AP Chemistry unit 5

Core Definitions in Chemical Kinetics

Spontaneous Reactions: Reactions that occur naturally, although kinetics cannot predict exactly when or how quickly they will happen.
Kinetics: The scientific study of the rates of chemical reactions.
Determining Stoichiometric Coefficients from Graphs: By analyzing a graph of concentration versus time, coefficients of a chemical equation can be determined using relative rates of disappearance and appearance. * Example: If the rate of disappearance of $A$ is equal to the rate of appearance of $B$ and $\frac{1}{2}$ the rate of appearance of $C$ .
Relative Reaction Speeds: To determine which reactant is consumed most quickly, one must examine the balanced equation and utilize Molar Ratios.

Collision Theory and Factors Affecting Reaction Rates

Collision Theory Principle: The rate of a reaction is influenced by any factor that changes the number or force of molecular collisions. An increase in collisions results in an increased reaction rate.
1. Reactant Concentration: * Increasing the concentration increases the frequency of collisions. * This leads to increased reaction rates for all orders except for zero-order reactions.
2. Temperature: * Increasing the temperature increases the frequency of collisions. * It also increases the average Kinetic Energy (KE) of the reactants. * Result: More particles meet the minimum activation energy ( $E_a$ ) required to react. * General Rule: Increased temperature equals an increased number of successful collisions.
3. Pressure: * Specifically for reactions involving gases, increasing the pressure increases the collision frequency between reactants.
4. Particle Size and Surface Area: * Smaller particle sizes possess more sides, which equals a larger surface area. * Increasing surface area (via methods like stirring or grinding) leads to more collisions.
5. Catalysts: * A catalyst changes the value of $k$ (the rate constant) by changing the activation energy ( $E_a$ ). * The value of the rate constant is explicitly dependent on the activation energy.
Inhibitor: A substance that decreases the reaction rate by increasing the activation energy ( $E_a$ ).

Rate Laws and the Rate Constant

Initial Rate Method: Refers to the "instantaneous rate" measured just after the reaction begins. This is usually the fastest part of the reaction.
Differential Rate Law Equation: $\text{Rate} = k[A]^m [B]^n$ * $k$ = rate constant. * $m$ and $n$ = rate orders.
The Rate Constant ( $k$ ): * Relates the reaction rate to the concentration of the reactants. * Represents the slope of the concentration/time relationship. * The value of $k$ is positive when representing the slope of products and negative for the slope of reactants. * The value of this constant is dependent only on temperature. * The units of $k$ reflect the overall reaction order.
General Rules for Writing Rate Laws: 1. The concentrations of products never appear in the rate law. 2. The values of rate orders ( $m$ and $n$ ) must be determined experimentally; they cannot be taken from the coefficients of a balanced overall equation (except in the case of elementary reactions).
Types of Rate Laws: * Differential Rate Law (The Rate Law): Expresses how the rate depends on the concentration of the reactants. * Integrated Rate Law: Expresses how the concentrations depend on time. This is used when asked to determine the amount of reactant or product present at a specific time.

Determining Reaction Orders and Rates

Finding Rate Law from Data Tables: 1. Locate two experiments where the concentration of one reactant changes while the other remains constant. 2. Use the ratio: $\frac{\text{Rate Formula 1}}{\text{Rate Formula 2}}$ 3. Solve for the individual rate orders. 4. Plug the determined orders back into the rate formula to solve for $k$ .
Overall Reaction Order: The sum of all individual rate orders. * Example: If $\text{Rate} = k[A]^1 [B]^2 [C]^3$ , the overall reaction order is $1 + 2 + 3 = 6$ .
Catalyst Surface Sites and Order: A reaction starting on a catalyst begins as first-order because the rate depends on the molecules adsorbed. Once all surface sites are fully occupied, the rate is no longer affected by concentration, and the reaction becomes zero-order.

Kinetics of Different Reaction Orders

Zero-Order Reactions

Definition: A change in reactant concentration has no effect on the rate.
Rate Law: $\text{Rate} = k$
Reactant Presence: The reactant does not appear in the rate law equation.
Integrated Rate Law: $[A]_t - [A]_0 = -kt$
Graphical Representation: A plot of $[reactant]$ versus time is a straight line.
Slope and k: $k = -(\text{slope})$ .

First-Order Reactions

Definition: Changing the reactant concentration has an identical effect on the reaction rate (proportional by the same factor).
Rate Law: $\text{Rate} = k[A]$
Integrated Rate Law: $\ln[A]_t - \ln[A]_0 = -kt$
Graphical Representation: A plot of $\ln[A]$ versus time creates a linear graph.
Graph Features: Slope is $-k$ and the y-intercept is $\ln[A]_0$ .
Half-Life ( $t_{1/2}$ ): The time required for a reactant to reach half of its original concentration. * The half-life of a 1st-order reaction depends only on $k$ (it is independent of concentration). * Formula: $t_{1/2} = \frac{0.693}{k}$ * A constant time is required to reduce concentration by half repeatedly. * All types of radioactive decay are first-order.

Second-Order Reactions

Definition: Changing reactant concentration affects the reaction rate to the square of the change.
Rate Law: $\text{Rate} = k[A]^2$
Integrated Rate Law: $\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt$
Graphical Representation: A plot of $\frac{1}{[A]}$ versus time is a straight line.
Slope and k: $k = \text{slope}$ .
Half-Life ( $t_{1/2}$ ): The half-life increases with time as the reaction proceeds. * Formula: $t_{1/2} = \frac{1}{k[A]_0}$ * As concentration decreases, it takes significantly more time for the concentration to halve again.

The Collision Model and Activation Energy

The Collision Model: The energy for breaking reactant bonds comes from the kinetic energies possessed by molecules before collision.
Requirements for a Successful Reaction: 1. Reactant molecules must collide. 2. They must collide with enough energy to overcome the activation energy barrier. 3. They must make contact at the correct molecular orientation to allow bonds to rearrange.
Orientation specifics: Orientation is critical for molecules but not for atoms.
Reaction Fraction: Only a small fraction of collisions produce a reaction because both energy ( $E \ge E_a$ ) and orientation must be correct. This is why the observed rate is much lower than the total number of collisions with minimum energy.
Energy Calculations: * Activation Energy ( $E_a$ ): $E_a = \text{Energy of Activated Complex} - \text{Energy of Reactants}$ (calculated as Peaks - line). * Transition State / Activated Complex: An unstable state where bonds are partially formed; located at the peak of the energy curve. * Activation Energy of Reverse Reaction: Equal to $\text{Forward Activation Energy} - \Delta H$ . * Enthalpy Change ($\Delta H$ or $\Delta E$): The total energy change or energy released. Calculated as $H_{\text{products}} - H_{\text{reactants}}$ ; visualized as the distance between products and reactants on the graph.

Reaction Mechanisms

Rate-Determining Step: The rate of a reaction can be no faster than its slowest step.
Elementary Reactions: * Unlike overall equations, the coefficients of an elementary reaction can be used to determine the rate orders.
Graphical Features of Mechanisms: * The slowest step (highest peak) has the highest activation energy. * The number of peaks on the graph represents the number of elementary steps. * Intermediates: Found at the minimums (valleys) of the graph. * Transition States: Found at the maximums (peaks) of the graph.

Determining Rate Laws for Multi-Step Mechanisms

Standard Procedure when the slow step is not the first step: 1. Write the rate law for the slow step. 2. For the preceding fast equilibrium step, set the rate of the forward reaction ( $k_1$ ) equal to the rate of the reverse reaction ( $k_{-1}$ ). 3. Solve that equilibrium expression for the concentration of the intermediate. 4. Substitute the intermediate concentration expression into the slow step's rate law.
Mechanism Example: * Step 1: $2A \rightleftharpoons X$ (fast); $\text{Rate} = k[A]^2$ * Step 2: $X + B \rightarrow Y$ (slow); $\text{Rate} = k[X][B]$ * Step 3: $Y + B \rightarrow D$ (fast); $\text{Rate} = k[Y][B]$ * Derivation: The expected rate law from the slow step is $\text{Rate} = k[X][B]$ . However, $X$ is an intermediate. Since Step 1 is in equilibrium, $[X]$ is equivalent to $[A]^2$ . Substituting results in: $\text{Rate} = k[A]^2[B]$ .

Catalysis and Energy Barriers

Catalyst Properties: * Increases reaction speed without being consumed. * Can be reused repeatedly. * Does not affect the free energy ($\Delta G$) or enthalpy ($\Delta H$) of a reaction. * Catalyzed reactions typically involve at least two steps. * Function: Lowers $E_a$ so more molecules have enough energy to react.
Types of Catalysts: * Homogeneous: Present in the same phase as the reactants. * Heterogeneous: Exists in a different phase (usually a solid, like a catalytic converter).
Steps of Heterogeneous Catalysis: 1. Adsorption (collection of a substance on a surface) and activation of reactants. 2. Migration of adsorbed reactants across the surface. 3. Reaction of the adsorbed substances. 4. Escape (desorption) of the products.
Mechanisms of Speeding up Reactions: * Increasing the rate constant ( $k$ ). * Lowering the $E_a$ barrier. * Forming a more stable activated complex. * Increasing collision frequency. * Improving orientation effects.
Activation Energy Barrier Details: * Total energy needed for reactants to reach the transition state and break bonds. * Often provided by environmental heat, but using high heat is disadvantageous in biological systems because it speeds up all reactions indiscriminately, denatures proteins, and kills cells. * The transition state is a highly reactive and unstable condition of the substrate after absorbing enough energy to start the reaction.