AP/IB Chemistry Chapter 5 Thermochemistry Handout
Thermodynamics is the study of energy and its transformations
Thermochemistry is the study of relationships between energy changes involving heat
and chemical reactions
Energy is the capacity to do work or to transfer heat
Heat is the energy transformed from one object to another because of a temperature
difference
A calorie is the amount of energy required to raise the temperature of one gram of
water by one Celsius degree. The Calorie is equivalent to kcal and is used to measure
food energy
A joule is the SI unit of energy. 1 cal = 4.184 J
Energy changes involve the transfer of heat between the system (the portion of the
universe that we are studying) and the surroundings (everything else)
The first law of thermodynamics, also called the law of the conservation of energy,
states that energy is conserved. The energy of the universe is constant. Energy can
neither be created nor destroyed, but it can be changed from one form to another
An exothermic process is a process that releases heat to the surroundings. Heat flows
out of the system and into the surroundings. The temperature of the surroundings
increases.
An endothermic process absorbs heat from the surroundings. Heat flows into the
system from the surroundings. The temperature of the surroundings decreases.
Enthalpy is the property of a system that accounts for heat flow between the system
and the surroundings at constant pressure. Enthalpy is a state function, a property of a
system that depends only on the present state of the system, not the path the system
took to reach that state. A state is determined by the system’s temperature, pressure.
Location, and other conditions that define the system.
Enthalpy change, ∆H, is the heat absorbed during a physical or chemical process
Enthalpy: accounts for heat flow in process occurring at constant pressure when no
forms of work are performed other than P-V work
represented by H
H=E+PV
∆H= ∆E+P ∆V
∆H is positive for an endothermic process, one that absorbs heat
∆H is negative for an exothermic process, one that releases heat
Calorimetry is the measurement of heat flow. A calorimeter is an apparatus that
measures heat flow
Heat capacity, C, is the heat required to raise the temperature of an object by 1K. The
units of C are J/K/
Molar heat capacity, Cmolar is the amount of heat absorbed by one mole of a substance
when it experiences a one degree temperature change. The units of Cmolar are J/molK.
Specific heat capacity is the heat capacity of one gram of a substance. The units are
J/gK.
The specific heat of water is worth remembering: 1cal/gK=4.184 J/gK
There are many different kinds of ∆H’s—so when a problem asks for one be sure you
know which value of ∆H to use (∆Hrxn, ∆Hcomb, ∆Hfus, ∆Hvap, ∆HBDE, ∆Hf, ∆Hsoln)
Hess’s Law: states that the reaction is carried out in a series of steps, ∆H of the overall
reation is equal to the sum of the ∆H’s for each individual step. Hess’ law is useful in
calculating enthalpies of reactions that are difficult or impossible to measure directly.
The formation of sulfur trioxide by the reaction of sulfur with oxygen, for example,
does not proceed directly under normal conditions, but the heat of the reaction can
be calculated from the heats of the individual steps that lead to the reaction.
A formation reaction is a reaction that produces only one mole of the substance from
its elements in their most stable thermodynamic state..
The standard heat of formation, ∆H°f, is the heat absorbed when one mole of a
substance is formed from its elements in their standard states at 25°C and 1 atm.
Q=mC∆T (Q= heat absorbed/released, m=mass in grams, ∆T=change in temperature)
*must be at STP=1atm + 273.15K so sometimes you must use Gas laws to get the
conditions correct (P1V1/T1= P2V2/T2)
Practice Problems:
Problem 1:
H2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ
Calculate the change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide
decomposes.
Problem 2
For the decomposition of hydrogen peroxide, it is known that:
H2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ
Using this information, determine ΔH for the reaction:
2 H2O(l) + O2(g) → 2 H2O2(l)
Problem 3
The bond energy (kJ) for H2, F2, and HF are 436, 158 and 568 kJ
respectively, calculate the enthalpy (energy) of the reaction,
H2(g) + F2(g) = 2 HF
Problem 4
Standard enthalpies of formation are: C2H5OH(l) -228, CO2 -394, and H2O(l) -286
kJ/mol. Calculate the enthalpy of the reaction,
C2H5OH + 3 O2
2 + 3 H2O
Problem 5 (endothermic or exothermic process)
a. mixing water with calcium chloride
b. nuclear fission
c. crystallizing liquid salts (as in sodium acetate in chemical handwarmers)
d. mixing water and strong acids
e. forming ion pairs
f. burning sugar
g. rusting iron
h. a candle flame
i. condensation of rain from water vapor
j. formation of snow in clouds
k. making ice cubes
Problem 6 (exothermic or endothermic processes)
a. melting solid salts
b. producing sugar by photosynthesis
c. cooking an egg
d. baking bread
e. evaporation of water
f. conversion of frost to water vapor
g. melting ice cubes
Answer KEY:
Problem 1:
The thermochemical equation tells us that ΔH for the decomposition of 1 mole of
H2O2 is -98.2 kJ, so this relationship can be used as a conversion factor. Using the
Periodic Table, the molecular mass of H2O2 is 34.0, which means that 1 mol H2O2 =
34.0 g H2O2.
Using these values:
ΔH = 1.00 g H2O2 x 1 mol H2O2 / 34.0 g H2O2 x -98.2 kJ / 1 mol H2O2
ΔH = -2.89 kJ
Answer: The change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide decomposes
= -2.89 kJ
Problem 2:
When looking at the second equation, we see it is double the first reaction and in the
opposite direction.
First change the direction of the first equation. When the direction of the reaction is
changed, the sign on ΔH changes for the reaction
H2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ
it becomes…
H2O(l) + 1/2 O2(g) → H2O2(l); ΔH = +98.2 kJ
Second, multiply this reaction by 2. When multiplying a reaction by a constant, the
ΔH is multiplied by the same constant.
2 H2O(l) + O2(g) → 2 H2O2(l); ΔH = +196.4 kJ
Answer: ΔH = +196.4 kJ for the reaction: 2 H2O(l) + O2(g) → 2 H2O2(l)
Problem 3:
Based on the bond energies given, we have
H2
D = 436 kJ/mol
F2
D = 158 kJ/mol
H = -568*2 kJ/mol
Adding all three equations and energies leads to the following
H2(g) + F2(g) = 2 HF
H = -542 kJ/equation
Problem 4:
From the definition of the enthalpy of formation, we have the following equations
and the energy changes of reactions.
C2H5
2(l) + 0.5 O2(g) H = 228 kJ/mol
2 C(graphite) + 2 O2
H = -394*2 kJ/mol
2(g)
3 H2(g) + 1.5 O2
H = -286*3 kJ/mol
2O(l)
Adding all three equations and energies leads to the following
C2H5OH(l) + 3 O2
Problem 5
All are exothermic processes
Problem 6
All are endothermic processes
2(g)
+ 3 H2O (l)
H = -1418 kJ/mol