Chemistry
Mrs. Cribb
Name ____________________
Chapter 13 notes
13A Thermochemistry -the branch of science that studies the transfer of energy during chemical
reactions or phase changes
13.2
Temperature – the AVERAGE KINETIC energy in the particles of a sample
Heat – measures the TOTAL amount of thermal energy transferred from one object
to another. It is expressed in Joules (J).
Joule is the standard SI unit for work and can be measured for ANY KIND OF
ENERGY. Use kJ ---kilojoules
Because it is IMPOSSIBLE to determine the absolute amount of heat in something,
chemists measure the CHANGE IN ENERGY that occurs during a CHEMICAL
REACTION.
ENTHALPY (H) – is the heat content of a system at a constant pressure.
Calorimeter – an insulated container, similar to a thermos, in which a thermometer
detects the temperature change that occurs during a chemical reaction. Used
to measure ENTHALPY. SEE FIGURE 13-1.
13.3
SENSIBLE HEAT – produces a temperature change (you can sense it happening)
LATENT HEAT – produces a phase change! FLAT on the graph. NO temperature
change is detected. See figure 13-2.
Illustration:
•
MOLAR ENTHALPY OF FUSION ( Δ H fus ) - the amount of heat
that has to be added to overcome the intermolecular forces so a solid can
MELT. This is a type of LATENT HEAT. Defined as – the quantity of heat
required to melt one mole of a solid to a liquid with NO TEMPERATURE
CHANGE. It is usually in kJ/mole.
•
MOLAR ENTHALPY OF VAPORIZATION (___________) - the amount
of heat required to convert a mole of a liquid at its BOILING POINT to its
vapor at he SAME temperature.
Practice: Refer to your reference sheet for

H fus and 
H vap values.
1. How many kJ does it take to melt 56.9 moles of ice?
2. How many kJ does it take to turn 34.0 g of water from a liquid to a solid?
3. How many kJ does it take to turn 34.0 g of water from a liquid to a gas?
13.4
Not all substances heat at the same rate. The amount of energy that raises 1g of
something by 1 ºC is the SPECIFIC HEAT. SEE CHART 13-2!!!
Substances with HIGH SPECIFIC HEAT will need LARGE amounts of heat for a
temperature change.
Q = _____________________
Q – Thermal energy put into a substance
m – mass of substance in grams
c sp - specific heat of substance. You get this off of a chart!
T – temperature in °C
Rates of COOLING are also related to specific heat. Substances with HIGH specific
heat values retain heat for longer periods.
The warming curve of a substance reveals its relative specific heat. If the temperature
rises quickly with the addition of heat, the specific heat must be LOW. This means a
SMALL amount of heat results in a LARGE change in temperature. A GENTLE slope
on the warming curve means has a HIGH specific heat because added heat does not
change the temperature very much.
High specific heat
Low Specific heat
Example 13-1 If 942 J of thermal energy are added to 50.0g of water at 25.1 ºC, what will its final
temperature be?
Example 13-2A 28g sample of silver is heated from 15ºC to 85ºC. How many joules were added to the
sample?
A complex thermodynamic Problem
An ice cube is heated until completely converted to steam at 373K (100 degrees C), how many
Joules of heat must be added? The mass of the ice cube is 5.0g.
Draw the warming curve
13.5
Enthalpy (heat) of reaction
• The change on enthalpy that occurs during a reaction is called the enthalpy (heat)
of reaction (______).
Formula: __________________________
2H 2 g  O 2 g  → 2H 2 O l 571.6 kJ
Thermochemical equation – shows where the heat is added or produced in kJ.
Draw exothermic reaction warming curve. These have negative values for ΔH.
Draw endothermic reaction warming curve. These have positive values for ΔH. The heat is added on
the reactant side in kJ.
Example reaction: 6 C(s)+3 H 2 (g)+49.1 kJ →C 6 H 6 (l)
Reverse reactions have the same magnitude, but different signs.
The endothermic reaction could be written the in opposite direction. Then the ΔH would be negative
and not positive.
Example:
13.6 Enthalpy (heat) of Formation ΔH f (Molar enthalpy of formation)
• The change in enthalpy that occurs when one mole of the compound is FORMED from its
elements.
ΔH f Of elements is ZERO because they are not being formed.
•
Example reaction: ______________________________________________________
ΔH f = _____________
•
Endothermic or Exothermic? _________________
The amount of enthalpy in a substance varies with temperature and pressure, so these conditions
must be specified.
◦ Standard state => ___________ (298 K) and 1 atm (_______ torr)
◦ Standard molar enthalpy of formation ( _____) is the enthalpy change that makes 1 mole
in its standard state from its elements in their standard states.
◦ See reference chart
Example 13-3 What is the ΔH f
for
N 2 +2O 2 →2 NO 2 ?
13.7 Enthalpy (heat) of Combustion ΔH c
• The energy released as heat by the complete burning of one mole of a substance in standard
conditions.
• Remember combustion reactions must contain oxygen
Example reaction: ______________________________________________________
ΔH c = _______________
• See reference chart or Chart 13-4 in your text book.
• All of these are exothermic because heat exits so all the values are negative.
13.8 Calculating Enthalpies of Reaction
Steps: 1. Each reactant breaks down to its elements in their standard state
2. Elements combine to form the products in their standard state
▪ Remember opposite processes like forming and breaking down have opposite signs on
their heats of formation
Hess's Law → The enthalpy change of a reaction equals the sum of the enthalpy changes for each step
of the process.
•
Mathematical version of this: ∑ΔH = ∑H (products)−∑H (reactants)
◦ Remember you can use any valid BALANCED equation, but change the sign of the values
if you are using the reverse of the equation.
◦ Multiply the ΔH when needed by the coefficients to account for each mole used.
Example 13-4 Estimate the
formation.
ΔH for the combustion of methane, using standard molar enthalpies of
Chapter 13B --- Honors Only
Chemical bonds STORE energy. When the bonds are broken energy is RELEASED. Bonds are broken
through VIBRATION.
The strength of the bond between any two atoms depends on the ELEMENTS that are bonding and on
the LENGTH of the bond. Bonded atoms are MORE STABLE than unbonded ones. Energy IS
ALWAYS REQUIRED to break bonds. Bond breaking is thus ALWAYS an ENDOTHERMIC
process. FORMING bonds is ALWAYS an EXOTHERMIC
process.
The STRONGER the bond the MORE ENERGY is released.
A reaction that breaks strong bonds and forms weak ones
REQUIRES energy.
(Think about it. Forming strong bonds releases a lot of
energy and thus requires a lot back to break them. Weak bonds
release a small amount of energy.)
stronger bonds → weaker bonds : ΔH › 0 (positive) (endothermic)
A reaction that breaks weak bonds and forms strong ones RELEASES energy.
(Think about it. Weak bonds release SMALL amounts of energy and thus require small amounts
to break them. Strong bonds release a large amount of energy!)
weaker bonds → stronger bonds : Δ H ‹ 0 (negative) (Exothermic)
See figure 13-12 in your text.
In most cases, exothermic processes are thermodynamically favorable (that means they are more likely
to happen automatically) and endothermic processes are thermodynamically unfavorable.
Endothermic reactions happen all the time like ice melting or cold packs becoming cold when you
break the seal and the chemical reaction happens. So, more than determining if something is
endothermic or exothermic is needed to know if something happens spontaneously.
Chemical processes tend to proceed naturally in the direction that INCREASES THE DISORDER of a
system. The SECOND LAW OF THERMODYNAMICS relates this when it says that natural
processes decrease the order found in the universe. It spreads out the energy of a system.
Which has more order?
Gas molecules < liquids < solids
(Think about how the molecules move and how much energy they have.)
ENTROPY (S) is a measure of RANDOMNESS or lack of orderliness in a system. (DiSSSSSSorder)
It is the measure of the dispersal of energy in a system. In natural processes, energy tends to spread
out, not concentrate. A hot pan cooling off on a stove top is an example of entropy in action. The heat
is leaving the pan and heating up the surrounding air giving the air more energy and more randomness
or disorder.
Which has more ENTROPY? Gas molecules > liquids > solids
The ENTROPY of a substance INCREASES with TEMPERATURE because things get more energy!
At absolute zero, theoretically all random movement of molecules or atoms stops and the ENTROPY is
zero ( S=O).
You calculate entropy just like enthalpy.
Formula:
∑ΔS=∑S( products)−∑S (reactants)
A reaction with a POSITIVE ΔSº INCREASES disorder, energy, temperature. The products
have more entropy than the reactants. A reactions with a NEGATIVE ΔSº DECREASES
disorder, lowers energy, lower temperature, increases order.
Example 13-5: Does the reaction NH 3 (g)+HCl(g)→NH 4 Cl (s) produce an increase or
decrease in entropy? See Reference chart for needed values. These are copied from your
book.
∑ΔS=∑S( products)−∑S (reactants)
All procresses INCREASE the entropy of the universe. A positive ΔSº means a reaction is
MORE FAVORABLE. A NEGATIVE ΔSº goes against this tendency and attempts to decrease
entropy.
Two tendencies drive reactions:
1. Decrease ENTHALPY (Exothermic)
2. INCREASE ENTROPY
See the four possible combinations.
1. Exothermic and increasing entropy.
(
)
2. Exothermic and decreasing entropy.
(
)
3. Endothermic and increasing entropy.
(
)
4. Endothermic and decreasing entropy.
(
)
J. Willard Gibbs formulated a single way of figuring out favoribility. He combined entropy and
enthalpy into a single term called FREE ENERGY, or GIBBS FREE ENERGY (G).
ΔG=ΔH−T ΔS
(Here T is in KELVINs.)
ΔG is the change in FREE ENERGY. It is the difference between the free energy of the
products and the free energy of the reactants. A NEGATIVE ΔG indicates a DECLINE in
free energy and signifies that a reaction MAY OCCUR NATURALLY and is
SPONTANEOUS.
See that? A -ΔG means it is SPONTANEOUS! A positive value is NOT spontaneous.
2 ways to calculate ΔG.
1. Products – reactants.
2.
∑ΔG=∑G(products)−∑G(reactants)
ΔG=ΔH f −T ΔS
Be sure to make sure units match up!
Example: 13-6 Calculate Gibbs free-energy change for the following reaction:
2 Ag(s)+S(s)→Ag 2 S(s) At 298K
Example: 13-7 Calculate Gibbs free-energy change for the following reaction:
3
2 Fe (s )+ O 2 (g)→Fe2 O 3 (s) At 298 K
2
Example: 13-8 Calculate Gibbs free-energy change for the following reaction:
2C (s)+H 2 (g)→C 2 H 2 (g) At 298 K
Example: 13-9 Calculate Gibbs free-energy change for the following reaction:
2C (s)+2 H 2(g)→C 2 H 4 (g) At 298K
Example: 13-10 Determine whether the reaction between ammonia and hydrogen chloride is
probable at 298K and at 1000. K according to table 13-7 on pages 344 and 345.
NH 3 (g)+HCl(g)→NH 4 Cl (s)