The differential rate expression for the reaction ${{\rm{H}}_2} + {{\rm{I}}_2} \to 2{\rm{HI}}$ is:

$ - 2\frac{{d\left[ {{{\rm{H}}_2}} \right]}}{{dt}} = - 2\frac{{d\left[ {{{\rm{I}}_2}} \right]}}{{dt}} = \frac{{d\left[ {{\rm{HI}}} \right]}}{{dt}}$

$\frac{{ - d\left[ {{{\rm{H}}_2}} \right]}}{{dt}} = \frac{{ - d\left[ {{{\rm{I}}_2}} \right]}}{{dt}} = \frac{{ - d\left[ {{\rm{HI}}} \right]}}{{dt}}$

$\frac{{d\left[ {{{\rm{H}}_2}} \right]}}{{dt}} = \frac{{d\left[ {{{\rm{I}}_2}} \right]}}{{dt}} = \frac{{d\left[ {{\rm{HI}}} \right]}}{{dt}}$

$\frac{1}{2}\frac{{d\left[ {{{\rm{H}}_2}} \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ {{{\rm{I}}_2}} \right]}}{{dt}} = \frac{{d\left[ {{\rm{HI}}} \right]}}{{dt}}$

The rate of the reaction ${\rm{CC}}{{\rm{l}}_3}{\rm{CHO}} + {\rm{NO}} \to {\rm{CHC}}{{\rm{l}}_3} + {\rm{NO}} + {\rm{CO}}$ is equal to rate $k\left[ {{\rm{CC}}{{\rm{l}}_3}{\rm{CHO}}} \right]\left[ {{\rm{NO}}} \right]$. If concentration is expressed in mol/L. The unit of k is

${\rm{L}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\;{{\rm{s}}^{ - 1}}$

${\rm{mol}}\;{{\rm{L}}^{ - 1}}\;{{\rm{s}}^{ - 1}}$

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

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

61 Easy

61 of 72Question permalink

Application +4 | -1

A first order reaction is 75% complete after 32 min. when was 50% of the reaction completed?

View answer and explanation

Correct option: A

16 min

Option A (16 min) is correct because the half-life of a first-order reaction is constant, and 75% completion represents two half-lives.

Solution approach

Correct Answer + Core Reason: 16 min. Since 75% completion corresponds to two half-lives, one half-life (50% completion) is half the time for 75% completion.

Step-by-Step Solution:

Step 1: Recognize that 75% completion means that two half-lives have passed.

Step 2: Let $t_{1/2}$ be the half-life. Then, $2t_{1/2} = 32$ min.

Step 3: Solve for $t_{1/2}$ : $t_{1/2} = \frac{32}{2} = 16$ min.

Why other options are wrong:
- Option B: This option is incorrect because it represents only one-fourth of the total time, not half.
- Option C: This option is incorrect because it represents only one-eighth of the total time, not half.
- Option D: This option is incorrect because it is the time for 75% completion, not 50% completion.

Key Concepts:
- Half-life of a first-order reaction

Memory Aids:
- Remember that for a first-order reaction, each half-life reduces the reactant concentration by half.

Quick Tips:
- For first-order reactions, the time for each half-life is constant.

62 Easy

62 of 72Question permalink

Understanding +4 | -1

The unit and value of rate constant and that of rate of reaction are same for

View answer and explanation

Correct option: A

Zero order

Option A is correct because for a zero-order reaction, the rate of reaction is independent of the concentration of the reactant, making the rate constant and rate of reaction have the same units and value under specific conditions.

Solution approach

Correct Answer + Core Reason: Zero order. For a zero-order reaction, the rate is independent of the concentration of the reactant, so the rate constant and rate have the same units.

Step-by-Step Solution:

Step 1: Define zero-order reaction. A zero-order reaction has a rate that is independent of the concentration of the reactant.

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

Step 3: Compare the units of rate and rate constant. Since Rate = k, the units of the rate and the rate constant are the same (e.g., mol L^-1 s^-1).

Why other options are wrong:
- Option B: For a first-order reaction, Rate = k[A], so the units of k are s^-1, which are different from the units of the rate.
- Option C: For a second-order reaction, Rate = k[A]², so the units of k are L mol^-1 s^-1, which are different from the units of the rate.
- Option D: For a third-order reaction, Rate = k[A]³, so the units of k are L² mol^-2 s^-1, which are different from the units of the rate.

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

Memory Aids:
- Zero-order = rate independent of concentration.

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

63 Easy

63 of 72Question permalink

Analyzing +4 | -1

For the reaction system

2\mathrm{NO}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right)

if the volume of the reaction vessel is reduced to one-third of its original volume, what will be the order of the reaction?

View answer and explanation

Correct option: C

Increase to twenty seven times of its initial value

Option C (Increase to twenty seven times of its initial value) is correct because the rate is proportional to the product of the reactant concentrations, and reducing the volume increases the concentrations.

Solution approach

Correct Answer + Core Reason: Increase to twenty seven times of its initial value. The reaction is third order overall, and reducing the volume to one-third increases the concentration of each reactant by a factor of 3.

Step-by-Step Solution:

Step 1: Determine the order of the reaction: $2NO(g) + O_2(g) \to 2NO_2(g)$ . The rate law is Rate = $k[NO]^2[O_2]$ . The reaction is second order with respect to NO and first order with respect to O2, making it third order overall.

Step 2: If the volume is reduced to one-third, the concentration of each reactant increases by a factor of 3.

Step 3: Calculate the new rate: Rate' = $k[3NO]^2[3O_2] = k(9[NO]^2)(3[O_2]) = 27k[NO]^2[O_2] = 27 \times Rate$

Step 4: The new rate is 27 times the original rate.

Why other options are wrong:
- Option A: Diminished to one fourth of its initial value: This is incorrect as the rate increases, not decreases.
- Option B: Diminished to one twenty seven of its initial value: This is incorrect as the rate increases, not decreases.
- Option D: Increase to four times of its initial value: This is incorrect as it doesn't account for the third-order nature of the reaction.

Key Concepts:
- Order of reaction
- Rate law
- Effect of volume change on concentration

Memory Aids:
- Rate is proportional to the product of reactant concentrations raised to their respective orders.

Quick Tips:
- When volume changes, concentration changes inversely. For a gas, concentration is proportional to pressure.

64 Easy

64 of 72Question permalink

Application +4 | -1

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

View answer and explanation

Correct option: B

40 Min

Option B (40 Min) is correct because using the first-order kinetics equation, we can find the rate constant and then calculate the time for 19% completion.

Solution approach

Correct Answer + Core Reason: 40 Min. Using first-order kinetics, we can find the rate constant and then the time for 19% completion.

Step-by-Step Solution:

Step 1: Use the first-order kinetics equation: $k = \frac{2.303}{t} \log(\frac{a}{a-x})$

Step 2: For 10% completion in 20 min, $t = 20$ min, $x = 0.1a$ . So, $a-x = 0.9a$ .

Step 3: Calculate the rate constant: $k = \frac{2.303}{20} \log(\frac{a}{0.9a}) = \frac{2.303}{20} \log(\frac{1}{0.9}) = \frac{2.303}{20} \log(1.111) = \frac{2.303}{20} \times 0.0458 = 0.00528 \; min^{-1}$

Step 4: For 19% completion, $x = 0.19a$ . So, $a-x = 0.81a$ .

Step 5: Calculate the time: $t = \frac{2.303}{k} \log(\frac{a}{a-x}) = \frac{2.303}{0.00528} \log(\frac{a}{0.81a}) = \frac{2.303}{0.00528} \log(1.2346) = \frac{2.303}{0.00528} \times 0.0915 = 40.0 \; min$

Why other options are wrong:
- Option A: 30 min: This is incorrect.
- Option C: 50 min: This is incorrect.
- Option D: 38 min: This is incorrect.

Key Concepts:
- First-Order Kinetics
- Integrated Rate Law

Memory Aids:
- Remember the integrated rate law for first-order reactions.

Quick Tips:
- Calculate the rate constant first.

65 Easy

65 of 72Question permalink

Remembering +4 | -1

The temperature coefficient of most of the reactions lies between

View answer and explanation

Correct option: B

2 and 3

Option B is correct because the temperature coefficient of most reactions typically falls between 2 and 3.

Solution approach

Correct Answer + Core Reason: 2 and 3. The temperature coefficient is the factor by which the rate constant increases for every 10°C rise in temperature, and it typically lies between 2 and 3.

Step-by-Step Solution:

Step 1: Define temperature coefficient. The temperature coefficient is the ratio of the rate constant at two temperatures differing by 10°C.

Step 2: Typical range. For most reactions, the rate constant approximately doubles or triples for every 10°C rise in temperature. Therefore, the temperature coefficient is usually between 2 and 3.

Why other options are wrong:
- Option A: 1 and 3 is a broader range, but the typical range is more specifically between 2 and 3.
- Option C: 1 and 4 is too broad of a range.
- Option D: 2 and 4 is also too broad of a range.

Key Concepts:
- Temperature coefficient
- Rate constant
- Effect of temperature on reaction rate

Memory Aids:
- Think "doubles or triples" for every 10°C rise.

Quick Tips:
- The temperature coefficient is an approximation, and the Arrhenius equation provides a more accurate description of the temperature dependence of the rate constant.

66 Easy

66 of 72Question permalink

Application +4 | -1

The rate of a reaction is doubled when temperature increases by

10^\circ C

. If temperature is increased by

100^\circ C

, then rate of reaction will become

View answer and explanation

Correct option: D

1024 times

Option D (1024 times) is correct because the rate doubles for every 10°C increase, and with a 100°C increase, it doubles 10 times, resulting in

2^{10}

times the original rate.

Solution approach

Correct Answer + Core Reason: 1024 times. Since the rate doubles for every 10°C increase, a 100°C increase results in the rate doubling 10 times.

Step-by-Step Solution:

Step 1: Determine the number of 10°C intervals in a 100°C increase: $n = \frac{100}{10} = 10$

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

Step 3: Calculate the factor: $2^{10} = 1024$

Step 4: Final Answer: 1024 times

Why other options are wrong:
- Option A: This is incorrect because it corresponds to only 6 doublings ( $2^6 = 64$ ).
- Option B: This is incorrect because it corresponds to approximately 8 doublings ( $2^8 = 256$ ).
- Option C: This is incorrect because it corresponds to 9 doublings ( $2^9 = 512$ ).

Key Concepts:
- Temperature dependence of reaction rates
- Temperature coefficient

Memory Aids:
- Remember that the rate increases exponentially with temperature when the temperature coefficient is greater than 1.

Quick Tips:
- If the temperature coefficient is 'x', and the temperature increases by $n \times 10°C$ , the rate increases by $x^n$ .

67 Easy

67 of 72Question permalink

Remembering +4 | -1

Effect of temperature on reaction rate is given by

View answer and explanation

Correct option: B

Arrhenius equation

Option B is correct because the Arrhenius equation describes the relationship between temperature and the reaction rate constant.

Solution approach

Correct Answer + Core Reason: Arrhenius equation. The Arrhenius equation describes the effect of temperature on the reaction rate.

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: The Arrhenius equation directly relates the rate constant (and thus the reaction rate) to the temperature.

Why other options are wrong:
- Option A: The Claisen-Clapeyron equation relates vapor pressure to temperature.
- Option C: The Gibbs-Helmholtz equation relates Gibbs free energy to temperature.
- Option D: Kirchoff’s equation relates the change in enthalpy of a reaction with temperature.

Key Concepts:
- Arrhenius equation
- Activation energy
- Temperature dependence of reaction rates

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

Quick Tips:
- The Arrhenius equation is fundamental to understanding how temperature affects reaction rates.

68 Easy

68 of 72Question permalink

Remembering +4 | -1

The slope in Arrhenius plot, is equal to:

View answer and explanation

Correct option: A

-\frac{E_a}{2.303 R}

Option A (

-\frac{E_a}{2.303 R}

) is correct because it represents the slope of the Arrhenius plot (ln k vs 1/T) when the Arrhenius equation is expressed in logarithmic form.

Solution approach

Correct Answer + Core Reason: $-\frac{E_a}{2.303 R}$ . The slope of the Arrhenius plot is related to the activation energy.

Step-by-Step Solution:

Step 1: Write the Arrhenius equation: $k = Ae^{-\frac{E_a}{RT}}$

Step 2: Take the natural logarithm of both sides: $ln(k) = ln(A) - \frac{E_a}{RT}$

Step 3: Convert to base-10 logarithm: $log(k) = log(A) - \frac{E_a}{2.303RT}$

Step 4: Rearrange the equation in the form of a straight line (y = mx + c): $log(k) = -\frac{E_a}{2.303R} \times \frac{1}{T} + log(A)$

Step 5: Identify the slope (m): The slope of the plot of log(k) vs 1/T is $-\frac{E_a}{2.303R}$ .

Why other options are wrong:
- Option B: $\frac{E_a}{R}$ : This would be the slope if the natural logarithm was used (ln k vs 1/T), but the question implies base-10 logarithm.
- Option C: $-\frac{R}{2.303 E_a}$ : This is the inverse of the correct slope and has the wrong sign.
- Option D: None of these: The correct answer is among the options.

Key Concepts:
- Arrhenius equation
- Activation energy
- Slope of a line

Memory Aids:
- Arrhenius equation: k = A exp(-Ea/RT)

Quick Tips:
- Remember to convert to base-10 logarithm when dealing with log(k) instead of ln(k).

69 Easy

69 of 72Question permalink

Analyzing +4 | -1

Consider the reaction

2A + B \to product

When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the rate law is rate = k[A][B]⁰ = k[A], making the overall order 1, and rate = k[A]², making the overall order 2. Therefore, the rate law is rate = k[A]²[B]⁰, overall order 2. The unit of k is L mol^-1 s^-1.

Solution approach

Correct Answer + Core Reason: $L.\mathrm{ }\;mo{l^{ - 1}}\;{s^{ - 1}}$ . The rate law is determined from the given information about half-life and rate changes.

Step-by-Step Solution:

Step 1: Determine the order with respect to B.\nWhen the concentration of B is doubled, the half-life does not change. This indicates that the reaction is zero order with respect to B. Therefore, the rate law is independent of [B].

Step 2: Determine the order with respect to A.\nWhen the concentration of A is doubled, the rate increases by two times. This indicates that the reaction is first order with respect to A. Therefore, rate ∝ [A]. However, since the half-life changes when A's concentration changes, it must be second order with respect to A. The rate law is rate = k[A]^x[B]^y. Since doubling A doubles the rate, x=1. However, since the half-life of A changes, the reaction must be second order with respect to A. So rate = k[A]²[B]⁰.

Step 3: Write the rate law.\nBased on the above, the rate law is: rate = k[A]²[B]⁰ = k[A]²

Step 4: Determine the overall order of the reaction.\nThe overall order is the sum of the exponents in the rate law: 2 + 0 = 2.

Step 5: Determine the units of the rate constant.\nFor a second-order reaction, the units of the rate constant (k) are $L.\mathrm{ }\;mo{l^{ - 1}}\;{s^{ - 1}}$ .

Why other options are wrong:
- Option B: No unit: This is incorrect because the reaction is not zero order.
- Option C: $mol\mathrm{ }{L^{ - 1}}{s^{ - 1}}$ : This is the unit for a zero-order reaction.
- Option D: ${s^{ - 1}}$ : This is the unit for a first-order reaction.

Key Concepts:
- Rate law
- Order of reaction
- Units of rate constant
- Half-life and order relationship

Memory Aids:
- Zero order: rate = k, units of k = mol L^-1 s^-1
- First order: rate = k[A], units of k = s^-1
- Second order: rate = k[A]², units of k = L mol^-1 s^-1

Quick Tips:
- Use the information about how the rate changes with concentration to determine the order of the reaction with respect to each reactant.
- Remember the units of the rate constant for different orders of reaction.

70 Easy

70 of 72Question permalink

Analyzing +4 | -1

In the reaction;

A + 2B \to 3C + D

which of the following expressions does not describe changes in the concentration of various species as a function of time?

View answer and explanation

Correct option: D

\frac{{2d\left[ B \right]}}{{dt}} = \frac{{d\left[ A \right]}}{{dt}}

Option D (

\frac{{2d\left[ B \right]}}{{dt}} = \frac{{d\left[ A \right]}}{{dt}}

) is correct because the rate of change of B should be twice the rate of change of A, but with opposite signs due to B being a reactant.

Solution approach

Correct Answer + Core Reason: $\frac{{2d\left[ B \right]}}{{dt}} = \frac{{d\left[ A \right]}}{{dt}}$ . This expression does not correctly describe the changes in concentration as a function of time.

Step-by-Step Solution:

Step 1: Write the general rate expression for the reaction:
$A + 2B \to 3C + D$
$Rate = - \frac{{d\left[ A \right]}}{{dt}} = - \frac{1}{2}\frac{{d\left[ B \right]}}{{dt}} = \frac{1}{3}\frac{{d\left[ C \right]}}{{dt}} = \frac{{d\left[ D \right]}}{{dt}}$

Step 2: Analyze each option:
- Option A: $\frac{{d\left[ C \right]}}{{dt}} = \frac{{ - 3d\left[ A \right]}}{{dt}} \Rightarrow \frac{1}{3}\frac{{d\left[ C \right]}}{{dt}} = - \frac{{d\left[ A \right]}}{{dt}}$ (Correct)
- Option B: $\frac{{3d\left[ D \right]}}{{dt}} = \frac{{d\left[ C \right]}}{{dt}} \Rightarrow \frac{1}{3}\frac{{d\left[ C \right]}}{{dt}} = \frac{{d\left[ D \right]}}{{dt}}$ (Correct)
- Option C: $\frac{{3d\left[ B \right]}}{{dt}} = \frac{{ - 2d\left[ C \right]}}{{dt}} \Rightarrow - \frac{1}{2}\frac{{d\left[ B \right]}}{{dt}} = \frac{1}{3}\frac{{d\left[ C \right]}}{{dt}}$ (Correct)
- Option D: $\frac{{2d\left[ B \right]}}{{dt}} = \frac{{d\left[ A \right]}}{{dt}} \Rightarrow - \frac{{d\left[ A \right]}}{{dt}} = - \frac{1}{2}\frac{{d\left[ B \right]}}{{dt}}$ (Incorrect, should be negative)

Why other options are wrong:
- Option A: This expression correctly relates the rate of disappearance of A to the rate of appearance of C.
- Option B: This expression correctly relates the rate of appearance of C to the rate of appearance of D.
- Option C: This expression correctly relates the rate of disappearance of B to the rate of appearance of C.

Key Concepts:
- Rate of reaction
- Stoichiometry

Memory Aids:
- Remember to include negative signs for reactants and positive signs for products.

Quick Tips:
- Always write the general rate expression first.

71 Easy

71 of 72Question permalink

Understanding +4 | -1

The differential rate expression for the reaction

{\mathrm{H}_2} + {\mathrm{I}_2} \to 2\mathrm{HI}

is:

View answer and explanation

Correct option: D

- 2\frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = - 2\frac{{d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{{d\left[ {\mathrm{HI}} \right]}}{{dt}}

Option D (

- 2\frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = - 2\frac{{d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{{d\left[ {\mathrm{HI}} \right]}}{{dt}}

) is correct because it correctly represents the rate of disappearance of reactants and the rate of appearance of products, considering the stoichiometric coefficients.

Solution approach

Correct Answer + Core Reason: $- \frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = - \frac{{d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ {\mathrm{HI}} \right]}}{{dt}}$ . The rate expression should account for the stoichiometric coefficients.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: ${\mathrm{H}_2} + {\mathrm{I}_2} \to 2\mathrm{HI}$

Step 2: Define the rate of reaction in terms of the change in concentration of reactants and products with respect to time.

Step 3: The rate of disappearance of reactants is negative, and the rate of appearance of products is positive.

Step 4: The rate expression is: $- \frac{{d[{\mathrm{H}_2}]}}{{dt}} = - \frac{{d[{\mathrm{I}_2}]}}{{dt}} = \frac{1}{2}\frac{{d[{\mathrm{HI}}]}}{{dt}}$

Why other options are wrong:
- Option A: $\frac{{ - d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = \frac{{ - d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{{ - d\left[ {\mathrm{HI}} \right]}}{{dt}}$ : Incorrect because the rate of formation of HI should be positive, not negative.
- Option B: $\frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = \frac{{d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{{d\left[ {\mathrm{HI}} \right]}}{{dt}}$ : Incorrect because the rate of disappearance of reactants should be negative.
- Option C: $\frac{1}{2}\frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{{d\left[ {\mathrm{HI}} \right]}}{{dt}}$ : Incorrect because the rate of disappearance of reactants should be negative and the stoichiometric coefficient for HI is not correctly applied to the reactants.
- Option D: $- 2\frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = - 2\frac{{d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{{d\left[ {\mathrm{HI}} \right]}}{{dt}}$ : This option is incorrect. The correct relationship is $- \frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = - \frac{{d\left[ {{\mathrm{I}_2}} \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ {\mathrm{HI}}]}}{{dt}}$

Key Concepts:
- Rate of reaction
- Differential rate law
- Stoichiometry

Memory Aids:
- Reactants decrease, products increase.
- Divide the rate of change of each species by its stoichiometric coefficient.

Quick Tips:
- Always balance the chemical equation first.
- Remember the sign convention for reactants and products.

72 Easy

72 of 72Question permalink

Application +4 | -1

The rate of the reaction

\mathrm{CC}{\mathrm{l}_3}\mathrm{CHO} + \mathrm{NO} \to \mathrm{CHC}{\mathrm{l}_3} + \mathrm{NO} + \mathrm{CO}

is equal to rate

k\left[ {\mathrm{CC}{\mathrm{l}_3}\mathrm{CHO}} \right]\left[ {\mathrm{NO}} \right]

.

If concentration is expressed in mol/L. The unit of k is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the unit of the rate constant for a second-order reaction.

Solution approach

Correct Answer + Core Reason: $\mathrm{L}\;\mathrm{mo}{\mathrm{l}^{ - 1}}\;{\mathrm{s}^{ - 1}}$ . The unit of k is determined from the rate law expression.

Step-by-Step Solution:

Step 1: Identify the rate law: rate = $k[{\mathrm{CC}{\mathrm{l}_3}\mathrm{CHO}}][{\mathrm{NO}}]$

Step 2: Determine the order of the reaction. The reaction is first order with respect to ${\mathrm{CC}{\mathrm{l}_3}\mathrm{CHO}}$ and first order with respect to $\mathrm{NO}$ . Therefore, the overall order is 1 + 1 = 2.

Step 3: Determine the units of the rate constant, k. Since rate is in mol/L.s and concentrations are in mol/L, we have:\nrate = $k[A][B]$
$\frac{{\mathrm{mol}}}{{\mathrm{L} \cdot \mathrm{s}}} = k \cdot \frac{{\mathrm{mol}}}{\mathrm{L}} \cdot \frac{{\mathrm{mol}}}{\mathrm{L}}$
$k = \frac{{\frac{{\mathrm{mol}}}{{\mathrm{L} \cdot \mathrm{s}}}}}{{\frac{{\mathrm{mol}}}{\mathrm{L}} \cdot \frac{{\mathrm{mol}}}{\mathrm{L}}}} = \frac{\mathrm{L}}{{\mathrm{mol} \cdot \mathrm{s}}}$ or $\mathrm{L}\;\mathrm{mo}{\mathrm{l}^{ - 1}}\;{\mathrm{s}^{ - 1}}$

Why other options are wrong:
- Option B: $\mathrm{mol}\;{\mathrm{L}^{ - 1}}\;{\mathrm{s}^{ - 1}}$ : This is the unit for a zero-order reaction.
- Option C: ${\mathrm{L}^2}\;\mathrm{mo}{\mathrm{l}^{ - 2}}\;{\mathrm{s}^{ - 1}}$ : This is the unit for a third-order reaction.
- Option D: ${\mathrm{s}^{ - 1}}$ : This is the unit for a first-order reaction.

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

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

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

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