The rate of first order reaction, $A \to \;products$, is $7.5 \times {10^{ - 4}}\;{\rm{mol}}\;{\rm{litr}}{{\rm{e}}^{ - 1}}\;{\sec ^{ - 1}}.$ If the concentration of $A\;is\;0.5\;{\rm{mol}}\;{\rm{litr}}{{\rm{e}}^{ - 1}}$ the rate constant is:

$3.75 \times {10^{ - 4}}\;{\sec ^{ - 1}}$

$2.5 \times {10^{ - 5}}\;{\sec ^{ - 1}}$

$8.0 \times {10^{ - 4}}\;{\sec ^{ - 1}}$

The chemical reaction $2{{\rm{O}}_3} \to 3{{\rm{O}}_2}$ proceeds as follows ${{\rm{O}}_3}\;\leftrightharpoons\;{{\rm{O}}_2} + {\rm{O}}(fast)$ ${\rm{O}} + {{\rm{O}}_3} \to 2{{\rm{O}}_2}(slow)$ The rate law expression should be

$r = k{\left[ {{{\rm{O}}_3}} \right]^2}{\left[ {{{\rm{O}}_2}} \right]^{ - 1}}$

$r = k{\left[ {{{\rm{O}}_3}} \right]^2}$

$r = k\left[ {{{\rm{O}}_3}} \right]\left[ {{{\rm{O}}_2}} \right]$

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

31 Medium

31 of 71Question permalink

Understanding +4 | -1

It rate constant

K = 2.3 \times {10^{ - 5}}L\,.\,mo{l^{ - 1}}.\,{S^{ - 1}}

, identify order of reaction

View answer and explanation

Correct option: D

2

Option D (2) is correct because the order of a reaction can be determined from the units of the rate constant, and the given units correspond to a second-order reaction.

Solution approach

Correct Answer + Core Reason: 2. The order of the reaction is determined by the units of the rate constant.

Step-by-Step Solution:

Step 1: Identify the given rate constant and its units: $K = 2.3 \times {10^{ - 5}}L.mo{l^{ - 1}}.{S^{ - 1}}$

Step 2: Recall the general form of the rate constant units for an nth order reaction: $mo{l^{1 - n}}.{L^{n - 1}}.{s^{ - 1}}$

Step 3: Compare the given units with the general form to find the order (n) of the reaction.

Step 4: In this case, we have $L.mo{l^{ - 1}}.{S^{ - 1}}$ , which corresponds to $mo{l^{1 - n}}.{L^{n - 1}}.{s^{ - 1}}$ . Therefore, $1-n = -1$ and $n-1 = 1$ .

Step 5: Solving for n, we get $n = 2$ .

Step 6: Therefore, the order of the reaction is 2.

Why other options are wrong:
- Option A (0): A zero-order reaction has units of $mol.L^{-1}.s^{-1}$ .
- Option B (1): A first-order reaction has units of $s^{-1}$ .
- Option C (3): A third-order reaction has units of $L^2.mol^{-2}.s^{-1}$ .

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

Memory Aids:
- Remember the general formula for the units of the rate constant: $mo{l^{1 - n}}.{L^{n - 1}}.{s^{ - 1}}$

Quick Tips:
- Always check the units of the rate constant to quickly determine the order of the reaction.

32 Medium

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Understanding +4 | -1

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

View answer and explanation

Correct option: C

Remains unchanged

Option C (Remains unchanged) is correct because a catalyst only affects the rate of reaction, not the enthalpy change.

Solution approach

Correct Answer + Core Reason: Remains unchanged. A catalyst lowers the activation energy but does not change the enthalpy of the reaction.

Step-by-Step Solution:

Step 1: Understand the role of a catalyst.

Step 2: A catalyst provides an alternative reaction pathway with a lower activation energy.

Step 3: The enthalpy change ( $\Delta H$ ) is a state function and depends only on the initial and final states, not the path taken. Therefore, a catalyst does not affect $\Delta H$ .

Why other options are wrong:
- Option A: Increases: This is incorrect because a catalyst does not increase the heat evolved or absorbed.
- Option B: Decreases: This is incorrect because a catalyst does not decrease the heat evolved or absorbed.
- Option D: May increase or decrease: This is incorrect because a catalyst has no effect on the heat evolved or absorbed.

Key Concepts:
- Catalyst
- Enthalpy Change
- Activation Energy

Memory Aids:
- Catalysts speed up reactions without being consumed and do not change the thermodynamics of the reaction.

Quick Tips:
- Remember that catalysts affect kinetics, not thermodynamics.

33 Medium

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Application +4 | -1

Observe the following reaction,

2A + B \to C

The rate of formation of C is

2.2 \times {10^{ - 3}}\;mol\;{L^{ - 1}}\;{\min ^{ - 1}}

What is the value of

- \frac{{d\left[ A \right]}}{{dt}}(mol\;{L^{ - 1}}\;\;{\min ^{ - 1}})

?

View answer and explanation

Correct option: C

4.4 \times {10^{ - 3}}

Option C (

4.4 \times {10^{ - 3}}

) is correct because the rate of disappearance of A is twice the rate of formation of C, based on the stoichiometry of the reaction.

Solution approach

Correct Answer + Core Reason: $4.4 \times {10^{ - 3}}$ . The rate of disappearance of A is twice the rate of formation of C, based on the stoichiometry.

Step-by-Step Solution:

Step 1: Write the rate expression for the reaction:\nRate = $-\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[C]}{dt}$

Step 2: We are given that $\frac{d[C]}{dt} = 2.2 \times 10^{-3} mol L^{-1} min^{-1}$

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

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

Step 5: $-\frac{d[A]}{dt} = 2 \times 2.2 \times 10^{-3} mol L^{-1} min^{-1} = 4.4 \times 10^{-3} mol L^{-1} min^{-1}$

Step 6: Final Answer: $4.4 \times 10^{-3}$

Why other options are wrong:
- Option A: This is incorrect because it is the rate of formation of C, not the rate of disappearance of A.
- Option B: This is incorrect because it is half the rate of formation of C.
- Option D: This is incorrect because it is 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.

34 Medium

34 of 71Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: C

Decomposition of

{N_2}O

Option C (Decomposition of

{N_2}O

) is correct because the unimolecular decomposition of nitrous oxide (

N_2O \rightarrow N_2 + \frac{1}{2}O_2

) follows first-order kinetics under certain conditions, particularly at high temperatures.

Solution approach

Correct Answer + Core Reason: Decomposition of ${N_2}O$ . This unimolecular decomposition often follows first-order kinetics.

Step-by-Step Solution:

Step 1: Define a true first-order reaction: A reaction where the rate is directly proportional to the concentration of only one reactant.

Step 2: Analyze the given options:
- Alkaline hydrolysis of ethyl acetate: This is a second-order reaction (SN2). It is bimolecular.
- Acid catalyst hydrolysis of ethyl acetate: This is a pseudo-first-order reaction because the concentration of water is in large excess and remains effectively constant.
- Decomposition of ${N_2}O$ : Under certain conditions (high temperature), this unimolecular decomposition follows first-order kinetics.
- Decomposition of gaseous ammonia on a hot platinum surface: This is a zero-order reaction under certain conditions (high pressure) because the surface is saturated with ammonia.

Why other options are wrong:
- Option A: Alkaline hydrolysis of ethyl acetate: This is a second-order reaction.
- Option B: Acid catalyst hydrolysis of ethyl acetate: This is a pseudo-first-order reaction, not a true first-order reaction.
- Option D: Decomposition of gaseous ammonia on a hot platinum surface: This is a zero-order reaction under certain conditions.

Key Concepts:
- First-order reaction
- Pseudo-first-order reaction
- Reaction order
- Unimolecular decomposition

Memory Aids:
- Remember that unimolecular decompositions can often be first order.

Quick Tips:
- Distinguish between true first-order and pseudo-first-order reactions.

35 Medium

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Analyzing +4 | -1

The reaction,

NO + (1/2){O_2}\xrightarrow{\,}N{O_2}

exhibits:

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because the reaction exhibits a small negative temperature coefficient, a decrease in the value of K with temperature, and a decrease in the value of the rate with temperature.

Solution approach

Correct Answer + Core Reason: All of the above. The reaction exhibits a small negative temperature coefficient, decrease in value of K with temperature, and decrease in value of rate with temperature.

Step-by-Step Solution:

Step 1: Analyze the reaction: $NO + (1/2){O_2} \xrightarrow{\}N{O_2}$ .

Step 2: Consider the temperature dependence of the reaction. This reaction is exothermic. For exothermic reactions, increasing the temperature decreases the equilibrium constant (K) and the rate of the reaction.

Step 3: A small negative temperature coefficient means that the rate constant decreases with increasing temperature.

Step 4: Therefore, all the given statements are correct.

Why other options are wrong:
- Option A: Small negative temperature coefficient: This is a correct statement, but not the only correct one.
- Option B: Decrease in value of K with temperature: This is a correct statement, but not the only correct one.
- Option C: Decrease in value of rate with temperature: This is a correct statement, but not the only correct one.

Key Concepts:
- Temperature Dependence of Reaction Rates
- Equilibrium Constant
- Exothermic Reactions

Memory Aids:
- For exothermic reactions, increasing temperature decreases K.

Quick Tips:
- Consider the reaction type (exothermic or endothermic) when analyzing temperature dependence.

36 Medium

36 of 71Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: C

Second

Option C (Second) is correct because for a second-order reaction, the half-life is inversely proportional to the initial concentration of the reactant.

Solution approach

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

Step-by-Step Solution:

Step 1: Recall the half-life equations for different reaction orders.

Step 2: For a zero-order reaction: ${t_{1/2}} = \frac{{{{\left[ R \right]}_0}}}{{2k}}$

Step 3: For a first-order reaction: ${t_{1/2}} = \frac{{0.693}}{k}$ (independent of concentration)

Step 4: For a second-order reaction: ${t_{1/2}} = \frac{1}{{k{{\left[ R \right]}_0}}}$

Step 5: Identify the order for which half-life is inversely proportional to the concentration of the reactant. From the above equations, it is the second-order reaction.

Why other options are wrong:
- Option A: For a zero-order reaction, half-life is directly proportional to the initial concentration.
- Option B: For a first-order reaction, half-life is independent of the initial concentration.
- Option D: Half order is not a standard order and not relevant to the question.

Key Concepts:
- Half-life of a reaction
- Reaction order
- Zero-order reaction
- First-order reaction
- Second-order reaction

Memory Aids:
- 0, 1, 2 order half-life equations.

Quick Tips:
- Memorize the half-life equations for common reaction orders.

37 Medium

37 of 71Question permalink

Understanding +4 | -1

The order of a reaction with rate equal to

k\;C_A^{3/2}\;C_B^{ - 1/2}

is

View answer and explanation

Correct option: A

1

Option A (1) is correct because the order of a reaction is the sum of the exponents of the reactant concentrations in the rate law.

Solution approach

Correct Answer + Core Reason: 1. The order is the sum of the exponents in the rate law.

Step-by-Step Solution:

Step 1: Identify the rate law: $rate = k \\;C_A^{3/2}\\;C_B^{-1/2}$

Step 2: Determine the exponents for each reactant: $C_A$ has an exponent of 3/2, and $C_B$ has an exponent of -1/2.

Step 3: Calculate the overall order by summing the exponents: Order = (3/2) + (-1/2) = 2/2 = 1

Step 4: Final Answer: 1

Why other options are wrong:
- Option B: This is incorrect because it only considers the exponent of B and doesn't sum the exponents of all reactants.
- Option C: This is incorrect because it calculates (3/2) - (-1/2) instead of (3/2) + (-1/2).
- Option D: This is incorrect because it calculates (3/2) * (-1/2) + (9/4) which is not the correct method to calculate the order of the reaction.

Key Concepts:
- Rate Law
- Order of Reaction

Memory Aids:
- Order = Sum of exponents in rate law

Quick Tips:
- Always sum the exponents, considering their signs.

38 Medium

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Application +4 | -1

If the rate constant of a first order reaction is

9.212 \times {10^{ - 3}}\,{s^{ - 1}}

, then the time required for the completion of 75% of the reaction will be

View answer and explanation

Correct option: B

150 s

Option B (150 s) is correct because it is the time calculated for 75% completion using the first-order rate constant.

Solution approach

Correct Answer + Core Reason: 150 s. This is the time required for 75% completion of a first-order reaction, calculated using the given rate constant.

Step-by-Step Solution:

Step 1: Identify the given information and what needs to be found.\nGiven: Rate constant (k) = $9.212 \times 10^{-3} s^{-1}$ . Find: Time for 75% completion (t).

Step 2: Recall the integrated rate law for a first-order reaction:
$t = \frac{{2.303}}{k} \log \frac{{[A]_0}}{{[A]_t}}$

Step 3: For 75% completion, $[A]_t = [A]_0 - 0.75[A]_0 = 0.25[A]_0$ . Therefore, $\frac{{[A]_0}}{{[A]_t}} = \frac{{[A]_0}}{{0.25[A]_0}} = 4$

Step 4: Substitute the values into the integrated rate law:
$t = \frac{{2.303}}{{9.212 \times 10^{-3}}} \log 4$
$t = \frac{{2.303}}{{9.212 \times 10^{-3}}} \times 0.6021$
$t = \frac{{1.3868}}{{9.212 \times 10^{-3}}} = 150.54 \approx 150 \text{{ s}}$

Why other options are wrong:
- Option A (250 s): This is incorrect. It might be obtained by using an incorrect logarithmic value or calculation error.
- Option C (450 s): This is incorrect. It might be obtained by using an incorrect logarithmic value or calculation error.
- Option D (500 s): This is incorrect. It might be obtained by using an incorrect logarithmic value or calculation error.

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

Memory Aids:
- Remember the 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.

39 Medium

39 of 71Question permalink

Understanding +4 | -1

For a general chemical change,

2A + 3B \to

Products, the rate of disappearance of A is

{r_2}.

The rates

{r_1}

and

{r_2}

are related as

View answer and explanation

Correct option: A

3{r_1} = 2{r_2}

Option A (

3{r_1} = 2{r_2}

) is correct because it correctly relates the rates of disappearance of A and B based on their stoichiometric coefficients.

Solution approach

Correct Answer + Core Reason: $3{r_1} = 2{r_2}$ . The rates of disappearance are related by the stoichiometric coefficients.

Step-by-Step Solution:

Step 1: The reaction is $2A + 3B \to Products$ .

Step 2: Let $r_1$ be the rate of disappearance of B, and $r_2$ be the rate of disappearance of A.

Step 3: The rate of the reaction can be expressed as: $rate = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt}$ .

Step 4: Given $r_1 = -\frac{d[B]}{dt}$ and $r_2 = -\frac{d[A]}{dt}$ , we have: $rate = \frac{1}{2} r_2 = \frac{1}{3} r_1$ .

Step 5: Therefore, $3r_1 = 2r_2$ .

Why other options are wrong:
- Option B: ${r_1} = {r_2}$ : This is incorrect because it does not account for the stoichiometric coefficients.
- Option C: $2{r_1} = 3{r_2}$ : This is incorrect because the coefficients are reversed.
- Option D: $r_1^2 = 2r_2^2$ : This is incorrect because the relationship is linear, not quadratic.

Key Concepts:
- Rate of reaction
- Stoichiometry

Memory Aids:
- Remember to divide the rate of change of concentration by the stoichiometric coefficient.

Quick Tips:
- Use the stoichiometric coefficients to relate the rates of different reactants and products.

40 Medium

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Understanding +4 | -1

If

X

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

N

molecules per unit volume is

View answer and explanation

Correct option: D

NX/2

Option D (

NX/2

) is correct because it accounts for the fact that each collision involves two molecules, thus the total collision frequency must be divided by 2 to avoid double-counting.

Solution approach

Correct Answer + Core Reason: $NX/2$ is the collision frequency of the gas, accounting for each collision involving two molecules.

Step-by-Step Solution:

Step 1: Define the terms. $X$ is the number of collisions a single gas molecule registers per unit time. $N$ is the number of molecules per unit volume.

Step 2: Calculate the total number of collisions registered by all molecules. This would be $NX$ .

Step 3: Account for double-counting. Each collision involves two molecules. Therefore, the total number of collisions is counted twice in the $NX$ value. To correct for this, divide by 2.

Step 4: The collision frequency is therefore $NX/2$ .

Why other options are wrong:
- Option A: $X/N$ would represent the average number of collisions per molecule divided by the total number of molecules, which is not the collision frequency.
- Option B: $NX$ counts each collision twice, as it doesn't account for the fact that each collision involves two molecules.
- Option C: $2NX$ overestimates the collision frequency even further by multiplying by 2 instead of dividing.

Key Concepts:
- Collision Theory
- Collision Frequency

Memory Aids:
- Remember to divide by 2 to avoid double-counting collisions.

Quick Tips:
- Always consider whether you are counting each event once or multiple times.
- Understand the definitions of the terms involved in collision theory.

41 Medium

41 of 71Question permalink

Remembering +4 | -1

A catalyst is used in a reaction to

View answer and explanation

Correct option: D

decrease the time required for the reaction

Option D is correct because a catalyst speeds up a reaction, thus decreasing the time required for the reaction to reach completion.

Solution approach

Correct Answer + Core Reason: decrease the time required for the reaction. A catalyst speeds up a reaction, reducing the time needed for it to occur.

Step-by-Step Solution:

Step 1: Define catalyst. A catalyst is a substance that speeds up a chemical reaction without being consumed in the reaction.

Step 2: Explain the effect of a catalyst on reaction time. Catalysts lower the activation energy of a reaction, allowing it to proceed faster. This results in a decrease in the time required for the reaction to reach completion.

Why other options are wrong:
- Option A: Catalysts do not change the nature of the products.
- Option B: Catalysts do not increase the reaction yield. They only affect the rate at which equilibrium is reached.
- Option C: Catalysts do not decrease the need for reactants. The stoichiometry of the reaction remains the same.

Key Concepts:
- Catalyst
- Activation energy
- Reaction rate

Memory Aids:
- Catalyst = faster reaction = less time.

Quick Tips:
- Catalysts do not change the equilibrium constant or the thermodynamics of a reaction, only the rate.

42 Medium

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Application +4 | -1

The reaction

2S{O_2}\left( g \right) + {O_2}\left( g \right)\;\leftrightharpoons\;2S{O_3}\left( g \right)

is carried out in a

1\;d{m^3}

vessel and

2\;d{m^3}

vessel separately. The ratio of the reaction velocities will be

View answer and explanation

Correct option: D

8:1

Option D (

8:1

) is correct because the reaction velocity is proportional to the concentration of reactants, and changing the volume affects the concentrations.

Solution approach

Correct Answer + Core Reason: $8:1$ . The ratio of reaction velocities is determined by the change in concentrations due to the change in volume.

Step-by-Step Solution:

Step 1: Assume the initial number of moles of $SO_2$ and $O_2$ are $n_{SO_2}$ and $n_{O_2}$ respectively.

Step 2: The initial concentrations in the $1 dm^3$ vessel are $[SO_2]_1 = n_{SO_2}/1$ and $[O_2]_1 = n_{O_2}/1$ .

Step 3: The concentrations in the $2 dm^3$ vessel are $[SO_2]_2 = n_{SO_2}/2$ and $[O_2]_2 = n_{O_2}/2$ .

Step 4: The rate of the reaction is given by $Rate = k[SO_2]^2[O_2]$ .

Step 5: The rate in the $1 dm^3$ vessel is $Rate_1 = k(n_{SO_2}/1)^2(n_{O_2}/1) = k n_{SO_2}^2 n_{O_2}$ .

Step 6: The rate in the $2 dm^3$ vessel is $Rate_2 = k(n_{SO_2}/2)^2(n_{O_2}/2) = k (n_{SO_2}^2/4)(n_{O_2}/2) = k n_{SO_2}^2 n_{O_2}/8$ .

Step 7: The ratio of the reaction velocities is $\frac{Rate_1}{Rate_2} = \frac{k n_{SO_2}^2 n_{O_2}}{k n_{SO_2}^2 n_{O_2}/8} = 8$ .

Step 8: Final Answer: $8:1$

Why other options are wrong:
- Option A: This is incorrect because the ratio is inverted.
- Option B: This is incorrect because it doesn't consider the square dependence on $SO_2$ .
- Option C: This is incorrect because it's the inverse of the correct ratio.

Key Concepts:
- Rate of reaction
- Effect of concentration on reaction rate

Memory Aids:
- Remember that the rate is proportional to the product of the concentrations of the reactants raised to their respective orders.

Quick Tips:
- When the volume increases, the concentration decreases proportionally.

43 Medium

43 of 71Question permalink

Application +4 | -1

If rate constant of a reaction is

1.155 \times {10^{ - 3}}{s^{ - 1}}

then the half life period for the reaction will be

View answer and explanation

Correct option: D

10 min

Option D (10 min) is correct because it is the half-life calculated from the given rate constant for a first-order reaction.

Solution approach

Correct Answer + Core Reason: 10 min. This is the half-life calculated for a first-order reaction using the provided rate constant.

Step-by-Step Solution:

Step 1: Identify the given information and what needs to be found.\nGiven: Rate constant (k) = $1.155 \times 10^{-3} s^{-1}$ . Find: Half-life ( $t_{1/2}$ ).

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

Step 3: Substitute the value of k into the formula:
$t_{1/2} = \frac{{0.693}}{{1.155 \times 10^{-3}}} \text{{ s}}$

Step 4: Calculate the half-life in seconds:
$t_{1/2} = 600 \text{{ s}}$

Step 5: Convert the half-life from seconds to minutes:
$t_{1/2} = \frac{{600}}{{60}} = 10 \text{{ min}}$

Why other options are wrong:
- Option A (20 min): This is incorrect. It might be obtained by using an incorrect formula or calculation error.
- Option B (15 min): This is incorrect. It might be obtained by using an incorrect formula or calculation error.
- Option C (5 min): This is incorrect. It might be obtained by using an incorrect formula or calculation error.

Key Concepts:
- First-order kinetics
- Half-life

Memory Aids:
- Remember the formula for the half-life of a first-order reaction: $t_{1/2} = \frac{{0.693}}{k}$

Quick Tips:
- Ensure the units of rate constant and half-life are consistent.
- Always double-check the calculations.

44 Medium

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Application +4 | -1

The rate constant for a first order reaction which has half-life 231 s is

View answer and explanation

Correct option: B

3 \times {10^{ - 3}}\,{s^{ - 1}}

Option B (

3 \times {10^{ - 3}}\,{s^{ - 1}}

) is correct because it is the rate constant calculated from the given half-life for a first-order reaction.

Solution approach

Correct Answer + Core Reason: $3 \times {10^{ - 3}}\,{s^{ - 1}}$ . This is the rate constant calculated from the half-life of a first-order reaction.

Step-by-Step Solution:

Step 1: Identify the given information and what needs to be found.\nGiven: Half-life ( $t_{1/2}$ ) = 231 s. Find: Rate constant (k).

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: Substitute the value of half-life into the formula:
$k = \frac{{0.693}}{{231}} \text{{ s}}^{-1}$

Step 4: Calculate the rate constant:
$k = 0.003 \text{{ s}}^{-1} = 3 \times 10^{-3} \text{{ s}}^{-1}$

Why other options are wrong:
- **Option A ( ${10^{ - 4}}\,{s^{ - 1}}$ ): This is incorrect. It might be obtained by using an incorrect formula or calculation error.
- Option C ( ${10^{ - 6}}\,{s^{ - 1}}$ ): This is incorrect. It might be obtained by using an incorrect formula or calculation error.
- Option D ( $3 \times {10^{ - 2}}\,{s^{ - 1}}$ ): This is incorrect. It might be obtained by using an incorrect formula or calculation error.

Key Concepts:
- First-order kinetics
- Half-life

Memory Aids:**
- Remember the relationship between half-life and rate constant for first-order reactions: $k = \frac{{0.693}}{{t_{1/2}}}$

Quick Tips:
- Ensure the units of rate constant and half-life are consistent.
- Always double-check the calculations.

45 Medium

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Analyzing +4 | -1

The rate constant is numerically the same for three reactions of first, second and third order respectively. Which one is true for rate of three reaction?

View answer and explanation

Correct option: D

All of these

Option D ((A)Only) is correct because mechanism A is consistent with the rate equation, as the slow step determines the rate, and its reactants match the rate law.

Solution approach

Correct Answer + Core Reason: (A)Only. Mechanism A is consistent with the rate equation.

Step-by-Step Solution:

Step 1: Analyze the given rate equation: $rate = k\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]$

Step 2: Analyze mechanism A:
- Slow step: $C{l_2} + {H_2}S\left( {aq} \right) \to {H^ + }C{l^ - } + C{l^ + } + H{S^ - }$
- The rate law for this slow step is: $rate = k\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]$
- This matches the given rate equation.

Step 3: Analyze mechanism B:
- Fast equilibrium: ${H_2}S \leftrightarrow {H^ + } + H{S^ - }$
- Slow step: $C{l_2} + H{S^ - } \to 2C{l^ - } + {H^ + }S$
- The rate law for this slow step is: $rate = k\left[ {C{l_2}} \right]\left[ {H{S^ - }} \right]$
- Since ${H_2}S \leftrightarrow {H^ + } + H{S^ - }$ is a fast equilibrium, we can write $K = \frac{{\left[ {{H^ + }} \right]\left[ {H{S^ - }} \right]}}{{\left[ {{H_2}S} \right]}}$
- Therefore, $\left[ {H{S^ - }} \right] = K\frac{{\left[ {{H_2}S} \right]}}{{\left[ {{H^ + }} \right]}}$
- Substituting this into the rate law for the slow step, we get: $rate = k\left[ {C{l_2}} \right]K\frac{{\left[ {{H_2}S} \right]}}{{\left[ {{H^ + }} \right]}} = k'\frac{{\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]}}{{\left[ {{H^ + }} \right]}}$
- This does not match the given rate equation.

Step 4: Conclusion: Only mechanism A is consistent with the given rate equation.

Why other options are wrong:
- Option B: Mechanism B is not consistent with the rate equation.
- Option C: Mechanism B is not consistent with the rate equation.
- Option A: Mechanism B is not consistent with the rate equation.

Key Concepts:
- Rate law
- Reaction mechanism
- Rate-determining step
- Equilibrium

Memory Aids:
- The rate law must match the rate-determining step.

Quick Tips:
- Identify the slow step in the mechanism and derive its rate law.

46 Medium

46 of 71Question permalink

Application +4 | -1

The rate of first order reaction,

A \to \;products

, is

7.5 \times {10^{ - 4}}\;\mathrm{mol}\;\mathrm{litr}{\mathrm{e}^{ - 1}}\;{\sec ^{ - 1}}.

If the concentration of

A\;is\;0.5\;\mathrm{mol}\;\mathrm{litr}{\mathrm{e}^{ - 1}}

the rate constant is:

View answer and explanation

Correct option: C

1.5 \times {10^{ - 3}}\;{\sec ^{ - 1}}

Option C (

1.5 \times {10^{ - 3}}\;{\sec ^{ - 1}}

) is correct because for a first-order reaction, k = rate/[A].

Solution approach

Correct Answer + Core Reason: $1.5 \times {10^{ - 3}}\;{\sec ^{ - 1}}$ . Rate constant is calculated by k = rate/[A].

Step-by-Step Solution:

Step 1: Identify the rate law for a first-order reaction: $Rate = k[A]$

Step 2: Rearrange the equation to solve for k: $k = \frac{Rate}{[A]}$

Step 3: Substitute the given values: $Rate = 7.5 \times 10^{-4} \\;mol \\;L^{-1} \\;sec^{-1}$ and $[A] = 0.5 \\;mol \\;L^{-1}$

Step 4: Calculate the rate constant: $k = \frac{7.5 \times 10^{-4}}{0.5} = 1.5 \times 10^{-3} \\;sec^{-1}$

Step 5: Final Answer: $1.5 \times {10^{ - 3}}\;{\sec ^{ - 1}}$

Why other options are wrong:
- Option A: This is incorrect because it divides the rate by 2 instead of dividing by 0.5.
- Option B: This is incorrect because it divides the rate by 30 instead of dividing by 0.5.
- Option D: This is incorrect because it multiplies the rate by 0.5 instead of dividing by 0.5.

Key Concepts:
- First-Order Reaction
- Rate Law

Memory Aids:
- k = Rate/[A] for first-order reactions

Quick Tips:
- Always remember the rate law for the given order of reaction and rearrange it to solve for the desired variable.

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Understanding +4 | -1

\frac{{{K_{{{35}^0}}}}}{{{K_{{{34}^0}}}}} > 1

, this means that

View answer and explanation

Correct option: A

Rate increases with rise in temperature

Option A (Rate increases with rise in temperature) is correct because a rate constant increasing with temperature indicates a faster reaction rate at higher temperatures.

Solution approach

Correct Answer + Core Reason: Rate increases with rise in temperature. A higher rate constant at a higher temperature indicates a faster reaction.

Step-by-Step Solution:

Step 1: The given condition is $\frac{{{K_{{{35}^0}}}}}{{{K_{{{34}^0}}}}} > 1$ .

Step 2: This means that $K_{35^0} > K_{34^0}$ .

Step 3: Since the rate constant $K$ is greater at $35^0$ than at $34^0$ , the rate of the reaction increases with an increase in temperature.

Why other options are wrong:
- Option B: Rate decreases with rise in temperature: This is incorrect because the rate constant increases with temperature.
- Option C: Rate does not change with rise in temperature: This is incorrect because the rate constant changes with temperature.
- Option D: None of the above: Option A is the correct explanation.

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

Memory Aids:
- Higher temperature usually means higher rate constant and faster reaction.

Quick Tips:
- Understand the relationship between temperature and rate constant.

48 Medium

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Analyzing +4 | -1

During the kinetic study of the reaction

2A + B \to C + D

following results were obtained.

\begin{array}{l} Run\;\;\;\;[A]\;\;in\;M\;\;\;\;[B]\;in\;M\;\;\;\;Initial\;rate\;of\;formation\;of\;D\;in\;M\;{s^{ - 1}}\\ I\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;6.0 \times {10^{ - 3}}\;\\ II\;\;\;\;\;\;\;\;\;\;\;0.3\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;7.2 \times {10^{ - 2}}\;\;\\ III\;\;\;\;\;\;\;\;\;\;0.3\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.4\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2.88 \times {10^{ - 1}}\;\;\;\;\\ IV\;\;\;\;\;\;\;\;\;\;0.4\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2.40 \times {10^{ - 2}}\;\;\;\; \end{array}

On the basis of above data which one is correct:

View answer and explanation

Correct option: D

\mathrm{r} = k\left[ A \right]{\left[ B \right]^2}

Option D (

\mathrm{r} = k[A][B]^2

) is correct because it is the rate law that fits the experimental data provided.

Solution approach

Correct Answer + Core Reason: $\mathrm{r} = k[A][B]^2$ . This rate law is determined by comparing the initial rates with the corresponding concentrations of A and B.

Step-by-Step Solution:

Step 1: Assume the rate law is of the form: Rate = $k[A]^x[B]^y$

Step 2: Compare runs I and IV: [A] changes from 0.1 to 0.4 (4 times), [B] is constant at 0.1, and the rate changes from $6.0 \times 10^{-3}$ to $2.40 \times 10^{-2}$ (4 times). This means: $4 = (4)^x$ , so $x = 1$ .

Step 3: Compare runs II and III: [A] is constant at 0.3, [B] changes from 0.2 to 0.4 (2 times), and the rate changes from $7.2 \times 10^{-2}$ to $2.88 \times 10^{-1}$ (4 times). This means: $4 = (2)^y$ , so $y = 2$ .

Step 4: Therefore, the rate law is: Rate = $k[A]^1[B]^2$

Step 5: Write the rate equation: $\mathrm{r} = k[A][B]^2$

Why other options are wrong:
- Option A: $\mathrm{r} = k{ A ]^2} B ]$ : This is incorrect because the order with respect to A is 1 and with respect to B is 2.
- Option B: $\mathrm{r} = k\left[ A \right]\left[ B \right]$ : This is incorrect because the order with respect to B is 2.
- Option C: $\mathrm{r} = k\left[ {{A^2}} \right]{\left[ B \right]^2}$ : This is incorrect because the order with respect to A is 1.

Key Concepts:
- Rate law
- Order of reaction

Memory Aids:
- The exponents in the rate law represent the order of the reaction with respect to each reactant.

Quick Tips:
- Use the given data to determine the order of the reaction with respect to each reactant.
- Compare runs where only one concentration changes to isolate the effect of that reactant.

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Analyzing +4 | -1

Consider the following statements

a) Increase in concentration of reactant increases the rate of zero order reaction

b) Rate constant K is equal to collision frequency A if

{E_a} = 0

c) Rate constant K is equal to collision frequency A if

{E_a} = \, \propto

d) ln K Vs T is a straight line

e) ln K Vs 1/T is a straight line

Correct statements are

View answer and explanation

Correct option: B

b and e

Option B (b and e) is correct because statement b is true when activation energy is zero, and statement e is the correct form of the Arrhenius equation.

Solution approach

Correct Answer + Core Reason: b and e. Statement b is correct based on the Arrhenius equation, and statement e represents the linear form of the Arrhenius equation.

Step-by-Step Solution:

Step 1: Analyze each statement individually.

Step 2: Statement a) is incorrect because a zero-order reaction's rate is independent of reactant concentration.

Step 3: Statement b) is correct. According to the Arrhenius equation, $k = A e^{-\frac{E_a}{RT}}$ . If $E_a = 0$ , then $k = A$ .

Step 4: Statement c) is incorrect. If $E_a = \infty$ , then $k = 0$ .

Step 5: Statement d) is incorrect. ln K vs T is not a straight line. The Arrhenius equation in logarithmic form is $\ln k = \ln A - \frac{E_a}{RT}$ .

Step 6: Statement e) is correct. ln K vs 1/T is a straight line according to the Arrhenius equation.

Why other options are wrong:
- Option A: Includes statement a (incorrect) and statement d (incorrect).
- Option C: Includes statement c (incorrect) and statement d (incorrect).
- Option D: Includes statement c (incorrect).

Key Concepts:
- Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$
- Zero-order reactions

Memory Aids:
- Remember the Arrhenius equation and its logarithmic form.

Quick Tips:
- Understand the relationship between activation energy and the rate constant.

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Understanding +4 | -1

A first order reaction is carried out with an initial concentration of 10 mole per litre and

80\%

of the reactant changes into the product. Now if the same reaction is carried out with an initial concentration of 5 mol per litre the percentage of the reactant changing to the product is:

View answer and explanation

Correct option: B

80

Option B (80) is correct because the percentage of reactant changing into product in a first-order reaction is independent of the initial concentration.

Solution approach

Correct Answer + Core Reason: 80. For a first-order reaction, the percentage of reactant converted to product depends only on time, not initial concentration.

Step-by-Step Solution:

Step 1: Recognize that this is a first-order reaction.

Step 2: For a first-order reaction, the rate constant $k$ is independent of the initial concentration.

Step 3: The percentage of reactant that changes into product depends only on the rate constant and time.

Step 4: Since the rate constant and time are the same in both cases, the percentage of reactant changing to the product will be the same.

Step 5: Therefore, if 80% of the reactant changes into the product with an initial concentration of 10 mol/L, then 80% will also change into the product with an initial concentration of 5 mol/L.

Why other options are wrong:
- Option A: 40: Incorrect, the percentage remains the same for first order reactions.
- Option C: 160: Incorrect, percentages cannot exceed 100.
- Option D: Cannot be calculated: Incorrect, the percentage can be calculated because it is a first-order reaction.

Key Concepts:
- First-order reaction
- Rate constant
- Initial concentration

Memory Aids:
- First-order reactions: Percentage conversion is independent of initial concentration.

Quick Tips:
- For first-order reactions, focus on the rate constant and time.

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Analyzing +4 | -1

Two reactions

A \to products

and

B \to products

have rate constants

{K_A}

and

{K_B}

at temperature

T

and activation energies

{E_A}

and

{E_B}

respectively. If

{K_A} > {K_B}

  and

{E_A} < {E_B}

and assuming that

A

for both the reactions is same, then:

View answer and explanation

Correct option: D

All of the above

Option D (All of these) is correct because the rates of the reactions depend on the concentrations of the reactants, and without knowing the concentrations, we cannot definitively compare the rates.

Solution approach

Correct Answer + Core Reason: All of these. The rates of the reactions depend on the concentrations of the reactants.

Step-by-Step Solution:

Step 1: Write the rate equations for each reaction:
- First order: ${r_1} = k[A]$
- Second order: ${r_2} = k[B]^2$
- Third order: ${r_3} = k[C]^3$

Step 2: Given that the rate constant k is numerically the same for all three reactions.

Step 3: Analyze the rates:\nThe rates depend on the concentrations of A, B, and C. Without knowing the concentrations, we cannot definitively compare the rates.

Step 4: Consider different scenarios:
- If [A] = [B]² = [C]³, then ${r_1} = {r_2} = {r_3}$
- If [A] > [B]² > [C]³, then ${r_1} > {r_2} > {r_3}$
- If [A] < [B]² < [C]³, then ${r_1} < {r_2} < {r_3}$

Step 5: Since all scenarios are possible, the answer is All of these.

Why other options are wrong:
- Option A: This is only true if [A] = [B]² = [C]³.
- Option B: This is only true if [A] > [B]² > [C]³.
- Option C: This is only true if [A] < [B]² < [C]³.

Key Concepts:
- Rate law
- Order of reaction

Memory Aids:
- Remember that the rate depends on both the rate constant and the concentrations of the reactants.

Quick Tips:
- Always consider the concentrations of the reactants when comparing rates.

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Understanding +4 | -1

The correct difference between first and second – order reactions is that

View answer and explanation

Correct option: D

The half-life of a first order reaction does not depend on

{\left[ \mathrm{A} \right]_\mathrm{0}}

; the half-life of a second order reaction does
depend on

{\left[ \mathrm{A} \right]_\mathrm{0}}

Option D (The half-life of a first order reaction does not depend on

{\left[ \mathrm{A} \right]_\mathrm{0}}

; the half-life of a second order reaction does depend on

{\left[ \mathrm{A} \right]_\mathrm{0}}

) is correct because this is the key difference in how half-life relates to initial concentration for first and second-order reactions.

Solution approach

Correct Answer + Core Reason: The half-life of a first order reaction does not depend on ${\left[ \mathrm{A} \right]_\mathrm{0}}$ ; the half-life of a second order reaction does depend on ${\left[ \mathrm{A} \right]_\mathrm{0}}$ . This is the defining difference in half-life behavior.

Step-by-Step Solution:

Step 1: Recall the half-life equation for a first-order reaction: ${t_{1/2}} = \frac{{0.693}}{k}$ . Notice that it does not depend on the initial concentration.

Step 2: Recall the half-life equation for a second-order reaction: ${t_{1/2}} = \frac{1}{{k{{\left[ A \right]}_0}}}$ . Notice that it is inversely proportional to the initial concentration.

Step 3: Compare the two half-life equations. The first-order half-life is independent of initial concentration, while the second-order half-life depends on the initial concentration.

Why other options are wrong:
- Option A: The rate of both first and second-order reactions depends on reactant concentrations.
- Option B: The rate of both first and second-order reactions depends on reactant concentrations.
- Option C: Catalysis can affect both first and second-order reactions.

Key Concepts:
- First-order reaction
- Second-order reaction
- Half-life
- Initial concentration

Memory Aids:
- First-order half-life is constant; second-order half-life depends on initial concentration.

Quick Tips:
- Memorize the half-life equations for first and second-order reactions.

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Understanding +4 | -1

Activation energy is given by

View answer and explanation

Correct option: A

The energy difference between activated complex and the energy reactant molecules

Option A (The energy difference between activated complex and the energy reactant molecules) is correct because activation energy is the energy required to transform reactants into the activated complex, representing the energy difference between these two states.

Solution approach

Correct Answer + Core Reason: The energy difference between activated complex and the energy reactant molecules. 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 activated complex and reactants: Activation energy is the difference in energy between the activated complex (transition state) and the reactant molecules.

Why other options are wrong:
- Option B: Always equal to energy of product molecules: This is incorrect; activation energy is related to the activated complex, not necessarily the product molecules.
- Option C: Always equal to energy of reactant molecules: This is incorrect; activation energy is the *difference* between the activated complex and reactant energies.
- Option D: Summation of threshold energy and energy of product molecules: This is incorrect; threshold energy is the sum of activation energy and reactant energy, not related to product energy in this way.

Key Concepts:
- Activation energy
- Activated complex
- Transition state

Memory Aids:
- Activation energy = Energy of activated complex - Energy of reactants

Quick Tips:
- Visualize the energy diagram to understand the relationship between these energies.

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Analyzing +4 | -1

For an elementary reaction

2 {A} \rightarrow {B},

consider following statements:

(a) Rate law can be written as rate

={k}[{A}]^2

(b) On increasing temperature, rate constant

k

decreases.

(c) On doubling the volume, rate becomes one fourth.

(d) On doubling the concentration, rate becomes one fourth.

Choose the correct answer from the options given below:

View answer and explanation

Correct option: A

(a), (c) only

Option A ((a), (c) only) is correct because for an elementary reaction 2A -> B, the rate law is indeed rate = k[A]², and doubling the volume quarters the rate.

Solution approach

Correct Answer + Core Reason: (a), (c) only. The rate law is correctly written, and doubling the volume quarters the rate.

Step-by-Step Solution:

Step 1: Analyze statement (a):\nFor an elementary reaction $2A \to B$ , the rate law is directly determined by the stoichiometry of the reaction. Therefore, $Rate = k[A]^2$ . This statement is correct.

Step 2: Analyze statement (b):\nIncreasing the temperature increases the rate constant $k$ , not decreases. This statement is incorrect.

Step 3: Analyze statement (c):\nIf the volume is doubled, the concentration is halved. If $[A]$ becomes $[A]/2$ , then the new rate is $Rate_{new} = k([A]/2)^2 = k[A]^2/4 = Rate/4$ . This statement is correct.

Step 4: Analyze statement (d):\nIf the concentration is doubled, $[A]$ becomes $2[A]$ , then the new rate is $Rate_{new} = k(2[A])^2 = 4k[A]^2 = 4 \times Rate$ . The rate becomes four times, not one-fourth. This statement is incorrect.

Why other options are wrong:
- Option B ((b), (d) only): Both statements are incorrect.
- Option C ((c), (d) only): Statement (d) is incorrect.
- Option D ((a), (b) only): Statement (b) is incorrect.

Key Concepts:
- Rate Law for Elementary Reactions
- Effect of Temperature on Rate Constant
- Effect of Concentration on Reaction Rate

Memory Aids:
- Remember that rate constants increase with temperature (Arrhenius equation).

Quick Tips:
- For elementary reactions, the rate law can be directly written from the stoichiometry.

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Analyzing +4 | -1

Rate of a reaction

\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}

is given below as a function of differential concentrations of

\mathrm{A}

and

\mathrm{B}

.

\begin{array}{cccccccccccccccccccc} {{\text{Exp}}{\text{.}}}&{\left[ {\text{A}} \right]}&{\left[ {\text{B}} \right]}&{{\text{Rate}}} \\ {}&{{\text{mol }}{{\text{L}}^{ - 1}}}&{{\text{mol }}{{\text{L}}^{ - 1}}}&{{\text{mol }}{{\text{L}}^{ - 1}}\,{\text{tim}}{{\text{e}}^{ - 1}}} \\ {1.}&{0.01}&{0.01}&{0.005} \\ {2.}&{0.02}&{0.01}&{0.010} \\ {3.}&{0.01}&{0.02}&{0.005} \end{array}

Determine the order of reaction with respect to

\mathrm{A}

and and also calculate the half-life of

\mathrm{A}

in the reaction.

View answer and explanation

Correct option: A

1.386

The correct answer is 1.386 because the reaction is first order with respect to A and zero order with respect to B, allowing calculation of the rate constant and subsequently the half-life.

Solution approach

Correct Answer + Core Reason: 1.386. The half-life of A is calculated based on the first-order kinetics.

Step-by-Step Solution:

Step 1: Determine the order with respect to A:\nComparing experiments 1 and 2, [B] is constant at 0.01 mol/L. When [A] doubles from 0.01 to 0.02 mol/L, the rate doubles from 0.005 to 0.010 mol/L * time^-1. Therefore, the reaction is first order with respect to A.

Step 2: Determine the order with respect to B:\nComparing experiments 1 and 3, [A] is constant at 0.01 mol/L. When [B] doubles from 0.01 to 0.02 mol/L, the rate remains constant at 0.005 mol/L * time^-1. Therefore, the reaction is zero order with respect to B.

Step 3: Write the rate law:
$Rate = k[A]^1[B]^0 = k[A]$

Step 4: Calculate the rate constant k:\nUsing experiment 1: $0.005 = k(0.01)$ , so $k = 0.005 / 0.01 = 0.5 \text{{ time}}^{-1}$

Step 5: Calculate the half-life of A:\nFor a first-order reaction, $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.5} = 1.386 \text{{ time units}}$ \nAssuming the time unit is minutes, $t_{1/2} = 1.386$ minutes.

Why other options are wrong:
- Option B (2.386): Incorrect calculation of half-life.
- Option C (3.586): Incorrect calculation of half-life.
- Option D (6.386): Incorrect calculation of half-life.

Key Concepts:
- Rate Law
- Order of Reaction
- Half-life of a First-Order Reaction

Memory Aids:
- For a first-order reaction, $t_{1/2} = \frac{0.693}{k}$

Quick Tips:
- Use the method of initial rates to determine the order of the reaction.
- Remember the half-life formula for first-order reactions.

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Analyzing +4 | -1

Which of the following statements are incorrect?

View answer and explanation

Correct option: B

Rate of the reaction involving the thermal decomposition of acetaldehyde =

k{\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}} \right]^{3/2}}

Option B (Rate of the reaction involving the thermal decomposition of acetaldehyde =

k{\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}} \right]^{3/2}}

) is correct because the rate law for the thermal decomposition of acetaldehyde is experimentally determined and not always 3/2.

Solution approach

Correct Answer + Core Reason: Rate of the reaction involving the thermal decomposition of acetaldehyde = $k{\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}} \right]^{3/2}}$ . The rate law is experimentally determined and may not always be 3/2.

Step-by-Step Solution:

Step 1: Analyze each statement:
- Option A: Rate of the reaction involving conversion of ortho hydrogen to para hydrogen $= - \frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = k{\left[ {{\mathrm{H}_2}} \right]^{3/2}}$ . This is a possible rate law.
- Option B: Rate of the reaction involving the thermal decomposition of acetaldehyde = $k{\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}} \right]^{3/2}}$ . This rate law is not always true. The order is experimentally determined.
- Option C: In the formation of phosgene from ${\mathrm{CO}}$ and ${\mathrm{C}{\mathrm{l}_2}}$ , the rate of the reaction $= k\left[ {\mathrm{CO}} \right]{\left[ {\mathrm{C}{\mathrm{l}_2}} \right]^{1/2}}$ . This is a possible rate law.
- Option D: In the decomposition of ${{\mathrm{H}_2}{\mathrm{O}_2}}$ , the rate of reaction $= k\left[ {{\mathrm{H}_2}{\mathrm{O}_2}} \right]$ . This is a possible rate law.

Step 2: Identify the incorrect statement. The rate law for the thermal decomposition of acetaldehyde is not always $k{\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}} \right]^{3/2}}$ .

Why other options are wrong:
- Option A: Rate of the reaction involving conversion of ortho hydrogen to para hydrogen $= - \frac{{d\left[ {{\mathrm{H}_2}} \right]}}{{dt}} = k{\left[ {{\mathrm{H}_2}} \right]^{3/2}}$ : This is a correct rate law.
- Option C: In the formation of phosgene from ${\mathrm{CO}}$ and ${\mathrm{C}{\mathrm{l}_2}}$ , the rate of the reaction $= k\left[ {\mathrm{CO}} \right]{\left[ {\mathrm{C}{\mathrm{l}_2}} \right]^{1/2}}$ : This is a correct rate law.
- Option D: In the decomposition of ${{\mathrm{H}_2}{\mathrm{O}_2}}$ , the rate of reaction $= k\left[ {{\mathrm{H}_2}{\mathrm{O}_2}} \right]$ : This is a correct rate law.

Key Concepts:
- Rate Law
- Order of Reaction

Memory Aids:
- The rate law is experimentally determined.

Quick Tips:
- Rate laws are determined experimentally and cannot be predicted from the balanced chemical equation.

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Analyzing +4 | -1

The initial rates of reaction

3 \mathrm{A}+2 \mathrm{B}+\mathrm{C} \longrightarrow

Products, at different initial concentrations are given below:

\begin{array}{cccccccccccccccccccc} {{\text{Initial rate,}}}&{{{\left[ {\text{A}} \right]}_0},{\text{M}}}&{{{\left[ {\text{B}} \right]}_0},{\text{M}}}&{{{\left[ {\text{C}} \right]}_0},{\text{M}}} \\ {{\text{M}}{{\text{s}}^{ - 1}}}&{}&{}&{} \\ {5.0 \times {{10}^{ - 3}}}&{0.010}&{0.005}&{0.010} \\ {5.0 \times {{10}^{ - 3}}}&{0.010}&{0.005}&{0.015} \\ {1.0 \times {{10}^{ - 2}}}&{0.010}&{0.010}&{0.010} \\ {1.25 \times {{10}^{ - 3}}}&{0.005}&{0.005}&{0.010} \end{array}

The order with respect to the reactants,

\mathrm{A}, \mathrm{B}

and

\mathrm{C}

are respectively

View answer and explanation

Correct option: D

2, 1,0

The correct answer is D (2, 1, 0) because the rate is second order with respect to A, first order with respect to B, and zero order with respect to C, based on the given experimental data.

Solution approach

Correct Answer + Core Reason: 2, 1, 0. The orders with respect to A, B, and C are 2, 1, and 0, respectively.

Step-by-Step Solution:

Step 1: Write the general rate law: $Rate = k[A]^x[B]^y[C]^z$

Step 2: Compare experiments 1 and 2 to find the order with respect to C:
$[A]_0$ and $[B]_0$ are constant, while $[C]_0$ changes from 0.010 to 0.015. The rate remains constant at $5.0 \times 10^{-3}$ . Therefore, the reaction is zero order with respect to C (z = 0).

Step 3: Compare experiments 1 and 3 to find the order with respect to B:
$[A]_0$ is constant, while $[B]_0$ changes from 0.005 to 0.010. The rate changes from $5.0 \times 10^{-3}$ to $1.0 \times 10^{-2}$ . Doubling $[B]_0$ doubles the rate, so the reaction is first order with respect to B (y = 1).

Step 4: Compare experiments 1 and 4 to find the order with respect to A:
$[B]_0$ and $[C]_0$ are constant, while $[A]_0$ changes from 0.010 to 0.005. The rate changes from $5.0 \times 10^{-3}$ to $1.25 \times 10^{-3}$ . Halving $[A]_0$ reduces the rate by a factor of 4 ( $5.0 \times 10^{-3} / 1.25 \times 10^{-3} = 4$ ), so the reaction is second order with respect to A (x = 2).

Step 5: Write the rate law: $Rate = k[A]^2[B]^1[C]^0$

Why other options are wrong:
- Option A (3,2,0): Incorrect order with respect to A and B.
- Option B (3,2,1): Incorrect order with respect to A, B, and C.
- Option C (2,2,0): Incorrect order with respect to B.

Key Concepts:
- Rate Law
- Order of Reaction
- Method of Initial Rates

Memory Aids:
- Use the method of initial rates to determine the order of reaction with respect to each reactant.

Quick Tips:
- Choose experiments where only one concentration changes to easily determine the order with respect to that reactant.

58 Medium

58 of 71Question permalink

Analyzing +4 | -1

The chemical reaction

2{\mathrm{O}_3} \to 3{\mathrm{O}_2}

proceeds as follows

{\mathrm{O}_3}\;\leftrightharpoons\;{\mathrm{O}_2} + \mathrm{O}(fast)

\mathrm{O} + {\mathrm{O}_3} \to 2{\mathrm{O}_2}(slow)

The rate law expression should be

View answer and explanation

Correct option: B

r = k{\left[ {{\mathrm{O}_3}} \right]^2}{\left[ {{\mathrm{O}_2}} \right]^{ - 1}}

Option B (

r = k{\left[ {{\mathrm{O}_3}} \right]^2}{\left[ {{\mathrm{O}_2}} \right]^{ - 1}}

) is correct because the rate law is determined by the slow step and the equilibrium of the fast step.

Solution approach

Correct Answer + Core Reason: $r = k{\left[ {{\mathrm{O}_3}} \right]^2}{\left[ {{\mathrm{O}_2}} \right]^{ - 1}}$ . The rate law is determined by the slow step and the equilibrium of the fast step.

Step-by-Step Solution:

Step 1: Identify the rate-determining step (slow step): $\mathrm{O} + {\mathrm{O}_3} \to 2{\mathrm{O}_2}$ . The rate law for this step is $r = k[\mathrm{O}][{\mathrm{O}_3}]$ .

Step 2: Express [O] in terms of $[{\mathrm{O}_3}]$ and $[{\mathrm{O}_2}]$ using the fast equilibrium step: ${\mathrm{O}_3} \leftrightharpoons {\mathrm{O}_2} + \mathrm{O}$ . The equilibrium constant $K = \frac{{[{\mathrm{O}_2}][\mathrm{O}]}}{{[{\mathrm{O}_3}]}}$ .

Step 3: Solve for [O]: $[\mathrm{O}] = K\frac{{[{\mathrm{O}_3}]}}{{[{\mathrm{O}_2}]}}$ .

Step 4: Substitute the expression for [O] into the rate law from Step 1: $r = k\left[ {K\frac{{[{\mathrm{O}_3}]}}{{[{\mathrm{O}_2}]}}} \right][{\mathrm{O}_3}] = kK\frac{{[{\mathrm{O}_3}]^2}}{{[{\mathrm{O}_2}]}}$ .

Step 5: Combine the constants $k$ and $K$ into a single rate constant $k'$ : $r = k'\frac{{[{\mathrm{O}_3}]^2}}{{[{\mathrm{O}_2}]}} = k'[{\mathrm{O}_3}]^2[{\mathrm{O}_2}]^{-1}$ .

Why other options are wrong:
- Option A: This option only considers the concentration of O3 squared, neglecting the effect of O2 which is present in the fast equilibrium step.
- Option C: This option incorrectly assumes a first-order dependence on both O3 and O2, which is not supported by the mechanism.
- Option D: The rate law can be predicted based on the given mechanism, so it is not unpredictable.

Key Concepts:
- Rate-determining step
- Chemical kinetics
- Equilibrium constant
- Rate law expression

Memory Aids:
- The rate law is determined by the slow step.
- Fast equilibrium steps can be used to express intermediates in terms of reactants and products.

Quick Tips:
- Identify the slow step first.
- Express the concentration of intermediates in terms of reactants and products using equilibrium expressions.

59 Medium

59 of 71Question permalink

Analyzing +4 | -1

The activation energies of two reactions are

{E_a}

and

{E'_a}

with

{E_a} > {E'_a}.

If the temperature of the reacting systems is increased from

{T_1}

to

{T_2}

, predict which alternative is correct.

K'

are rate constants at higher temperature. Assume

A

being same for both the reactions:

View answer and explanation

Correct option: B

{K_1} < {K_2}

and

{K'_2} < {K'_1}

Option B (

{K_1} < {K_2}

and

{K'_2} < {K'_1}

) is correct because the reaction with the lower activation energy will have a greater increase in rate constant with increasing temperature.

Solution approach

Correct Answer + Core Reason: ${K_1} < {K_2}$ and ${K'_2} < {K'_1}$ . The reaction with the lower activation energy will have a greater increase in rate constant with increasing temperature.

Step-by-Step Solution:

Step 1: Consider the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$ .

Step 2: When the temperature increases from $T_1$ to $T_2$ , the rate constant increases. The increase is more significant for the reaction with lower activation energy.

Step 3: Since ${E_a} > {E'_a}$ , the reaction with activation energy ${E'_a}$ will have a greater increase in the rate constant.

Step 4: Therefore, $\frac{{{{K'}_2}}}{{{K_2}}} > \frac{{{{K'}_1}}}{{{K_1}}}$ . This implies that ${K_1} < {K_2}$ and ${K'_2} < {K'_1}$ .

Why other options are wrong:
- Option A: $\frac{{{{K'}_1}}}{{{K_1}}} = \frac{{{{K'}_2}}}{{{K_2}}}$ : This is incorrect because the increase in rate constant depends on the activation energy.
- Option C: ${K_1} > {K_2}$ and ${K'_1} > {K'_2}$ : This is incorrect because the rate constants increase with temperature.
- Option D: $\frac{{{{K'}_1}}}{{{K_1}}} < \frac{{2{{K'}_2}}}{{{K_2}}}$ : This is not necessarily true. The factor of 2 is arbitrary.

Key Concepts:
- Arrhenius Equation
- Activation Energy
- Temperature Dependence of Reaction Rates

Memory Aids:
- Lower activation energy means a greater increase in rate constant with temperature.

Quick Tips:
- Use the Arrhenius equation to analyze the temperature dependence of reaction rates.

60 Medium

60 of 71Question permalink

Understanding +4 | -1

For the reaction,

{\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{B}{\mathrm{r}_2}\left( \mathrm{g} \right) \to 2\mathrm{HBr}\left( \mathrm{g} \right)

The experimental data suggest

\mathrm{rate} = k\left[ {{\mathrm{H}_2}} \right]{\left[ {\mathrm{B}{\mathrm{r}_2}} \right]^{1/2}}

the molecularity and order of the reaction are respectively

View answer and explanation

Correct option: D

2,\frac{3}{2}

Option D (2,

\frac{3}{2}

) is correct because molecularity is the number of reacting molecules (2), and the order is the sum of the exponents in the rate law (1 + 1/2 = 3/2).

Solution approach

Correct Answer + Core Reason: $2,\frac{3}{2}$ . Molecularity is the number of molecules involved in an elementary step, and the order is the sum of the exponents in the rate law.

Step-by-Step Solution:

Step 1: Write the reaction: ${\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{B}{\mathrm{r}_2}\left( \mathrm{g} \right) \to 2\mathrm{HBr}\left( \mathrm{g} \right)$

Step 2: The rate law is given as: $\mathrm{rate} = k\left[ {{\mathrm{H}_2}} \right]{\left[ {\mathrm{B}{\mathrm{r}_2}} \right]^{1/2}}$

Step 3: Molecularity is determined by the elementary step. Since the reaction involves one molecule of H2 and one molecule of Br2, the molecularity is 2.

Step 4: The order of the reaction is the sum of the exponents in the rate law: $1 + \frac{1}{2} = \frac{3}{2}$

Why other options are wrong:
- Option A: $1,\frac{1}{2}$ : Incorrect molecularity.
- Option B: $1,1$ : Incorrect order and molecularity.
- Option C: $\frac{3}{2},\frac{3}{2}$ : Incorrect molecularity.

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

Memory Aids:
- Molecularity: Number of reacting molecules.
- Order: Sum of exponents in the rate law.

Quick Tips:
- Molecularity is a theoretical concept, while the order is experimentally determined.

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