State where the rates of the forward and reverse reactions are equal, causing the concentrations of reactants and products to remain constant.
The concentrations or pressures are not always equal at equilibrium, but they will remain constant at equilibrium. Remember that the reaction never stops. Source: telgrus.co.uk
Reversible reactions
Reactions that can go both forward and backward at the same time.
reactants ⇌ products
Left-to-right arrow describes the forward reaction Reactants are being consumed to form products
Right-to-left arrow describes the reverse reaction Products are being consumed to form reactants.
Molar Solubility
Number of moles of a substance that can be dissolved per liter of solution before the solution becomes saturated.
7.2 Direction of Reversible Reactions
Favored Direction
Happens when the reaction rate in one direction is greater than the other.
Forward Reaction Favored
When the rate of the forward reaction is greater than the rate of the reverse reaction, this means more reactants are being converted to products than products to reactants.
Reverse Reaction Favored
When the rate of the reverse reaction is greater than the rate of the forward reaction, this means more products are being converted to reactants than reactants to products.
7.3 Reaction Quotient and Equilibrium Constant
The Reaction Quotient (Q)
Q is the reaction quotient that gives the relative amount of reactants and products at any time in the reaction (different from K).
The value gives the direction the reaction needs to shift to reach equilibrium.
For reaction aA + bB ⇌ cC + dD
The subscript c stands for concentration (molarity). The [ ] must be used instead of parenthesis.
The equilibrium constant can be written in terms of partial pressures for reactions that occur in the gas phase.
The subscript p stands for pressure. The ( ) must be used instead of brackets.
Equilibrium Constant (K)
K is the equilibrium constant that gives the relative amount of reactants and products when the equation is at equilibrium.
The value identifies whether more products or more reactants are in the reaction mixture once equilibrium is reached.
For reaction aA + bB ⇌ cC + dD
The equilibrium constant can be written in terms of partial pressures for reactions that occur in the gas phase.
IMPORTANT Solids and liquids do not have concentrations and partial pressures, meaning they cannot be included in Q or K calculations. Instead, they are replaced with the number one. Always check the states of matter of each reaction.
Example: 2SO2 (g) + O2 (g) ⇌ 2SO3 (l)
Shifting Direction to Reach Equilibrium
Q > K
The concentration of products is greater than the concentration of reactants relative to K. The reaction will shift in the reverse direction and consume products to form reactants.
Q = K
The reaction is already at equilibrium.
Q < K
The concentration of reactants is greater than the concentration of products relative to K. The reaction will shift in the forward direction and consume reactants to form products.
7.4 Calculating the Equilibrium Constant
K (or Keq) is a constant only affected by temperature, which is usually provided but not needed actually to solve for the value of K.
7.5 Magnitude of the Equilibrium Constant
K = [products]/[reactants]
K > 1
The reaction would be product-favored, with more products than reactants at equilibrium.
If K is extremely large (108 or more), the forward reaction essentially proceeds to completion, with the amount of reactants negligible.
K = 1
Equal amounts of products and reactants are at equilibrium.
K < 1
The reaction would be reactant-favored, with more reactants than products at equilibrium.
If K is extremely small (10-8 or more), the forward reaction essentially doesn’t proceed at all, with the amount of reactants being the vast majority.
7.6 Properties of the Equilibrium Constant
Hess’s Law
Reversed reaction = inversed K
Equation • n = Kn
Added reactions = multiply individual K
(Also applies to ∆H, Kp, Kc, Qp, Qc)
7.7 Calculating Equilibrium Concentrations
ICE Tables
As problems become more mathematically complex, a method known as the ICE table becomes very helpful.
ICE Stands for Initial, Change, and Equilibrium table, each in Molarity (convert moles to molarity).
5% rule
If the actual equilibrium concentrations differ from the approximated values by no more than 5%, then the approximation is valid.
You do not have to approximate, but it could make solving easier since the College Board wants to test your chemistry knowledge, not math.
7.8 Representations of Equilibrium
Particulate Models
Visual representation of equilibrium that represents matter as discrete particles.
Example: R ⇌ G, where R is red, and G is green
Since the fourth and fifth models have the same number of green (4) and red (6) particles, equilibrium is reached in the fourth model.
Throughout, it shifts in the forward direction from reactants to products. This means the equilibrium constant is increasing (more products, less reactants).
7.9 Introduction to Le Chatelier’s Principle
Le Chatlier’s Principle
Le Chatlier’s Principle states that when stress is applied to an equilibrium system, the system will shift in a direction to reduce that stress and re-establish equilibrium.
Concentration Change
The system will shift away from the added substance or toward the removed substance.
In the image below, H2 is added to the equation H2 + I2 ⇌ 2HI, forming more HI and less I2. Without the values given, you cannot determine the change to H2.
Diluting Reaction
When a solution is diluted, the concentration of all the reactants and products decreases, shifting to the side with more aqueous species.
Pressure/Volume Change
If the pressure increases (or the volume decreases), the reaction will shift to the side with fewer moles of gas and vice versa.
The “fewer moles of gas” refers to the side where the total stoichiometric coefficients of gaseous species are smaller in the balanced equation. Remember that only gaseous species are considered.
Suppose the moles of gas in the reactants and products are equal. In that case, changing the pressure or volume will not affect the equilibrium constant.
Inert gases, a gas that is not in the equilibrium constant expression, will have no impact on equilibrium.
Temperature Change
The only stress that can affect the equilibrium constant. Heat is thought of as a reactant/product, following the same rules as concentration.
Tip: Write “+ heat” to the equations yourself to help visualize how temperature affects equilibrium shifts.
If ∆H˚ < 0 (exothermic)
A + B ⇌ C + heat
Increased temperature = shifts left
Decreased temperature = shifts right
If ∆H˚ > 0 (endothermic)
Heat + A + B ⇌ C
Increased heat = shift right
Decreased temperature = shift left
7.10 Reaction Quotient and Le Chatelier’s Principle
A disturbance to a system at equilibrium causes Q to differ from K. The reaction will shift to bring Q back to equal K
7.11 Introduction to Solubility Equilibria
Solubility-Product Constant
Ksp, a constant at a given temperature that allows for the comparison of the solubility of solids.
If Ksp > 1, the ionic compound is soluble (dissolves easily)
If Ksp >> 1, it could spontaneously dissolve
If Ksp < 1, the ionic compound is not very soluble
In ICE tables, the solubility (s) is the value of x. I find it helpful to use the variable “s” instead.
Molar solubility
The number of moles of a solid that dissolve to form a liter of the saturated solution is expressed as moles/liter or M.
Saturated Solution
A solution that has dissolved the maximum amount of solute.
7.12 Common Ion Effect
Common Ion Effect
Describes how a common ion can suppress/reduce the solubility of an ionic compound with a common ion is added to a solution containing a salt of that ion.
Example: Calculate the solubility of solid Mg3(PO4)2 (Ksp = 1.3 • 10-32) in a solution of 0.20M Na3PO4.
Identify the reaction: Mg3(PO4)2 (s) ⇌ 3Mg2+ (aq) + 2PO43- (aq)
Identify the Solubility-Constant Constant: Ksp = [Mg2+]3[PO43-]2
Solve using an ICE Table: Since the solid is now dissolved in a solution with PO4, the reaction would start with an initial concentration of 0.2 because there is one mole of PO4
Mg3(PO4)2
Mg2+
PO43-
I
—
0
0.20
C
—
+3s
+2s
E
—
3s
2s+0.20
Ksp = [3s]3[2s+0.20]2 = 1.3 • 10-32 s = 1.1 • 10-11 M (two sig figs) Answer: 1.1 • 10-11 M
