Class 12 Chemistry Chapter 3 Chemical Kinetics

Key Concept: Average Rate Calculation

a) $0.001 mol L^{-1}s^{-1}$

[Solution Description] The average rate of reaction is calculated using the formula:

$$\text{Average Rate} = -\frac{\Delta[\text{A}]}{\Delta t} = -\frac{(0.03 – 0.05)\, \text{M}}{20\, \text{s}} = \frac{0.02\, \text{M}}{20\, \text{s}} = 0.001\, \text{mol L}^{-1}\text{s}^{-1}.$$

Key Concept: Basic Definition

a) The time taken for the concentration of a reactant to reduce to one half of its initial concentration.
[Solution Description]
The half-life $t_{1/2}$ is defined as the time required for the concentration of a reactant to reduce to one half of its initial concentration.

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Key Concept: Orientation and Energy, Theoretical Limitations, Molecular Structure

d) Assertion is false, but Reason is true.

[Solution Description] In collision theory, while $Z$ (collision frequency) and $E_a$ (activation energy) are important factors in determining the reaction rate, the theory does not fully account for all aspects due to its limitations. One crucial limitation is that it ignores the molecular structure which affects the orientation during collisions; proper orientation is also required besides sufficient kinetic energy. Hence, proper orientation is a significant factor that collision theory fails to completely address. Therefore, both assertions about collision frequency and activation energy alone determining reaction rates are false under complete explanation standards, but the necessity of proper orientation as stated in the reason is true.

Key Concept: Factors Affecting Rate

d) Color
[Solution Description] The factors affecting the rate of a chemical reaction typically include concentration, temperature, and catalysts. The color of substances is not a factor that affects the reaction rate.

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Key Concept: Basic Concept

c) The number of collisions per second per unit volume
[Solution Description] The collision frequency Z refers to the number of collisions that occur per second per unit volume in a reaction mixture.

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Key Concept: Role of Steric Factor

c) Adjusts for molecular orientation in collisions
[Solution Description] The steric factor P accounts for the proper orientation of molecules during a collision. In collision theory, effective collisions require not only sufficient kinetic energy but also proper molecular orientation. $P$ adjusts the calculated collision frequency to account for this requirement.

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Key Concept: Arrhenius Equation Basics

b) Frequency factor
[Solution Description] The Arrhenius equation is used to describe how the rate constant $k$ of a reaction depends on the temperature $T$ and activation energy $E_a$. The frequency factor, represented by $A$, is a pre-exponential factor that indicates the number of times that reactants approach the transition state per unit time.

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Key Concept: Collision Theory, Real-world Application

b) Increases effective collisions
[Solution Description]
According to collision theory, a catalyst provides an alternate reaction pathway that lowers the activation energy. This increases the number of effective collisions because more molecules possess the required energy for the reaction. Therefore, the catalyst increases the frequency of successful molecular collisions without changing the orientation requirement or the equilibrium position.

Thus, the correct statement is: “Increases effective collisions.”

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Key Concept: Units of Rate

b) s$^{-1}$

[Solution Description]

In a first-order reaction, the rate law is expressed as:

$$\text{Rate} = k[A]$$

The unit of rate is mol L$^{-1}$s$^{-1}$ and the unit of concentration is mol L$^{-1}$. Therefore, the units of $k$ will be:

$$\text{Unit of } k = \frac{\text{mol L}^{-1}\text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}$$

Key Concept: Intercept Understanding

b) $ln A$
[Solution Description] The intercept of the ln k vs. 1T graph represents $ln A$, which is a constant related to the frequency factor in the Arrhenius equation. It shows how often molecules collide with the correct orientation and enough energy to react.

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Key Concept: Activation Energy

c) It increases
[Solution Description] The Arrhenius equation is given by $$k = Ae^{-Ea/RT}$$, where $k$ is the rate constant, $Ea$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature. When the activation energy $(Ea)$ is decreased, the exponent $$-Ea/RT$$ becomes less negative, thus increasing the value of $$ e^{-Ea/RT} $$, leading to a higher rate constant $k$. Hence, the rate of the reaction increases.

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Key Concept: Molecularity Concept

c) 3
[Solution Description] Molecularity refers to the number of molecules colliding in an elementary step. For the reaction $2NO + O_2 \rightarrow 2NO_2$, three molecules $(2NO)$ and $(1O_2)$ are involved, making it a trimolecular reaction.

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Key Concept: Activation Energy

b) A catalyst decreases the activation energy required for a reaction.
[Solution Description] A catalyst provides an alternate pathway for the reaction with a lower activation energy. This means that it reduces the height of the energy barrier that must be overcome for reactants to form products.

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Key Concept: Integrated Rate Equation

b) $ln[R] = -kt + ln[R]_0$
[Solution Description] For a first-order reaction, the integrated rate equation is often represented as: $ln[R] = -kt + ln[R]_0$ where [R] is the concentration of reactant at time t, k is the rate constant, and [R]_0 is the initial concentration. Hence, option b) correctly represents a first-order reaction.

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Key Concept: First Order Independence

a) Independent of initial concentration
[Solution Description] In a first order reaction, the half-life is independent of the initial concentration. This means it remains constant regardless of how much reactant you start with. Therefore, the correct option is that the half-life is independent of initial concentration.

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Key Concept: Overall Order Calculation

c)$\frac{7}{3}$
[Solution Description] The overall order of a reaction is calculated by summing the exponents in the rate law expression. Here, the exponents are $\frac{1}{3}$ for $[A]$ and $2$ for $[B]$. Thus, the overall order is: $\frac{7}{3}$ Therefore, the overall order of the reaction is $\frac{7}{3}$.

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.

Key Concept: Multi-Step Analysis, Real-World Application

d) Third order
[Solution Description] The units of the rate constant $k_1$ suggest that the first step is third order $(2 + 1 = 3)$, since: $L^2 mol^{-2} s^{-1} = \frac{1}{(mol \, L^{-1})^2} \times s^{-1}$ Given that the first step is the rate-determining step, the overall order is determined by the first step’s rate constant. Thus, the overall order is third order.

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Key Concept: Basic Concept

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] In zero order reactions, the rate does not depend on the concentration of the reactant because it is fixed by some other factor, such as a surface or catalyst. The rate equation $Rate = k$ signifies that the rate is constant and equal to the rate constant $k$. Thus, both statements are true and Reason explains why Assertion is correct.

Key Concept: Pseudo First Order

b) Hydrolysis of ethyl acetate

[Solution Description] A pseudo-first-order reaction is one that appears first order because one of the reactants is present in large excess. Hydrolysis of ethyl acetate in the presence of water is a common example where water is in excess, making it effectively first order with respect to ethyl acetate.

Key Concept: Calculation of Ea

c) 25.55,kJ/mol
[Solution Description] To find the activation energy $(Ea)$, we use the Arrhenius equation in its logarithmic form:

$$ln \frac{k_2}{k_1} = -\frac{Ea}{R} \left( \frac{1}{T_2} – \frac{1}{T_1} \right)$$

We have:
$$k_1 = 1.5 \times 10^{-3}\,s^{-1}, T_1 = 350\,K$$,
$$k_2 = 4.5 \times 10^{-3}\,s^{-1},T_2 = 400\,K$$,
$$ R = 8.314\,J/(mol\,K)$$.

Substituting these values, we get:

$$ln \frac{4.5 \times 10^{-3}}{1.5 \times 10^{-3}} = -\frac{Ea}{8.314} \left( \frac{1}{400} – \frac{1}{350} \right)$$

Simplifying,

$$ln 3 = -\frac{Ea}{8.314} \left( \frac{1}{400} – \frac{1}{350} \right)$$

$$1.0986 = -\frac{Ea}{8.314} \left( -0.000357 \right)$$

$$Ea = \frac{1.0986 \times 8.314}{0.000357} = 25549.56\,J/mol$$

The activation energy is approximately $25.55\,kJ/mol$.

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Key Concept: Zero Order Formula

a) $t_{1/2} = \frac{[R]_0}{2k}$
[Solution Description] The formula for the half-life in a zero order reaction is derived from the rate law where the concentration decreases linearly with time. It is given by $t_{1/2} = \frac{[R]_0}{2k}$, where $[R]_0$ is the initial concentration and $k$ is the rate constant.

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Key Concept: Units of Rate Constant

c) Second-order
[Solution Description] To determine the order from the units of the rate constant, we use the formula for the unit of rate constant $k$, which is dependent on the overall reaction order $n$. The general unit for $k$ in terms of molarity (M) and time (s) is $$\text{(concentration)}^{1-n} \text{time}^{-1}$$. Given the unit as $$\text{mol}^{-1}\,\text{L}\,\text{s}^{-1}$$, it implies that $k$ corresponds to a second-order reaction where $n = 2$.

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Key Concept: Rate Law Derivation, Order and Mechanism

b) Exponents do not match stoichiometric coefficients
[Solution Description] The rate law $Rate = k [A]^3$ shows that the order with respect to $A$ is 3. The stoichiometric coefficient for $A$ is 2, which differs from the exponent in the rate law, indicating that the reaction is not straightforward and might proceed via a different mechanism than directly represented by the balanced chemical equation.

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Key Concept: Catalyst Definition

a) A substance that increases the rate of a reaction without undergoing permanent change
[Solution Description]
A catalyst is a substance that increases the rate of a chemical reaction without undergoing any permanent chemical change itself. It provides an alternative pathway with a lower activation energy for the reaction to occur.

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Key Concept: First Order Reaction

b) $ln[R] = -kt + ln[R]_0$
[Solution Description] The integrated rate equation for a first order reaction is given by $ln[R] = -kt +ln[R]_0 $. This shows that the natural logarithm of the concentration of reactant $[R]$ linearly depends on time with slope $-k$.

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Key Concept: Reaction Pathway

c) Lowers the activation energy needed for the reaction.
[Solution Description] A catalyst provides an alternate reaction pathway with a lower activation energy, thereby increasing the reaction rate. The key to understanding catalysis is recognizing that it reduces the energy barrier required for the reaction to proceed.

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Key Concept: Complex Example Identification

b) $H_2 + I_2 \rightarrow 2HI$
[Solution Description] A bimolecular reaction involves two reacting species (atoms, ions, or molecules). Examining each option:

Option a) $2NO_2 \rightarrow 2NO + O_2$ has only one type of molecule on the reactant side being decomposed into products; it is not bimolecular.

Option b) $H_2 + I_2 \rightarrow 2HI$ involves two different molecules $(H_2 and I_2)$ coming together to form the product; this is a bimolecular reaction.

Option c) $2H_2O_2 \rightarrow 2H_2O + O_2$ involves the decomposition of one molecule type $(H_2O_2)$, meaning it is unimolecular with respect to each $H_2O_2$ unit.

Option d) $N_2O_5 \rightarrow 2NO_2 + 1/2 O_2$ involves a single molecule decomposing into multiple products; it is unimolecular.

Therefore, the correct answer is b).

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Key Concept: Real-World Application, Reaction Mechanism Analysis

d) It offers a surface for reactants to form transient intermediates, lowering overall energy barriers.
[Solution Description] In catalytic hydrogenation, the metal surface enables the adsorption of both ethylene and hydrogen, facilitating their interaction through formation of temporary bonds. This creates a lower-energy pathway compared to the uncatalyzed reaction, thereby increasing the reaction rate. The catalyst itself is not consumed in the overall reaction.

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Key Concept: Catalyst Role

d) It reduces the activation energy, thereby speeding up the reaction.

[Solution Description] A catalyst provides an alternative reaction pathway with a lower activation energy, allowing more reactant molecules to have enough energy to overcome this new barrier and thus increasing the rate of reaction without being consumed.

Key Concept: Temperature Effect

c) Increases the rate
[Solution Description] According to the Arrhenius equation, increasing temperature increases the rate of reaction because it provides more energy for particles to surpass the activation energy barrier, resulting in more effective collisions.

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Key Concept: Average vs Instantaneous Rate

c) Average rate analyzes overall speed over time; instantaneous rate focuses on specific moments.
[Solution Description] The average rate is used to determine the rate over a finite interval of time by calculating the change in concentration divided by the change in time. It gives a general idea of the reaction’s speed over that duration. On the other hand, the instantaneous rate is determined at a specific point in time and is calculated from the slope of the tangent line on a concentration vs. time graph. This provides a precise rate at that exact moment.

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Key Concept: Frequency Factor

b) Collision frequency
[Solution Description] The frequency factor A in the Arrhenius equation accounts for the collision frequency and the proper orientation needed for molecules to undergo a successful reaction.

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Key Concept: Reaction Order Determination, Real World Application, Integrated Rate Laws

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that for a first-order reaction, the half-life $(t_{1/2})$ is constant and does not depend on the initial concentration. This is true, as the formula for the half-life of a first-order reaction is derived from the integrated rate law: $t_{1/2} = \frac{\ln 2}{k}$ where $k$ is the rate constant. It shows that $$t_{1/2}$$ depends only on $k$, thus is constant regardless of initial concentration. The reason relates to the application of this concept in real-world scenarios like carbon dating, where the decay of carbon-14, a first-order process, is used. Since the half-life is constant, it allows for accurate measurement of time elapsed since the death of an organism by measuring remaining carbon-14, hence enabling archaeologists to determine the age of finds. Both Assertion and Reason are true, and Reason correctly explains the Application aspect of Assertion, providing its relevance in the real world. Therefore, the correct answer is (a).

Key Concept: Rate Law Application

d) It reduces to one-fourth.

[Solution Description]

In a first-order reaction, the rate is directly proportional to the concentration of the reactant. The rate law can be expressed as:

$$\text{Rate} = k[\text{Reactant}]$$

When the concentration is reduced to one-fourth, the new rate becomes:

$$\text{New Rate} = k\left(\frac{[\text{Reactant}]}{4}\right) = \frac{1}{4}(k[\text{Reactant}])$$

Thus, the rate of the reaction also reduces to one-fourth.

Key Concept: Activation Energy Concept

c) The energy required to form an activated complex
[Solution Description] Activation energy $Ea$ is the minimum energy that the reactants must possess in order to transform into an activated complex, which then leads to the formation of products.

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Key Concept: Concentration Effect

c) Concentration
[Solution Description] Increasing the concentration of reactants typically increases the rate of a chemical reaction because there are more molecules available to collide with each other. With more frequent collisions, the likelihood of effective collisions that lead to reactions also increases.

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Key Concept: Complex Rate Law, Reaction Order Determination

b) m = 2, n = 1
[Solution Description] To determine the order with respect to each reactant, we analyze how changes in concentration affect the rate. Doubling [A] causing a fourfold increase indicates that the rate is proportional to $[A]^2$. Therefore, $m=2$. Similarly, doubling [B] causes the rate to double, indicating first-order dependence on [B], so $n=1$. Thus, the orders are $m=2$ and $n=1$.

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Key Concept: Average Rate Calculation

c) 0.0025
[Solution Description] To calculate the average rate of the reaction, we use the formula:
$$\text{Average rate} = \frac{\Delta [A]}{\Delta t} = \frac{[A]_{\text{final}} – [A]_{\text{initial}}}{t_{\text{final}} – t_{\text{initial}}}$$.
Substituting the values given:
$$= \frac{0.025\, \text{M} – 0.05\, \text{M}}{10\, \text{s}} = \frac{-0.025\, \text{M}}{10\, \text{s}} = -0.0025\, \text{mol L}^{-1} \text{s}^{-1}.$$
Since the concentration decreases, the negative sign indicates the decrease and is not required for the magnitude of rate.

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Key Concept: Rate Law and Stoichiometry

d) The rate-determining step defines the rate law for the overall reaction.
[Solution Description] For a complex reaction involving multiple steps, the rate law is determined by the rate-determining step, which is the slowest step in the sequence. This step controls the overall reaction rate because it serves as a bottleneck.

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Key Concept: Intermediate Complex Theory

b) It forms intermediate complexes with lower energy.
[Solution Description]
A catalyst increases the rate of a chemical reaction by providing an alternate pathway with lower activation energy. This is often achieved by forming intermediate complexes with the reactants, which require less energy to convert into products. By lowering the activation energy barrier, more molecules have sufficient energy to undergo effective collisions and form products, thus increasing the reaction rate.

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Key Concept: Integrated Rate Laws, Real World Application

c) 20 minutes
[Solution Description] To solve this problem, we start by using the integrated rate law for a first-order reaction: $$ln[R] = -kt + ln[R]_0$$. From the initial data, we can calculate the rate constant $$k$$ when the concentration changes from 1.0 M to 0.5 M:

$$
ln(0.5) = -k \times 20 + ln(1.0)
$$

Solving for $$k$$, we find:

$$
k = \frac{ln(1.0) – ln(0.5)}{20} = \frac{0.693}{20} = 0.03465 \, min^{-1}
$$

Next, we use this $$k$$ value to find the time required for the concentration to drop from 0.5 M to 0.25 M:

$$
ln(0.25) = -0.03465 \, t + ln(0.5)
$$

Simplifying and solving for $$t$$:

$$
t = \frac{ln(0.5) – ln(0.25)}{0.03465} = \frac{0.693}{0.03465} = 20 \, minutes
$$

Therefore, it will take another 20 minutes for the concentration to decrease from 0.5 M to 0.25 M.

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Key Concept: Order Determination from Rate Data, Real-world Application

c) Order = 2
[Solution Description] Let’s assume the rate expression is $$\text{Rate} = k[X]^n$$. If tripling [X] results in the rate increasing by nine times, then:

$$9 \times \text{original rate} = k(3[X])^n$$

Simplifying, we have: $9 = 3^n$ Solving for $n$, we take the logarithm base 3 of both sides: $n = log_3{9} = 2$ Thus, the order with respect to X is 2.

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Key Concept: Complex Reaction Analysis, Zero Order Reaction Understanding

c) Assertion is true, but Reason is false.

[Solution Description] To determine the truth of the assertion and reason:

– **Assertion**: This statement is true as per the definition of a zero-order reaction, where the rate is independent of reactant concentration. In practical scenarios like the decomposition of $O_3$, certain surface-catalyzed reactions can indeed exhibit zero-order kinetics since the reaction occurs at a constant rate when the catalyst’s active sites are fully occupied.

– **Reason**: This statement is false. By the definition of zero-order reactions, the rate is actually independent of the concentrations of reactants. Therefore, increasing the concentration does not affect the rate of zero-order reactions.

Thus, the Assertion is true, but the Reason is false.

Key Concept: Simple Calculation

c) 1
[Solution Description] To find the overall order of the reaction, sum the exponents of the concentration terms in the rate expression. Here, the exponent of $[C]$ is $\frac{3}{4}$
and the exponent of $[D]$ is $\frac{1}{4}$ . Adding these gives $\ce{\frac{3}{4} + \frac{1}{4} = 1}$ Thus, the overall order is 1.

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Key Concept: Formula Identification

c) $[R] = [R]_0 e^{-kt}$
[Solution Description] For a first order reaction, the integrated rate equation can be expressed in terms of the initial concentration $[R]_0$ and the concentration at any time $[R]$. It is given by the expression $[R] = [R]_0 e^{-kt}$, where $k$ is the rate constant and $t$ is time.

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Key Concept: Reaction Feasibility

c) When $\Delta G < 0$ [Solution Description] A chemical reaction is considered thermodynamically feasible if the change in Gibbs free energy $(\Delta G)$ is negative, which indicates that the reaction can proceed spontaneously under the given conditions. Click here to explore notes for this chapter or topic

Key Concept: Tangent Slope

b) By drawing a tangent at a specific time and calculating its slope
[Solution Description] The instantaneous rate of reaction at any point on a concentration vs. time curve can be found by drawing a tangent to the curve at that specific point and calculating the slope of this tangent line. The slope represents the rate of change of concentration with respect to time.

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Key Concept: Activation Energy

b) Activation Energy
[Solution Description] The activation energy $E_a$ is the least amount of energy necessary for reactants to transform into products in a chemical reaction.

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Key Concept: Rate Law Application

c) Rate = $$k[A]^2$$

[Solution Description] The rate of reaction depends on the concentration of reactants raised to their respective powers. Given that doubling $$[A]$$ quadruples the rate, the order with respect to $$A$$ is 2. Doubling $$[B]$$ has no effect, so the order with respect to $$B$$ is 0. Thus, the rate law is $$\text{Rate} = k[A]^2[B]^0 = k[A]^2$$.