MCAT Study Guide Chemistry Ch. 10 – Equilibrium 2017-08-15T06:45:06+00:00

I.          10.1:  EQUILIBRIUM

The balance of forward and reverse reactions


1.     aA + bB ⇋ cC + dD

2.     Keq = (

[C]c[D]d)/([A]a[B]b)  → known as mass-action ratio

a)     Each reaction has its own Keq for a given temperature

b)     Keq is constant for a given reaction at a given temperature

c)     Keq < 1 → reaction favors reactants

d)     Keq > 1 → reaction favors products

e)     K is often given a subscript (Ka for acids, Kb for base, Ksp for solubility product)

3.     EX:  2NO ⇋ N2 + O2

a)     Keq = ([N2][O2])/[NO]2  → don’t forget to use the quantity in the exponent!

B.     How Keq is obtained

1.     Products are in numerator, reactants are in denominator → brackets indicate concentration

2.     The coefficient of each species becomes an exponent on its concentration

3.     Solids (s) and pure liquids (l) are not included, because their concentrations don’t change; dilute solvents are also omitted

4.     Aqueous dissolved particles are included

5.     If the reaction is gaseous, we can use the partial pressure of each gas as its concentration (Kp); note that the value of the Kp will be different with partial pressure because units are different


A.     Reaction quotient (Q)

This is the same formula as Ke, but it is when the reaction is not at equilibrium

1.     Q = ([C]c[D]d)/([A]a[B]b)

2.     Comparing Q with Ke will tell whether the reaction will proceed forwards or backwards

3.     Q < Ke → reaction will proceed forward (to increase products and decrease reactants)

4.     Q > Ke → reaction will proceed backwards (to decrease products and increase reactants)



A system at ⇋ will try to neutralize any imposed change (or stress) in order to reestablish ⇋

A.     EX:  N2(g) + 3H2(g) ⇋ 2 NH3(g) + heat

1.     ADDING AMMONIA → by adding more product, Q would increase, so the reverse reaction would be favored, to decrease Q back towards Keq

2.     REMOVING AMMONIA → removing product, Q would decrease, so the forward reaction would be favored, increasing Q back towards Keq

3.     ADDING HYDROGEN → adding reactant will decrease Q, causing the forward reaction to be favored; the reaction “shifts to the right” to reach ⇋ again

4.     REMOVING NITROGEN → removing reactant will increase Q, causing reverse reaction to be favored; the reaction “shifts to the left” to reach ⇋ again


a)     Increasing volume decreases partial pressures

b)     The total number of moles in reactants vs products needs to be examined

c)     Summary:

(1)   If V ↓, the side of the reaction with the least number of moles is favored

(2)   If V ↑, the side of the the reaction with the most number of moles is favored


a)     Heat is treated like a reactant or product; endothermic reactions need heat on the left side, exothermic reactions release heat on the right side

b)     General rules:

(1)   Lowering the temperature favors exothermic reactions (ΔH < 0)

(2)   Raising the temperature favors endothermic reactions (ΔH > 0)

c)     This only applies when a reaction is at ⇋! Remember, all reactions proceed faster at ↑ T

7.     ADDING AN INERT (OR NON-REACTIVE) GAS – if it does not participate in reaction, there is no effect on the reaction

8.     ADDING A CATALYST – no effect on ⇋! Just changes reaction rate to get there




1.     Solution – formed when a substance dissolves in another forming a homogenous mixture

a)     Can involve any of the three phases of matter (two gases, gas in liquid, two liquids, solid in liquid, two solids (called an alloy)

b)     Ideal solution – when the ΔH = 0; this is when the energies bonds that break and the bonds that form are the same, so no net change in enthalpy (ΔG = -TΔS)

c)     Aqueous solution – solution with water as the solvent

d)     Rule:  like dissolves like

(1)   Solutes will dissolve best in solvents where the intermolecular forces being broken in the solute are being replaced by equal (or stronger) intermolecular forces between the solvent and the solute

2.     Dissolution – the process of dissolving

3.     Solute – the smaller proportion of the substance in a solution

4.     Solvent – the larger proportion of the substance in a solution

5.     Solvation – the process that occurs when solvent molecules surround a solute molecule

6.     Hydration – solvation when the solvent is water

a)     Hydration number → the number of water molecules that bond to an ion in aqueous solution


Free ions in solution, called electrolytes because they can conduct electricity

1.     Strong electrolytes – solutes that dissociate completely in solution (like ionic substances)

a)     Note that dissociation is not equivalent to solubility!!

2.     Weak electrolytes – solutes that dissociate, but remain ion-paired to some extent

3.     Nonelectrolytes – covalent compounds that do not dissociate into ions (like sugar; it will dissolve, but not dissociate)

4.     van’t Hoff (ionizability) factor (i) – tells us how many ions one unit of a substance will produce in solution

a)     C6H12O6 → nonionic, does not dissociate; i = 1

b)     NaCl → i = 2

c)     HNO3 → i = 2

d)     CaCl→ i = 3

5.     Dynamic equilibrium – when the molar solubility is reaches, and the dissolving of products occurs at the same rate as precipitation


Refers to the amount of a solute that will saturate a particular solvent; is specific for the type of solute and solvent

1.     Phase Solubility Rules

a)     The solubility of solids in liquids tends to ↑ with ↑ T

b)     The solubility of gases in liquids tends to ↓ with ↑ T

c)     The solubility of gases in liquids tends to ↑ with ↑ P

2.     Salt Solubility Rules (in water)

a)     All Group I (Li+, Na+, K+, Rb+, Cs+) and ammonium (NH4+) salts are soluble

b)     All nitrate (NO3), perchlorate (ClO4), and acetate (C2H3O2) salts are soluble

c)     All silver (Ag+), lead (Pb2+/Pb 4+), and mercury (Hg22+/Hg 2+) salts are insoluble, except for their nitrates, perchlorates, and acetates


1.     Solubility product constant (Ksp) – the extent at which a salt will dissolve in water (is just another equilibrium constant)  → solution is saturated!

a)     The larger the Ksp, the more soluble the salt is

2.     EX:  Mg(OH)2(s) ⇋ Mg2+(aq) + 2OH(aq)

a)     Ksp = [Mg2+][OH]2  →  note that the MgOH(s) is left out because is is a pure solid


How to calculate molar solubility (# of moles that will dissolve in water)

1.     First, figure out the number of moles of each ion:

a)     Mg(OH)2(s) ⇋ Mg2+(aq) + 2OH(aq)

b)     x ⇋ x + 2x

2.     Next, replace the ions with “x” in solubility constant:

a)     Ksp = [Mg2+][OH]2

b)     Ksp = [2x]2 = 4x3

3.     If Ksp for Mg(OH)2 at 25 degrees is 1.6*10-11, solve for x:

a)     1.6*10-11 = 4x3 → x ≈ 1.4*10-4


1.     Homogenous mixtures – particles are > 500 nm, will sink to the bottom

a)     EX:  paint, choc milk

2.     Colloid – particles are 2-500 nm, will stay suspended

a)     EX:  jello, fog, smoke

3.     Solution – particles < 2 nm (most important in chemistry)


V.          10.5:  ION PRODUCT

A.     Ion product (Qsp)

The reaction quotient for a solubility reaction (does not have to be at ⇋)

1.     Qsp < Ksp  → more salt can be dissolved

2.    Qsp = Ksp  → solution is saturated

3.     Qsp > Ksp  → excess salt will precipitate


VI.          10.6:  THE COMMON-ION EFFECT

A.     Consider a saturated solution of magnesium hydroxide:

1.     Mg(OH)2(s) ⇋ Mg2+(aq) + 2OH(aq)

B.     Add another solute with OH:

1.     NaOH(s) ⇋ Na+(aq) + OH(aq)

C.    NaOH is very soluble, so there is therefore an excess of OH ions

D.    This disturbs the Keq of Mg(OH)2 and shifts it to the left, causing it to precipitate out

E.     This is an example of the common ion effect



A.     Complex ions

Consist of metal ions surrounded by 2, 4, or 6 ligands (aka Lewis bases)

1.     These ligands substantially ↑ the solubility of usually insoluble metal salts

2.     EX:

a)    AgCl(s) ⇋ Ag+(aq) + Cl(aq) → Ksp = 1.6 x 10-10

b)    Ag+(aq) + 2NH3(aq) ⇋ [Ag(NH3)2]+(aq) → Ksp = 1.5 x 107

c)    AgCl(s) + 2NH3(aq) ⇋ [Ag(NH3)2]+(aq) + Cl(aq) → Ksp ≈ 10-3  (overall)



A.     A reaction will proceed until ΔG = 0 and the products and reactants reach ⇋; this implies there must be a relationship between ΔG and the reaction quotient (and Keq)

1.     ΔG = ΔG° + RTlnQ

2.     0 = ΔG° + RTlnKeq  → ΔG° = –RTlnKeq

a)     Remember, ΔG is a statement about spontaneity of a reaction (or tendency to go one way or another), and ΔG° is the change in energy at standard conditions

b)     ΔG will change with the changing reaction composition until it reaches 0, but ΔG° is a constant

(1)  ΔG° < 0; Keq > 1→ products are favored at ⇋

(2)   ΔG° = 0; Keq = 1 → products and reactants are about equal at ⇋

(3)   ΔG° > 1; Keq < 1 → reactants are favored at ⇋

IX.          10.9:  Naming ionic compounds

A.     Roman numerals – used for atoms that have more than 1 oxidation state

1.     Fe2+ Iron (II) → aka ferrous (-ous is lesser charge)

2.     Fe3+ Iron (III) → aka ferric (-ic is greater charge)

B.     -ide – added to the name of a monoatomic ion of an element

1.     NaH – sodium hydride

C.    Oxyanions – suffixes are relative to the total number of anions formed;

1.     -ite – if 2 oxyanions are formed, this is the name given to the one with the lesser # of O’s

a)     Nitrite → NO2

b)     Sulfite → SO3

2.     -ate – if 2 oxyanions are formed this is the name given to the one with the greater # of O’s

a)     Nitrate → NO32-

b)     Sulfate → SO32-

D.    Oxyanions – prefixes are added to the suffixes if there are 4 oxyanions

1.     hypo- – used with “-ite”, indicates 1 less O than the “-ite”

2.     per- – used with “-ate”, indicates 1 more O than the “-ate”

a)     ClO → hypochlorite

b)     ClO2 → chlorite

c)     ClO3 → chlorate

d)     ClO4 → perchlorate

Chapter 10 Summary

  • The ⇋ constant dictates the relative ratios of products to reactants when a system is at ⇋
  • For aA + bB ⇋ cC + dD: Keq = ([C]c[D]d)/([A]a[B]b)
  • Pure solids and pure liquids are not included in the equilibrium expression
  • If K > 1, products are favored; if K < 1, reactants are favored
  • The reaction quotient (Q) is a ratio of products to reactants with the same form as K, but can be used when the reaction is not at ⇋. If Q < K, the reaction will proceed in the forward direction. If Q > K, the reaction will proceed in the reverse direction until ⇋ is achieved
  • The only factor that changes the Kqe is T
  • Changing the concentrations of the products or reactants of a reaction at ⇋ will force the system to shift according to Le Chateleir’s principle
  • ↑ T at ⇋ favors the products in an endothermic rxn and the reactants in an exothermic rxn
  • ↓ T at ⇋ favors the reactants in an endothermic rxn and the products in an exothermic rxn
  • In a gaseous rxn, ↑ P by ↓ V favors the side of the reaction with fewer moles of gas. ↓ P has the opposite effect
  • An electrolyte is a solute that produces free ions in solution. Strong electrolytes produce more ions in solution than weak ones
  • The van’t Hoff (ionizability) factor, i, tells us how many ions one unit of a substance will produce
  • All group I, ammonium, nitrate, perchlorate, and acetate salts are completely soluble. All silver lead, and mercury salts are insoluble, except when they are paired with nitrate, perchlorate, and acetate
  • The solubility of solids in liquids ↑ with ↑ T
  • The solubility of gases in liquids ↓ with ↑ T, and ↑ with ↑ P
  • The amount of salt that can be dissolved in a solute is given by its solubility product constant (Ksp)
  • For a reaction at ⇋ under ΔG°, ΔG° = RTlnK
  • For a reaction under nonstandard conditions, ΔG can be calculated by using ΔG = ΔG° + RTlnK

MCAT Study Guide Chemistry - Kim Matsumoto

More MCAT Study Guide Chemistry


Ch. 3 Chemistry Basics


Ch. 4 Atomic Structure


Ch. 5 Chemical Bonds


Ch. 6 Enthalpy + Entropy


Ch. 7 Calorimetry + Phase Diagrams


Ch. 8 Ideal Gas Law


Ch. 9 Rate Laws


Ch. 10 Equilibrium


Ch. 11 Acids and Bases

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