Equilibrio quimico

1. Chem II Review
KARELYS ROMAN

2. Equilibrium
A chemical equilibrium occurs when the rates of the forward and reverse reactions are the same
and the concentrations for both reactants and products remain constant. The equilibrium
constant is a mathematical expression of the law of mass action for both the forward and
reverse reaction, this equilibrium constant (K) will not change for the equilibrium unless the
temperature change.
𝑎𝐴 + 𝑏𝐵 ⇄ 𝑐𝐶 + 𝑑𝐷 𝐾 =
[𝐶]𝑐[𝐷]𝑑
[𝐴]𝑎 𝐵 𝑏
* Important: pure solids and liquids will not be included in the expression of K.
Example, Ni(s) + 4CO (g) ⇄ Ni(CO)4 (g)

4. Kc and Kp
The equilibrium constant can be expressed in using concentrations in molarity (M) and that is Kc,
or it can be expressed using pressure (atm) and that is Kp, however, both expressions are related
to each other since they are expressions for the same equilibrium
𝐾𝑃 = 𝐾𝑐(𝑅𝑇)∆𝑛 where Δn = moles gas products – moles of gas reactants
For the equilibrium, PCl5 g ⇄ PCl3 g + Cl2(g)

7. Operations with equilibriums
If A + B ⇄ C + D K
Multiply by a factor
2A + 2B ⇄ 2C + 2D K’ = K2
Reverse the equilibrium
For C + D ⇄ A + B K’ = 1/K
Addition of multiple equilibriums
𝐾𝑐 = 𝐾𝑐
′𝐾𝑐
′′

9. Predict the direction
Determine a reaction coefficient (Q) which will have the same form of K but taking into
consideration only the initial concentrations, then comparing the values of Qc and Kc we can
determine in which direction the reaction will proceed.
Qc < Kc more reactants then products, forward direction will be favored
Qc = Kc the concentrations have the same ratio than equilibrium. System is at equilibrium.
Qc > Kc more products than reactants, reverse direction will be favored

11. Solve equilibrium problems
To solve the equilibrium, first we need to determine the initial concentrations of the substances
present, then stablish a ICE table and use the value of K to determine concentrations at
equilibrium
A + 2B ⇄ 2C + 2D K = 120
I 0.5 1.5
C
E

13. Le Chatelier’s Principle
When a external stress is applied to a system at equilibrium, the system will adjust in such way that
the stress is partially offset as the system reach a new equilibrium position.
N2 g + 3 H2 g ⇄ 2 NH3 g ∆H < 0
Changes in concentrations (additions or removal of reactant or product)
Changes in volume
An increase in volume will lower the pressure, the system will favor the reaction that increase the
number of moles of gas.
Changes in temperature
An increase in temperature will favor the endothermic reaction
*Addition of inert gas
*Presence of a catalyst

17. Acid – Base equilibrium
Acid
For Arrhenius, anything that generates H+ ions in solution
For Bronsted-Lowry, a substance that donate a H+ ion
For Lewis, a substance that can receive a pair of electrons.
Bases
For Arrhenius, a substance that generates OH- ions in solution (like hydroxides, NaOH)
For Bronsted-Lowry, a substance that can receive a H+ ion
For Lewis, a substance that can donate a pair of electrons.

18. Bronsted-Lowry
𝐶𝐻3𝐶𝑂𝑂𝐻 𝑎𝑞 + 𝐻2𝑂 𝑙 ⇄ 𝐶𝐻3𝐶𝑂𝑂− 𝑎𝑞 + 𝐻3𝑂+(𝑎𝑞)
Acid Base conjugate base conjugate acid
𝑁𝐻3 𝑎𝑞 + 𝐻2𝑂 𝑙 ⇄ 𝑁𝐻4
+
𝑎𝑞 + 𝑂𝐻−(𝑎𝑞)
Base acid conjugate acid conjugate base
Water can act as an acid or base, it is amphoteric.

20. Autoionization of water
Water can autoionize to form H3O+ and OH-
2H2O l ⇄ H3O+ aq + OH−(aq) Kw = 1.4x10-14
Relationship between Ka and Kb
CH3COOH aq + H2O l ⇄ CH3COO− aq + H3O+(aq) Ka =
CH3COO− [H3O+]
[CH3COOH]
CH3COO− aq + H2O l ⇄ CH3COOH aq + OH− aq Kb =
CH3COOH [OH−]
[CH3COO−]
Ka*Kb = Kw
The stronger the acid, the weaker the conjugate base.

22. pH
The pH is a measure of the acidity of a solution
pH = − log H+
In the same way, pOH = − log OH− , in consequence, pH + pOH = 14

27. Percent Ionization
The percent ionization is a measure of how much of the substance have ionize
%𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 =
𝑥
[ ]0
× 100

28. Buffer
A buffer solution contain both species of the conjugate pair, it could be: acid/conjugate base or
base/conjugate acid.
Example: a solution that contain 0.5 M CH3COOH and 0.4 M CH3COONa
Another example: solution that contains 0.3 M NH3 and 0.3 M NH4Cl
We can use the Henderson-Hasselbach equation
pH = pKa + log
[A−]
[HA]
The perfect buffer is when both concentrations are the same ([HA] = [A-]), at this point the
buffering capacity is highest; also at this point pH = pKa

32. Solubility equilibria
Saturated solution
PbI2(s) ⇄ Pb2+(aq) + 2I−(aq)
Solubility product constant (Kps)
Kps = Pb2+ I− 2

33. Kps and Molar solubility
Molar solubility: moles of salt soluble in 1 liter of saturated solution.
AgCl s ⇄ Ag+ + Cl−
S S
Ksp = Ag+ Cl− = S S = S2
PbI2(s) ⇄ Pb2+(aq) + 2I−(aq)
S 2S
Kps = Pb2+
I− 2
= 𝑆 2𝑆 2
= 4𝑆3