Order & Molecularity of a Reaction — JEE Main Chemistry MCQs with Solutions

The rate constant of a reaction has the units $L^{2}\,mol^{-2}\,s^{-1}$. The order of the reaction is:

A reaction has the rate law $Rate=k[A]^{3/2}$. If the concentration of $A$ is increased to four times its initial value, the rate becomes:

For a reaction the half-life is $100\,s$ when the initial concentration is $0.2\,M$ and $200\,s$ when the initial concentration is $0.1\,M$. The order of the reaction is:

The half-life of a reaction becomes one-half of its original value when the initial concentration is increased four-fold. The order of the reaction is:

For $A + B \rightarrow$ products the initial-rate data are: (i) $[A]=0.1,\,[B]=0.1$, rate $=1.0\times10^{-3}$; (ii) $[A]=0.2,\,[B]=0.1$, rate $=4.0\times10^{-3}$; (iii) $[A]=0.1,\,[B]=0.4$, rate $=2.0\times10^{-3}$ (all in $mol\,L^{-1}s^{-1}$). The overall order of the reaction is:

For $A \rightarrow$ products, the rate is $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ when $[A]=0.1\,M$ and $4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ when $[A]=0.4\,M$. The order of the reaction with respect to $A$ is:

A reaction $A + B \rightarrow P$ occurs by a fast pre-equilibrium $A + B \rightleftharpoons C$ (equilibrium constant $K$) followed by a slow step $C + B \rightarrow P$ (rate constant $k_2$). The rate law for the reaction is:

Which of the following statements is INCORRECT?

The acid hydrolysis of ethyl acetate, $CH_3COOC_2H_5 + H_2O \rightarrow CH_3COOH + C_2H_5OH$, carried out in a large excess of water, follows first-order kinetics even though two species react. This is because:

For $2NO_2 + F_2 \rightarrow 2NO_2F$ the accepted mechanism has a slow step $NO_2 + F_2 \rightarrow NO_2F + F$ followed by a fast step $NO_2 + F \rightarrow NO_2F$. The molecularity of the rate-determining step and the overall order of the reaction are, respectively:

A reaction has the rate law $Rate=k[A]^{2}[B]^{1/2}[C]^{-1}$. The overall order of the reaction is:

For a certain reaction the half-life is found to be $40\,min$ irrespective of the initial concentration of the reactant. The order of the reaction is:

For $A + B \rightarrow$ products, tripling $[A]$ (with $[B]$ constant) leaves the rate unchanged, while doubling $[B]$ (with $[A]$ constant) makes the rate $8$ times larger. The overall order of the reaction is:

A reaction has the rate law $Rate=k[A]^{2}[B]^{-1}$. If the concentrations of both $A$ and $B$ are doubled, the rate becomes:

The rate constant of a reaction carries the units $mol^{-3/2}\,L^{3/2}\,s^{-1}$. The order of the reaction is:

The reaction $2NO + Br_2 \rightarrow 2NOBr$ proceeds by a fast pre-equilibrium $NO + Br_2 \rightleftharpoons NOBr_2$ (constant $K$) followed by a slow step $NOBr_2 + NO \rightarrow 2NOBr$ (rate constant $k_2$). The rate law is:

For $CO + Cl_2 \rightarrow COCl_2$ the mechanism is: fast equilibrium $Cl_2 \rightleftharpoons 2Cl$ (constant $K$); slow step $Cl + CO \rightarrow COCl$; fast step $COCl + Cl_2 \rightarrow COCl_2 + Cl$. The overall order of the reaction is:

A gas-phase decomposition $A \rightarrow$ products follows the Lindemann mechanism: $A + A \rightleftharpoons A^{*} + A$ (forward $k_1$, reverse $k_{-1}$) then $A^{*} \rightarrow$ products ($k_2$). Applying the steady-state approximation to $A^{*}$, the order of the reaction at very low pressure (low $[A]$, where $k_2 \gg k_{-1}[A]$) is:

For $A + B \rightarrow$ products the initial-rate data are: (i) $[A]=0.1,\,[B]=0.1$, rate $=4.0\times10^{-3}$; (ii) $[A]=0.4,\,[B]=0.1$, rate $=8.0\times10^{-3}$; (iii) $[A]=0.1,\,[B]=0.2$, rate $=1.6\times10^{-2}$ (all in $mol\,L^{-1}s^{-1}$). The rate law is:

The decomposition $2O_3 \rightarrow 3O_2$ proceeds as: fast equilibrium $O_3 \rightleftharpoons O_2 + O$ (constant $K$); slow step $O + O_3 \rightarrow 2O_2$. The rate law is:

The reaction $2NO + O_2 \rightarrow 2NO_2$ follows the rate law $Rate=k[NO]^{2}[O_2]$. If the concentration of $NO$ is halved while that of $O_2$ is doubled, the rate becomes:

A reaction is first order in $A$ and its overall order is $3/2$. If $[A]$ is doubled and $[B]$ is increased four-fold at the same time, the rate increases by a factor of:

The reaction $4HBr + O_2 \rightarrow 2H_2O + 2Br_2$ has a slow (rate-determining) step $HBr + O_2 \rightarrow HOOBr$. The overall order of the reaction is:

For a reaction $A \rightarrow$ products, the successive times taken for the concentration to fall from $0.8\,M$ to $0.4\,M$, from $0.4\,M$ to $0.2\,M$, and from $0.2\,M$ to $0.1\,M$ are $20\,min$, $40\,min$ and $80\,min$ respectively. The order of the reaction is:

For the acid-catalysed reaction $S + H^{+} \rightarrow$ products, the mechanism is a fast pre-equilibrium $S + H^{+} \rightleftharpoons SH^{+}$ followed by a slow step $SH^{+} \rightarrow$ products. The order of the reaction is best described as:

An enzyme-catalysed reaction proceeds as $E + S \rightleftharpoons ES$ (fast) followed by $ES \rightarrow E + P$ (slow). At very high substrate concentration the order of the reaction with respect to $S$ is:

The rate constant of a reaction $A \rightarrow$ products has the units $L\,mol^{-1}\,s^{-1}$ and the value $k=2.0\times10^{-3}$. When the initial concentration of $A$ is $0.5\,M$, the initial rate of the reaction is:

A reaction $2A + B \rightarrow$ products occurs by: fast $A + B \rightleftharpoons C$ ($K_1$); fast $C + A \rightleftharpoons D$ ($K_2$); slow $D \rightarrow$ products ($k_3$). The rate law is:

For a reaction the half-life is found to be inversely proportional to the square of the initial concentration of the reactant. The order of the reaction is:

Consider the assertion (A) and reason (R). A: A reaction may have an order of zero, but its molecularity can never be zero. R: Order is an experimentally determined quantity that may be zero or fractional, whereas molecularity counts the species reacting in an elementary step and is always a positive integer. Which is correct?