Chem 30 – Unit 1-4 Formula Review – Mr. Chin’s class
Unit 1 - Thermochemistry:
Thermal Energy Changes
∆t = tf - ti
Q = mct
Thermal Energy Mass
Specific heat capacity (usually 4.19 J/(goC) for water and solutions)
E rH
Enthalpy Changes
H = n rH
(kJ)
(mol) (kJ/mol)
Enthalpy
Molar Enthalpy
Calorimetry
∆Egain = - ∆Elost
Example- Exothermic:
∆Egain by surroundings = - ∆Elost by system
As the reaction proceeds, it is giving off energy, and the environment is gaining E
(getting hotter)
Example- Endothermic:
∆Egain by system = - ∆Elost by surroundings
As the reaction proceeds, it is absorbing energy, and the environment is losing E (getting
cooler).
Flame Calorimetry
Egain = - Elost
Tip: There is a list of
common specific heat
capacities of metals
and other substances
in your data booklet
m c t + m c t = - n cH
water
metal
fuel
For example, to calculate cH,
they must give you the amount
of water, mass of metal,
specific heat capacity of the
metal, temperature change
and the amount of fuel used.
(Only 1 unknown variable)
Solution Calorimetry
-
Generally, we just use Q = mct
Remember, if you dissolve 5 g of something into 100 mL of water, we assume both the volume is
105 mL (and thus the mass is 105 g).
We use water’s specific heat capacity for solutions
Bomb Calorimetry
Q = C t
Tip: Remember, bomb calorimetry
uses HEAT CAPACITY (in kJ/oC)
Egain = - Elost
C t = - n cH
calorimeter
fuel
Hess’s Law 1 – Adding Multiple Steps
-Rearrange the equations so that they will cancel out to give the balanced target equation
-Then, add the enthalpies for each step together to get the overall enthalpy
Hess’s Law 2 – Using Enthalpies of Formation
rHo = fHo products - fHo reactants
rHo = (n fHo products) - (n fHo reactants)
Tip: Standard Enthalpies of Formation are listed in your data booklet.
-You will have to usually have to solve for one of these:
1) The rHo by using all of the fHo for all products and reactants
2) Solve for a fHo that is not listed in your book. They must give you rHo
Energy Efficiency
Useful Energy



energy delivered to consumer in usable form
actual work done
calculate by Q = mcT or H = n rH
Efficiency 
useful energy
x 100%
total available energy
Total Available Energy


ideal energy content of fuel
calculate using Hess’s Law and H = n rH
Unit 2 - Electrochemistry:
Oxidizing and Reducing Agents (I know, not really a formula)
Example:
Copper metal with an aqueous silver solution
Cu(s)  Cu+2(aq) + 2e2[Ag+(aq) + e-  Ag(s)]
2 electrons lost
2 electrons gained
Tip: electrons on right side =
oxidized, so Cu(s) = SRA
(thus SOA = Ag+)
Cu(s) +2Ag+(aq)  Cu+2(aq) + 2 Ag(s)
Oxidation Numbers
{RA} Oxidation (Lost electrons)
Example:
+6
-2
+1
-2 +1 -2 +1
+3
+1 -2
0 +1 0-2-2+1
Cr2O72-(aq) + H+(aq) C2H5OH(aq)  Cr3+ (aq) + H2O(l) + CH3COOH(aq)
{OA} Reduction (Gain electrons)
Tip: To balance this, you would multiply the OA and RA by numbers that would give the
lowest common multiple (so total electrons gained and lost will balance). These become
the coefficients. Ex. Cr gained 6e- and C lost 4e-, so Cr x 2 and C x 3 would give 2 Cr2O72-(aq)
and 3 CH3COOH(aq). Balance the rest by inspection.
Redox Stoichiometry


A balanced net ionic equation can be used to identify the correct ratio between the various
reactants and products.
This can be used along with the equations:
n
m
M
ncV
Cell Notation
anode | anode electrolyte | | cathode electrolyte | cathode
Cell Potential


standard reduction potential (Eo): the ability of a half-cell to gain electrons
In a voltaic cell, the half-cell that contains the SOA will always have a greater reduction
potential than the other half-cell. As a result, the SOA always gains electrons.

standard cell potential (Eocell): the difference in reduction potentials between the cathode
and the anode
Eocell = Eo cathode - Eo anode
Faraday’s Law

current: the continuous flow of electrons
 the amount of charge (Q) that passes any given point in a period of time (t)
I
Amperes (C/s)

Q
t
Coulombs (C)
Seconds (s)
Faraday’s Law: The amount of a substance produced or consumed in an electrochemical
reaction is directly proportional to the amount of charge that flows through the circuit.
n e 

F represents the Faraday constant, F

I is in amperes (A) and t is in seconds (s)!!!
It
F
= 9.65 x 104 C/mol
Tip: you will usually
need to write the
balanced half rxn, so
that you can determine
the moles of ELECTRONS
transferred
Unit 3 – Organic Chemistry
1)
Combustion Reactions
Balanced Chemical Equation:
Ex. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
CH3
CH2
CH3
+
O

O
O
C
O
+
H
O
H
2) Addition Reactions
C
C
+
X
X
X
X
C
C
3) Elimination Reactions
Y
Z
C
C
C
C
H
X
C
C
+
Y
Z
4) Substitution Reactions
H
H
UV
C
C
+ X
X
+
H
X
5) Esterification Reactions (Alcohol + Carboxylic acid = Ester! Yay!)
O
O
H2SO4
R
C
OH
carboxylic acid
+
HO
R’
alcohol
R
C
ester
O
R’
+
HO
water
H
Unit 4 - Equilibrium:
Equilibrium Law
aA + bB
cC + dD
the equilibrium law is:
𝐾𝑐 =
Equilibrium constant
[𝐶 ]𝑐 [𝐷]𝑑
[𝐴]𝑎 [𝐵]𝑏
*Never include pure solids or liquids in the equilibrium law expression!
*Kc smaller than 1 means reactants are favored (<50% yield)
* Kc larger than 1 means products are favored (>50% yield)
*The only Le Chatelier change that will change Kc is temperature
Water Equilibrium (Kw)
KW = [H3O+(aq)] [OH-(aq)] = 1.00 x 10-14 at 298.15 K
In other words, if we know the concentration of [H3O+(aq)] or [OH-(aq)], we can figure out the
other.
Always 1.00x10-14
Ex.
[𝐻3 𝑂
+
(𝑎𝑞) ]
=
𝐾𝑊
[𝑂𝐻 − (𝑎𝑞) ]
pH and pOH
pH = - log [H3O+(aq)]
pOH = - log[OH-(aq)]
[H3O+(aq)] = 10-pH
[OH-(aq)] = 10-pOH
pH + pOH = 14
Acid Ionization Constant (Ka)

The Ka value can be used to predict the [H3O+(aq)] and pH of a solution.
Hint: For strong acids (that ionize completely), the [H3O+(aq)] will be the same as the
concentration of the acid
-For weak acids, we can’t predict the [H3O+(aq)] so easily, so we use:
HA(aq)
+
H3O+(aq)
H2O(l)
𝐾𝑎 =
+
[𝐻3 𝑂+ (𝑎𝑞) ][𝐴− (𝑎𝑞) ]
[𝐻𝐴(𝑎𝑞) ]
Tip: You may need to set up an ICE
table to solve for x2
A-(aq)
Same concentration [x][x] or
[x]2 (get these x’s from ICE
table)
This will usually be [HA – x]
(from ICE table). X can
often be dropped using the
approximation method
Tip: Ka is often listed in data book if
solving for a concentration
Base Ionization Constant (Kb)
*Same guidelines as acids
Tip: Weak bases we use the following (If solving a concentration, Kb can be calculated based off
the Ka value listed in data booklet):
[𝐻𝐵(𝑎𝑞)][𝑂𝐻 − (𝑎𝑞)]
𝐾𝑏 =
[𝐵− (𝑎𝑞)]
Ka x Kb = Kw
Always 1.00x10-14