simultaneously in the elementary step,which is $2$ ($RCl$ and $H_2O$).
Therefore,the molecularity is $2$ and the order of reaction is $1$.

(B) For a $1^{st}$ order reaction,the rate is given by $Rate = k[A]$.
In Run $1$,the concentration of $A$ is $10 \ M$.
In Run $2$,the volume is doubled,but the concentration of $A$ remains $10 \ M$,so the rate is the same as Run $1$.
In Run $3$,$100 \ mL$ of $H_2O$ is added to $100 \ mL$ of $10 \ M$ solution of $A$,which dilutes the solution to $5 \ M$.
Since the rate depends on the concentration of $A$,the rate in Run $3$ will be lower than in Run $1$ and Run $2$.
Therefore,the correct order is $Run \ 3 < Run \ 1 = Run \ 2$.

(B) The general unit for the rate constant $k$ of an $n^{th}$ order reaction is given by the formula: $\text{mol}^{1-n} \text{L}^{n-1} \text{s}^{-1}$.
Given the unit of the rate constant is $\text{L}^2 \text{mol}^{-2} \text{s}^{-1}$.
Comparing the exponent of $\text{L}$ (liters) in the given unit with the general formula: $n - 1 = 2$.
Solving for $n$,we get $n = 3$.
Therefore,the order of the reaction is $3$.

(B) The graph shows $[R]$ vs $t$. Reaction $1$,$2$,and $3$ represent decreasing concentrations with time.
For zero-order reactions,$[R] = [R]_{0} - kt$,which is a straight line. As the order increases,the curves become more convex.
Reaction $1$ is a straight line (zero-order),Reaction $2$ is first-order,and Reaction $3$ is second-order.
The unit of the rate constant for zero-order is $mol \text{ L}^{-1} \text{ s}^{-1}$. Thus,option $C$ is incorrect.
For a fixed initial concentration,the rate constant order is $k_{3} > k_{2} > k_{1}$ to maintain the decay profiles,as reaction $3$ decays faster than $2$. Therefore,option $B$ is correct.

(B) The general unit for the rate constant $k$ for a reaction of order $n$ is given by the formula: $k = (mol \text{ } L^{-1})^{1-n} s^{-1}$.
For $n=0$ (Zero order): $k = (mol \text{ } L^{-1})^{1-0} s^{-1} = mol \text{ } L^{-1} s^{-1}$ (Matches $IV$).
For $n=1$ (First order): $k = (mol \text{ } L^{-1})^{1-1} s^{-1} = s^{-1}$ (Matches $III$).
For $n=2$ (Second order): $k = (mol \text{ } L^{-1})^{1-2} s^{-1} = mol^{-1} L s^{-1}$ (Matches $I$).
For $n=3$ (Third order): $k = (mol \text{ } L^{-1})^{1-3} s^{-1} = mol^{-2} L^{2} s^{-1}$ (Matches $II$).
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.