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 $A-IV, B-III, C-I, D-II$.

(D) Step $1$: The rate law is expressed as $\text{Rate} = k[A]^x[B]^y$. Here, $k$ is the rate constant.
Step $2$: $k$ is independent of the initial concentration of reactants, but it depends on temperature (Arrhenius equation: $k = Ae^{-E_a/RT}$).
Step $3$: $A$ larger value of $k$ indicates a faster reaction rate for a given set of concentrations. Therefore, the statement that a greater value of $k$ makes the reaction slower is incorrect.

(C) Step $1$: For a zero order reaction, the rate is constant and independent of the initial concentration, so $(A)$ is incorrect.
Step $2$: The decomposition of acetaldehyde $(CH_3CHO \rightarrow CH_4 + CO)$ is a second order reaction, so $(B)$ is incorrect.
Step $3$: For a first order reaction, the half-life is given by $t_{1/2} = \frac{0.693}{k}$, which is independent of the initial concentration $[R]_0$. Thus, $(C)$ is correct.
Step $4$: For a zero order reaction, the half-life is $t_{1/2} = \frac{[R]_0}{2k}$, which depends on the initial concentration. Thus, $(D)$ is incorrect.

(B) The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
Given rate law: $\text{rate} = k[H_2O_2]^1$.
The power of the concentration term $[H_2O_2]$ is $1$.
Therefore, the order of the reaction is $1$.

(A) The general unit for the rate constant $k$ of a reaction of order $n$ is given by the formula: $(\text{mol dm}^{-3})^{1-n} \text{s}^{-1}$.
For a zero-order reaction, $n = 0$.
Substituting $n = 0$ into the formula: $(\text{mol dm}^{-3})^{1-0} \text{s}^{-1} = \text{mol dm}^{-3} \text{s}^{-1}$.
Therefore, the reaction is of zero order.

(A) Given rate law: $\text{rate} = k[A][B]^2$
Substitute the given values: $0.25 \text{ Ms}^{-1} = k(1 \text{ M})(0.2 \text{ M})^2$
$0.25 = k(1)(0.04)$
$k = \frac{0.25}{0.04}$
$k = 6.25 \text{ M}^{-2}\text{s}^{-1}$

(B) The rate law for the reaction is given by: $\text{Rate} = k[A]^1[B]^0 = k[A]$.
Given: $\text{Rate} = 6 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}$ and $[A] = 0.3 \text{ mol dm}^{-3}$.
Substituting the values into the rate law: $6 \times 10^{-4} = k \times 0.3$.
Solving for $k$: $k = \frac{6 \times 10^{-4}}{0.3} = 20 \times 10^{-4} \text{ s}^{-1} = 2 \times 10^{-3} \text{ s}^{-1}$.

(B) The rate law for the reaction is given by: $\text{Rate} = k[A]^1[B]^2$.
Given: $\text{Rate} = 4 \times 10^{-2} \text{ mol dm}^{-3} \text{ s}^{-1}$, $[A] = 0.2 \text{ mol dm}^{-3}$, $[B] = 0.1 \text{ mol dm}^{-3}$.
Substituting the values into the rate law: $4 \times 10^{-2} = k(0.2)^1(0.1)^2$.
$4 \times 10^{-2} = k(0.2)(0.01)$.
$4 \times 10^{-2} = k(0.002)$.
$k = \frac{4 \times 10^{-2}}{2 \times 10^{-3}} = 2 \times 10^1 = 20 \text{ mol}^{-2} \text{ dm}^6 \text{ s}^{-1}$.

(B) For a first-order reaction, the rate law is given by: $\text{Rate} = k[N_2O_5]$.
Given: $\text{Rate} = 2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$ and $k = 3.0 \times 10^{-5} \text{ s}^{-1}$.
Substituting the values into the rate equation: $2.4 \times 10^{-5} = (3.0 \times 10^{-5}) \times [N_2O_5]$.
Solving for $[N_2O_5]$: $[N_2O_5] = \frac{2.4 \times 10^{-5}}{3.0 \times 10^{-5}}$.
$[N_2O_5] = 0.8 \text{ mol L}^{-1}$.

(A) Step $1$: Identify the given values.
Rate constant $k = 6.0 \text{ M}^{-2} \text{s}^{-1}$.
Concentration $[A] = 1 \text{ M}$.
Concentration $[B] = 0.2 \text{ M}$.
Step $2$: Use the rate law expression.
$\text{Rate} = k[A]^2[B]$
Step $3$: Substitute the values into the equation.
$\text{Rate} = 6.0 \times (1)^2 \times (0.2)$
$\text{Rate} = 6.0 \times 1 \times 0.2$
$\text{Rate} = 1.2 \text{ M s}^{-1}$.
Note: The calculated value is $1.2 \text{ M s}^{-1}$. Since this is not among the options, the question options are incorrect.

(C) Step $1$: Analyze the rate laws for the given reactions.
Step $2$: The reaction $2 NO_2(g) + F_2(g) \rightarrow 2NO_2F(g)$ follows the rate law $Rate = k[NO_2][F_2]$. This is a second-order reaction (first order with respect to $NO_2$ and first order with respect to $F_2$, total order $1+1=2$).
Step $3$: $2 H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)$ is a first-order reaction.
Step $4$: $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$ is a fractional order reaction (order $1.5$).
Step $5$: $2 NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(l)$ is a third-order reaction.
Step $6$: Therefore, option $C$ is the correct answer.

(A) Step $1$: Identify the given values: $k = 1.62 \times 10^{-2} \text{ L mol}^{-1} \text{ s}^{-1}$ and $[NOBr] = 2 \times 10^{-3} \text{ mol L}^{-1}$.
Step $2$: Use the rate law expression $r = k[NOBr]^2$.
Step $3$: Substitute the values into the equation: $r = (1.62 \times 10^{-2}) \times (2 \times 10^{-3})^2$.
Step $4$: Calculate the square of the concentration: $(2 \times 10^{-3})^2 = 4 \times 10^{-6} \text{ mol}^2 \text{ L}^{-2}$.
Step $5$: Multiply by the rate constant: $r = 1.62 \times 10^{-2} \times 4 \times 10^{-6} = 6.48 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}$.

(C) Step $1$: Analyze the rate laws for the given reactions.
Step $2$: The decomposition of acetaldehyde, $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$, follows the rate law $Rate = k[CH_3CHO]^{1.5}$.
Step $3$: The overall order of a reaction is the sum of the powers of the concentration terms in the rate law expression.
Step $4$: For the decomposition of acetaldehyde, the order is $1.5$.
Step $5$: Other reactions listed are typically first or second order. Thus, option $C$ is correct.

(A) Given:
Rate law: $r = k[NOBr]^2$
Rate constant $k = 1.62 \text{ M}^{-1} \text{ s}^{-1}$
Concentration $[NOBr] = 5 \times 10^{-3} \text{ M}$
Substitute the values into the rate law:
$r = (1.62 \text{ M}^{-1} \text{ s}^{-1}) \times (5 \times 10^{-3} \text{ M})^2$
$r = 1.62 \times 25 \times 10^{-6} \text{ M s}^{-1}$
$r = 40.5 \times 10^{-6} \text{ M s}^{-1}$
$r = 4.05 \times 10^{-5} \text{ M s}^{-1}$

(C) Step $1$: Analyze the order of the given reactions.
Step $2$: The decomposition of acetaldehyde $(CH_3CHO)$ follows a chain mechanism where the rate law is given by $\text{Rate} = k[CH_3CHO]^{3/2}$.
Step $3$: Since the exponent of the concentration term is $3/2$ (which is $1.5$), this reaction is of fractional order.
Step $4$: The other reactions listed are typically elementary or follow integer order kinetics under standard conditions.
Step $5$: Therefore, option $C$ is the correct example of a fractional order reaction.

(A) An elementary reaction is a single-step reaction that occurs in one go without any intermediate.
Option $A$: The decomposition of ozone $O_3(g) \rightarrow O_2(g) + O(g)$ is a single-step unimolecular elementary reaction.
Option $B$, $C$, and $D$ are complex reactions involving multiple steps and intermediates.
Therefore, the correct answer is $A$.

(A) $1$. The order of a reaction is the sum of the powers of the concentration terms in the rate law expression. Here, the rate $= k[NO_2]^2$, so the order is $2$.
$2$. The molecularity of an elementary reaction is the number of reacting species (atoms, ions, or molecules) taking part in the reaction. In the reaction $2NO_2(g) \rightarrow 2NO(g) + O_2(g)$, two molecules of $NO_2$ are involved, so the molecularity is $2$.
$3$. $A$ reaction with molecularity $2$ is called bimolecular.
$4$. Therefore, the reaction is second order and bimolecular.