Representations of Equilibrium - AP Chemistry

Higher $K_{sp}$ indicates greater solubility. Larger $K_{sp}$ values correspond to more soluble compounds at equilibrium.

Equilibrium shifts right, producing more products. Increasing reactant concentration makes the forward reaction more favorable.

Equilibrium shifts right, producing more $NH_3$. Adding reactants increases their concentration, driving the reaction forward.

A catalyst does not affect the position of equilibrium. Catalysts only speed up the rate of reaching equilibrium but don't change the final position.

Shifts towards the side with fewer moles of gas. System responds to pressure increase by reducing the total number of gas molecules.

Products are favored at equilibrium. Large equilibrium constants mean the forward reaction is thermodynamically favorable.

$K_{eq} > 1$ indicates a spontaneous reaction. Large equilibrium constants indicate thermodynamically favorable forward reactions.

No effect on the equilibrium position. Solid concentrations don't appear in equilibrium expressions and don't affect position.

Neither reactants nor products are favored. Equal concentrations of reactants and products at equilibrium when $K_{eq} = 1$.

Le Chatelier's Principle. States that systems respond to stress by shifting to counteract the disturbance.

Shifts the equilibrium to the left (toward reactants). Heat is treated as a product in exothermic reactions, so adding heat shifts left.

No effect if $H_2O$ is not a reactant or product in gaseous form. Liquid water doesn't affect gas-phase equilibria unless it participates as a gas.

$K_c$ decreases. For exothermic reactions, higher temperature makes products less stable, decreasing $K_c$.

Solubility decreases due to the common ion effect. Increased concentration of a common ion shifts equilibrium toward the undissolved solid.

Shifts equilibrium to the left (toward reactants). Heat is treated as a reactant in endothermic reactions, so removing heat shifts left.

Equilibrium shifts toward the side with fewer moles of gas. System reduces stress by shifting to the side that occupies less volume.

No effect on the equilibrium position. Inert gases don't participate in the reaction and don't change partial pressures.

$K_{sp} = [Ca^{2+}][F^-]^2$. Products of ion concentrations raised to their stoichiometric coefficients for the dissolution.

The reaction favors reactants at equilibrium. Small $K_c$ values mean the equilibrium position lies far to the left.

Pressure changes do not affect $K_c$.. Only temperature changes affect $K_c$; pressure and concentration changes don't.

Equilibrium shifts right to produce more product. Removing products decreases $Q$, making the forward reaction more favorable.

M^{-2}. Units are $\frac{M^2}{M \cdot M^3} = M^{-2}$ based on the concentration terms.

Shifts towards the side with more moles of gas. Increasing volume decreases pressure, favoring the side with more gas molecules.

$Q = \frac{[B][C]}{[A]^2} = \frac{1 \times 3}{2^2} = 0.75$. Using $Q = \frac{[B][C]}{[A]^2}$ with the given concentrations.

$K_c = \frac{[HI]^2}{[H_2][I_2]}$. Products raised to stoichiometric coefficients over reactants raised to their coefficients.

No change in equilibrium position. At constant volume, adding inert gas doesn't change partial pressures of reactants.

Shifts equilibrium to the left (toward reactants). Heat is treated as a reactant in endothermic reactions, so removing heat shifts left.

$K_{sp} = [Ca^{2+}][F^-]^2$. Products of ion concentrations raised to their stoichiometric coefficients for the dissolution.

The reaction favors reactants at equilibrium. Small $K_c$ values mean the equilibrium position lies far to the left.