Chem 109 - General Chemistry - Spring 2015

Lecture Notes 19: 6 March

Hydrogen/Oxygen gas stoichiometry demonstration

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 =

H =

E - w =

E - P

V = q @ constant P

where P

V is the pressure-volume work

Enthalpy is often approximately = E for chemical processes, since little or no work is usually done in solution chemistry (both P and V are constant).

Calorimetry

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

For solution and solid state reactions at constant pressure there is no significant change in volume, so we can assume no work: therefore heat = E.
If a reaction takes place in a rigid container, there can be no volume change, and no work can have occurred: therefore heat = E.

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?

joule (SI unit): It takes 4.184 J to raise 1 g of water at 20 °C 1 °C.
calorie (metric, non-SI unit): It takes 1 cal to raise 1 g of water at 20 °C 1°C. Thus 1 cal = 4.184 J (defined).
Calorie (large Calorie = nutritional Calorie): 1 Calorie = kilocalorie. [2000 Calorie burger is 8.4 million juoles!]

Specific 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 = mC_sp

where m is the mass of substance and C_sp 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 = mC_spT

Rearranging equations gives:

T = q/ (mC_sp)

Substituting values into the equations get:

T = 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 = 27.5 °C

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