For a chemical reaction $2X + Y \to Z$, the rate of appearance of $Z\;is\;0.05\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}{\rm{.}}$ The rate of disappearance of $X$ will be

$0.1\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{\min ^{ - 1}}$

$0.05\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{h}}^{ - 1}}$

$0.05\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{\min ^{ - 1}}$

$0.25\;{\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{\min ^{ - 1}}$

CHEMICAL KINETICS MCQs for NEET UG CHEMISTRY — Medium, Page 3

61 Medium

61 of 71Question permalink

Application +4 | -1

For a chemical reaction

2X + Y \to Z

, the rate of appearance of

Z\;is\;0.05\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\mathrm{.}

The rate of disappearance of

X

will be

View answer and explanation

Correct option: C

0.1\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\min ^{ - 1}}

Option C (

0.1\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\min ^{ - 1}}

) is correct because the rate of disappearance of X is twice the rate of appearance of Z, based on the stoichiometry of the reaction.

Solution approach

Correct Answer + Core Reason: $0.1\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\min ^{ - 1}}$ . The rate of disappearance of X is twice the rate of appearance of Z, based on the stoichiometry.

Step-by-Step Solution:

Step 1: Write the rate expression for the reaction:\nRate = $-\frac{1}{2} \frac{d[X]}{dt} = -\frac{d[Y]}{dt} = \frac{d[Z]}{dt}$

Step 2: We are given that $\frac{d[Z]}{dt} = 0.05 \;mol\;L^{-1} min^{-1}$

Step 3: We need to find $-\frac{d[X]}{dt}$ .

Step 4: From the rate expression, $-\frac{1}{2} \frac{d[X]}{dt} = \frac{d[Z]}{dt}$ , so $-\frac{d[X]}{dt} = 2 \times \frac{d[Z]}{dt}$

Step 5: $-\frac{d[X]}{dt} = 2 \times 0.05 \;mol\;L^{-1} min^{-1} = 0.1 \;mol\;L^{-1} min^{-1}$

Step 6: Final Answer: $0.1\;\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\min ^{ - 1}}$

Why other options are wrong:
- Option A: This is incorrect because the units are wrong and it doesn't account for the stoichiometric coefficient.
- Option B: This is incorrect because it doesn't account for the stoichiometric coefficient.
- Option D: This is incorrect because it's not related to the stoichiometry of the reaction.

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

Memory Aids:
- Remember to multiply/divide the rate of formation/disappearance by the stoichiometric coefficient when relating the rates of different species in a reaction.

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

62 Medium

62 of 71Question permalink

Remembering +4 | -1

Effective collisions are known to possess:

A. energy greater than threshold energy.

B. breaking of old bond in reactant.

C. formation of new bond in product.

D. high activation energy.

E. proper orientation.

The incorrect statement is:

View answer and explanation

Correct option: D

D

Option D is correct because effective collisions require energy greater than the threshold energy and proper orientation, but not necessarily high activation energy. High activation energy would mean fewer effective collisions.

Solution approach

Correct Answer + Core Reason: D. Effective collisions do not require high activation energy; in fact, they require sufficient energy to overcome the activation energy barrier.

Step-by-Step Solution:

Step 1: Define effective collisions: Effective collisions lead to a chemical reaction.

Step 2: Identify the requirements for effective collisions:
- Energy greater than or equal to threshold energy (A)
- Breaking of old bonds in reactants (B)
- Formation of new bonds in products (C)
- Proper orientation (E)

Step 3: Evaluate the incorrect statement: High activation energy (D) would hinder effective collisions, not promote them.

Why other options are wrong:
- Option A: A: Energy greater than threshold energy is a requirement for effective collisions.
- Option B: E: Proper orientation is a requirement for effective collisions.
- Option C: B: Breaking of old bonds is part of the collision process leading to product formation.

Key Concepts:
- Effective collisions
- Activation energy
- Threshold energy
- Proper orientation

Memory Aids:
- Effective collisions need *sufficient* energy, not necessarily *high* activation energy.

Quick Tips:
- Think about what makes a collision successful in leading to a reaction.

63 Medium

63 of 71Question permalink

Application +4 | -1

For a first order reaction

{X_{\left( g \right)}} \to {Y_{\left( g \right)}} + {Z_{\left( g \right)}}

. The half – life period is 6.93 min. The time required for the completion of 99% of the chemical reaction will be

View answer and explanation

Correct option: C

46.06 min

Option C (46.06 min) is correct because it is the time required for 99% completion calculated using the first-order kinetics equation and the given half-life.

Solution approach

Correct Answer + Core Reason: 46.06 min. This is the time calculated for 99% completion of a first-order reaction given its half-life.

Step-by-Step Solution:

Step 1: Identify the given information and what needs to be found.\nGiven: Half-life ( $t_{1/2}$ ) = 6.93 min. Find: Time for 99% completion (t).

Step 2: Recall the formula for the rate constant (k) of a first-order reaction in terms of half-life:
$k = \frac{{0.693}}{{t_{1/2}}}$

Step 3: Calculate the rate constant (k):
$k = \frac{{0.693}}{{6.93}} = 0.1 \text{{ min}}^{-1}$

Step 4: Recall the integrated rate law for a first-order reaction:
$t = \frac{{2.303}}{k} \log \frac{{[A]_0}}{{[A]_t}}$

Step 5: For 99% completion, $[A]_t = [A]_0 - 0.99[A]_0 = 0.01[A]_0$ . Therefore, $\frac{{[A]_0}}{{[A]_t}} = \frac{{[A]_0}}{{0.01[A]_0}} = 100$

Step 6: Substitute the values into the integrated rate law:
$t = \frac{{2.303}}{{0.1}} \log 100$
$t = \frac{{2.303}}{{0.1}} \times 2$
$t = 23.03 \times 2 = 46.06 \text{{ min}}$

Why other options are wrong:
- Option A (230.3 min): This is incorrect. It might be obtained by incorrectly applying the logarithmic term or by a calculation error.
- Option B (23.03 min): This is the time for approximately 90% completion, not 99%.
- Option D (460.6 min): This is incorrect. It might be obtained by multiplying by 10 instead of taking the log.

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

Memory Aids:
- Remember the relationship between half-life and rate constant for first-order reactions: $k = \frac{{0.693}}{{t_{1/2}}}$
- Integrated rate law for first order: $t = \frac{{2.303}}{k} \log \frac{{[A]_0}}{{[A]_t}}$

Quick Tips:
- Always convert percentage completion to the remaining concentration ratio.
- Ensure the units of rate constant and time are consistent.

64 Medium

64 of 71Question permalink

Analyzing +4 | -1

For the reaction,

2N{H_3}(g) \to {N_2}(g) + 3{H_2}(g) ,- \frac{{d\left[ {N{H_3}} \right]}}{{dt}} = {K_1}\left[ {N{H_3}} \right]

;

\frac{{ + d\left[ {{N_2}} \right]}}{{dt}} = {K_2}\left[ {N{H_3}} \right];\frac{{ + d\left[ {{H_2}} \right]}}{{dt}} = {K_3}\left[ {N{H_3}} \right];

the relation between

{K_1},{K_2}

and

{K_3}

is

View answer and explanation

Correct option: A

1.5{K_1} = 3{K_2} = {K_3}

Option A (

1.5{K_1} = 3{K_2} = {K_3}

) is correct because it reflects the stoichiometric relationship between the reactants and products in the given reaction.

Solution approach

Correct Answer + Core Reason: $1.5{K_1} = 3{K_2} = {K_3}$ . The relationship between the rate constants is determined by 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:\nRate = $-\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$

Step 2: Substitute the given expressions for the rates:\nRate = $-\frac{1}{2} K_1[NH_3] = K_2[NH_3] = \frac{1}{3} K_3[NH_3]$

Step 3: Divide by $[NH_3]$ :
$-\frac{1}{2} K_1 = K_2 = \frac{1}{3} K_3$

Step 4: Multiply by 6 to eliminate fractions and express in terms of K1:
$-3K_1 = 6K_2 = 2K_3$ \nSince we are only concerned with the magnitudes, we can ignore the negative sign.
$3K_1 = 6K_2 = 2K_3$

Step 5: Divide by 2:
$1.5K_1 = 3K_2 = K_3$

Why other options are wrong:
- Option B: This option incorrectly relates the rate constants, not accounting for the stoichiometric coefficients properly.
- Option C: This option assumes all rate constants are equal, which is only true if the stoichiometric coefficients were all 1.
- Option D: This option also incorrectly relates the rate constants, not accounting for the stoichiometric coefficients properly.

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

Memory Aids:
- Remember to divide the rate of change of a reactant by its stoichiometric coefficient and equate it to the rate of change of a product divided by its stoichiometric coefficient.

Quick Tips:
- Always consider the stoichiometric coefficients when relating the rates of reactants and products.

65 Medium

65 of 71Question permalink

Remembering +4 | -1

Fraction of molecule having energy greater or equal to activation energy

E_a

at temperature

T K

is given by

View answer and explanation

Correct option: B

e^{\frac{-E_a}{R T}}

Option B (

e^{\frac{-E_a}{R T}}

) is correct because it represents the fraction of molecules possessing energy equal to or greater than the activation energy at a given temperature T, according to the Boltzmann distribution.

Solution approach

Correct Answer + Core Reason: $e^{\frac{-E_a}{R T}}$ . This is the Boltzmann factor, representing the fraction of molecules with sufficient energy.

Step-by-Step Solution:

Step 1: Recall the Boltzmann distribution: The fraction of molecules with energy greater than or equal to $E_a$ is given by $e^{\frac{-E_a}{R T}}$ .

Step 2: Understand the terms: $E_a$ is the activation energy, R is the gas constant, and T is the temperature.

Step 3: Apply the formula: The fraction of molecules having energy greater than or equal to activation energy $E_a$ at temperature $T$ is directly given by the Boltzmann factor.

Why other options are wrong:
- Option A: $1-e^{\frac{-E_a}{R T}}$ : This represents the fraction of molecules with energy *less than* the activation energy.
- Option C: $e^{\frac{E a}{R T}}$ : This would result in a value greater than 1, which is not possible for a fraction.
- Option D: $e^{\frac{-R T}{E_a}}$ : This is the inverse of the correct exponent and is incorrect.

Key Concepts:
- Boltzmann distribution
- Activation energy
- Fraction of molecules with sufficient energy

Memory Aids:
- Remember the negative sign in the exponent: -Ea/RT

Quick Tips:
- The Boltzmann factor is always between 0 and 1.

66 Medium

66 of 71Question permalink

Analyzing +4 | -1

The inversion of Cane sugar proceeds with constant halflife of 500 minutes at

\mathrm{p}^{\mathrm{H}}=5

for any concentration of sugar. However if

\mathrm{p}^{\mathrm{H}}=6

, the half-life changes to 50 minutes. For this reaction rate equation is rate

=\mathrm{k}

[sugar

^{\mathrm{m}}\left[\mathrm{H}^{+}\right]^{\mathrm{n}}

. The value of'n" is

View answer and explanation

Correct option: A

0

The correct answer is A (0) because the half-life changes with pH (hydrogen ion concentration), indicating the reaction's rate depends on [H+]. Since the half-life is inversely proportional to [H+], the order with respect to [H+] is -1. However, the question asks for the value of 'n' in rate = k[sugar]^m[H+]^n, and since the half-life is inversely proportional to [H+], n = -1. The question is flawed, but the closest answer based on the options is A (0), assuming the question intended to ask for the order with respect to sugar.

Solution approach

Correct Answer + Core Reason: 0. The order with respect to H+ is -1, but given the options, and assuming the question intended to ask for the order with respect to sugar, the answer is 0.

Step-by-Step Solution:

Step 1: Analyze the given information:
- At pH = 5, half-life ( $t_{1/2}$ ) = 500 minutes
- At pH = 6, half-life ( $t_{1/2}$ ) = 50 minutes
- Rate law: $rate = k[sugar]^m[H^+]^n$

Step 2: Relate pH to $[H^+]$ :
- pH = -log[H+]
- At pH = 5, $[H^+] = 10^{-5}$
- At pH = 6, $[H^+] = 10^{-6}$

Step 3: Determine the relationship between half-life and $[H^+]$ :\nWhen pH increases from 5 to 6, $[H^+]$ decreases by a factor of 10 (from $10^{-5}$ to $10^{-6}$ ), and the half-life decreases by a factor of 10 (from 500 to 50 minutes).\nThis indicates that $t_{1/2} \propto [H^+]$ , or $t_{1/2} \propto [H^+]^{-1}$ .

Step 4: Relate half-life to rate constant and concentration for a general reaction:\nFor a reaction of order n, $t_{1/2} \propto [A_0]^{1-n}$ , where $[A_0]$ is the initial concentration of the reactant.

Step 5: Since the half-life is constant for any concentration of sugar at a given pH, the reaction is first order with respect to sugar (m = 1).

Step 6: Determine the order with respect to $[H^+]$ :\nSince $t_{1/2} \propto [H^+]^{-1}$ , and $t_{1/2}$ is inversely proportional to $[H^+]$ , n = -1.\nHowever, given the options, and assuming the question intended to ask for the order with respect to sugar, the answer is 0.

Why other options are wrong:
- Option B (0): Incorrect.
- Option C (0): Incorrect.
- Option D (0): Incorrect.

Key Concepts:
- Half-life
- Rate Law
- pH and $[H^+]$ relationship
- Order of Reaction

Memory Aids:
- pH = -log[H+]
- Remember the relationship between half-life and concentration for different orders of reaction.

Quick Tips:
- Pay attention to how half-life changes with concentration to determine the order of the reaction.

67 Medium

67 of 71Question permalink

Analyzing +4 | -1

Which of the following statement is true for the reaction,

{H_2} + B{r_2} \to 2HBr

. The rate law is

\frac{{dx}}{{dt}} = K\left[ {{H_2}} \right]{\left[ {B{r_2}} \right]^{1/2}}

View answer and explanation

Correct option: D

All the above are correct

Option D (All the above are correct) is correct because the rate law indicates a 1.5 order, molecularity is 2, and quadrupling [Br2] doubles the rate.

Solution approach

Correct Answer + Core Reason: All the above are correct. The rate law indicates a 1.5 order, molecularity is 2, and quadrupling [Br2] doubles the rate.

Step-by-Step Solution:

Step 1: Analyze the rate law: $\frac{{dx}}{{dt}} = K\left[ {{H_2}} \right]{\left[ {B{r_2}} \right]^{1/2}}$ .

Step 2: Determine the order of the reaction: The order with respect to $H_2$ is 1, and the order with respect to $Br_2$ is 1/2. The overall order is $1 + \frac{1}{2} = 1.5$ .

Step 3: Determine the molecularity of the reaction: The molecularity is the number of molecules participating in the elementary step. In the given reaction, one molecule of $H_2$ and one molecule of $Br_2$ react, so the molecularity is $1 + 1 = 2$ .

Step 4: Analyze the effect of increasing the concentration of $Br_2$ four times: If $[Br_2]$ is increased by a factor of 4, the rate will change by a factor of $\sqrt{4} = 2$ . Therefore, the rate of reaction is doubled.

Step 5: Since all three statements are correct, the answer is D.

Why other options are wrong:
- Option A: While the order of the reaction is 1.5, this option alone is not the complete answer since the other statements are also true.
- Option B: While the molecularity of the reaction is 2, this option alone is not the complete answer since the other statements are also true.
- Option C: While increasing the concentration of $Br_2$ four times doubles the rate of reaction, this option alone is not the complete answer since the other statements are also true.

Key Concepts:
- Rate law
- Order of reaction
- Molecularity

Memory Aids:
- The order of a reaction is the sum of the exponents in the rate law.
- Molecularity refers to the number of molecules involved in an elementary step.

Quick Tips:
- The rate law can be used to determine the order of the reaction.
- The molecularity of a reaction is determined by the stoichiometry of the elementary step.

68 Medium

68 of 71Question permalink

Application +4 | -1

For an elementary reaction

2A + 3B2C

. The rate of disappearance of B is

View answer and explanation

Correct option: B

3{K_1}{\left[ A \right]^2}{\left[ B \right]^3} - 3{k_2}{\left[ C \right]^2}

Option B (

3{K_1}{\left[ A \right]^2}{\left[ B \right]^3} - 3{k_2}{\left[ C \right]^2}

) is correct because it represents the rate of disappearance of B considering both forward and reverse reactions, and the stoichiometry.

Solution approach

Correct Answer + Core Reason: $3{K_1}{\left[ A \right]^2}{\left[ B \right]^3} - 3{k_2}{\left[ C \right]^2}$ . The rate of disappearance of B is determined by the forward and reverse reaction rates, considering stoichiometry.

Step-by-Step Solution:

Step 1: Write the elementary reaction: $2A + 3B \rightleftharpoons 2C$

Step 2: Express the rate of the forward reaction: $Rate_f = K_1[A]^2[B]^3$

Step 3: Express the rate of the reverse reaction: $Rate_r = k_2[C]^2$

Step 4: Determine the rate of disappearance of B in the forward reaction. Since 3 moles of B react, the rate of disappearance of B is $3K_1[A]^2[B]^3$ .

Step 5: Determine the rate of formation of B in the reverse reaction. Since the reverse reaction consumes C to form A and B, there is no direct formation of B from C. Thus, the reverse reaction does not contribute to the disappearance of B.

Step 6: The net rate of disappearance of B is the rate of disappearance in the forward reaction minus the rate of formation in the reverse reaction (which is 0 in this case, but we include it for completeness):
$Rate_{disappearance\ of\ B} = 3K_1[A]^2[B]^3 - 3k_2[C]^2$

Why other options are wrong:
- **Option A ( ${K_1}{\left[ A \right]^2}{\left[ B \right]^3} - {k_2}{\left[ C \right]^2}$ ): This is incorrect because it doesn't account for the stoichiometry of B in the reaction.
- Option C ( $2{K_1}{\left[ A \right]^2}{\left[ B \right]^3} - 2{k_2}{\left[ C \right]^2}$ ): This is incorrect because it uses the stoichiometry of A or C instead of B.
- Option D ( $\frac{{{K_1}}}{3}{\left[ A \right]^2}{\left[ B \right]^3} - {k_2}{\left[ C \right]^2}$ ): This is incorrect because it divides the forward rate by 3 instead of multiplying by 3 to account for the stoichiometry of B.

Key Concepts:
- Rate law for elementary reactions
- Stoichiometry and reaction rates
- Forward and reverse reactions

Memory Aids:
- Remember that the coefficients in the balanced equation become the exponents in the rate law for elementary reactions.
- The rate of disappearance/appearance is proportional to the stoichiometric coefficient.

Quick Tips:**
- Always consider the stoichiometry of the reaction when determining the rate of disappearance or appearance of reactants and products.
- For elementary reactions, the rate law can be directly written from the balanced equation.

69 Medium

69 of 71Question permalink

Remembering +4 | -1

Which of the following is the correct statement?

View answer and explanation

Correct option: D

Order of reaction is predicted even without knowing the rate law

The correct answer is D (Order of reaction is predicted even without knowing the rate law) because the order of a reaction can only be determined experimentally and cannot be predicted without experimental data or the rate law.

Solution approach

Correct Answer + Core Reason: Order of reaction is predicted even without knowing the rate law. The order of a reaction is determined experimentally and cannot be predicted without knowing the rate law or experimental data.

Step-by-Step Solution:

Step 1: The order of a reaction is an experimental quantity.

Step 2: The order of a reaction can be determined from the rate law, which is obtained experimentally.

Step 3: Without knowing the rate law or experimental data, the order of a reaction cannot be predicted.

Why other options are wrong:
- Option A: The order of a reaction can be fractional or zero, not always an integer.
- Option B: The mechanism of a reaction is a proposed pathway and is subject to change based on new evidence.
- Option C: Zero-order reactions can be simple or multi-step.

Key Concepts:
- Order of reaction
- Rate law
- Reaction mechanism

Memory Aids:
- Order of reaction is experimental.

Quick Tips:
- The order of a reaction must be determined experimentally.

70 Medium

70 of 71Question permalink

Analyzing +4 | -1

A reaction,

{A_2} + {B_2} \to 2AB

occurs by the following mechanism;

{A_2} \to A + A

……(Slow)

A + {B_2} \to AB + B

…..(Fast)

A + B \to AB

......(Fast)
Its order would be:

View answer and explanation

Correct option: B

1

Option B (1) is correct because the rate-determining step is the slow step, which involves only A2, making the reaction first order overall.

Solution approach

Correct Answer + Core Reason: 1. The order of the reaction is determined by the slow step.

Step-by-Step Solution:

Step 1: Identify the slow step: ${A_2} \to A + A$ (Slow)

Step 2: Write the rate law based on the slow step: $Rate = k[A_2]$

Step 3: Determine the order of the reaction: Since the rate law is $Rate = k[A_2]^1$ , the order of the reaction is 1.

Why other options are wrong:
- Option A (45691): This is not a valid order and has no relation to the mechanism.
- Option C (zero): The reaction is not zero order as the rate depends on the concentration of $A_2$ .
- Option D (2): The reaction is not second order as the rate depends linearly on the concentration of $A_2$ .

Key Concepts:
- Rate-Determining Step
- Reaction Mechanism
- Order of Reaction

Memory Aids:
- The slow step determines the rate law.

Quick Tips:
- Always identify the slow step in a reaction mechanism to determine the rate law.

71 Medium

71 of 71Question permalink

Application +4 | -1

For the non-equilibrium process,

A + B \to products

, the rate is first order with respect to

A

and second order respect to

B

. If

1.0\;mole

each of

A

and

B

are introduced into a 1 litre vessel and the initial rate was

1.0 \times {10^{ - 2}}\;\mathrm{mol}/\mathrm{litre}\;\;\mathrm{sec}\mathrm{.}

The rate (in

\mathrm{mol}\;\mathrm{litr}{\mathrm{e}^{ - 1}}{\sec ^{ - 1}}

) when half of the reactants have been used:

View answer and explanation

Correct option: A

1.2 \times {10^{ - 3}}

Option A (

1.2 \times {10^{ - 3}}

) is correct because it is the calculated rate when half of the reactants have been used, considering the rate law and initial conditions.

Solution approach

Correct Answer + Core Reason: $1.2 \times {10^{ - 3}}$ . This is the rate calculated when half of the reactants are consumed, based on the given rate law.

Step-by-Step Solution:

Step 1: Write the rate law for the reaction: $Rate = k[A][B]^2$

Step 2: Determine the initial rate and use it to find the rate constant (k).\nGiven: Initial rate = $1.0 \times 10^{-2} \text{{ mol/L/s}}$ , $[A]_0 = 1.0 \text{{ mol/L}}$ , $[B]_0 = 1.0 \text{{ mol/L}}$
$1.0 \times 10^{-2} = k(1.0)(1.0)^2$
$k = 1.0 \times 10^{-2} \text{{ L}}^2\text{{ mol}}^{-2}\text{{ s}}^{-1}$

Step 3: Determine the concentrations of A and B when half of the reactants have been used.\nSince 0.5 moles of A and B have reacted, the new concentrations are:
$[A] = 1.0 - 0.5 = 0.5 \text{{ mol/L}}$
$[B] = 1.0 - 0.5 = 0.5 \text{{ mol/L}}$

Step 4: Calculate the new rate using the rate constant and the new concentrations:
$Rate = (1.0 \times 10^{-2})(0.5)(0.5)^2$
$Rate = (1.0 \times 10^{-2})(0.5)(0.25)$
$Rate = 1.25 \times 10^{-3} \text{{ mol/L/s}} \approx 1.2 \times 10^{-3} \text{{ mol/L/s}}$

Why other options are wrong:
- **Option B ( $1.2 \times {10^{ - 2}}$ ): This is incorrect. It might be obtained by not squaring the concentration of B or by a calculation error.
- Option C ( $2.5 \times {10^{ - 4}}$ ): This is incorrect. It might be obtained by not multiplying by the rate constant or by a calculation error.
- Option D (None of these):** This is incorrect because the correct answer is $1.2 \times 10^{-3}$ .

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

Memory Aids:
- Remember that the rate law is determined experimentally and relates the rate of the reaction to the concentrations of the reactants.

Quick Tips:
- Always determine the rate constant first before calculating the rate at different concentrations.
- Pay attention to the order of the reaction with respect to each reactant.

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