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

Decrease in the concentration of oxygen

Providing inert atmosphere for burning

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

Increasing the concentration of ${\rm{NO}}$

Increasing the concentration of the ${{\rm{C}}{{\rm{l}}_2}}$

CHEMICAL KINETICS MCQs for NEET UG CHEMISTRY — Easy, Page 2

31 Easy

31 of 72Question permalink

Analyzing +4 | -1

Which of the following statement is not correct for the reaction:

4A + B \to 2C + 2D

View answer and explanation

Correct option: A

The rate of disappearance of ‘B’ is twice the rate of appearance of ‘C’

Option A (The rate of disappearance of ‘B’ is twice the rate of appearance of ‘C’) is correct because the stoichiometry of the reaction dictates that the rate of disappearance of B is half the rate of appearance of C.

Solution approach

Correct Answer + Core Reason: The rate of disappearance of ‘B’ is twice the rate of appearance of ‘C’. This statement is not correct based on the stoichiometry of the reaction.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $4A + B \to 2C + 2D$

Step 2: Express the rate of the reaction in terms of the change in concentration of each species:
$Rate = - \frac{1}{4}\frac{{d\left[ A \right]}}{{dt}} = - \frac{{d\left[ B \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ C \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ D \right]}}{{dt}}$

Step 3: Analyze each statement:
- Option A: The rate of disappearance of ‘B’ is twice the rate of appearance of ‘C’. This is incorrect. $\frac{{d\left[ B \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ C \right]}}{{dt}}$ . The rate of disappearance of B is half the rate of appearance of C.
- Option B: The rate of disappearance of B is one fourth the rate of disappearance of A. $\frac{{d\left[ B \right]}}{{dt}} = \frac{1}{4}\frac{{d\left[ A \right]}}{{dt}}$ . This is correct.
- Option C: The rate of formation of D is one half the rate of consumption of A. $\frac{1}{2}\frac{{d\left[ D \right]}}{{dt}} = \frac{1}{4}\frac{{d\left[ A \right]}}{{dt}}$ . This is correct.
- Option D: The rate of formation of C and D are equal. $\frac{1}{2}\frac{{d\left[ C \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ D \right]}}{{dt}}$ . This is correct.

Why other options are wrong:
- Option B: This statement is correct based on the stoichiometry of the reaction.
- Option C: This statement is correct based on the stoichiometry of the reaction.
- Option D: This statement is correct based on the stoichiometry of the reaction.

Key Concepts:
- Rate of reaction
- Stoichiometry

Memory Aids:
- Remember to consider the stoichiometric coefficients when relating rates.

Quick Tips:
- Always balance the chemical equation before relating the rates of reaction.

32 Easy

32 of 72Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: A

{\mathrm{s}^{ - 1}},\mathrm{M}\;{\mathrm{s}^{ - 1}}

Option A (

{\mathrm{s}^{ - 1}},\mathrm{M}\;{\mathrm{s}^{ - 1}}

) is correct because the rate constant units are derived from the rate law for first and zero-order reactions, respectively.

Solution approach

Correct Answer + Core Reason: ${\mathrm{s}^{ - 1}},\mathrm{M}\;{\mathrm{s}^{ - 1}}$ . The units of the rate constant depend on the order of the reaction.

Step-by-Step Solution:

Step 1: Define the rate law for a zero-order reaction: Rate = $k[A]^0 = k$ , where k is the rate constant and [A] is the concentration of reactant A.

Step 2: Determine the units of the rate constant for a zero-order reaction. Since Rate has units of M/s (Molarity per second) and $[A]^0$ is dimensionless, the units of k are M/s or $\mathrm{M}\;{\mathrm{s}^{ - 1}}$ .

Step 3: Define the rate law for a first-order reaction: Rate = $k[A]^1 = k[A]$ , where k is the rate constant and [A] is the concentration of reactant A.

Step 4: Determine the units of the rate constant for a first-order reaction. Since Rate has units of M/s and [A] has units of M, then M/s = k * M. Solving for k, we get k = (M/s) / M = 1/s or ${\mathrm{s}^{ - 1}}$ .

Step 5: Combine the units for zero and first-order reactions: Zero-order: $\mathrm{M}\;{\mathrm{s}^{ - 1}}$ , First-order: ${\mathrm{s}^{ - 1}}$ .

Step 6: State the final answer: ${\mathrm{s}^{ - 1}},\mathrm{M}\;{\mathrm{s}^{ - 1}}$ .

Why other options are wrong:
- Option B: This is incorrect because it only provides the correct unit for the first-order reaction but not the zero-order reaction.
- Option C: This is incorrect because it reverses the order of the units for the zero and first-order reactions.
- Option D: This is incorrect because it provides incorrect units for both zero and first-order reactions.

Key Concepts:
- Rate Laws
- Order of Reactions
- Units of Rate Constants

Memory Aids:
- Remember that the overall order of a reaction is the sum of the exponents in the rate law.

Quick Tips:
- To find the units of the rate constant, rearrange the rate law to solve for k and then substitute the units of rate and concentration.

33 Easy

33 of 72Question permalink

Analyzing +4 | -1

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:

View answer and explanation

Correct option: A

2{K_1} = {K_2} = 4{K_3}

Option A (

2{K_1} = {K_2} = 4{K_3}

) is correct because it correctly relates the rate constants based on the stoichiometry of the reaction.

Solution approach

Correct Answer + Core Reason: $2{K_1} = {K_2} = 4{K_3}$ . The rate constants are related based on the stoichiometry of the reaction.

Step-by-Step Solution:

Step 1: Write the rate of the reaction in terms of the change in concentration of each species:
$- \frac{{d\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]}}{{dt}} = {K_1}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]$
$\frac{{d\left[ {\mathrm{N}{\mathrm{O}_2}} \right]}}{{dt}} = {K_2}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]$
$\frac{{d\left[ {{\mathrm{O}_2}} \right]}}{{dt}} = {K_3}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]$

Step 2: Relate the rates based on the stoichiometry of the reaction:
${\mathrm{N}_2}{\mathrm{O}_5} \to 2\mathrm{N}{\mathrm{O}_2} + \frac{1}{2}{\mathrm{O}_2}$
$- \frac{{d\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ {\mathrm{N}{\mathrm{O}_2}} \right]}}{{dt}} = 2\frac{{d\left[ {{\mathrm{O}_2}} \right]}}{{dt}}$

Step 3: Substitute the rate expressions:
${K_1}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right] = \frac{1}{2}{K_2}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right] = 2{K_3}\left[ {{\mathrm{N}_2}{\mathrm{O}_5}} \right]$

Step 4: Simplify to find the relationship between the rate constants:
${K_1} = \frac{1}{2}{K_2} = 2{K_3}$
$2{K_1} = {K_2} = 4{K_3}$

Why other options are wrong:
- Option B: This answer is incorrect because it does not account for the stoichiometry of the reaction.
- Option C: This answer is incorrect because it does not account for the stoichiometry of the reaction.
- Option D: This answer is incorrect because it does not account for the stoichiometry of the reaction.

Key Concepts:
- Rate of reaction
- Stoichiometry
- Rate constants

Memory Aids:
- Pay attention to the stoichiometric coefficients when relating rates.

Quick Tips:
- Always balance the chemical equation before relating the rates of reaction.

34 Easy

34 of 72Question permalink

Application +4 | -1

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

View answer and explanation

Correct option: D

{\mathrm{s}^{ - 1}}\;\mathrm{mo}{\mathrm{l}^{ - 2}}\;{\mathrm{L}^2}

Option D (

{\mathrm{s}^{ - 1}}\;\mathrm{mo}{\mathrm{l}^{ - 2}}\;{\mathrm{L}^2}

) is correct because it represents the correct unit for a third-order reaction rate constant.

Solution approach

Correct Answer + Core Reason: ${\mathrm{s}^{ - 1}}\;\mathrm{mo}{\mathrm{l}^{ - 2}}\;{\mathrm{L}^2}$ . The unit of the rate constant for a third-order reaction is derived from the rate law.

Step-by-Step Solution:

Step 1: General rate law for a third-order reaction: Rate = $k[A]^3$ or Rate = $k[A][B][C]$ or Rate = $k[A]^2[B]$

Step 2: Units of rate: mol L $^{-1}$ s $^{-1}$ \nUnits of concentration: mol L $^{-1}$

Step 3: Determine the units of k for Rate = $k[A]^3$ :\nmol L $^{-1}$ s $^{-1}$ = $k$ (mol L $^{-1}$ ) $^3$
$k$ = (mol L $^{-1}$ s $^{-1}$ ) / (mol L $^{-1}$ ) $^3$
$k$ = L $^2$ mol $^{-2}$ s $^{-1}$ or ${\mathrm{s}^{ - 1}}\;\mathrm{mo}{\mathrm{l}^{ - 2}}\;{\mathrm{L}^2}$

Why other options are wrong:
- Option A: $\mathrm{mo}{\mathrm{l}^{ - 1}}\;\mathrm{L}\;{\mathrm{s}^{ - 1}}$ : This is the unit for a second-order reaction.
- Option B: $\mathrm{mo}{\mathrm{l}^{ - 1}}\;{\mathrm{s}^{ - 2}}$ : This unit is not generally applicable to any simple order reaction.
- Option C: $\mathrm{mol }\;\mathrm{L}$ : This is not a unit for a rate constant.

Key Concepts:
- Rate law
- Order of reaction
- Units of rate constant

Memory Aids:
- For a reaction of order n, the units of k are (mol/L) $^{1-n}$ s $^{-1}$

Quick Tips:
- Remember that the rate always has units of concentration per time (e.g., mol L $^{-1}$ s $^{-1}$ ).

35 Easy

35 of 72Question permalink

Remembering +4 | -1

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

View answer and explanation

Correct option: D

Threshold energy

Option D (Threshold energy) is correct because threshold energy is the minimum energy required for a molecule to have a successful collision and undergo a chemical reaction.

Solution approach

Correct Answer + Core Reason: Threshold energy is the minimum energy required for a fruitful collision.

Step-by-Step Solution:

Step 1: Define the terms.

Step 2: Reaction energy is the energy change during a chemical reaction (difference between reactants and products).

Step 3: Collision energy is the kinetic energy of the colliding molecules.

Step 4: Activation energy is the energy required to reach the transition state.

Step 5: Threshold energy is the minimum energy required for a reaction to occur.

Why other options are wrong:
- Option A: Reaction energy is the overall energy change of the reaction, not the minimum energy for a collision.
- Option B: Collision energy is the energy the molecules have when they collide, but it might not be enough for a reaction.
- Option C: Activation energy is the energy difference between the reactants and the transition state, not the minimum energy the molecules must possess.
- Option D: Threshold energy is the minimum energy required for a successful reaction.

Key Concepts:
- Threshold energy
- Activation energy
- Collision theory

Memory Aids:
- Threshold = minimum

Quick Tips:
- Understand the difference between threshold energy and activation energy.

36 Easy

36 of 72Question permalink

Application +4 | -1

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:

View answer and explanation

Correct option: D

512 times

Option D (512 times) is correct because the rate doubles for every 10°C increase, and there are nine 10°C intervals between 10°C and 100°C, so the rate increases by a factor of 2⁹.

Solution approach

Correct Answer + Core Reason: 512 times. The rate doubles for every 10°C increase, so the rate increases by $2^9$ from 10°C to 100°C.

Step-by-Step Solution:

Step 1: Determine the number of 10°C intervals between 10°C and 100°C.

Step 2: The temperature increase is $100^\circ C - 10^\circ C = 90^\circ C$ .

Step 3: The number of 10°C intervals is $\frac{90^\circ C}{10^\circ C} = 9$ .

Step 4: Since the rate doubles for every 10°C increase, the rate will increase by a factor of $2^9$ .

Step 5: Calculate $2^9 = 512$ .

Why other options are wrong:
- Option A: 64 times: This corresponds to $2^6$ , which would be a temperature increase of 60°C.
- Option B: 128 times: This corresponds to $2^7$ , which would be a temperature increase of 70°C.
- Option C: 256 times: This corresponds to $2^8$ , which would be a temperature increase of 80°C.

Key Concepts:
- Temperature Dependence of Reaction Rates
- Zero-Order Reactions

Memory Aids:
- Remember that the rate doubles for every 10°C increase.

Quick Tips:
- Calculate the number of 10°C intervals carefully.

37 Easy

37 of 72Question permalink

Understanding +4 | -1

Activation energy of a reaction is:

View answer and explanation

Correct option: C

Minimum amount of energy needed to overcome the potential barrier of reaction

Option C (Minimum amount of energy needed to overcome the potential barrier of reaction) is correct because activation energy is the minimum energy required for reactants to transform into products by overcoming the energy barrier.

Solution approach

Correct Answer + Core Reason: Minimum amount of energy needed to overcome the potential barrier of reaction. This is the definition of activation energy.

Step-by-Step Solution:

Step 1: Define activation energy: The minimum energy required for a reaction to occur.

Step 2: Relate activation energy to the potential energy barrier: Activation energy is the energy needed to overcome the potential energy barrier between reactants and products.

Why other options are wrong:
- Option A: The energy released during the reaction: This refers to the enthalpy change of the reaction, not the activation energy.
- Option B: The energy evolved when activated complex is formed: This is related to the stability of the activated complex, not the activation energy itself.
- Option D: The energy needed to form one mole of the product: This is related to the enthalpy of formation, not the activation energy.

Key Concepts:
- Activation energy
- Potential energy barrier
- Activated complex

Memory Aids:
- Activation energy is the "energy of activation" needed to start the reaction.

Quick Tips:
- Visualize the potential energy diagram to understand the concept of activation energy.

38 Easy

38 of 72Question permalink

Understanding +4 | -1

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:

View answer and explanation

Correct option: B

Powdering the lumps of coal

Option B (Powdering the lumps of coal) is correct because increasing the surface area by powdering the coal increases the rate of reaction.

Solution approach

Correct Answer + Core Reason: Powdering the lumps of coal. Increasing the surface area increases the rate of reaction.

Step-by-Step Solution:

Step 1: The reaction is $\mathrm{C}(s) + \mathrm{O}_2(g) \to \mathrm{CO}_2(g)$ .

Step 2: Consider the factors affecting the rate of reaction: concentration, temperature, surface area, and catalysts.

Step 3: Powdering the lumps of coal increases the surface area available for the reaction with oxygen.

Step 4: Increasing the surface area increases the number of collisions between carbon and oxygen, thus increasing the rate of reaction.

Why other options are wrong:
- Option A: Decrease in the concentration of oxygen: Decreasing the concentration of oxygen will decrease the rate of reaction.
- Option C: Decreasing the temperature: Decreasing the temperature will decrease the rate of reaction.
- Option D: Providing inert atmosphere for burning: Providing an inert atmosphere will decrease the concentration of oxygen, thus decreasing the rate of reaction.

Key Concepts:
- Factors affecting reaction rate
- Surface area and reaction rate

Memory Aids:
- Increasing surface area increases reaction rate.

Quick Tips:
- Consider the factors that affect reaction rate when analyzing the options.

39 Easy

39 of 72Question permalink

Understanding +4 | -1

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:

View answer and explanation

Correct option: A

Increasing the temperature

Option A (Increasing the temperature) is correct because increasing the temperature increases the rate constant, as described by the Arrhenius equation.

Solution approach

Correct Answer + Core Reason: Increasing the temperature. The rate constant is temperature-dependent according to the Arrhenius equation.

Step-by-Step Solution:

Step 1: The rate equation is given as $rate = k[\mathrm{NO}]^2[\mathrm{Cl}_2]$ .

Step 2: The rate constant $k$ is affected by temperature according to the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$ .

Step 3: Increasing the temperature $T$ increases the rate constant $k$ .

Step 4: Changing the concentrations of $\mathrm{NO}$ and $\mathrm{Cl}_2$ will change the rate of the reaction, but it will not change the value of the rate constant $k$ .

Why other options are wrong:
- Option B: Increasing the concentration of $\mathrm{NO}$ : Increasing the concentration of $\mathrm{NO}$ will increase the rate of the reaction, but it will not change the value of the rate constant $k$ .
- Option C: Increasing the concentration of the ${\mathrm{Cl}_2}$ : Increasing the concentration of ${\mathrm{Cl}_2}$ will increase the rate of the reaction, but it will not change the value of the rate constant $k$ .
- Option D: Doing all of these: Only increasing the temperature will increase the rate constant.

Key Concepts:
- Arrhenius equation
- Rate constant
- Temperature dependence of reaction rates

Memory Aids:
- Remember that the rate constant is temperature-dependent.

Quick Tips:
- Understand the difference between factors affecting the rate of reaction and the rate constant.

40 Easy

40 of 72Question permalink

Remembering +4 | -1

A zero order reaction is one:

View answer and explanation

Correct option: C

Whose rate does not change with time

The correct answer is C (Whose rate does not change with time) because the rate of a zero-order reaction is independent of the concentration of the reactants.

Solution approach

Correct Answer + Core Reason: Whose rate does not change with time. The rate of a zero-order reaction is constant and independent of the concentration of the reactants.

Step-by-Step Solution:

Step 1: Define a zero-order reaction: A reaction whose rate is independent of the concentration of the reactants.

Step 2: Rate law for a zero-order reaction: Rate = k, where k is the rate constant.

Step 3: Since the rate is equal to the rate constant, it does not change with time.

Why other options are wrong:
- Option A: Reactants do react in a zero-order reaction, but the rate is independent of their concentration.
- Option B: While having one reactant in large excess can sometimes lead to pseudo-order kinetics, it doesn't define a zero-order reaction in general.
- Option D: The rate of a zero-order reaction is constant, not increasing with time.

Key Concepts:
- Zero-order reaction
- Rate law
- Rate constant

Memory Aids:
- Zero-order means rate is independent of concentration.

Quick Tips:
- The rate of a zero-order reaction is constant and equal to the rate constant.

41 Easy

41 of 72Question permalink

Understanding +4 | -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.

View answer and explanation

Correct option: D

Pseudo first order

Option D (Pseudo first order) is correct because the reaction is first order with respect to the ester when water is in large excess.

Solution approach

Correct Answer + Core Reason: Pseudo first order. The reaction is actually bimolecular but behaves as first order because the concentration of one of the reactants (water) is in large excess and remains almost constant.

Step-by-Step Solution:

Step 1: Identify the reaction: $\mathrm{C}{\mathrm{H}_3}\mathrm{COO}{\mathrm{C}_2}{\mathrm{H}_5} + {\mathrm{H}_2}\mathrm{O}\xrightarrow{{\mathrm{H} + }}\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} + {\mathrm{C}_2}\mathrm{H}_5\mathrm{OH}$

Step 2: The rate law for this reaction is: rate = $k[{\mathrm{CH}_3}\mathrm{COO}{\mathrm{C}_2}{\mathrm{H}_5}][{\mathrm{H}_2}\mathrm{O}]$

Step 3: Since water is present in large excess, its concentration remains nearly constant during the reaction. Therefore, the rate law can be approximated as: rate = $k'[{\mathrm{CH}_3}\mathrm{COO}{\mathrm{C}_2}{\mathrm{H}_5}]$ , where $k' = k[{\mathrm{H}_2}\mathrm{O}]$

Step 4: This simplified rate law indicates a first-order reaction.

Why other options are wrong:
- Option A: Zero: The reaction is not zero order because the rate depends on the concentration of the ester.
- Option B: Second: The reaction is bimolecular, but due to the excess of water, it behaves as first order.
- Option C: Third: The reaction is not third order.

Key Concepts:
- Order of reaction
- Rate law
- Pseudo first-order reactions

Memory Aids:
- Pseudo first order: One reactant is in large excess.

Quick Tips:
- Look for reactions where one reactant is in large excess.

42 Easy

42 of 72Question permalink

Application +4 | -1

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

View answer and explanation

Correct option: C

32 times

Option C (32 times) is correct because the rate doubles for every 10°C rise, so for a 50°C rise, the rate increases by

2^5 = 32

times.

Solution approach

Correct Answer + Core Reason: 32 times. The rate doubles for every 10°C rise, so for a 50°C rise, the rate increases by $2^5 = 32$ times.

Step-by-Step Solution:

Step 1: Determine the number of 10°C intervals in a 50°C rise: $\frac{{50}}{{10}} = 5$

Step 2: Since the rate doubles for every 10°C rise, the rate increases by $2^5$ times.

Step 3: Calculate $2^5 = 32$

Step 4: Therefore, the rate of the reaction increases by 32 times.

Why other options are wrong:
- Option A (10 times): This is incorrect because it assumes a linear relationship between temperature and rate, rather than an exponential one.
- Option B (24 times): This is incorrect and doesn't follow the doubling pattern.
- Option D (64 times): This is incorrect because it corresponds to a 60°C rise ( $2^6 = 64$ ).

Key Concepts:
- Temperature dependence of reaction rates
- Exponential increase

Memory Aids:
- Remember that the rate doubles for every 10°C rise.

Quick Tips:
- Calculate the number of 10°C intervals and then raise 2 to that power.

43 Easy

43 of 72Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because if the rate of a gaseous reaction remains the same upon doubling the volume, it implies that the rate is independent of the concentration of the reactants, which is characteristic of a zero-order reaction.

Solution approach

Correct Answer + Core Reason: Zero. The rate is independent of concentration changes.

Step-by-Step Solution:

Step 1: Define the relationship between rate and volume: Doubling the volume halves the concentration of the reactants.

Step 2: Analyze the given information: The rate remains the same even when the volume is doubled (concentration is halved).

Step 3: Determine the order of the reaction: If the rate is unaffected by changes in concentration, the reaction is zero order.

Why other options are wrong:
- Option B: One: For a first-order reaction, the rate would halve when the volume is doubled.
- Option C: Two: For a second-order reaction, the rate would decrease by a factor of four when the volume is doubled.
- Option D: Three: For a third-order reaction, the rate would decrease by a factor of eight when the volume is doubled.

Key Concepts:
- Reaction order
- Rate law
- Effect of concentration on rate

Memory Aids:
- Zero order: Rate = k (independent of concentration)

Quick Tips:
- Relate the change in volume to the change in concentration and then to the change in rate for different reaction orders.

44 Easy

44 of 72Question permalink

Application +4 | -1

The time taken for the completion of

3/4

of a first order reaction is

View answer and explanation

Correct option: B

\left( {2.303/k} \right)\log 4

Option B (

\left( {2.303/k} \right)\log 4

) is correct because it represents the time taken for 3/4 completion of a first-order reaction based on the integrated rate law.

Solution approach

Correct Answer + Core Reason: $\left( {2.303/k} \right)\log 4$ . This is the correct expression for the time taken for 3/4 completion of a first-order reaction.

Step-by-Step Solution:

Step 1: Recall the integrated rate law for a first-order reaction: $t = \frac{{2.303}}{k}\log \frac{{[A]_0}}{{[A]_t}}$ , where $[A]_0$ is the initial concentration and $[A]_t$ is the concentration at time t.

Step 2: If 3/4 of the reaction is complete, then the remaining concentration is 1/4 of the initial concentration. So, $[A]_t = \frac{1}{4}[A]_0$ .

Step 3: Substitute $[A]_t = \frac{1}{4}[A]_0$ into the integrated rate law: $t = \frac{{2.303}}{k}\log \frac{{[A]_0}}{{\frac{1}{4}[A]_0}}$ .

Step 4: Simplify the expression: $t = \frac{{2.303}}{k}\log 4$ .

Why other options are wrong:
- Option A: This option is incorrect because it has $\log 3/4$ instead of $\log 4$ .
- Option C: This option is incorrect because it has $\log 1/4$ which would result in a negative time value.
- Option D: This option is incorrect because it has $\left( {2.303/0.75} \right)$ which is not the correct form of the integrated rate law.

Key Concepts:
- Integrated rate law for first-order reactions

Memory Aids:
- Remember the integrated rate law and how to relate the fraction of completion to the remaining concentration.

Quick Tips:
- Always use the integrated rate law when dealing with time and concentrations in reaction kinetics.

45 Easy

45 of 72Question permalink

Analyzing +4 | -1

The following equilibrium constants are given

{N_2} + 3{H_2} \rightleftarrows 2N{H_3};{K_1}

{N_2} + {O_2} \rightleftarrows 2NO;{K_2}

{H_2} + \frac{1}{2}{O_2} \rightleftarrows {H_2}O;{K_3}

The equilibrium constant for the oxidation of

2\,{\text{mol}}\,N{H_3}

by oxygen to

NO

is

View answer and explanation

Correct option: A

\frac{{{K_2}K_3^3}}{{{K_1}}}

Option A (

\frac{{{K_2}K_3^3}}{{{K_1}}}

) is correct because it represents the equilibrium constant for the oxidation of ammonia to NO, derived from the given equilibrium constants.

Solution approach

Correct Answer + Core Reason: $\frac{{{K_2}K_3^3}}{{{K_1}}}$ . This is derived by manipulating the given equilibrium reactions to obtain the desired reaction and combining their equilibrium constants.

Step-by-Step Solution:

Step 1: Write the desired reaction: $2N{H_3} + \frac{5}{2}{O_2} \rightleftharpoons 2NO + 3{H_2}O$ ; K

Step 2: Write the given reactions and their equilibrium constants:
- ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3};{K_1}$
- ${N_2} + {O_2} \rightleftharpoons 2NO;{K_2}$
- ${H_2} + \frac{1}{2}{O_2} \rightleftharpoons {H_2}O;{K_3}$

Step 3: Manipulate the given reactions to obtain the desired reaction:
- Reverse the first reaction: $2N{H_3} \rightleftharpoons {N_2} + 3{H_2};\frac{1}{{{K_1}}}$
- Keep the second reaction as is: ${N_2} + {O_2} \rightleftharpoons 2NO;{K_2}$
- Multiply the third reaction by 3: $3{H_2} + \frac{3}{2}{O_2} \rightleftharpoons 3{H_2}O;K_3^3$

Step 4: Add the manipulated reactions:
$2N{H_3} + {N_2} + {O_2} + 3{H_2} + \frac{3}{2}{O_2} \rightleftharpoons {N_2} + 3{H_2} + 2NO + 3{H_2}O$
$2N{H_3} + \frac{5}{2}{O_2} \rightleftharpoons 2NO + 3{H_2}O$

Step 5: Multiply the equilibrium constants of the manipulated reactions to get the equilibrium constant for the desired reaction:
$K = \frac{1}{{{K_1}}} \times {K_2} \times K_3^3 = \frac{{{K_2}K_3^3}}{{{K_1}}}$

Why other options are wrong:
- Option B: $\frac{{K_2^2{K_3}}}{{{K_3}}}$ : This is incorrect due to incorrect manipulation of the equilibrium constants.
- Option C: $\frac{{{K_1}{K_2}}}{{{K_3}}}$ : This is incorrect due to incorrect manipulation of the equilibrium constants.
- Option D: $\frac{{{K_2}K_3^3}}{{{K_1}}}$ : This is the same as the correct answer, but it was already identified as the correct answer.

Key Concepts:
- Equilibrium constant
- Hess's law of equilibrium

Memory Aids:
- When reversing a reaction, take the inverse of K. When multiplying a reaction by a factor, raise K to that power.

Quick Tips:
- Manipulate the given reactions to match the desired reaction and combine their equilibrium constants accordingly.

46 Easy

46 of 72Question permalink

Application +4 | -1

In a first order reaction the

a/\left( {a - x} \right)

was found to be 8 after 10 minute. The rate constant is:

View answer and explanation

Correct option: A

\left( {2.303 \times 3\log 2} \right)/10

Option A (

\left( {2.303 \times 3\log 2} \right)/10

) is correct because it correctly applies the integrated rate law for a first-order reaction to find the rate constant.

Solution approach

Correct Answer + Core Reason: $\left( {2.303 \times 3\log 2} \right)/10$ . This is the correct expression for the rate constant based on the given information.

Step-by-Step Solution:

Step 1: Recall the integrated rate law for a first-order reaction: $k = \frac{{2.303}}{t}\log \frac{a}{{a - x}}$ , where a is the initial concentration and (a-x) is the concentration at time t.

Step 2: We are given that $\frac{a}{{a - x}} = 8$ and $t = 10$ minutes.

Step 3: Substitute the given values into the integrated rate law: $k = \frac{{2.303}}{{10}}\log 8$ .

Step 4: Simplify the expression: $k = \frac{{2.303}}{{10}}\log {2^3} = \frac{{2.303}}{{10}} \times 3\log 2$ .

Why other options are wrong:
- Option B: This option is incorrect because it has $2\log 3$ instead of $3\log 2$ .
- Option C: This option is incorrect because the 10 is multiplied instead of divided.
- Option D: This option is incorrect because the 10 is multiplied instead of divided.

Key Concepts:
- Integrated rate law for first-order reactions

Memory Aids:
- Remember the integrated rate law and how to use the given ratio of concentrations.

Quick Tips:
- Always use the integrated rate law when dealing with time and concentrations in reaction kinetics.

47 Easy

47 of 72Question permalink

Remembering +4 | -1

The temperature coefficient of a reaction is:

View answer and explanation

Correct option: C

The ratio of rate constant differing by

10^\circ C

, preferably

35^\circ C\;and\;25^\circ C

Option C is correct because the temperature coefficient is defined as the ratio of rate constants at two temperatures differing by 10°C.

Solution approach

Correct Answer + Core Reason: The ratio of rate constant differing by $10^\circ C$ , preferably $35^\circ C\;and\;25^\circ C$ . The temperature coefficient describes how the reaction rate changes with a 10°C temperature difference.

Step-by-Step Solution:

Step 1: Define the temperature coefficient.

Step 2: The temperature coefficient is given by the ratio of the rate constant at two temperatures differing by 10°C.

Step 3: Mathematically, it can be expressed as: Temperature Coefficient = $\frac{{k_{{T+10}}}}{{k_T}}$ , where k is the rate constant and T is the temperature.

Why other options are wrong:
- Option A: The rate constant at a fixed temperature is simply the rate constant, not the temperature coefficient.
- Option B: The ratio of rate constant at two temperature is a general statement, but the temperature coefficient specifically refers to a 10°C difference.
- Option D: Option C is the correct definition, so "None of the above" is incorrect.

Key Concepts:
- Chemical kinetics
- Rate constant
- Temperature coefficient

Memory Aids:
- Remember that the temperature coefficient relates rate constants at temperatures 10°C apart.

Quick Tips:
- The temperature coefficient is usually around 2 to 3, meaning the reaction rate doubles or triples for every 10°C increase in temperature.

48 Easy

48 of 72Question permalink

Remembering +4 | -1

In the respect of the equation

k = A{e^{ - {E_a}/RT}}

in chemical kinetics, which one of the following statements is correct?

View answer and explanation

Correct option: C

{E_a}

is energy of activation

The correct answer is C (

{E_a}

is energy of activation) because

E_a

represents the activation energy in the Arrhenius equation.

Solution approach

Correct Answer + Core Reason: ${E_a}$ is energy of activation. In the Arrhenius equation, $k = A{e^{ - {E_a}/RT}}$ , $E_a$ represents the activation energy.

Step-by-Step Solution:

Step 1: Recall the Arrhenius equation: $k = A{e^{ - {E_a}/RT}}$

Step 2: Identify each term:\nk = rate constant\nA = pre-exponential factor (frequency factor)
$E_a$ = activation energy\nR = gas constant\nT = temperature (in Kelvin)

Step 3: From the equation, it is clear that $E_a$ represents the energy of activation.

Why other options are wrong:
- Option A: k is the rate constant, not the equilibrium constant.
- Option B: A is the pre-exponential factor or frequency factor, not the adsorption factor.
- Option D: R is the gas constant, not the Rydberg constant.

Key Concepts:
- Arrhenius equation
- Activation energy
- Rate constant

Memory Aids:
- Remember the Arrhenius equation: $k = A{e^{ - {E_a}/RT}}$

Quick Tips:
- Know the meaning of each term in the Arrhenius equation.

49 Easy

49 of 72Question permalink

Understanding +4 | -1

The activation energy of a reaction is zero. The rate constant for the reaction

View answer and explanation

Correct option: D

Is nearly independent of temp

Option D (Is nearly independent of temp) is correct because when the activation energy is zero, the rate constant is independent of temperature according to the Arrhenius equation.

Solution approach

Correct Answer + Core Reason: Is nearly independent of temp. When activation energy is zero, the rate constant is independent of temperature.

Step-by-Step Solution:

Step 1: Recall the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$ , where $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature.

Step 2: If the activation energy $E_a = 0$ , then the equation becomes $k = A e^0 = A$ .

Step 3: Since $k = A$ , the rate constant $k$ is equal to the pre-exponential factor $A$ , which is independent of temperature.

Step 4: Therefore, the rate constant is nearly independent of temperature when the activation energy is zero.

Why other options are wrong:
- Option A: Decreases with decrease of temp: This is incorrect because if $E_a = 0$ , the rate constant does not depend on temperature.
- Option B: Increases with increase of temp: This is incorrect because if $E_a = 0$ , the rate constant does not depend on temperature.
- Option C: Decreases with increase of temp: This is incorrect because if $E_a = 0$ , the rate constant does not depend on temperature.

Key Concepts:
- Arrhenius equation
- Activation energy
- Rate constant

Memory Aids:
- Remember the Arrhenius equation and how activation energy affects the rate constant.

Quick Tips:
- If the activation energy is zero, the rate constant is independent of temperature.

50 Easy

50 of 72Question permalink

Remembering +4 | -1

For the first order reaction with the rate constant

k

, which expression gives the rate half-life period?

(Initial\mathrm{ }conc.\mathrm{ } = a)

View answer and explanation

Correct option: D

\frac{{0.693}}{k}

Option D (

\frac{{0.693}}{k}

) is correct because it is the standard formula for the half-life of a first-order reaction.

Solution approach

Correct Answer + Core Reason: $\frac{{0.693}}{k}$ . This is the standard formula for the half-life of a first-order reaction.

Step-by-Step Solution:

Step 1: Recall the integrated rate law for a first-order reaction: $[A]_t = [A]_0 e^{-kt}$ , where $[A]_t$ is the concentration at time t, $[A]_0$ is the initial concentration, k is the rate constant, and t is time.

Step 2: Define half-life ( $t_{1/2}$ ) as the time it takes for the concentration to reduce to half of its initial value: $[A]_{t_{1/2}} = \frac{1}{2}[A]_0$ .

Step 3: Substitute this into the integrated rate law: $\frac{1}{2}[A]_0 = [A]_0 e^{-kt_{1/2}}$ .

Step 4: Divide both sides by $[A]_0$ : $\frac{1}{2} = e^{-kt_{1/2}}$ .

Step 5: Take the natural logarithm of both sides: $\ln{\frac{1}{2}} = -kt_{1/2}$ .

Step 6: Simplify: $-\ln{2} = -kt_{1/2}$ .

Step 7: Solve for $t_{1/2}$ : $t_{1/2} = \frac{\ln{2}}{k} = \frac{0.693}{k}$ .

Why other options are wrong:
- **Option A ( $\frac{3}{2}k{a^2}$ ): This expression is not related to the half-life of a first-order reaction.
- Option B ( $\frac{1}{{ka}}$ ): This expression is related to the half-life of a second-order reaction.
- Option C ( $\frac{1}{k}$ ): This expression is not the correct half-life for a first-order reaction.

Key Concepts:
- First-order kinetics
- Half-life of a reaction
- Integrated rate law

Memory Aids:
- Remember that the half-life of a first-order reaction is independent of the initial concentration.

Quick Tips:**
- Memorize the formula for the half-life of a first-order reaction.

51 Easy

51 of 72Question permalink

Application +4 | -1

For the reaction,

2{N_2}{O_5}\left( g \right) \to 4N{O_2}\left( g \right) + {O_2}\left( g \right)

If the concentration of

N{O_2}

increase by

5.2 \times {10^{ - 3}}\;M

in 100 s then the rate of the reactions

View answer and explanation

Correct option: A

1.3 \times {10^{ - 5}}\;M\;{s^{ - 1}}

Option A (

1.3 \times {10^{ - 5}}\;M\;{s^{ - 1}}

) is correct because it is the rate of the reaction calculated from the rate of formation of

NO_2

considering the stoichiometry.

Solution approach

Correct Answer + Core Reason: $1.3 \times {10^{ - 5}}\;M\;{s^{ - 1}}$ . The rate of reaction is calculated based on the stoichiometry of the reaction and the given rate of formation of $NO_2$ .

Step-by-Step Solution:

Step 1: Write the rate expression for the reaction:\nRate = $-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$

Step 2: We are given that $\frac{d[NO_2]}{dt} = \frac{5.2 \times 10^{-3} M}{100 s} = 5.2 \times 10^{-5} M/s$

Step 3: We need to find the rate of the reaction, which is $\frac{1}{4} \frac{d[NO_2]}{dt}$ .

Step 4: Rate = $\frac{1}{4} \times 5.2 \times 10^{-5} M/s = 1.3 \times 10^{-5} M/s$

Step 5: Final Answer: $1.3 \times 10^{-5} M/s$

Why other options are wrong:
- Option B: This is incorrect because it's not calculated using the correct stoichiometric coefficient.
- Option C: This is incorrect because it's not calculated using the correct stoichiometric coefficient.
- Option D: This is incorrect because it's not calculated using the correct stoichiometric coefficient.

Key Concepts:
- Rate of reaction
- Stoichiometry and rate expression

Memory Aids:
- Remember to divide the rate of formation/disappearance by the stoichiometric coefficient when calculating the rate of reaction.

Quick Tips:
- Always write the balanced chemical equation and the corresponding rate expression before solving the problem.

52 Easy

52 of 72Question permalink

Application +4 | -1

For the reaction,

2A + B \to products

, the active mass of B is kept constant, and that of A is doubled. The rate of reaction will be then

View answer and explanation

Correct option: C

Increase 4 times

Option C (Increase 4 times) is correct because if the rate law is rate = k[A]²[B], doubling [A] while keeping [B] constant will quadruple the rate.

Solution approach

Correct Answer + Core Reason: Increase 4 times. The rate of reaction is proportional to the square of the concentration of A.

Step-by-Step Solution:

Step 1: Assume the rate law is of the form: Rate = $k[A]^m[B]^n$ .

Step 2: Since the active mass of B is kept constant, we can focus on the effect of changing the active mass of A.

Step 3: If the active mass of A is doubled, the new rate will be: Rate' = $k[2A]^m[B]^n = 2^m k[A]^m[B]^n = 2^m \times Rate$ .

Step 4: Since the stoichiometry is $2A + B \to products$ , it is likely that the reaction is second order with respect to A. Therefore, $m=2$ .

Step 5: Rate' = $2^2 \times Rate = 4 \times Rate$ .

Why other options are wrong:
- Option A: Decrease 4 times: This is incorrect because the rate increases when the concentration of A increases.
- Option B: Decrease 2 times: This is incorrect because the rate increases when the concentration of A increases, and it increases by a factor of 4, not 2.
- Option D: Increase 2 times: This would be correct if the reaction was first order with respect to A.

Key Concepts:
- Rate Law
- Order of Reaction

Memory Aids:
- The rate is proportional to the concentration of the reactants raised to their respective orders.

Quick Tips:
- Determine the order of the reaction with respect to each reactant.

53 Easy

53 of 72Question permalink

Remembering +4 | -1

The unit of rate constant for a zero order reaction is

View answer and explanation

Correct option: A

mol.{L^{ - 1}}.{S^{ - 1}}

Option A (

mol.{L^{ - 1}}.{S^{ - 1}}

) is correct because it represents the unit of the rate constant for a zero-order reaction, where the rate is independent of reactant concentration.

Solution approach

Correct Answer + Core Reason: $mol.{L^{ - 1}}.{S^{ - 1}}$ . The rate of a zero-order reaction is independent of the concentration of the reactants.

Step-by-Step Solution:

Step 1: Define zero-order reaction: Rate = k[A]⁰ = k

Step 2: Determine the units of the rate constant (k): Since Rate = k, the units of k are the same as the units of Rate.

Step 3: Units of Rate: Rate is typically expressed as change in concentration per unit time, so the units are $mol.{L^{ - 1}}.{S^{ - 1}}$ .

Step 4: Therefore, the unit of the rate constant for a zero-order reaction is $mol.{L^{ - 1}}.{S^{ - 1}}$ .

Why other options are wrong:
- Option B: $L.mo{l^{ - 1}}.{S^{ - 1}}$ : This is the unit for the rate constant of a first-order reaction.
- Option C: ${L^2}mo{l^{ - 2}}{S^{ - 1}}$ : This is not the correct unit for any common reaction order.
- Option D: ${S^{ - 1}}$ : This is the unit for the rate constant of a first-order reaction when dealing with unimolecular gas-phase reactions where concentration is expressed in pressure units.

Key Concepts:
- Zero-order reaction
- Rate constant
- Units of rate and rate constant

Memory Aids:
- Zero order = rate independent of concentration

Quick Tips:
- Remember the units for different reaction orders. For a reaction of order n, the units of k are $mol^{1-n} L^{n-1} s^{-1}$

54 Easy

54 of 72Question permalink

Remembering +4 | -1

The branch of chemistry which deals with the reaction rates and reaction mechanism is called:

View answer and explanation

Correct option: D

Chemical kinetics

Option D (Chemical kinetics) is correct because chemical kinetics is the branch of chemistry specifically focused on reaction rates and reaction mechanisms.

Solution approach

Correct Answer + Core Reason: Chemical kinetics deals with reaction rates and mechanisms.

Step-by-Step Solution:

Step 1: Define the terms.

Step 2: Thermochemistry deals with heat changes in chemical reactions.

Step 3: Photochemistry deals with chemical reactions initiated by light.

Step 4: Analytical chemistry deals with the identification and quantification of substances.

Step 5: Chemical kinetics deals with the rates and mechanisms of chemical reactions.

Why other options are wrong:
- Option A: Thermochemistry focuses on the heat involved in chemical reactions, not the rates or mechanisms.
- Option B: Photochemistry focuses on reactions initiated by light, not general reaction rates and mechanisms.
- Option C: Analytical chemistry focuses on identifying and quantifying substances, not reaction rates and mechanisms.
- Option D: Chemical kinetics is the study of reaction rates and mechanisms.

Key Concepts:
- Chemical kinetics
- Thermochemistry
- Photochemistry
- Analytical chemistry

Memory Aids:
- Kinetics = movement/rate

Quick Tips:
- Remember the specific focus of each branch of chemistry.

55 Easy

55 of 72Question permalink

Remembering +4 | -1

The Arrhenius equation expressing the effect of temperature on the rate constant of reaction is:

View answer and explanation

Correct option: B

K = A{e^{ - {E_a}/RT}}

Option B is correct because the Arrhenius equation is mathematically represented as

K = A{e^{ - {E_a}/RT}}

.

Solution approach

Correct Answer + Core Reason: $K = A{e^{ - {E_a}/RT}}$ . This is the standard form of the Arrhenius equation.

Step-by-Step Solution:

Step 1: Recall the Arrhenius equation. The Arrhenius equation is $k = A e^{-E_a/RT}$ , where k is the rate constant, A is the pre-exponential factor, $E_a$ is the activation energy, R is the gas constant, and T is the temperature.

Step 2: Identify the correct form. The correct form of the equation is $K = A{e^{ - {E_a}/RT}}$ .

Why other options are wrong:
- Option A: $K = \frac{{{E_a}}}{{RT}}$ is not the Arrhenius equation.
- Option C: $K = {\log _e}\frac{{{E_a}}}{{RT}}$ is not the Arrhenius equation.
- Option D: $K = {e^{ - Ea/RT}}$ is missing the pre-exponential factor A.

Key Concepts:
- Arrhenius equation
- Rate constant
- Activation energy

Memory Aids:
- Remember the exponential term and the pre-exponential factor.

Quick Tips:
- The Arrhenius equation is essential for understanding the temperature dependence of reaction rates.

56 Easy

56 of 72Question permalink

Analyzing +4 | -1

Which one of the following is wrongly matched?

View answer and explanation

Correct option: D

Combination of

{H_2}

and

{B_{r2}}\;

to give

HBr

-first order reaction

Option D (Combination of

{H_2}

and

{B_{r2}}\;

to give

HBr

- first order reaction) is wrongly matched because the reaction is not first order.

Solution approach

Correct Answer + Core Reason: Combination of ${H_2}$ and ${B_{r2}}\;$ to give $HBr$ - first order reaction. This reaction has a complex rate law and is not simply first order.

Step-by-Step Solution:

Step 1: Analyze each option:
- A: Saponification of $C{H_3}COO{C_2}{H_5}$ - second order reaction: This is generally true as saponification involves the reaction of an ester with a base, which is second order.
- B: Hydrolysis of $C{H_3}COOC{H_3}$ - pseudo unimolecular: This is true because the concentration of water is in large excess, making it appear as a first-order reaction.
- C: Decomposition of ${H_2}{O_2}$ - first order reaction: This is generally true as the decomposition of hydrogen peroxide follows first-order kinetics.
- D: Combination of ${H_2}$ and ${B_{r2}}$ to give $HBr$ - first order reaction: The actual rate law for this reaction is complex and involves fractional orders. It is not a simple first-order reaction.

Step 2: The rate law for the reaction ${H_2} + B{r_2} \to 2HBr$ is given by $\frac{{d[HBr]}}{{dt}} = \frac{{k[H_2][B{r_2}]^{1/2}}}{{1 + k'[HBr]/[B{r_2}]}}$ . This is not a first-order reaction.

Why other options are wrong:
- Option A: Saponification of $C{H_3}COO{C_2}{H_5}$ - second order reaction: This is a correct match.
- Option B: Hydrolysis of $C{H_3}COOC{H_3}$ - pseudo unimolecular: This is a correct match.
- Option C: Decomposition of ${H_2}{O_2}$ - first order reaction: This is a correct match.

Key Concepts:
- Order of reaction
- Rate law
- Pseudo unimolecular reaction

Memory Aids:
- Remember common reaction orders for typical reactions.

Quick Tips:
- Be aware of the complex rate laws for certain reactions, such as the formation of HBr.

57 Easy

57 of 72Question permalink

Application +4 | -1

If the rate of reaction

A \to B

doubles on increasing the concentration of A by 4 times, the order of the reaction is

View answer and explanation

Correct option: C

\frac{1}{2}

Option C (

\frac{1}{2}

) is correct because the rate doubles when the concentration increases by a factor of 4, indicating a square root relationship.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}$ . The order of the reaction is 1/2 because rate is proportional to the square root of the concentration.

Step-by-Step Solution:

Step 1: Let the rate law be: Rate = $k[A]^n$ , where n is the order of the reaction.

Step 2: We are given that when the concentration of A is increased by 4 times, the rate doubles. So, if $[A]_2 = 4[A]_1$ , then Rate $_2 = 2$ Rate $_1$ .

Step 3: Write the rate laws for both conditions: Rate $_1 = k[A]_1^n$ and Rate $_2 = k[A]_2^n$ .

Step 4: Substitute the given relationships: $2$ Rate $_1 = k(4[A]_1)^n$ .

Step 5: Divide the second equation by the first: $\frac{{2\mathrm{Rate}{\mathrm{ }_1}}}{{\mathrm{Rate}{\mathrm{ }_1}}} = \frac{{k{{\left( {4{{\left[ A \right]}_1}} \right)}^n}}}{{k{{\left[ A \right]}_1}^n}}$ .

Step 6: Simplify: $2 = 4^n$ .

Step 7: Solve for n: $n = \frac{1}{2}$ .

Why other options are wrong:
- Option A: This option is incorrect because if the order was 2, the rate would increase by a factor of 16.
- Option B: This option is incorrect because if the order was 1, the rate would increase by a factor of 4.
- Option D: This option is incorrect because if the order was 4, the rate would increase by a factor of 256.

Key Concepts:
- Rate law and order of reaction

Memory Aids:
- Remember that the order of the reaction determines how the rate changes with concentration.

Quick Tips:
- Use the given information to set up a ratio of rates and concentrations to solve for the order of the reaction.

58 Easy

58 of 72Question permalink

Remembering +4 | -1

The rate at which a substance reacts depends on its

View answer and explanation

Correct option: D

Active mass

Option D is correct because the rate of a reaction is directly proportional to the active mass (effective concentration) of the reactants.

Solution approach

Correct Answer + Core Reason: Active mass. The rate of a reaction depends on the effective concentration, or active mass, of the reactants.

Step-by-Step Solution:

Step 1: Define rate of reaction. The rate of a reaction is the change in concentration of reactants or products per unit time.

Step 2: Relate rate to active mass. The rate law expresses the relationship between the rate of a reaction and the concentrations of the reactants. The active mass is a measure of the effective concentration of a substance in a reaction mixture.

Step 3: Atomic weight, atomic number, and molecular weight are properties of the substance but do not directly determine the rate at which it reacts.

Why other options are wrong:
- Option A: Atomic weight is a property of the substance but doesn't directly determine the reaction rate.
- Option B: Atomic number is a property of the substance but doesn't directly determine the reaction rate.
- Option C: Molecular weight is a property of the substance but doesn't directly determine the reaction rate.

Key Concepts:
- Rate of reaction
- Active mass
- Rate law

Memory Aids:
- Rate depends on concentration (active mass).

Quick Tips:
- The rate law is experimentally determined and relates rate to reactant concentrations.

59 Easy

59 of 72Question permalink

Understanding +4 | -1

The half time of a second order reaction is:

View answer and explanation

Correct option: B

Inversely proportional to the initial concentration of the reactants

Option B (Inversely proportional to the initial concentration of the reactants) is correct because the half-life of a second-order reaction is inversely proportional to the initial concentration.

Solution approach

Correct Answer + Core Reason: Inversely proportional to the initial concentration of the reactants. The half-life of a second-order reaction is inversely proportional to the initial concentration.

Step-by-Step Solution:

Step 1: Recall the rate law for a second-order reaction: Rate = $k[A]^2$ , where k is the rate constant and [A] is the concentration of the reactant.

Step 2: The integrated rate law for a second-order reaction is: $\frac{1}{{[A]_t}} = \frac{1}{{[A]_0}} + kt$ , where $[A]_t$ is the concentration at time t, $[A]_0$ is the initial concentration, k is the rate constant, and t is time.

Step 3: Define half-life ( $t_{1/2}$ ) as the time it takes for the concentration to reduce to half of its initial value: $[A]_{t_{1/2}} = \frac{1}{2}[A]_0$ .

Step 4: Substitute this into the integrated rate law: $\frac{1}{{\frac{1}{2}[A]_0}} = \frac{1}{{[A]_0}} + kt_{1/2}$ .

Step 5: Simplify: $\frac{2}{{[A]_0}} = \frac{1}{{[A]_0}} + kt_{1/2}$ .

Step 6: Solve for $t_{1/2}$ : $t_{1/2} = \frac{1}{{k[A]_0}}$ .

Step 7: Conclusion: The half-life is inversely proportional to the initial concentration of the reactants.

Why other options are wrong:
- Option A (Inversely proportional to the square of the initial concentration of the reactants): This is incorrect for a second-order reaction.
- Option C (Proportional to the initial concentration of reactants): This is incorrect for a second-order reaction.
- Option D (Independent of the initial concentration of reactants): This is true for a first-order reaction, not a second-order reaction.

Key Concepts:
- Second-order kinetics
- Half-life of a reaction
- Integrated rate law

Memory Aids:
- Remember the relationship between half-life and initial concentration for different reaction orders.

Quick Tips:
- Memorize the formula for the half-life of a second-order reaction.

60 Easy

60 of 72Question permalink

Understanding +4 | -1

Which is correct about zero order reaction?

View answer and explanation

Correct option: B

Rate of reaction is independent of concentration.

Option B (Rate of reaction is independent of concentration) is correct because, by definition, a zero-order reaction's rate is not influenced by the reactant concentration.

Solution approach

Correct Answer + Core Reason: Rate of reaction is independent of concentration for a zero-order reaction.

Step-by-Step Solution:

Step 1: Define a zero-order reaction. A zero-order reaction is one where the rate of the reaction is independent of the concentration of the reactants.

Step 2: Express the rate law for a zero-order reaction: Rate = k[A]⁰ = k, where k is the rate constant and [A] is the concentration of the reactant.

Step 3: Analyze the rate law. Since the rate is equal to the rate constant, it does not depend on the concentration of the reactant.

Why other options are wrong:
- Option A: Rate of reaction depends on decay constant. While the rate is equal to the rate constant, this statement is misleading as it doesn't highlight the independence from concentration.
- Option C: Unit of rate constant is $conc^{-1}$ . This is incorrect; the unit of the rate constant for a zero-order reaction is concentration/time.
- Option D: Unit of rate constant is $conc^{-1} time^{-1}$ . This is the unit for a second-order reaction, not a zero-order reaction.

Key Concepts:
- Zero-order reaction
- Rate law
- Rate constant

Memory Aids:
- Zero-order means zero dependence on concentration.
- Rate = k for zero-order reactions.

Quick Tips:
- Know the rate laws and units of rate constants for different reaction orders.
- Remember that zero-order reactions are rare and often occur on surfaces or with catalysts.

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