Chemistry Equilibrium and Equilibrium Constants Study Notes

Equilibrium constants, denoted as K, represent the ratio of concentrations of products to reactants in a chemical reaction at equilibrium.
These constants only account for gases and aqueous species, excluding solids and liquid solvents, notably water.

Solids are excluded from equilibrium expressions because they have a constant concentration that does not change during the reaction.
Water (as a liquid) is also excluded from these expressions since its concentration remains effectively constant (approximately 55 mol/L).
If the reaction involves water in a gaseous form (vapor), it is included in the equilibrium expression.
Example provided:
- For the reaction where ammonia reacts with water to form ammonium hydroxide:
- K = ( \frac{[\text{NH}_4^+] [\text{OH}^-]}{[\text{NH}_3]} )
- The concentration of water is not included because it is a solvent and considered constant.

In exothermic reactions, products are favored over reactants; thus, K > 1.
Example:
- If the ratio of concentration of products to reactants yields a number greater than one (e.g., K might be ( 12 / 2 = 6 )).
Large K values (as much as ( 10^{12} )) indicate that the reaction heavily favors the products.

In endothermic reactions, reactants are favored; thus, K < 1.
Example:
- If the ratio of concentration of products to reactants yields a number less than one (e.g., K might be ( 2 / 12 = 0.167 )).
K cannot be negative--negative K values imply impossible negative concentrations.

Catalysts speed up the reaction but do not alter the equilibrium constant K.
They accelerate both the forward and reverse reactions equally, leaving the ratio of concentrations at equilibrium unchanged.

Perturbations to a system at equilibrium will induce shifts to counterbalance the effect of the perturbation.
Types of perturbations:
1. Adding reactant: Shifts equilibrium to the right to produce more product.
2. Removing reactant: Shifts equilibrium to the left to produce more reactant.
3. Adding product: Shifts equilibrium to the left to produce more reactant.
4. Removing product: Shifts equilibrium to the right to produce more product.

Assume a 50/50 equilibrium between A and B, both at 1 mol/L initially.
Adding reactant A results in an imbalance—equilibrium will shift right to restore balance (more B will be produced).
If product B is added, equilibrium shifts left to create more reactant A.

Chemical equation for the reaction of acetic acid (( ext{CH}_3 ext{COOH} )) with water:
- ( ext{CH}_3 ext{COOH} + ext{H}_2 ext{O} \rightleftharpoons ext{H}_3 ext{O}^+ + ext{CH}_3 ext{COO}^- )
Equilibrium expression:
- K = ( \frac{[\text{H}_3 ext{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3 ext{COOH}]} )
- Note: Water is not included (it is the solvent).
Adding acetic acid will shift equilibrium to the right, increasing H3O+ concentrations.

Van Hoff's Principle: Heat can act as a reactant in endothermic reactions and as a product in exothermic reactions.
Example of cooling an endothermic reaction leads to a shift to the left (removing heat induces a response to generate warmth).

Acetic acid dissociating in water, with K = 1.8 x 10⁻⁵.
If initial concentrations are given and defined as:
- Initial concentrations: [Acetic Acid] = 1 M, [H3O+] = 0, [Acetate-] = 0.
At equilibrium, relationships can be expressed as:
- [H3O+] = x, [Acetate-] = x, [Acetic Acid] = 1 - x.
The equilibrium expression resolves to:
- K = ( \frac{x^2}{1-x} = 1.8 \times 10^{-5} )
- For simplification, assume 1 - x is approximately 1 (due to small x), leading to:
- ( x^2 = 1.8 \times 10^{-5} ), solving x gives the concentration of H3O+ formed.

Equilibrium constants are essential for understanding the dynamics of reversible reactions.
Manipulation of equilibrium through stressors has predictable outcomes based on established principles like Le Chatelier's principle.
Understanding the constants and their implications allows for deeper insight into chemical behavior under varying conditions.