THE THREE LAWS OF THERMODYNAMICS -It tells whether a process or chemical reaction can occur.

-It states that the entropy of the universe increases in a

The first law of thermodynamics says that energy can be spontaneous process and remains unchanged in an
converted from one form to another, but it cannot be equilibrium process.
created or destroyed.
Expressions:
Amount of heat given off or absorbed by a system during For a spontaneous process: ΔSuniv = ΔSsys + ΔSsurr > 0
a constant-pressure process, which we define as a For an equilibrium process: ΔSuniv = ΔSsys + ΔSsurr = 0
change in enthalpy (ΔH).
● For a spontaneous process, ΔSuniv must be
SPONTANEOUS PROCESS greater than zero, but it does not place a restriction
-a reaction that does occur under the given set of on either ΔSsys or ΔSsurr.
conditions. ● Thus, it is possible for either ΔSsys or ΔSsurr to be
NON SPONTANEOUS PROCESS negative, as long as the sum of these two
-a reaction does not occur under specified conditions. quantities is greater than zero.
● For an equilibrium process, ΔSuniv is zero. In this
NOTE: It is possible for an endothermic reaction to be case, ΔSsys and ΔSsurr must be equal in
spontaneous, it is possible for an exothermic reaction to magnitude, but opposite in sign.
be nonspontaneous.
Entropy Changes in the System
POSITIVE=NON SPONTANEOUS
NEGATIVE= SPONTANEOUS To calculate ΔSuniv, suppose that the system is
represented by the following reaction
ENTROPY
- a thermodynamic quantity often described as a measure aA + bB → cC + dD
of how spread out or dispersed the energy of a system.
-the greater the dispersal, the greater is the entropy. Standard entropy of reaction ΔS°rxn is given by the
- the entropy of the system increases because closely difference in standard entropies
spaced energy levels lead to a greater dispersal among
the energy levels. ΔS°rxn = [cS°(C) + dS°(D)] - [aS°(A) + bS°(B)] or
ΔS°rxn = ΣnS°(products) - ΣmS°(reactants)
THE SECOND LAW OF THERMODYNAMICS
The standard entropy values of a large number of Entropy Changes in the Surroundings
compounds have been measured in J/K · mol. To -when an exothermic process takes place in the system ,
calculate ΔS°rxn or ΔS° (which is ΔSsys) the entropy of the surroundings increases.
- endothermic process in the system absorbs heat from
Example 1: Calculate ΔS° for the reaction: the surroundings and so decreases the entropy of the
H2(g) + I2(s) → 2HI(g) surroundings.

Solution: Balance the equation first before using the For constant-pressure processes, the heat change is
formula. equal to the enthalpy change of the system(ΔHsys).
Therefore, the change in entropy of the surroundings,
H2(g) + I2(s) → 2HI(g) (ΔSsurr) is proportional to ΔHsys:
130.6 116.7 206.3
(From the Thermodynamic Data Table, S°(J/K·mol) ΔSsurr α – ΔHsys

ΔS°rxn = ΣnS°(products) - ΣmS°(reactants) IMPORTANT NOTE: MUST READ!

= [(2)So HI] – [(1)SoH2 + (1) So I2] ● Minus sign is used.
= [(2) (206.3 J/K·mol)] – [(1) (130.6 J/K·mol) + (1) (116.7 ● Exothermic process - increase in entropy
J/K·mol)] ● Endothermic process- decrease in entropy
= +165.3 J/K·mol ● If the temperature is high, the system will have
relatively little impact on molecular motion and
General Rules for Predicting Entropy Change of the increase in entropy is small.
System ● If the temperature is low, the system will more
drastically increase in molecular motion and larger
1. If the reaction produces more gas molecules than it increase in entropy
consumes, ΔS° is positive.
2. If the total number of gas molecules diminishes, ΔS° is From the inverse relationship between ΔSsurr and
negative. temperature T (in kelvins):
3. If there is no net change in the total number of gas
molecules. ΔSsurr = −ΔHsys
T
N2(g) + 3H2(g) → 2NH3(g) ΔH°rxn = -92.6 kJ/mol
Convert first -92.6 kj/mol to J G = H – TS

From Example 2 we have ΔSsys = -199 J/K·mol, and The change in free energy (ΔG) of a system for a
ΔHsys = -92.6 kJ/mol, we obtain constant-temperature process is ΔG = ΔH – TΔS.

ΔSsurr = −(− 92.6kJ mol ) ΔG less than 0:The reaction is spontaneous in the forward
298 K direction.
ΔSsurr = −(−92600 J mol) ΔG greater than 0:The reaction is nonspontaneous. The
298 K reaction is spontaneous in the opposite direction.
= 311 J/K·mol ΔG = 0 :The system is at equilibrium. There is no net
change.
The change in entropy of the universe is ΔSuniv
Standard Free-Energy Changes
= ΔSsys + ΔSsur = -199 J/K·mol + 311 J/K·mol ΔG°rxn = ΣnΔG°f(products) - ΣmΔG°f(reactants)
= 112 J/K·mol
Example 1: The old camera flash bulb used Mg metal
THE THIRD LAW OF THERMODYNAMICS AND ABSOLUTE sealed in a bulb with oxygen. Is the reaction spontaneous
ENTROPY at STP?
-the entropy of a perfect crystalline substance is zero at The reaction is:
the absolute zero of temperature. Mg(s) + ½ O2(g) →MgO(s) ΔS° (J/K·mol):
-As the temperature increases, the freedom of motion 32.7 205.0 26.9
increases. ΔHo f (kJ/mol): 0 0 - 601.2
-the entropy of any substance at a temperature above 0 K
is greater than zero. Solution: First, calculate ΔS and ΔH.
ΔS° = ΣnS°(products) − ΣmS°(reactants)
The change in entropy, ΔS, is given by: = [(1)SoMgO] – [(1)SoMg + (½)SoO2]
ΔS = Sf - Si ΔS = Sf = [(1) (26.9)] – [(1) (32.7) + (½) (205.0)] ΔS°
= - 108.3 J/K mol

GIBB’S FREE ENERGY ΔH° = ΣnΔH°f(products) – ΣmΔH°f (reactants)

-spontaneous reaction increases the entropy of the = [(1)ΔH°fMgO] – [(1)ΔH°fMg + (½) ΔH°f O2]
universe = [(1) (- 601.2) ] – [(1) (0) + (½) (0)] ΔH°
= - 601.2 kJ/mol -The reactants and products exist in different phases in a
heterogeneous equilibrium. (more on gas)
CHEMICAL EQUILIBRIUM
-Equilibrium is a word that indicates balance or Example 4: Breakdown of calcium carbonate (CaCO3) into
steadiness. calcium oxide (CaO) and carbon dioxide (CO2).

LAW OF MASS ACTION AND EQUILIBRIUM CONSTANT CaCO3(s) ⇌ CaO(s) + CO2(g)

EXPRESSION ft. Reversible reaction
-At equilibrium, no apparent change in the state of the Answer: K = [CO2]
system is observed
-rates of the forward and reverse reactions become equal, THE EQUILIBRIUM CONSTANT
the concentrations of the reactants and products remain - The speed of a chemical reaction is determined by
constant over time chemical kinetics.

: aA + bB ⇌ cC + dD or K = [C] c [D] d If K > 1, the equilibrium lies to the right and the mixture
[A] a [B] b contains mostly products.
If K < 1, the equilibrium lies to the left and the mixture
Homogeneous Equilibria contains mostly reactants.
-homogeneous equilibrium involves reactants and If K = 1, the amounts of reactants and products will be the
products that are present in only one phase. same.

Example 1: For the reaction: N2O4(g) ⇌ NO2(g) whose Calculating K when equilibrium concentrations/partial
equilibrium constant was experimentally determined to be equal pressures are known :
to 4.6 x 10-3 at 25oC, the equilibrium constant expression is
written as: Example 1: Gaseous hydrogen iodide partially
decomposes into hydrogen and iodine in a closed
Answer: K = [NO2 ] container at 425°C. At equilibrium, the concentration of
2 [N2O4 ] hydrogen iodide [HI] is 3.53 × 10–3 M, [H2] is 4.79 × 10–4
= 4.6 x 10-3 M and [I2] is 4.79 × 10–4 M. What is the value of K at this
temperature? 2
Heterogeneous Equilibria
Answer: K = [H2][I2]
[HI] 2
K = [4.79 × 10–4 𝑀][4.79 × 10–4 𝑀] [3.53 × 10–3 𝑀] 2
K = 0.0184

Calculating equilibrium concentrations/ partial pressures

from initial concentrations/ partial pressures and K values

Example 5: For the equilibrium Br2(g) + Cl2(g) ⇌ 2BrCl(g), the

equilibrium constant K is 7.00 at 400 K. If a cylinder is inserted
with BrCl(g) at an initial pressure of 1.00 atm, what is the
equilibrium partial pressure of BrCl at
equilibrium?

Solution: Let x = amount of BrCl that changes to attain

equilibrium.

ICE Table

The Reaction Quotient, Q

-value obtained when product and reactant
concentrations or partial pressures at any point of the
reaction are plugged in the equilibrium constant
expression.

Q = [C] c [D] d
[A] a [B] b

If Q = K, the reaction is in equilibrium.

If Q > K, favors the reverse reaction.
If Q < K, favors the forward reaction.
Example 7: A container initially contains N2, H2, and NH3 gasses
with concentrations of 0.82 M, 0.54 M, and 0.31 M, respectively. To illustrate Brønsted Acid and Base, examine the equation:
The mixture was heated to 375oC and allowed to attain equilibrium, HNO3 + H2O → H3O+ + NO3 -
if K is 1.2 at this temperature, to which direction the reaction will
proceed to reach equilibrium? N2(g) + 3H2(g) ⇌ 2NH3(g) TAKE NOTE: Conjugate base always has one fewer H
atom and one more negative charge (or one fewer positive
Answer: Q = [NH3 ] 2 charge) than the formula of the corresponding acid.
[N2 ] [H2 ] 3
Q = [0.31 𝑀] 2 Question: Which reactant loses a proton? Which reactant
[0.82 𝑀] [0.54 𝑀] 3 gains a proton?
= 0.74 Since, Q < K, the reaction proceeded forward until it
reached equilibrium. Answer: Nitric acid is the Brønsted acid because it loses
one proton. Water is the Brønsted base because it gains
THE Le CHATELIER’S PRINCIPLE one proton.
-when a change is imposed on a system at equilibrium,
the position of the equilibrium shifts in a direction that Explanation: Acids and bases in the Brønsted model exist
tends to reduce the effect of that change. as conjugate pairs whose formulas are related by the gain
or loss of a hydrogen ion. NO3 - is the conjugate base
Changes in Concentration formed from the acid HNO3, and H3O+ (hydronium ion) is
-When the concentration of either a reactant or a product the conjugate acid of the base H2O
is increased, equilibrium shifts into the direction of the
added component. The Acid-Base Properties of Water
-If the concentration is decreased, then the equilibrium -act either as an acid or as a base
shifts into the direction of the lost component.
H2O + H2O(l) ↔ H3O+ (aq) + OH- (aq)
ACIDS AND BASES -the reaction is called autoionization of water
-Brønsted acid as a substance capable of donating a
proton, and a Brønsted base as a substance that can
accept a proton.
-The conjugate base of a Brønsted acid is the species that Power of Hydrogen, pH: A Measure of Acidity
remains when one proton has been removed from the pH = - log [H+ ]
acid. pOH= pH-14 [H- ]
Concentration= shift+(-log), insert pH or Answer: Sodium acetate is a strong electrolyte, so it
antilog =[OH- ] = 10 -pOH dissociates completely in solution:

[H+ ]- Hydrogen ion CH3COONa(aq) → Na+ (aq) + CH3COO- (aq)

The initial concentrations, changes, and final concentrations of

the species involved in the equilibrium are; CH3COOH(aq) ↔
H+ (aq) + CH3COO- (aq)

pH increases as [H+ ] decreases.

pH decreases as [H+ ] increases.

REMEMBER: If the equation both contains negative from

products and reactants side, the reactant base which
contains the negative sign is paired with the product that
also contains negative signs.

Henderson-Hasselbalch Equation
-acid-base properties of a solution with two dissolved
solutes that contain the same ion (cation or anion), called
the common ion.
-common ion suppresses the ionization of a weak acid or
a weak base
BUFFER SOLUTION
EXPRESSION: pH = pKa + log [conjugate base] / [acid] -solution of a weak acid or a weak base and its salt; both
components must be present
Example 1: What is the pH of a solution containing both
0.20 M CH3COOH and 0.30 M CH3COONa? The Ka of
CH3COOH is 1.8 x 10-5 .
 2. The buffer resists changes in pH by reacting with any
added H+ or OH- so that these ions do not accumulate.
3. Any added H+ reacts with the base A- . H+ (aq) + A - (aq) →
HA(aq)
4. Any added OHreacts with the weak acid HA. OH- (aq) +
HA(aq) → H2O(l) A- (aq)

RedOx Reaction
-Electrochemistry is the branch of chemistry that deals
with the interconversion of electrical energy and chemical
Example 2: What buffer system should you use if the energy.
desired pH is 6? -movement of electrons or electrons from one particle to
Answer: Carbonic acid–hydrogen carbonate ion pair another.
(H2CO3–HCO3 - ) for desired pH is 6. -There is a change in oxidation number

Example 3: How would you prepare an ammonium- NonRedOx

ammonia buffer solution at pH 9? -No change in oxidation number

Answer: Use Henderson-Hasselbalch equation. OXIDATION

pH = pKa + log [NH3] / [NH4 + ] -losing of electron
rearrange the equation log: REDUCTION
[NH3] / [NH4 + ] = pH - pKa -gaining of electron
[NH3] / [NH4 + ] = antilog (9-9.25)
[NH3] / [NH4 + ] = antilog (-0.25) Question 1: What happens when sodium metal (Na) and
[NH3] / [NH4 + ] = 10-0.25 chlorine gas (Cl2) react to form table salt, NaCl. Which
= 0.56 atom was oxidized? Which atom was reduced?

Answer: Na atom lost electron; it was oxidized. Cl atom gained

electron; it was reduced. The equation is 2Na + Cl2 → 2NaCl.
Characteristics of a Buffer
1. The solution contains a weak acid HA and its conjugate
base A- .
 Simple way of determining if a reaction is redox or not is For example, in the ammonium ion, NH4 + , the oxidation
by assigning oxidation numbers to the elements involved number of N is -3 and that of H is +1. Thus, the sum of the
in the reaction. oxidation numbers is -3 + 4(+1) = +1, which is equal to the
net charge of the ammonium ion.
Rules for Assigning Oxidation Numbers (ON) 7. Oxidation numbers do not have to be integers. For
1. In free elements (that is, in the uncombined state), each example, the oxidation number of O in the superoxide ion,
atom has an oxidation number of zero. O2
For example, each atom in H2, Br2, Na, Be, K, O2, and P4
has the same oxidation number: zero. Question 4: Is the neutralization reaction between HCl and
2. For ions composed of only one atom (that is, NaOH a redox reaction?
monatomic ions), the oxidation number is equal to the
charge on the ion. Answer: No. To show that this reaction is non redox reaction, we
For example, Li+ ion has an oxidation number of +1; Ba2+ assign oxidation number to all elements present in the chemical
ion has +2; Fe3+ ion has +3; and so on. reaction. +1 -1 +1 -2 +1 +1 -2 +1 -1 HCl + NaOH → H2O + NaCl
3. The oxidation number of HYDROGEN is +1, except No change in oxidation number (ON) can be seen from reactant
when it is bonded to metals in binary compounds. side to product side for all the elements involved. Therefore, it is
For example, in LiH, NaH, or CaH, its oxidation number is a non-redox reaction.
-1.
4. The oxidation number of OXYGEN in most compounds Mnemonic Device in Identifying the Key Reactants in a
(for example, MgO and HO) is -2, but in hydrogen peroxide Redox Reaction
(H2O2) and peroxide ion (O2 -2 ), it is -1. LEORAION for Oxidation Reaction
5. FLUORINE has an oxidation number of -1 in all its
compounds. Other halogens (Cl, Br, and I) have negative L Loss of E Electrons
oxidation numbers when they occur as halide ions in their O OXIDATION
compounds. When combined with oxygen—for example R Reducing
in oxoacids and oxoanions — they have positive A Agent
oxidation numbers. I Increase in
6. In a neutral molecule, the sum of the oxidation numbers O Oxidation
of all the atoms must be zero. In a polyatomic ion, the sum N Number
of oxidation numbers of all the elements in the ion must
be equal to the net charge of the ion. GEROADON for Reduction Reaction
G Gain of
E Electrons
R REDUCTION
O Oxidizing
A Agent
D Decrease in
O Oxidation N Number

Balancing Redox Equations by the Change in Oxidation

Number Method

Question 8: The equation below is obviously not balanced. Is the

equation can be balanced? If yes, how? MnO2 + Cl- → Mn2+ +
Cl2
Answer: Yes. The reaction above is an example of Redox
reaction in its net ionic form.
Step 1: SPLIT the equation. MnO2 → Mn2+ Cl- → Cl2
Step 2: BALANCE the two half-reactions. In balancing the half-
reactions, we use H2O, H+ , e - . 2e- + 4H+ + MnO2 → Mn2+ +
2H2O 2Cl- → Cl2 + 2e
Step 3: COMBINE the reactants of two half-reactions as well as
the products. 2e- + 4H+ + MnO2 + 2Cl- → Mn2+ + 2H2O + Cl2
+ 2e Check Redox file for example:
Step 4: SIMPLIFY the equation by removing appropriate
numbers of substances that appear on both sides. 2e- + 4H+ +
MnO2 + 2Cl- → Mn2+ + 2H2O + Cl2 + 2e
4H+ + MnO2 + 2Cl- → Mn2+ + 2H2O + Cl2

Take note that this type of balancing is for reactions in

acidic condition/medium where H+ ions appear
somewhere in the equation.