AP CHEMISTRY – CHATER 6 THERMOCEMISTRY – STUDENT NOTES
 After this chapter, you should be able to solve problems relating you:
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 Definitions:
o Thermodynamics
o Energy
o Kinetic energy
o Potential energy
o Heat
o State function
o System
o Surrounds
o Exothermic
o Endothermic
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Work
o Force = mass x acceleration = kg * m/s2
o Work = force x distance = kg * m/s2 * m
o 1 Joule = 1 kg m2/s2
Sign Conventions
o When the system expands, it is doing ________________ work on the
surroundings, therefore it is doing ________________ work on the
system.
o When the system contracts, the surrounds have done work on the system,
there is ________________ work done on the system.
o W = -P∆V
o The system compression is positive work and system expansion is
negative work.
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The First Law of Thermodynamics
o The law of conservation of energy
o Energy can be converted from one form to another, but can neither be
created nor destroyed.
o Energy of the universe is constant
o Energy can be changed through work. Chemical bonds are made and
broken; energy is converted between the potential energy and thermal
energy.
o The change in the internal energy of the system, which is equal in size but
opposite in sign of that of the surroundings, is equal to the sum of the heat
and work.
o ∆E=q+w
o ∆ E = negative = system loses energy
o ∆ E = positive = system gains energy
Example (The First Law of Thermodynamics)
o Calculate the change in energy the system if 38.9 J of work is done by the
system with an associated heat loss of 16.2 J.
Example (Practice with Heat and Work)
o A piston is compressed from a volume of 8.3 L to 2.8 L against a constant
pressure of 1.9 atm. In the process, there is a heat gain by the system of
350J
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Enthalpy and Calorimetry
o Enthalpy (H) is a state function
o ∆ Hystem = H products – H reactants
o The change in enthalpy (∆ H) of the system is equal to the energy flow as
heat at a constant pressure
o ∆ H = qpressure
o if ∆ H > 0, the reaction is endothermic
o if ∆ H < 0, the reaction is exothermic
Example (Enthalpy)
o Upon adding solid potassium hydroxide pellets to water, the following
reaction takes place:
o For this reaction at constant pressure, ∆ H = -43 kJ/mol. Answer the
following questions:
 Does the beaker get warmer or colder?
 Is the reaction exo or endothermic?
 What is the enthalpy change for the dissolution.
Calorimetry
o Experimental technique used to determine the heat exchange (q)
associated with a reaction.
o The amount of heat exchanged in a reaction depends upon:
 The net temperature change during the reaction
 The amount of substance
 The heat capacity (C) of a substance
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There are three ways of expressing heat capacity:
 heat capacity = J/ C
 specific heat capacity = J/ C * g
 molar heat capacity = J/ C * mo
Example (Specific Heat Capacity)
o Based on the values for specific heat capacity, which conducts heat better,
water or aluminum? Why is this important for cooking?
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Example (Constant Pressure Calorimetry)
o The enthalpy change (∆ H) per mole of NaOH is -43 kJ/mol when
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o If 10.0 g of solid NaOH is added to 1.00 L of water at 25.0 C in a constant
pressure calorimeter, what will be the final temperature of the solution?
The density of the final solution is 1.05 g/mL
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Constant volume calorimetry is discussed. The bomb calorimeter is used for this
application. In this case, the change in volume is zero so no work is done and
∆ E = qw in units of joules. Each bomb calorimeter is different. The heat capacity
(J/ C) of the bomb and its parts must be determined using a known substance
before the energy (or heat) of combustion can be determined.
Example (Constant Volume Calorimetery)
o The heat of comubustion for glucose is 2800 kJ/mol. A sample of glucose
weighing 5.00 grams was burned with excess oxygen in a bomb
calorimeter. The temperature of the bomb rose 2.40 C. What is the heat
capacity of the calorimeter?
o A 4.40 gram sample of glucose was then buned with excess oxygen in the
same bomb calorimeter. The temperature of the bomb increased 6.85 C.
Calculate ∆ Ecombustion of propane.
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Hess’s Law
o Enthalpy changes are state functions
o It does not matter if ∆ H for a reaction is calculated in one step or a series
of steps.
o By using the known values of ∆ H of known reactions, we can use Hess’s
law to solve for enthalpies of reactions whose values we do not know.
Example (Hess’s Law)
o Given the following reactions and ∆ H values:
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o Calculate the ∆ H for
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Example (Hess’s Law)
o Given the following reactions and ∆ H values; calculate the ∆ H.
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Standard Enthalpies of Formation
o There is a table with the standard enthalpy of formation ∆ Hf0
o ∆ Hf0 is always given per mole of compound formed.
o ∆ Hf0 involves formation of a compound from its elements with the
substances in their standard states
o ∆ Hf0 for an element in its standard state equals 0.
Example (Standard Enthalpies of Formation)
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The key to calculate standard enthalpy changes in reactions is to remember
products minus reactants.
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Example (Calculating Standard Enthalpies of Formation)
o Calculate the ∆ H0 for the following reaction:
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Example (Practice with Standard Enthalpies)
o The thermite reaction is discussed in your textbook. A variation of this
reaction is:
o Calculate the ∆ H0 for this reaction.