Chemical Kinetics and Equilibrium: Mechanisms, Rate Laws, and Catalysis

Chemical Kinetics

Rate Laws and Reaction Progress

Chemical kinetics studies the speed of chemical reactions and the factors that affect them. The rate law expresses how the rate depends on the concentration of reactants.

• Rate Law: For a reaction A → B, the rate can be written as $\text{rate}_f = k_f [A]$ for the forward reaction and $\text{rate}_r = k_r [B]$ for the reverse reaction.

• Equilibrium Constant: $K = \frac{k_f}{k_r}$, where $k_f$ and $k_r$ are the rate constants for the forward and reverse reactions, respectively.

• Interpretation: If $k_f > k_r$, then $K > 1$ and products are favored at equilibrium. If $k_f < k_r$, then $K < 1$ and reactants are favored.

• Activation Energy ($E_a$): The minimum energy required for a reaction to occur. Lower $E_a$ means a faster reaction.

• Arrhenius Equation: $k = Ae^{-E_a/RT}$, where $A$ is the frequency factor, $E_a$ is activation energy, $R$ is the gas constant, and $T$ is temperature.

Example: Increasing temperature increases the rate constant $k$ and decreases $E_a$'s effect, leading to a faster reaction.

Effect of Temperature on Dissolution and Crystallization

When a saturated solution of NaCl is heated from 20 °C to 40 °C:

• The rate of dissolution increases with temperature.

• At higher temperature, the rate of dissolution is faster than the rate of crystallization.

Example: Heating a saturated NaCl solution causes more salt to dissolve than to crystallize.

Reaction Mechanisms and Rate Laws

Elementary Steps and Intermediates

Complex reactions often occur in multiple steps, each called an elementary reaction. Intermediates are species formed in one step and consumed in another.

• Elementary Reaction: A single molecular event.

• Intermediate: A species that appears in the mechanism but not in the overall reaction.

• Transition State: High-energy state between reactants and products.

Example Mechanism: 2 NO(g) + Br2(g) → 2 NOBr(g)

• Step 1: NO(g) + Br2(g) → NOBr2(g) (slow)

• Step 2: NOBr2(g) + NO(g) → 2 NOBr(g) (fast)

Rate Law if Step 1 is Slow:

• $\text{rate} = k_{f,1} [NO][Br_2]$

Rate Law if Step 2 is Slow:

• $\text{rate} = k_{f,2} [NOBr_2][NO]$

• If NOBr2 is an intermediate, substitute its steady-state concentration using equilibrium from step 1:

• $[NOBr_2] = K_1 [NO][Br_2]$

• So, $\text{rate} = k_{f,2} K_1 [NO][Br_2][NO] = k_{obs} [NO]^2 [Br_2]$

Graphical Representation: Reaction progress diagrams show energy changes, intermediates, and transition states.

Energy Profiles and Catalysis

Energy diagrams illustrate the energy changes during a reaction, including activation energy and the effect of catalysts.

• With Catalyst: The activation energy is lowered, making the reaction faster.

• Multiple Steps: Each step has its own transition state and intermediate.

Example: Adding a catalyst lowers the energy barrier, as shown by a lower peak in the energy diagram.

Chemical Equilibrium

Definition and Characteristics

Chemical equilibrium occurs when the forward and reverse reactions proceed at equal rates, resulting in constant concentrations of reactants and products.

• Dynamic Equilibrium: Both reactions continue, but net concentrations remain unchanged.

• Equilibrium Constant ($K$): Quantifies the ratio of product to reactant concentrations at equilibrium.

Equilibrium Constant Expressions

For a general reaction $aA + bB \rightleftharpoons cC + dD$:

• $K = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

• For gases, partial pressures can be used: $K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$

• $K_p = K_c (RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas.

Example: For $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $K = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Effect of Initial Conditions

Equilibrium is achieved regardless of whether the system starts with reactants or products; the same proportions are reached at equilibrium.

Practice Problems and Mechanism Analysis

Mechanism Identification

Given a reaction and its experimentally determined rate law, proposed mechanisms can be evaluated for consistency.

• Example: For $2H_2(g) + 2NO(g) \rightarrow N_2(g) + 2H_2O(g)$, if the rate law is $\text{rate} = k[H_2][NO]^2$, only mechanisms producing this rate law are valid.

Questions on Reaction Coordinate Diagrams

• Number of Intermediates: Count the valleys between peaks.

• Number of Elementary Reactions: Count the steps in the mechanism.

• Number of Transition States: Count the peaks in the energy diagram.

• Rate Determining Step: The slowest step (highest energy barrier).

• Fastest Step: The step with the lowest energy barrier.

• Exothermic vs. Endothermic: Compare energy of products and reactants.

Catalysis

Enzymatic Catalysts

Enzymes are biological catalysts that speed up reactions by lowering activation energy. The substrate fits into the enzyme's active site, forming an enzyme-substrate complex, much like a key fits into a lock.

• Enzyme: Biological catalyst.

• Substrate: Reactant molecule acted upon by the enzyme.

• Active Site: Region of the enzyme where the substrate binds.

Example: The breakdown of hydrogen peroxide by catalase.

Summary Table: Key Terms and Concepts

Term	Definition	Example/Application
Rate Law	Mathematical expression for reaction rate	$\text{rate} = k[NO]^2[Br_2]$
Equilibrium Constant ($K$)	Ratio of product to reactant concentrations at equilibrium	$K = \frac{[NH_3]^2}{[N_2][H_2]^3}$
Activation Energy ($E_a$)	Minimum energy required for reaction	Lower $E_a$ means faster reaction
Intermediate	Species formed and consumed in mechanism	NOBr2 in NO + Br2 mechanism
Transition State	High-energy state during reaction	Peak in energy diagram
Catalyst	Substance that increases reaction rate	Enzyme, platinum in hydrogenation

Additional info: Some diagrams and tables were inferred from context and standard chemistry knowledge to provide a complete study guide.