CHEMISTRY SEMESTER
REVIEW
LAST CALL FOR QUESTIONS
Do you remember how to identify
the type of reaction?
• 2 H2O → 2 H2 + O2
– What type of reaction is this?
Decomposition Reaction
• C10H8 + 12 O2 → 10 CO2 + 4 H2O
– What type of reaction is this?
Combustion Reaction
GRAMS to MOLECULES
• How many molecules are there in 250g of
NH3?
• What do we do first?
• First thing to do is find the molar mass of
the compound. Molar mass is the sum
total of every atom’s mass in the
compound.
What's the molar mass of NH3?
NITROGEN(N) = 14.01g/mol
HYDROGEN(H) = 1.008g/mol
What's the molar mass of NH3?
• N = 14.01 g
• H = 1.008 g (times 3 because there are
three hydrogen atoms in NH3)
• The molar mass is 17.034 g/mole (or
17.034 grams per 1 mole of NH3)
Next Step (250g of NH3)
• Next, once you know the total mass of your
substance, you can convert to moles by
dividing the total mass of NH3 (250 g) by the
molar mass (17.034 g/mol).
250 g
1 mol (of NH3)
14.68 mol =
17.034 g
Finally, the number of molecules
can be calculated
To do this you must convert moles to
molecules using Avogadro’s #, which is
6.02 X 1023
14.68 mol
8.837 x
1024 =
molecules
6.02 x 1023
molecules
1 mol
TYPES OF SYSTEMS
• open systems can exchange mass and energy
• closed systems allow the transfer of energy
(heat) but not mass
• isolated systems do not allow transfer of either
mass or energy
HYDROGEN EXPLOSION - combustion
• Is this reaction taking place in an open, closed or
isolated system?
2 H 2 ( g )  O2  2 H 2O(l )
• the reacting mixture is the system (hydrogen, oxygen,
and water molecules)... ...everything else is the
surroundings
• This is exothermic because it is a combustion reaction
(thermal energy leaves during this and all other
exothermic reaction -- a lot of thermal energy is released
in this example...）
open system
What's an example of an isolated system?
Calorimetry - "the measurement of heat change"
•
a "calorimeter" is an
insulated, closed container
that creates an ISOLATED
SYSTEM
• a specific quantity of water
surrounds a system carrying
out a reaction
• during the reaction, heat
leaves the inner container
and is absorbed by the
surrounding water
• by recording temperature
change, the heat generated
by a reaction can be
calculated
• Heat Capacity
– heat needed to raise a certain quanity of a
substance 1 degree (celsius)
• Specific Heat
– the heat needed to raise a 1 gram of a
specific substance 1 degree (celsius)
If volume and pressure remain constant, then q = H
• Either heat capacity or specific heat must be
applied to find a substances change in heat.
• Remember...
heat capacity
specific heat
Δt
°C
x
(C)
= m
x (s) x Δt
J

c
J

g c
g
q  Ct
q  mst
°C
Practice Question
• A quantity of 1.435 g naphthalene (C10H8) is
burned in a calorimeter. The water temp. rises
from 20.28 to 25.95 degrees celsius. If the heat
capacity (C) is 10.17 kJ/celsius, calculate the
molar heat combustion of C10H8.
• What is being asked here:
• How much heat per mole of napthalene is released into the
calorimeter?
To do this:
• You need to figure out how much heat is generated by the
combustion
• You need to convert this total heat to q per mole (kJ/mol)
solving this problem...
qcal  ccal t
qcal  (10.17 kJ / C )(25.95 C  20.28 C )

qcal  57.66kJ
qrxn  qcal  qsys
qsys  0
qcal  qrxn
qrxn  57.66kJ


We are not done...
• The question asks us to calculate "the
molar heat of combustion."
• So, you have to find the molar mass of
naphthalene (C10H8) have to use periodic table...
• molar mass of C10H8 is 128.2 g
• now we can convert from kJ/1.435 g to
kJ/mol
How to convert...
128.2 g C10 H 8
- 57.66kJ
molar heat of combustion 
x
1.435 g C10 H 8 1 mol C10 H 8
 5.151x10 kJ/mol
3
INTENSIVE and EXTENSIVE PROPERTIES
• Extensive Properties - are properties that
depend on how much matter is being considered
– for example: volume
• the property of space a substance takes up is a value
dependent on how much of the substance there is.
– Can you think of others...
INTENSIVE and EXTENSIVE PROPERTIES
• Intensive Properties - properties that do NOT
depend on how much matter is being considered
– for example: boiling point
• the boiling point of liquid water into water vapor is the same
regardless of how much water there is.
– can you think of others...
MOLE vs. MOLARITY
• mole (mol)
• molarity (M)
• 1 mol = 6.02 x 10^23 • molarity or molar
particles of a
concentration is the
particular substance
number of moles of
solute per liter of
the dissolved
A solvent is
solution
substance
what a solute
dissolves in
to make a
solution
molarity is defined as...
molarity = moles of solute / liters of solution
So, is molarity an intensive or extensive property?
intensive property
Change in Enthalpy of a Reaction
H

rxn
Standard Enthalpy of Formation
• DEFINITION:
– the enthalpy of a reaction carried out at 1 atm and at 25
degrees C
– the enthalpy value relates to the heat absorbed or released
during the formation of 1 mole of a specific compound
The symbol for this value is represented as:

H f  Standared enthalpy of formation
Once we know these values, we can calculate the change in

enthalpy of a reaction, or...
H
rxn
Standard Enthalpy of a Reaction
H

rxn
The generalized equation is...
H

rxn
  n H f (products)   n H f (reactants)


m
• n = the coefficient of the products
• m = the coefficient of the reactants
• Σ = "the sum of"
• f = formation
• ° = standard state conditions (1 atm and 25 degrees C)
The most stable forms of substances will = 0
EXAMPLE
Given the equation 3 O2(g)→2 O3(g) ΔH = +284.4 kJ,
calculate ΔH for the following reaction. 3/2 O2(g) → O3(g).
REMEMBER...
The generalized equation is...
H  rxn   n H  f (products)   n H  f (reactants)
O2 is the most stable form of oxygen. So, it is equal to 0.
Therefore, the the change in enthalpy is all about the product.
(x products) -i (0 reactants) = H  rxn = X products = +284.4 kJ
You can check using
the table on P. 247
P. 247
• The ΔH°f for O3(g) is +142.2
• The ΔH°f for O2(g) is 0
Given the equation 3 O2(g)→2 O3(g) ΔH = +284.4 kJ
There are two O3 in the equation (142.2 kJ x 2 = 284.4 kJ)
– we must remember the coefficients!!!!
• Then, the question asks us to calculate ΔH for the
following reaction. 3/2 O2(g) → O3(g).
• There is only one O3.
• Since 3/2 O2(g) → O3(g) is ½ of 3 O2(g)→2 O3(g)
the enthalpy of the reaction will be ½ as well:
½ (+284.4kJ)
...or +142.2kJ
Change of Enthalpy of Reaction
The combustion of thiophene, C4H4S(l), a compound
used in the manufacture of pharmaceuticals, produces
carbon dioxide and sulfur dioxide gases and liquid water.
The enthalpy change in the combustion of one mole of
C4H4S(l) is -2523kJ. Use this information and data from
Table 6.4 (p.247) to establish ΔH°f for C4H4S(l).
STEP I:
WRITE THE BALANCED EQUATION
*To carry any combustion reaction you need oxygen as a
reactant. Liquid water will always be included as one of the
products.
1 C4H4S(l) + ___
4 CO2(g) + ___
1 S02(g) + ___
6 O2(g) → __
2
__
H20(l)
Change of Enthalpy of Reaction
The combustion of thiophene, C4H4S(l), a compound
used in the manufacture of pharmaceuticals, produces
carbon dioxide and sulfur dioxide gases and liquid water.
The enthalpy change in the combustion of one mole of
C4H4S(l) is -2523kJ. Use this information and data from
Table 6.4 (p.247) to establish ΔH°f for C4H4S(l).
STEP II:
USE THE COEFFICIENTS FROM YOUR BALANCED
EQUATION TO DETERMINE THE TOTAL CHANGE IN
IN THE FORMATION OF EVERY SUBSTANCE
iENTHALPY
USED IN THE REACTION. USE THIS INFORMATION TO
WRITE YOUR ΔH°RXN EQUATION.
H  rxn   n H  f (products)   n H  f (reactants)
Change of Enthalpy of Reaction
The combustion of thiophene, C4H4S(l), a compound used in the
manufacture of pharmaceuticals, produces carbon dioxide and
sulfur dioxide gases and liquid water. The enthalpy change in the
combustion of one mole of C4H4S(l) is -2523kJ. Use this
information and data from Table 6.4 (p.247) to establish ΔH°f for
C4H4S(l).
STEP II:
PRODUCTS
REACTANTS
-393.5
4 (________kJ/mol
CO2(g))
i
2 (________kJ/mol
H2O(l))
-285.8
1 (________kJ/mol
SO2(g))
-296.8
X kJ/mol C4H4S(l)
0.00
6 (________kJ/mol
O2(g))
Change of Enthalpy of Reaction
The combustion of thiophene, C4H4S(l), a compound used in the
manufacture of pharmaceuticals, produces carbon dioxide and
sulfur dioxide gases and liquid water. The enthalpy change in the
combustion of one mole of C4H4S(l) is -2523kJ. Use this
information and data from Table 6.4 (p.247) to establish ΔH°f for
C4H4S(l).
STEP II:
[4(-393.5) + 1(-296.8) + 2(-285.8)] – [1(ΔH) + 6(0)] = -2523
PRODUCTS － REACTANTS = ΔHRXN
i
Change of Enthalpy of Reaction
The combustion of thiophene, C4H4S(l), a compound used in the
manufacture of pharmaceuticals, produces carbon dioxide and
sulfur dioxide gases and liquid water. The enthalpy change in the
combustion of one mole of C4H4S(l) is -2523kJ. Use this
information and data from Table 6.4 (p.247) to establish ΔH°f for
C4H4S(l).
STEP III:
SOLVE for "X"
-2442.4 - ΔH(reactants) = -2523
i
ΔHC4H4S(l) = 81.3kJ/mol
CHANGE IN H
ENDOTHERMIC : H  0
EXOTHERMIC : H  0
Change in Enthalpy in a multi-step reaction
Change in Enthalpy in a multi-step reaction
CHEMICAL EQUILIBRIA
Things to remember
This is how you convert an Equilibrium
constant from Kc to Kp:
Kp = Kc (0.0821 T)^Δn
T is the temperature during the
reaction in Kelvin. Remember that
Kelvin has exactly the same degree
of difference between integers as
celsius, but it does have a different
starting point. Kelvin's zero is
273° less than celsius. For
example, 2° C is equal to 275° K.
27° C is equal to 300° K.
Δn is the sum of all stoichiometric
coefficients belonging to products
minus the sum of all stoichiometric
coefficients belonging to reactants
Qc and Kc Question
h
EQUILIBRIUM of 2 or more RXN's
If a reaction can be expressed as the sum of
two or more reactions, the equilibrium constant
for the overall reaction is given by the product
of the equilibrium of the individual reactions.
Kc = (K'c) (K"c)
Let me elaborate...
1st Step Reaction
n
2nd Step Reaction
n
n
n
[E] [F ]
[C ] [ D]
K
"

K 'c 
c
n
n
n
n
[C ] [ D]
[ A] [ B]
Overall
Reaction
A B  C  D
K'c
CD EF
K"c
__________________
A B  E  F
Kc
To find the overall equilibrium of a
multi-step reaction
n
n
n
n
[C ] [ D] [ E ] [ F ]
K 'c K "c 
x

K
c
n
n
n
n
[ A] [ B] [C ] [ D]
2 step reaction
1st step reaction
N2(g) + O2(g) ↔ 2 NO(g) Kc1 = 2.3 x 10^-19
2nd step reaction
2 NO(g) + O2(g) ↔ 2 NO2(g) Kc2 = 3 x 10^6
Write the equilibrium equation for this multi-step reaction:
Kc = Kc1 x Kc2 = (2.3 x 10^-19)(3 x 10^6)
Kc = 7 x 10^-13
CHEMICAL KINETICS
Reaction rates...
N2 + 3H2 → 2NH3
Molecular hydrogen reacts at a rate of 0.074 M/s
• Suppose that at a particular moment during the
reaction molecular hydrogen is reacting at the
rate of 0.074 M/s.
a) At what rate is ammonia being formed?
b) At what rate is molecular nitrogen reacting?
 1  N 2  1  H 2  1  NH 3
  
 
 
 1  t
 3  t
 2  t
N2 + 3H2 → 2NH3
Molecular hydrogen reacts at a rate of 0.074 M/s
• At what rate is
ammonia being formed?
H 2  0.074 Ms
t
GIVEN
 1  N 2  1  H 2  1  NH 3
  
 
 
 1  t
 3  t
 2  t

NH
2


 1  H 2  1  NH 3
3
M
_

0
.
074







 
s
3
t


 3  t
 2  t
_  1
 1  NH 3
M
NH 3
  0.074 s   
M
0.049 s 
 3
 2  t
t
N2 + 3H2 → 2NH3
Molecular hydrogen reacts at a rate of 0.074 M/s
• At what rate is molecular
nitrogen reacting?
H 2  0.074 Ms
t
GIVEN
 1  N 2  1  H 2  1  NH 3
  
 
 
 1  t
 3  t
 2  t
1
N 2
 1  H 2  1  N 2

M
  
 
 (1)  0.074 s 
 3  t
 1  t
t
3
 1
 1  N 2
M
    0.074 s    
2
M
 3
 1  t
s
 0.025
N

t
Describe the difference between
Heat of solution and Heat of
dilution. Give an example of
each one.
Heat of solution is the amount of heat
absorbed or released during the act of
combining a solute with a solvent. An example
is adding salt to water. There are two steps
to this process: the Lattice Energy (breaking
down the molecules) and the heat of
hydration (where the ionically charged
particles dissolve by attracting to the dipole
water molecules.
Heat of dilution refers to the amount of
heat absorbed or released when water is
added to a solution already made. In this latter
case, if the act of making a solution is
exothermic, adding more water will make it
even more exothermic (hotter). For example,
as you add more and more water to a
solution of sulfuric acid and water, the
more heat it releases (the more it burns).
Neutralization
• Explain how you would neutralize an
acid and a base:
Mix an acidic solution with a basic solution
wherein the ratio of H+ ions to OH- ions
equal to 1.
Cabbage Juice as a pH indicator
•
Explain how cabbage juice works as a pH indicator:
There is anthocyanin in cabbage juice. Anthocyanin reacts
with hydroxyl ions (OH-). Hydroxyl ions jump off
anthocyanin when in an acidic solution, thereby changing
the chemical composition of the anthocyanin (the pH
indicator). This will cause the juice to react with light
differently than if these OH- ions had not jumped off the
anthocyanin (makes the juice turn a reddish color).
Conversely, Hydroxyl ions will jump from the solution onto
the anthrocyanin molecules when there is an excess of
them in the solution (such as when cabbage juice is mixed
with a basic solution). This causes the juice (the pH
indicator) to turn a yellowish color.
NOT mentioned...
•
•
•
•
•
•
Balancing redox reactions
Kw
Ka
Kb
pH
the galvanic cell
• KNOW THESE CONCEPTS!!!!!!!!!!!