Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 18: 2 March

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Thermochemistry, cont.

State Functions

Kinetic energy, potential energy, pressure, and volume are all examples of State Functions. They are all properties that depend only on the current state - they are all independent of the path used to reach this state.

The First Law of Thermodynamics

The First Law of Thermodynamics says that the energy of the Universe is constant. Thus it is another name for the law of conservation of energy. Symbolically it is written:

E = q + w

where E is energy, q is heat, and w is work.

Note that according to this law we can still do things with energy, its just that they are always compensated. (Thus as the Universe expands, work is done against gravity and the heat in the Universe decreases as manifested by a decreasing average temperature.)

Generally in thermodynamics we refer to systems. A system is simply a portion of the universe we wish to work with. For the expression

DeltaE = q + w

where E is the internal energy (the total KE and PE) of the system.

q = the quantity of heat exchanged by the system:

Notice in each case endo- and exo- are in respect to the system, not the surroundings. For example, a fire is exothermic, because heat comes out of the fire - the fuel loses heat, even though you (part of the surroundings) may gain some of it.

Keep in mind that heat always flows naturally from hotter to cooler systems. Energy must be used up to move heat in the opposite direction, as in a refrigerator.

w (in chemistry) = the work done by or on the system:

  • w = positive when work is done on the system (e.g. as gas is compressed)
  • w = negative when the system does work on its surroundings (e.g. a gas moves a piston by expansion - notice that an ideal gas expanding in space does no work!).

Note that if no heat is transferred to or from a system (it is isolated in a "thermos"), then all energy must appear as work. On the other hand, if no work is done, then all energy must appear as heat (this is utilized in calorimetry which is discussed below).

Example: A quantity of air in a cylinder expands against a piston doing 4.5 kJ of work while 10.0 kJ of heat is added. How much has the energy of the air changed?

DeltaE = q + w

So DeltaE = 10.0 kJ + (- 4.5 kJ) = 5.5 kJ

Enthalpy & Calorimetry

Most chemistry is done under conditions of constant pressure or constant volume (e.g. all of your body chemistry occurs at about atmospheric pressure - no pressure changes occur within single cells doing chemistry). Thus it is convenient to define a term for the heat involved in processes occurring with no change in pressure:

Enthalpy = DeltaH = DeltaE - w = DeltaE - PDeltaV = q @ constant P

where PDeltaV is the pressure-volume work

Enthalpy is often approximately = DeltaE for chemical processes, since little or no work is usually done in solution chemistry.

Calorimetry

Calorimetry is the science of measuring heat. It is particularly useful because under two readily achievable laboratory conditions heat = DeltaE.

Heat

Heat is a measure of energy transferred between objects of different temperatures. We are already familiar with the units of temperature, what are the units of heat?

Specific Heat is the amount of heat it takes to raise 1 g of a specific substance 1 °C. Specific heats for other substances are relative to water, so no units (comparing results in canceling out units).

The heat transferred in a process (q) is summarized in the equation:

Heat = q = mCspDeltaT

where m is the mass of substance and Csp is the specific heat of the substance.

Example: 750 calories of heat is transferred to 100.0 g of water at 20.00 °C. What will the new temperature of the water be assuming no heat is lost to the container or the surroundings?

Known: heat capacity of water = 1 cal / (g°C) [assume exact for problem]; q = mCspDeltaT

Rearranging equations gives: DeltaT = q/ (mCsp)

Substituting values into the equations get: DeltaT = 750 cal / {(100.0 g)(1 cal / (g°C)} = 7.50 °C

Adding the difference to the original temperature gives: 20.00 °C + 7.50 °C = 27.50 °C

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© R A Paselk

Last modified 28 February 2011