Chapter 4 Chemical Kinetics

1. Explain the effect of temperature on the rate of a reaction. Derive the Arrhenius equation and discuss its significance.

Answer:
The rate of a reaction increases with an increase in temperature. This is explained by the collision theory, which states that as the temperature rises, the molecules gain more energy and collide more frequently and forcefully. The Arrhenius equation is given as:

$k = A e^{-\frac{E_a}{RT}}$

Where:

• $k$ is the rate constant,
• $A$ is the pre-exponential factor (frequency of collisions),
• $E_a$ is the activation energy,
• $R$ is the gas constant (8.314 J/mol·K),
• $T$ is the temperature in Kelvin.

By taking the natural logarithm of both sides:

$$\ln k = \ln A - \frac{E_a}{RT}$$

This is the linear form of the Arrhenius equation, which shows that a plot of $\ln k$ vs $\frac{1}{T}$ will give a straight line with a slope of $-\frac{E_a}{R}$ . The equation shows how the rate constant $k$ changes with temperature. A higher temperature lowers the exponential factor, leading to a significant increase in the rate constant.

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2. Discuss the concept of half-life and its significance in first-order reactions. Derive the expression for the half-life of a first-order reaction.

Answer:
In chemical kinetics, half-life refers to the time required for the concentration of a reactant to reduce to half of its initial value. It is a critical concept in determining the speed of a reaction. For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant.

The rate equation for a first-order reaction is:

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

The integrated rate law for a first-order reaction is:

$$\ln \frac{[A]_0}{[A]} = kt$$

Where:

• $[A]_0$ is the initial concentration,
• $[A]$ is the concentration of reactant at time $t$ ,
• $k$ is the rate constant.

At half-life ($t_{1/2}$ ), the concentration of the reactant becomes half of its initial value. Therefore,

$$\ln \frac{[A]_0}{[A]_{t_{1/2}}} = \ln 2 = kt_{1/2}$$

Thus, the half-life expression for a first-order reaction is:

$$t_{1/2} = \frac{0.693}{k}$$

This shows that for a first-order reaction, the half-life is independent of the initial concentration of the reactant and depends only on the rate constant $k$ .

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3. For a reaction $A \rightarrow B$ , the rate constant $k$ is given by $k = 2.0 \times 10^{-3} \, s^{-1}$ . If the initial concentration of A is 0.5 M, calculate the concentration of A after 200 seconds.

Answer:
This is a first-order reaction, and the integrated rate law is:

$$\ln \frac{[A]_0}{[A]} = kt$$

Substitute the given values:

• $[A]_0 = 0.5 \, \text{M}$ ,
• $k = 2.0 \times 10^{-3} \, s^{-1}$ ,
• $t = 200 \, s$ .

The equation becomes:

$$\ln \frac{0.5}{[A]} = (2.0 \times 10^{-3})(200)$$ $$\ln \frac{0.5}{[A]} = 0.4$$ $$\frac{0.5}{[A]} = e^{0.4} \approx 1.4918$$ $$[A] = \frac{0.5}{1.4918} \approx 0.335 \, \text{M}$$

So, the concentration of A after 200 seconds is approximately 0.335 M.

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4. Derive the rate law for a second-order reaction and calculate the half-life for a second-order reaction given the rate constant $k = 4 \times 10^{-2} \, \text{L/mol·s}$ and initial concentration $[A]_0 = 0.2 \, \text{mol/L}$ .

Answer:
For a second-order reaction, the rate law is:

$$\text{Rate} = k[A]^2$$

The integrated rate law for a second-order reaction is:

$$\frac{1}{[A]} - \frac{1}{[A]_0} = kt$$

To find the half-life, we use the formula:

$$t_{1/2} = \frac{1}{k[A]_0}$$

Substituting the given values:

$$t_{1/2} = \frac{1}{(4 \times 10^{-2})(0.2)} = \frac{1}{0.008} = 125 \, \text{seconds}$$

Thus, the half-life for this second-order reaction is 125 seconds.

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5. Explain the concept of activation energy. How does the presence of a catalyst affect the activation energy of a reaction?

Answer:
Activation energy ($E_a$ ) is the minimum energy required by the reactants to undergo a chemical reaction. It is the energy barrier that must be overcome for the reactants to be converted into products. Activation energy is critical in determining the rate of a reaction. Higher activation energy means fewer molecules have sufficient energy to react, leading to a slower reaction.

A catalyst provides an alternative reaction pathway with a lower activation energy. It does not affect the final products of the reaction but increases the rate of reaction by lowering the energy barrier. This results in a faster reaction because more reactant molecules can achieve the required energy for the reaction to proceed.

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6. What is the difference between zero-order, first-order, and second-order reactions in terms of their rate laws and integrated rate laws?

Answer:

• Zero-Order Reactions:
  Rate law: $\text{Rate} = k$ .
  Integrated rate law:
  $[A] = [A]_0 - kt$
  Half-life:
  $t_{1/2} = \frac{[A]_0}{2k}$
  The concentration decreases linearly over time.

• First-Order Reactions:
  Rate law: $\text{Rate} = k[A]$ .
  Integrated rate law:
  $\ln [A] = \ln [A]_0 - kt$
  Half-life:
  $t_{1/2} = \frac{0.693}{k}$
  The concentration decreases exponentially.

• Second-Order Reactions:
  Rate law: $\text{Rate} = k[A]^2$ .
  Integrated rate law:
  $\frac{1}{[A]} = \frac{1}{[A]_0} + kt$
  Half-life:
  $t_{1/2} = \frac{1}{k[A]_0}$
  The concentration decreases according to a reciprocal relationship.

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7. A reaction follows the first-order rate law with a rate constant $k = 3.0 \times 10^{-2} \, \text{min}^{-1}$ . If the initial concentration of the reactant is 0.6 M, calculate the time required for the concentration to reduce to 0.15 M.

Answer:
For a first-order reaction, the integrated rate law is:

$$\ln \frac{[A]_0}{[A]} = kt$$

Substitute the values:

• $[A]_0 = 0.6 \, \text{M}$ ,
• $[A] = 0.15 \, \text{M}$ ,
• $k = 3.0 \times 10^{-2} \, \text{min}^{-1}$ .

The equation becomes:

$$\ln \frac{0.6}{0.15} = (3.0 \times 10^{-2}) \cdot t$$ $$\ln 4 = 0.4 = (3.0 \times 10^{-2}) \cdot t$$ $$t = \frac{0.4}{3.0 \times 10^{-2}} = 13.33 \, \text{min}$$

Thus, the time required is 13.33 minutes.

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8. What is a rate-determining step in a reaction mechanism? How is it related to the overall rate of reaction?

Answer:
The rate-determining step is the slowest step in a reaction mechanism. It controls the overall rate of the reaction because the entire reaction cannot proceed faster than the slowest step. In a multi-step reaction, even if the other steps are fast, the rate of the entire reaction will be governed by the slowest step.

This step often involves the highest activation energy, and it is the bottleneck of the reaction mechanism. The overall rate law is determined by the rate-determining step, and it involves the concentration of reactants involved in this step.

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9. Explain the term "order of reaction" and how it is determined experimentally. Provide a detailed example.

Answer:
The order of a reaction is the sum of the exponents of the concentrations of the reactants in the rate law. It indicates how the rate of the reaction depends on the concentration of the reactants.

Experimental determination of the order involves:

• Method of Initial Rates: Measure the initial rate of the reaction at different initial concentrations of reactants and deduce the order by comparing how the rate changes with concentration.
• Integrated Rate Law Method: Use concentration data over time, plot graphs, and deduce the order from the shape of the graph.

Example: For a reaction $A + B \rightarrow C$ , if changing the concentration of $A$ affects the rate more than $B$ , the order with respect to $A$ is higher than that of $B$ .

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10. For a reaction $A \rightarrow B$ , the rate constant $k$ is $1.0 \times 10^{-3} \, \text{s}^{-1}$ , and the initial concentration of A is 0.2 M. Calculate the time required for the concentration of A to reduce to 0.05 M.

Answer:
For a first-order reaction, the integrated rate law is:

$$\ln \frac{[A]_0}{[A]} = kt$$

Substituting the values:

• $[A]_0 = 0.2 \, \text{M}$ ,
• $[A] = 0.05 \, \text{M}$ ,
• $k = 1.0 \times 10^{-3} \, \text{s}^{-1}$ .

$$\ln \frac{0.2}{0.05} = (1.0 \times 10^{-3}) \cdot t$$ $$\ln 4 = 0.4 = (1.0 \times 10^{-3}) \cdot t$$ $$t = \frac{0.4}{1.0 \times 10^{-3}} = 400 \, \text{seconds}$$

Thus, the time required is 400 seconds.

11. Explain the term 'collision theory of reaction rates'. How does the orientation of reacting molecules affect the reaction rate?

Answer: The collision theory states that for a chemical reaction to occur, the reactant molecules must collide with sufficient energy and proper orientation. The rate of reaction is directly proportional to the frequency of effective collisions. These effective collisions are those where molecules have enough energy to overcome the activation energy barrier and are properly oriented for bond breaking and formation.

The energy needed for a collision to be effective is called the activation energy ($E_a$ ). Molecules collide more effectively when they are oriented correctly. For example, in a reaction between two molecules, if the reactive parts of the molecules do not align correctly, the collision, even if energetic enough, will not result in a reaction. Therefore, both the energy and orientation of the molecules are crucial in determining the reaction rate. A catalyst works by lowering the activation energy or providing an alternate pathway for the reaction with a lower activation energy.

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12. What is the integrated rate law for a zero-order reaction? How can you determine the rate constant $k$ for a zero-order reaction experimentally?

Answer: For a zero-order reaction, the rate of the reaction is independent of the concentration of the reactant. The rate law for a zero-order reaction is:

$$\text{Rate} = k$$

The integrated rate law for a zero-order reaction is derived from the rate law:

$$\frac{d[A]}{dt} = -k$$

Integrating this with respect to time gives:

$$[A] = [A]_0 - kt$$

Where:

• $[A]_0$ is the initial concentration,
• $[A]$ is the concentration at time $t$ ,
• $k$ is the rate constant.

To determine $k$ experimentally, you can monitor the concentration of the reactant over time and plot a graph of $[A]$ vs $t$ . For a zero-order reaction, this graph will be a straight line, and the rate constant $k$ can be calculated from the slope of the line (negative slope).

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13. For a reaction, the rate law is given by $\text{Rate} = k[A]^2[B]$ . If the concentration of $A$ is doubled and the concentration of $B$ is halved, how does the rate of reaction change?

Answer: For the rate law $\text{Rate} = k[A]^2[B]$ , we can analyze the effect of changing the concentrations of $A$ and $B$ .

Let the initial concentrations be $[A]_0$ and $[B]_0$ . The initial rate of the reaction is:

$$\text{Rate}_0 = k[A]_0^2[B]_0$$

Now, when the concentration of $A$ is doubled and the concentration of $B$ is halved, the new concentrations become $2[A]_0$ and $\frac{1}{2}[B]_0$ . The new rate of reaction is:

$$\text{Rate}_{\text{new}} = k(2[A]_0)^2 \left(\frac{1}{2}[B]_0\right)$$

Simplifying:

$$\text{Rate}_{\text{new}} = k \cdot 4[A]_0^2 \cdot \frac{1}{2}[B]_0 = 2k[A]_0^2[B]_0$$

Thus, the rate of the reaction becomes twice the original rate. Therefore, the reaction rate increases by a factor of 2.

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14. Describe the concept of a reaction mechanism. Explain the terms 'elementary step' and 'intermediate' with examples.

Answer: A reaction mechanism is a detailed step-by-step description of how a chemical reaction occurs at the molecular level. It breaks down the overall reaction into individual elementary steps. Each elementary step involves the collision of a small number of molecules or atoms.

• Elementary Step: An elementary step is a single molecular event that represents one reaction step in the mechanism. It describes a direct collision between reactants to form products. For example, in the reaction $A + B \rightarrow C$ , this could be an elementary step.

• Intermediate: An intermediate is a species that is formed in one elementary step and consumed in a subsequent step. It is not present in the overall reaction. For instance, in the reaction sequence $A + B \rightarrow C$ and $C + D \rightarrow E$ , the species $C$ is an intermediate because it is formed in the first step and consumed in the second.

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15. A reaction follows the second-order rate law with respect to reactant $A$ . If the concentration of $A$ is reduced to half in 20 minutes, how much time will it take for $A$ to reduce to one-fourth of its original concentration?

Answer: For a second-order reaction, the integrated rate law is:

$$\frac{1}{[A]} - \frac{1}{[A]_0} = kt$$

Let the initial concentration of $A$ be $[A]_0$ . When the concentration is reduced to half, we have:

$$\frac{1}{\frac{[A]_0}{2}} - \frac{1}{[A]_0} = k \cdot 20$$

This simplifies to:

$$\frac{2}{[A]_0} - \frac{1}{[A]_0} = k \cdot 20$$ $$\frac{1}{[A]_0} = k \cdot 20$$

Thus, $k = \frac{1}{20[A]_0}$ .

Now, to find the time when the concentration is reduced to one-fourth of its original value, use:

$$\frac{1}{\frac{[A]_0}{4}} - \frac{1}{[A]_0} = k \cdot t$$

Simplifying:

$$\frac{4}{[A]_0} - \frac{1}{[A]_0} = k \cdot t$$ $$\frac{3}{[A]_0} = k \cdot t$$

Substitute $k = \frac{1}{20[A]_0}$ :

$$\frac{3}{[A]_0} = \frac{1}{20[A]_0} \cdot t$$ $$t = 60 \, \text{minutes}$$

Thus, it will take 60 minutes for the concentration of $A$ to reduce to one-fourth of its original concentration.

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16. Explain the concept of the 'rate constant' and its units for different orders of reactions. How do you determine the order of reaction experimentally?

Answer: The rate constant ($k$ ) is a proportionality constant that relates the rate of a chemical reaction to the concentrations of the reactants, as per the rate law. Its value depends on the nature of the reaction and temperature. The units of $k$ vary with the order of the reaction.

• Zero-order reaction:
  Rate law: $\text{Rate} = k$
  Units of $k$ : $\text{mol/L·s}$

• First-order reaction:
  Rate law: $\text{Rate} = k[A]$
  Units of $k$ : $\text{s}^{-1}$

• Second-order reaction:
  Rate law: $\text{Rate} = k[A]^2$
  Units of $k$ : $\text{L/mol·s}$

To determine the order of the reaction experimentally, the method of initial rates is commonly used. This involves measuring the initial rate of the reaction at different initial concentrations of reactants and analyzing how the rate changes. The order can be determined by comparing the changes in concentration and rate.

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17. For a reaction, the rate constant $k$ is $1.0 \times 10^{-3} \, \text{s}^{-1}$ . If the concentration of the reactant is $0.2 \, \text{M}$ , how long will it take for the concentration to reduce to $0.05 \, \text{M}$ in a first-order reaction?

Answer: For a first-order reaction, the integrated rate law is:

$$\ln \frac{[A]_0}{[A]} = kt$$

Substitute the given values:

• $[A]_0 = 0.2 \, \text{M}$ ,
• $[A] = 0.05 \, \text{M}$ ,
• $k = 1.0 \times 10^{-3} \, \text{s}^{-1}$ .

The equation becomes:

$$\ln \frac{0.2}{0.05} = (1.0 \times 10^{-3}) \cdot t$$ $$\ln 4 = 0.4 = (1.0 \times 10^{-3}) \cdot t$$ $$t = \frac{0.4}{1.0 \times 10^{-3}} = 400 \, \text{seconds}$$

Thus, the time required is 400 seconds.

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18. Explain the Arrhenius equation and how activation energy affects the rate of reaction.

Answer: The Arrhenius equation describes the relationship between the rate constant $k$ and temperature $T$ :

$$k = A e^{-\frac{E_a}{RT}}$$

Where:

• $A$ is the pre-exponential factor,
• $E_a$ is the activation energy,
• $R$ is the universal gas constant,
• $T$ is the temperature in Kelvin.

The activation energy $E_a$ is the minimum energy required for a reaction to occur. As the activation energy increases, the rate constant $k$ decreases, meaning the reaction rate slows down. Conversely, lower activation energy results in a higher rate constant, increasing the reaction rate. The Arrhenius equation shows that a small increase in temperature can significantly increase the reaction rate, as $e^{-\frac{E_a}{RT}}$ increases.

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19. What is the significance of the collision frequency and steric factor in the context of chemical reactions?

Answer: The collision frequency refers to the number of collisions that occur per unit time per unit volume of reactants. It depends on the concentration and temperature of the reactants. For a reaction to occur, reactant molecules must collide with sufficient energy and in a favorable orientation.

The steric factor (also called the orientation factor) takes into account the geometry of the molecules involved in the collision. Even if two molecules collide with sufficient energy, if they are not aligned in the correct orientation, no reaction will occur. The steric factor quantifies this effect, and its value typically lies between 0 and 1. For reactions that require specific molecular orientations for effective collisions, the steric factor can significantly affect the reaction rate.

20. Explain the concept of a catalyst. How does a catalyst affect the rate of reaction and the activation energy?

Answer: A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. It works by providing an alternative reaction pathway that has a lower activation energy compared to the uncatalyzed reaction. This results in more molecules having sufficient energy to react, thus increasing the rate of the reaction.

There are two types of catalysts:

1. Homogeneous Catalyst: The catalyst is in the same phase as the reactants (e.g., liquid-phase reactions).
2. Heterogeneous Catalyst: The catalyst is in a different phase from the reactants (e.g., solid catalysts in gas or liquid reactions).

The presence of a catalyst lowers the activation energy ($E_a$ ) of the reaction, which means fewer molecules need high energy to collide effectively. The catalyst does not alter the final products or the overall enthalpy change ($\Delta H$ ) of the reaction. However, the reaction occurs more quickly due to the lower activation energy.

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21. A reaction follows first-order kinetics. If the initial concentration of reactant A is 0.2 M, and after 10 minutes the concentration is reduced to 0.1 M, calculate the half-life of the reaction.

Answer: For a first-order reaction, the integrated rate law is:

$$\ln \frac{[A]_0}{[A]} = kt$$

Where:

• $[A]_0 = 0.2 \, \text{M}$ (initial concentration),
• $[A] = 0.1 \, \text{M}$ (concentration after 10 minutes),
• $t = 10 \, \text{minutes}$ ,
• $k$ is the rate constant.

Substituting the known values into the rate law:

$$\ln \frac{0.2}{0.1} = k \cdot 10$$ $$\ln 2 = k \cdot 10$$ $$0.693 = k \cdot 10$$ $$k = \frac{0.693}{10} = 0.0693 \, \text{min}^{-1}$$

Now, using the formula for the half-life of a first-order reaction:

$$t_{1/2} = \frac{0.693}{k}$$

Substituting the value of $k$ :

$$t_{1/2} = \frac{0.693}{0.0693} = 10 \, \text{minutes}$$

Thus, the half-life of the reaction is 10 minutes.

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22. What is the effect of temperature on the rate constant $k$ of a reaction? Explain the significance of the Arrhenius equation in this context.

Answer: Temperature plays a significant role in determining the rate constant $k$ of a reaction. According to the Arrhenius equation:

$$k = A e^{-\frac{E_a}{RT}}$$

Where:

• $k$ is the rate constant,
• $A$ is the pre-exponential factor (frequency factor),
• $E_a$ is the activation energy,
• $R$ is the universal gas constant,
• $T$ is the absolute temperature in Kelvin.

As the temperature increases, the rate constant $k$ increases exponentially because the factor $e^{-\frac{E_a}{RT}}$ becomes larger. This is because at higher temperatures, more molecules possess the necessary energy to overcome the activation energy barrier, leading to more effective collisions and faster reactions.

The Arrhenius equation quantifies this relationship, allowing for the calculation of how the rate constant changes with temperature. The equation implies that even a small increase in temperature can significantly increase the reaction rate, which is particularly evident in many chemical processes.

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23. Discuss the different types of rate laws and how they can be determined experimentally. Provide an example for each type of reaction order (zero, first, and second order).

Answer: The rate law of a reaction expresses the rate of reaction as a function of the concentrations of reactants raised to certain powers (the order of the reaction). The order of reaction can be determined experimentally by measuring the rate at various concentrations of the reactants.

1. Zero-order Reaction: For a zero-order reaction, the rate is independent of the concentration of the reactant:

   $$\text{Rate} = k$$

   The integrated rate law is:

   $$[A] = [A]_0 - kt$$

   To determine the rate constant $k$ , you can plot $[A]$ vs $t$ . The slope of the line gives $-k$ .

   Example: A decomposition reaction like $2N_2O_5 \rightarrow 4NO_2 + O_2$ at a high temperature.

2. First-order Reaction: For a first-order reaction, the rate is directly proportional to the concentration of the reactant:

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

   The integrated rate law is:

   $$\ln [A] = \ln [A]_0 - kt$$

   To determine the rate constant $k$ , plot $\ln [A]$ vs $t$ . The slope of the line gives $-k$ .

   Example: Radioactive decay of isotopes, where the rate depends on the amount of the substance remaining.

3. Second-order Reaction: For a second-order reaction, the rate is proportional to the square of the concentration of the reactant:

   $$\text{Rate} = k[A]^2$$

   The integrated rate law is:

   $$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$$

   To determine the rate constant $k$ , plot $\frac{1}{[A]}$ vs $t$ . The slope of the line gives $k$ .

   Example: The reaction between hydrogen and iodine to form hydrogen iodide, $H_2 + I_2 \rightarrow 2HI$ .

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24. A reaction is first order with respect to $A$ and second order with respect to $B$ . If the initial concentrations of $A$ and $B$ are both doubled, by what factor will the rate of reaction change?

Answer: The rate law for this reaction is:

$$\text{Rate} = k[A]^1[B]^2$$

Let the initial concentrations of $A$ and $B$ be $[A]_0$ and $[B]_0$ . The initial rate of the reaction is:

$$\text{Rate}_0 = k[A]_0[B]_0^2$$

Now, when the concentrations of both $A$ and $B$ are doubled, the new concentrations become $2[A]_0$ and $2[B]_0$ . The new rate of the reaction is:

$$\text{Rate}_{\text{new}} = k(2[A]_0)(2[B]_0)^2$$

Simplifying:

$$\text{Rate}_{\text{new}} = k(2[A]_0)(4[B]_0^2) = 8k[A]_0[B]_0^2$$

Thus, the rate increases by a factor of 8 when the concentrations of both $A$ and $B$ are doubled.

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25. What is the concept of reaction order and how does it differ from molecularity? Discuss with examples.

Answer: Reaction order is the sum of the powers of the concentrations of the reactants in the rate law expression. It indicates how the rate of reaction is affected by the concentration of reactants.

• Zero-order reaction: The rate is independent of the concentration of the reactant. For example, the decomposition of ammonia on a hot platinum surface.

• First-order reaction: The rate is directly proportional to the concentration of one reactant. For example, radioactive decay follows first-order kinetics.

• Second-order reaction: The rate is proportional to the square of the concentration of one reactant or the product of the concentrations of two reactants. For example, the reaction between hydrogen and iodine.

On the other hand, molecularity refers to the number of reactant molecules involved in an elementary step of the reaction. Molecularity is always a whole number and refers to a specific step, while order is determined experimentally and refers to the overall reaction.

For example, in the reaction:

$$2A \rightarrow B + C$$

• The molecularity is 2 (since two molecules of $A$ are involved in the elementary step).
• The order could be zero, first, or second depending on the rate law derived experimentally.