Chemical kinetics practice · Mixed difficulty

Assertion (A): Molecularity of a reaction cannot be more than three.
Reason (R): Probability of simultaneous collision between more than three particles is very less.

In the following reaction, $A(g) \stackrel{k_1}{\underset{k_2}{\rightleftharpoons}} B(g)$
$\frac{-d[A]}{d t}=k_1[A]$ for the forward reaction
$\frac{-d[B]}{d t}=k_2[B]$ for the backward reaction
Half-life period of the forward reaction is 10 min and that of backward reaction is 1 min. Thus, equilibrium constant is

The rate of a chemical reaction doubles for every $10\,^\circ C$ rise of temperature. If the temperature is raised by $50\,^\circ C$, the rate of the reaction increases by about

Half life period of a reaction is found to be inversely proportional to the cube of the initial concentration. The order of reaction is

The slope of the line obtained on plotting $\log _{10} \mathrm{k}$ against $\frac{1}{\mathrm{T}}$ is equal to

A first order reaction is 10% complete in 20 min. The time taken for 19% completion is

The activation energies of two reactions are ${E_a}$ and ${E'_a}$ with ${E_a} > {E'_a}.$ If the temperature of the reacting systems is increased from ${T_1}$ to ${T_2}$, predict which alternative is correct. $K'$ are rate constants at higher temperature. Assume $A$ being same for both the reactions:

In the reversible reaction, $2N{O_2}\,\underset{{{k_2}}}{\overset{{{k_1}}}{\leftrightharpoons}}\,{N_2}{O_4};$ the rate of disappearance of $N{O_2}$ is equal to

What will happen to the rate of nitration of benzene if large amount of $\mathrm{KHSO}_4$ is added to the mixture responsible for nitration?

Assertion (A):The acid hydrolysis of ester takes place more rapidly in $\mathrm{D}_2 \mathrm{O}$ than $\mathrm{H}_2 \mathrm{O}$

Reason (R): $\mathrm{D}_3 \mathrm{O}^{+}$ is stronger acid than $\mathrm{H}_3 \mathrm{O}^{+}$ or $\mathrm{H}_2 \mathrm{D}^{+} \mathrm{O}$

In the reaction $a \mathrm{~A}+b \mathrm{~B}+c \mathrm{C} \rightarrow$ Products,

(I) if concentration of $\mathrm{A}$ is doubled, keeping concentration of $\mathrm{B}$ and $\mathrm{C}$ constant, the rate of reaction becomes doubled.
(II) if concentration of $B$ is halved, concentration of $A$ and $C$ constant, the rate of reaction remains unaffected.

(III) if concentration of $\mathrm{C}$ is made 1.5 times, the rate of reaction becomes 2.25 times.

The order of reaction is

A kinetic study of the reaction $\mathrm{A} \rightarrow \text{products}$ provides the data $t=0 \mathrm{~s},[\mathrm{A}]=2.00 \mathrm{~M} ;$ $500 \mathrm{~s},$ $1.00 \text{ M};$ $1500 \mathrm{~s},$ $0.50 \mathrm{M} ;$ $3500 \mathrm{~s},$ $0.25 \mathrm{~M}.$ In the simplest possible way determine whether this reaction is of

Assertion: Two different reaction can never have same rate of reaction

Reason: Rate of reaction always depends only on frequency of collision and Arrhenius factor

The order of a gaseous phase reaction for which rate becomes half if volume of container having same amount of reactant is doubled is:

A flask contains a mixture of compounds $A$ and $B.$ Both compounds decompose by first-order kinetics. The half-life for $A$ and $B$ are 300 s and 180 s, respectively. If the concentrations of $A$ and $B$ are equal initially, the time required for the concentration of $A$ to be four times that of $B$ (in s) is (Use ln $2 = 0.693)$

We are given with following information about concentration of reactant with initial rate of reaction.

$\begin{array}{cccccccccccccccccccc} {{\text{Initial concentration}}}&{{\text{Initial rate}}} \\ {0.005{\text{ M}}}&{7.5 \times {{10}^{ - 4}}} \\ {0.02{\text{ M}}}&{3 \times {{10}^{ - 3}}} \end{array}$

Find out order of reaction with respect to that reactant.

The rate of the reaction $2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2$ can be written in three ways:

$\begin{aligned} & \frac{-\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}={k}\left[\mathrm{~N}_2 \mathrm{O}_5\right] ; \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}={k}^{\prime}\left[\mathrm{N}_2 \mathrm{O}_5\right] \\ & \frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}={k}^{\prime \prime}\left[\mathrm{N}_2 \mathrm{O}_5\right] \end{aligned}$

The relationship between ${k}$ and ${k}^{\prime}$ and between ${k}$ and ${k}^{\prime \prime}$ are:

$\mathrm{T}_{50}$ of a first order reaction is $10 \mathrm{~min}$. Starting with $10 \mathrm{~mol} \mathrm{~L}^{-1}$, rate after $20 \mathrm{~min}$ is

Select the incorrect statement among the following

The rate equation for an autocatalytic reaction

$\mathrm{A}+\mathrm{R} \xrightarrow{k} \mathrm{R}+\mathrm{R} \text { is } r_{\mathrm{A}}=-\frac{\mathrm{d} C_{\mathrm{A}}}{\mathrm{dt}}=k C_{\mathrm{A}} C_{\mathrm{R}}$

The rate of disappearance of reactant $\text{A}$ is maximum when

For the reaction $A \to B,$ the rate expression is $r = k{\left[ A \right]^n}.$ When the concentration of $A$ is doubled, the rate of reaction is quadrupled. The value of $n$ is

What is the role of a catalyst in a catalysed
reaction?

A flask contains a mixture of compounds $A$ and $B$ initially
$\begin{aligned} & {[A]_0=[B]_0} \\ & A \xrightarrow{t_{50}=54 \mathrm{~min}} \text { product, }-\frac{d[A]}{d t}=k[A] \\ & B \xrightarrow{t_{50}=18 \mathrm{~min}} \text { product, }-\frac{d[B]}{d t}=k[B] \end{aligned}$
After 54 minutes, $[A]=x[B]$ what is the value of $x$ ?

The rate of first order reaction are $0.8 \mathrm{~M} \mathrm{~min}^{-1}$ and $0.08\mathrm{~M} \mathrm{~min}^{-1}$ at 40 min and 60 min respectively. Then half life is:

Half-life of two samples is 0.1 and 0.8 s. Their respective concentration is 400 and 50 respectively.

The order of reaction is

Assertion (A):Emission of light as a result of exposure of $P$ to air in night is called phosphorescence.

Reason (R): Phosphorus burns in $\mathrm{O}_2$ to give $\mathrm{P}_2 \mathrm{O}_5$ and conversion of chemical energy into light energy producing cold light and the phenomenon is called chemi-luminescence.

The rate constant $\left( {k'} \right)$ of one of the reaction is found to be double that of the rate constant $\left( {k''} \right)$ of another reaction. Then the relationship between the corresponding activation energies of the two reaction $\left( {{{E'}_a}\;\mathrm{and}\;{{E''}_a}} \right)$ can be represented as

The reaction of $(+)$ 2-iodobutane and $\mathrm{NaI}^*$ ( $\mathrm{I}^*$ is radioactive isotope of iodine) in acetone was studied by measuring the rate of racemization $\left(\mathrm{k}_{\mathrm{r}}\right)$ and the rate of inversion $\left(\mathrm{k}_{\mathrm{i}}\right)$.

$(+) \mathrm{CH}_3 \mathrm{CH}(\mathrm{I}) \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{NaI}^* \rightarrow \mathrm{CH}_3 \mathrm{CH}\left(\mathrm{I}^*\right) \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{NaI}$

For this reaction, the relationship between $\mathrm{k}_{\mathrm{r}}$ and $\mathrm{k}_{\mathrm{i}}$

STATEMENT-1 : If the activation energy of reaction is zero, temperature will have no effect on the rate constant.

STATEMENT-2 : Lower the activation energy fasten is the reaction.

Which one of the following is wrongly matched?