Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107

Fundamentals of Chemistry

Fall 2009

Lecture Notes: 1 October

© R. Paselk 2005


Energy Changes in Chemical Reactions

The First Law of Thermodynamics

In technology based systems (engines etc.) work is often done by expanding/contracting gases, e.g. piston engines in cars. For these systems work, 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 is the science of measuring heat. It is particularly useful because under two readily achievable laboratory conditions heat = DeltaE.


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

How about sig figs? Heat capacity of water is assumed to be exact for this problem. Three measurements: 750, 20.00°C, 100.0g. Thus the answer will have only three sig figs since two sig figs for 750 since using multiplication, but tenths place for addition of 7.5 to 20.00.

27.50 °C = 27.5 °C

In fact a slight error is introduced in this calculation because heat capacity changes slightly with temperature.

For constant volume calorimetry a bomb calorimeter is used. In this instrument a sample is burned in a heavy, rigid container which is in turn submerged in an insulated vat of water.

Note that for gases at constant V, no work can be done, so q = DeltaE.


Example: 1.40 g of vegetable oil is placed in a bomb calorimeter with excess oxygen and ignited with a spark. If the calorimeter temperature changes from 20.000 °C to 21.195 °C, find the energy released per gram of oil . The calorimeter contains 2.50 kg of water. The calorimeter without water has a heat capacity of 1.00 kJ°C-1.

q = nCpDeltaT, where Cp is the molar heat capacity at constant pressure (= 75.3 J C-1mol-1 for water).

q= qwater + qcalorimeter

qwater= {(2.50 kg H2O)(1000g/kg) / (18.02 g H2O/mole)}{75.3 J C-1mol-1}{1.195 °C} = 1.25 x 104J

qcalorimeter = CDeltaT = (1.00 x 103J°C-1)(1.195 °C) = 1.195 x 103J

qtot= 1.25 x 104J + 1.195 x 103J = 1.369 x 104J

E/g = (1.369 x 104J) / 1.40 g = 9.78 kJ/g

Notice that this is now the energy released, and it will also be the energy you could potentially get from consuming this much oil, since we are working with state functions, and the pathway (fire or metabolism) doesn't matter.

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Last modified 1 October 2009