Chemical kinetics practice · Mixed difficulty

Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively

For the reaction, ${\mathrm{N}_2}{\mathrm{O}_5} \to 2\mathrm{N}{\mathrm{O}_2} + \frac{1}{2}{\mathrm{O}_2}$

Given, $- \frac{{d\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]}}{{dt}} = {K_1}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]$

$\frac{{d\left[ {\mathrm{N}{\mathrm{O}_2}} \right]}}{{dt}} = {K_2}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]$

$\mathrm{and}\;\;\frac{{d\left[ {{\mathrm{O}_2}} \right]}}{{dt}} = {K_3}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]$

The relation in between ${K_1},{K_2}\;and\;{K_3}$ is:

The rate of reaction depends upon the

The unit of rate constant of a third order chemical reaction is

For a first order reaction, $\left( A \right) \to products$, the concentration of A changes from $0.1\;M$ to $0.025\;M$ in 40 minutes. The rate of reaction when the concentration of A is $0.01\;M$, is:

The minimum energy a molecule should posses in order to enter into a fruitful collision is known as

It rate constant $K = 2.3 \times {10^{ - 5}}L\,.\,mo{l^{ - 1}}.\,{S^{ - 1}}$, identify order of reaction

In a zero-order reaction for every $10^\circ C$ rise of temperature, the rate is doubled. If the temperature is increased from $10^\circ C\;to\;100^\circ C$, the rate of the reaction will become:

Activation energy of a reaction is:

The burning of coal represented by the equation; $\mathrm{C}\left( s \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)\mathrm{.}$ The rate of this reaction is increased by:

The rate of reaction:
$2\mathrm{NO} + \mathrm{C}{\mathrm{l}_2} \to 2\mathrm{NOCl}$ is given by the rate, equation $rate\; = k{\left[ {\mathrm{NO}} \right]^2}\left[ {\mathrm{C}{\mathrm{l}_2}} \right].$ The value of the rate constant can be increased by:

The rate of first order reaction is $1.5 \times {10^{ - 2}}\,mol\,{L^{ - 1}}\,\,{\min ^{ - 1}}$ at 0.5 M concentration of the reactant. The half-life of reaction is

In the presence of a catalyst, the heat evolved or absorbed during the reaction _______

Two substances A and B are present such that $\left[ A \right] = 4\left[ B \right]$ and half-life of $A$ is 5 minute and of $B$ is 15 minute. If they start decaying at the same time following first order, how much time later will the concentration of both of them would be same?

A zero order reaction is one:

Observe the following reaction,

$2A + B \to C$

The rate of formation of C is $2.2 \times {10^{ - 3}}\;mol\;{L^{ - 1}}\;{\min ^{ - 1}}$

What is the value of $- \frac{{d\left[ A \right]}}{{dt}}(mol\;{L^{ - 1}}\;\;{\min ^{ - 1}})$?

$\mathrm{C}{\mathrm{H}_3}\mathrm{COO}{\mathrm{C}_2}{\mathrm{H}_5} + {\mathrm{H}_2}\mathrm{O}\xrightarrow{{\mathrm{H} + }}\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} + {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ is an example of …… order.

Which of the following reactions is a true, first order reaction

At $325^{\circ} \mathrm{C},$ 1,3-butadiene dimerizes according to the equation: $2 \mathrm{C}_4 \mathrm{H}_6(\mathrm{g}) \rightarrow \mathrm{C}_8 \mathrm{H}_{12}(\mathrm{g}).$ It is found that the partial pressure $P,$ of the reactant obeys the law: $\frac{1}{P}=k t+\frac{1}{P_0}.$ The order of the reaction is

The reaction, $NO + (1/2){O_2}\xrightarrow{\,}N{O_2}$ exhibits:

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

For which order reaction, half-life is inversely proportional to the concentration of reactant?

The order of a reaction with rate equal to $k\;C_A^{3/2}\;C_B^{ - 1/2}$ is

For an elementary chemical reaction, ${A_2}\underset{{{k_{ - 1}}}}{\overset{{{k_1}}}{\leftrightharpoons}}2A$, the expression for $\frac{{d\left[ A \right]}}{{dt}}$ is

In on doubling the volume of reaction vessel, the rate of gaseous reaction remains same then the order o the reaction is

If the rate constant of a first order reaction is $9.212 \times {10^{ - 3}}\,{s^{ - 1}}$, then the time required for the completion of 75% of the reaction will be

A drop of a solution $\left( {\mathrm{volume} = 0.05\;\mathrm{mL}} \right)$ contains $6 \times {10^{ - 7}}\;mol\;of\;{\mathrm{H}^ + }.$ If the rate of disappearance of ${\mathrm{H}^ + }\;is\;6.0 \times {10^5}\;\mathrm{mol}/\mathrm{L} \times \mathrm{s},$ how long will it take for ${\mathrm{H}^ + }$ to disappear from the drop

For a general chemical change, $2A + 3B \to$ Products, the rate of disappearance of A is ${r_2}.$ The rates ${r_1}$ and ${r_2}$ are related as

If $X$ is the total number of collisions which a gas molecule register with others per unit time under particular conditions, then the collision frequency of the gas containing $N$ molecules per unit volume is

For the reaction:
${\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }} + {\mathrm{H}_2}\mathrm{O}\;\leftrightharpoons\;{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}{\mathrm{H}_2}\mathrm{O}} \right]^{2 + }} + \mathrm{N}{\mathrm{H}_3},$ the net rate of reaction at any time is given by, net $rate = 2.0 \times {10^{ - 4}}\;{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}\left[ {{\mathrm{H}_2}\mathrm{O}} \right] - 3.0 \times {10^5}{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}{\mathrm{H}_2}\mathrm{O}} \right]^{2 + }} \cdot \left[ {\mathrm{N}{\mathrm{H}_3}} \right]$
Then correct statement is/are: