Chemical Equilibria01: 9/09

If a system at equilibrium is disturbed (stressed), the system shifts its position to counteract the disturbance, thereby re-establishing a new equilibrium state.
Why it works: The system seeks to minimize the effect of the stress and return to a stable state where the net rates of forward and reverse reactions are equal. The magnitude of Q relative to K dictates the direction of this shift.

General reaction: aA + bB \rightleftharpoons cC + dD
Reaction Quotient: Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}
Q can be measured at any point during the reaction.
At equilibrium, Q = K. Both Q and K are ratios of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients.

The terms look similar, but they are not the same in general: Q \neq K except at equilibrium where Q = K.
K is a constant for a given temperature, reflecting the ratio of products to reactants at equilibrium; Q depends on the actual concentrations at any moment.
Why they work: Comparing Q to K tells us the direction the reaction must proceed to reach equilibrium:
- If Q < K: The ratio of products to reactants is too small, so the reaction proceeds to the right (shifts toward products) to increase product concentrations and decrease reactant concentrations until Q = K.
- If Q > K: The ratio of products to reactants is too large, so the reaction proceeds to the left (shifts toward reactants) to decrease product concentrations and increase reactant concentrations until Q = K.
- If Q = K: The system is already at equilibrium, and there is no net change.

When a system at equilibrium is stressed, it shifts to relieve the stress. This shift occurs because the stress causes Q to temporarily no longer equal K. The system adjusts its concentrations until Q once again equals K.
Stresses include:
- Addition of a reaction component
- Removal of a reaction component
- Change in pressure (for gaseous systems)
- Change in temperature

Example: aA + bB \rightleftharpoons cC + dD
Add more of A; the instantaneous concentration of A increases. This makes the denominator of Q larger, so Q becomes smaller than K (assuming it was at equilibrium). To re-establish Q=K, the equilibrium shifts to form more products (toward the right) to consume the added A and increase product concentrations.

If a substance is removed, its concentration decreases instantaneously. This makes the numerator of Q smaller, so Q becomes smaller than K. To compensate, equilibrium shifts toward producing more products (to replace the removed product) until Q=K.
General rule: Removal of a reactant shifts the equilibrium toward the reactants; removal of a product shifts toward the products to replace it. In both cases, the shift aims to increase the concentration of the removed component.

Affects only gaseous systems because solids and liquids are largely incompressible.
Increasing pressure (e.g., by decreasing volume) shifts the equilibrium toward the side with fewer moles of gas.
Why it works: The system tries to reduce the total number of gas particles to counteract the increased pressure, thereby reducing the frequency of collisions with the container walls and lowering the pressure.
Decreasing pressure shifts the equilibrium toward the side with more moles of gas to increase the number of gas particles and increase the pressure.

Depends on whether the reaction is exothermic or endothermic. Heat can be treated as a reactant or product.
Exothermic reactions: Heat is considered a product (A \rightleftharpoons B + \text{heat}). Increasing temperature is like adding a product;
- Why it works: The system shifts left (toward reactants) to consume the excess heat, making Q < K and then re-establishing equilibrium.
Endothermic reactions: Heat is considered a reactant (A + \text{heat} \rightleftharpoons B). Increasing temperature is like adding a reactant;
- Why it works: The system shifts right (toward products) to consume the added heat, making Q < K and then re-establishing equilibrium.

The value of K changes with temperature because the relative rates of the forward and reverse reactions are affected differently by temperature changes, leading to a new equilibrium composition.
Exothermic reactions: Adding energy (increasing temperature) shifts the equilibrium left; the equilibrium constant decreases because the products-to-reactants ratio at equilibrium becomes smaller.
Endothermic reactions: Adding energy (increasing temperature) shifts the equilibrium right; the equilibrium constant increases because the products-to-reactants ratio at equilibrium becomes larger.

Catalysts speed up both forward and reverse reactions equally by lowering the activation energy for both directions.
Why they don't affect equilibrium: Since both rates are increased proportionally, the point at which the forward and reverse rates become equal (equilibrium) is reached faster but does not change its position or the value of K.

For a saturated solution containing excess MX(s):
K_{sp} = [M^+][X^-] (solubility product constant)
K_{sp} is a constant at a given temperature and can be calculated from solubility. It represents the maximum product of ion concentrations in a saturated solution