Aubrey High School  AP Chemistry
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Period ___ Date ___/___/___
6  Thermodynamics
6.1 Lesson – Enthalpy
Change in Internal Energy (∆Eint)
It’s important to define the system − that at which we’re looking, usually a
container or a physical or chemical process − and the surroundings – the rest of
the universe, so we can study the movement of energy between them.
Open System
Movement of Matter between
System and Surroundings
Movement of Energy between
System and Surroundings
Closed System
Isolated System
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The internal energy (Eint) of a system is the sum of all of the kinetic energy (which contributes to the
temperature) and potential energy (from attractions between particles).
The First Law of Thermodynamics states that in an isolated system (one in which no matter or energy can be
exchanged), the internal energy of the system is conserved (∆Eint = 0): energy can be converted from one form or
another (such as from kinetic energy to potential energy), but the total energy cannot increase or decrease.
However, no system is truly isolated, so the internal energy of a closed system can change in two ways: (1) heat
flow or (2) work.
∆Eint = q + w
 Heat (q) is the flow of thermal energy between the system and the surroundings from the object of higher
temperature to that of lower temperature.
q = m·c·∆T
 Work (w) can be measured by the energy from the expansion or compression of a gas. w = −P·∆V
Example 1: A 20.0 g sample of Ar (g) in a cylinder with a piston at 30°C. Heat flows into the cylinder and
increases the temperature to 40°C. The cylinder expands against a constant pressure of 95.5 kPa from 21.5 L to
23.5 L. By how much has the internal energy of the system changed? Ar has a specific heat of 0.52 J/g·°C.
(1 J = 1 L·kPa).
Ideas about Heat (q)
Example 2: A student dissolves 10.0 g of NaOH in 100.0 mL of distilled water, and notices
the temperature of the mixture increase from 25.0°C to 51.6°C.
(a) The student states:
The process of dissolving NaOH is endothermic, because the temperature
increases (i.e. ∆T > 0) so q > 0.
Do you agree or disagree? Why? Use system, surroundings, heat, endothermic, and
exothermic in your explanation.
(b) Explain in terms of molecular motion what is happening when the temperature of the water is increasing.
(c) Describe (pictorially) the substances before (NaOH (s) and H2O (ℓ)) and after the dissolution (NaOH (aq)).
Identify the inter-particle forces involved in the substances before and after the dissolution. Do the substances
before the dissolution have less or more potential energy than those after the dissolution?
BEFORE:
AFTER:
(d) Describe the process of breaking and forming inter-particle forces/attractions in the
dissolution of NaOH. Draw a potential energy diagram for each step of the process.
Then explain why the temperature lowers when dissolving a different salt, NH4Cl (s).
(e) Calculate the heat flow (q) for this process. The specific heat of water is 4.18 J/g·°C.
(f) Calculate the ∆Hdiss for this process: NaOH (s) → NaOH (aq)
Enthalpy of Reaction (∆Hrxn)
Example 3: Consider the reaction:
2 SO2 (g) + O2 (g) → 2 SO3 (g)
(a) Experimental Approach to Determining ∆Hrxn
A 9.71 g sample of SO2 (g) reacts with excess O2 releasing 15.0 kJ of energy. Calculate ∆Hrxn.
(b) Products – Reactants Approach
Using the ∆Hrxn from part (a), determine the value of ∆Hf for SO2 (g). ∆Hf for SO3 (g) is −396 kJ/mol.
(c) Hess’s Law
Calculate the value of ∆Hrxn using the following reactions. Compare the result with that found in part (a).
2 H2S + 3 O2 → 2 H2O + 2 SO2
∆H1 = −1124 kJ/mol
(1)
(2)
H2O + SO3 → H2SO4
∆H2 = −132 kJ/mol
(3)
H2S + 2 O2 → H2SO4
∆H3 = −793 kJ/mol
(d) Bond Energy Approach
Using the following structures and bond energies,
calculate the ∆Hrxn. Compare the result with that
found in part (a) (and explain any difference).
Lewis Structures:
SO2
Bond Energies
S=O
O=O
Draw the graph of the potential energy vs. inter-nuclear distance for both the
S=O and O=O bonds to show why the bond energy for S=O is greater than
that of O=O.
SO3
522 kJ/mol
495 kJ/mol