High School Chemistry Review: Kinetics and Equilibrium Study Guide

The rate of a chemical reaction is influenced by various factors including the physical state (surface area) and the concentration of the reactants.
Collision Theory states that for a reaction to occur, reactant particles must collide with sufficient energy (activation energy) and proper orientation.
Example Analysis of Reaction Rates:
- Comparison of systems involving Zinc ( $Zn$ ) and Hydrochloric Acid ( $HCl$ ):
  - Option a: $0.5\,g$ zinc strip and $1.0\,M\,HCl$
  - Option b: $0.5\,g$ zinc strip and $3.0\,M\,HCl$
  - Option c: $0.5\,g$ zinc powder and $1.0\,M\,HCl$
  - Option d: $0.5\,g$ zinc powder and $3.0\,M\,HCl$
- Analysis: Option (d) has the highest reaction rate. Zinc powder provides a much greater surface area than a zinc strip, increasing the frequency of collisions. A higher concentration of $HCl$ ( $3.0\,M$ vs $1.0\,M$ ) increases the number of solute particles in a given volume, further increasing the frequency of effective collisions.

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process.
Mechanism: It functions by providing an alternative reaction pathway with a lower activation energy ( $E_a$ ).
By lowering the energy barrier required for a successful collision, a larger fraction of reactant particles possesses the necessary energy to react at a given temperature.

Enthalpy Change ( $\Delta H$ ): The expression representing the change in enthalpy for a chemical reaction in terms of the potential energy ( $PE$ ) of its components is:
- $\Delta H = PE_{\text{products}} - PE_{\text{reactants}}$
Haber Process Example:
- Reaction: $N_2 (g) + 3H_2 (g) \rightarrow 2NH_3 (g) + 91.8\,kJ$
- Conditions: $101.3\,kPa$ and $298\,K$ .
- Analysis: Since heat energy ( $91.8\,kJ$ ) is written on the product side, the reaction is exothermic. For exothermic reactions, energy is released to the surroundings, and $\Delta H$ is negative. Thus, $\Delta H = -91.8\,kJ$ .
Potential Energy (PE) Diagrams:
- Activation Energy ( $E_a$ ): On a PE diagram, this is represented by the energy difference between the peak of the curve (the activated complex) and the energy of the reactants.
- Heat of Reaction ( $\Delta H$ ): This is represented by the interval between the potential energy of the products and the potential energy of the reactants.
- Endothermic vs. Exothermic Identification:
  - If the products are at a higher energy level than the reactants, the reaction is endothermic (\Delta H > 0).
  - If the products are at a lower energy level than the reactants, the reaction is exothermic (\Delta H < 0).

Definition of Dynamic Equilibrium: A state where the forward and reverse processes of a reversible reaction occur at the same rate. At this point, the concentrations of reactants and products remain constant, though the molecules continue to react.
Phase Change Equilibrium:
- Example: $C_2H_5OH (l) \rightleftharpoons C_2H_5OH (g)$
- In this system, the forward process (evaporation) and the reverse process (condensation) happen at the same rate.

For a general gaseous reaction $aA + bB \rightleftharpoons cC + dD$ , the equilibrium constant expression is:
- $K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}$
Calculation Example 1:
- Reaction: $A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$
- Concentrations: $[A_2] = 3.45\,M$ , $[B_2] = 5.67\,M$ , $[AB] = 0.67\,M$
- Expression: $K_{eq} = \frac{[AB]^2}{[A_2][B_2]}$
- Numerical Value: $K_{eq} = \frac{(0.67)^2}{(3.45)(5.67)}$
Calculation Example 2 (Stoichiometry and Equilibrium):
- Reaction: $2AB_3(g) \rightleftharpoons A_2(g) + 3B_2(g)$
- Initial Conditions: $0.87\,moles$ of $AB_3$ in a $5.0\,L$ container at $25^{\circ}C$ .
- Initial Concentration of $AB_3$ : $[AB_3]_i = \frac{0.87\,mol}{5.0\,L} = 0.174\,M$
- At equilibrium, it is found that $[A_2] = 0.070\,M$ .
- Using a RICE table (Reaction, Initial, Change, Equilibrium):
  - Change in $[A_2]$ is $+0.070\,M$ .
  - By stoichiometry, change in $[B_2]$ is $3 \times 0.070\,M = +0.210\,M$ .
  - By stoichiometry, change in $[AB_3]$ is $-2 \times 0.070\,M = -0.140\,M$ .
- Final Equilibrium Concentrations:
  - $[AB_3] = 0.174\,M - 0.140\,M = 0.034\,M$
  - $[A_2] = 0.070\,M$
  - $[B_2] = 0.210\,M$

Chemical reactions often occur through a series of elementary steps known as a mechanism.
Example Mechanism:
- Step 1: $NO + NO \rightarrow N_2O_2$ (fast)
- Step 2: $N_2O_2 + H_2 \rightarrow N_2O + H_2O$ (slow)
- Step 3: $N_2O + H_2 \rightarrow N_2 + H_2O$ (fast)
Overall Balanced Equation: Summing the steps and canceling species that appear on both sides:
- $2NO + 2H_2 \rightarrow N_2 + 2H_2O$
Intermediates vs. Catalysts:
- Intermediates: Species produced in one step and consumed in a subsequent step. In this mechanism, $N_2O_2$ and $N_2O$ are intermediates.
- Catalysts: Species present at the start and regenerated at the end (not present in this specific mechanism).
Rate-Determining Step: The slowest step in a reaction mechanism (Step 2) determines the overall rate of the reaction. However, because Step 2 depends on $N_2O_2$ , and $N_2O_2$ is produced by $NO$ in Step 1, adding more $NO$ will increase the concentration of $N_2O_2$ , thereby increasing the overall reaction rate.

Le Chatelier's Principle states that if stress is applied to a system at equilibrium, the system will shift in a direction that tends to counteract the stress.
Haber Reaction shifts ( $N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g)$ ):
- To favor the formation of ammonia ( $NH_3$ ):
  - Increase in pressure: Shifting the equilibrium to the side with fewer moles of gas (from 4 moles on the left to 2 moles on the right).
  - Adding Reactants: Adding $N_2$ or $H_2$ .
  - Removing Product: Removing $NH_3$ .
Effect of Adding Concentration:
- System: $POCl_3 (g) + \text{energy} \rightleftharpoons 2PCl_3 (g) + O_2 (g)$
- If $O_2 (g)$ is added:
  - The system shifts to the left to consume the excess oxygen.
  - Consequently, the concentration of $PCl_3$ decreases as it reacts with the added $O_2$ .