Thermochemistry

Definitions

• Thermochemistry – study of energy changes that occur during phase changes and chemical reactions.

– Energy – capacity for doing work or supplying heat.

– Chemical (Potential) Energy – energy stored in chemical bonds.

Example

5473 kJ/mol

Energy difference.Lots of energy

stored in bonds!

Little energy stored in bonds.

Heat

• Heat - Energy that transfers from one object to another because of a temperature difference between them.

– Represented by the symbol q

• Heat flows from warm cool until the two objects are at the same temperature– heat = amount of energy– temperature = average kinetic energy

Exothermic vs. Endothermic

• Exothermic – energy is released– Surroundings are warmed up– q is negative because the system is losing heat.

• Endothermic – energy is absorbed– surroundings are cooled down.–q is positive because the system is gaining heat.

Potential Energy Diagram of Ice Melting at 0ºC.

Time

Potential E

nergy

Ice

Water

Is the melting of ice an endothermic or an exothermic process? How can you tell?

Measuring Heat Flow

• SI Unit of heat flow: Joule (J)

• Common unit used in chemistry: calorie (cal) -amount of heat needed to raise 1 gram of water by 1ºC.

– 1 cal = 4.184 J

• Food Calorie (capital “C”) = 1000 cal, or 1 kilocalorie = 4184 J

Heat Capacity

• Amount of heat needed to raise an object’s temperature by 1°C.– Depends on the chemical composition and the

mass(grams) of the object.

• EXAMPLE: 1 gram of water requires 1 cal to raise its temperature by 1°C.– 100. g of water require 100. cal to raise the

temp. by 1°C.

Specific Heat (c)

• Amt. of heat needed to raise 1 gram of a substance’s temperature by 1ºC.– Expressed in J/g ºC

• The higher a substance’s specific heat, the more energy it takes to heat it.– Water: 4.18 J/gºC

Energy of Temperature Change

• q = mcΔT– q = heat energy in Joules

– m = mass in grams

– c = specific heat(for water, it’s 4.18 J/gºC)

– ΔT = change in temperature

Example Problem

• What is the energy required to heat a 120.0 gram glass of water from 25.0ºC to 100.0ºC?– SOLUTION: q = m c ΔT

• q = (120.0 g) (4.18 J/gºC) (100.0ºC – 25.0ºC)• q = (120.0 g) (4.18 J/gºC) (75.0ºC)• q = 37620 J

Example Problem

• 1000. Joules of energy is added to a 50.0 gram sample of water at 25.0ºC. What is the change in temperature? – SOLUTION: q = m c ΔT– 1000. J = (50.0 g) (4.18 J/gºC) ΔT– 1000. J = (209 J) ΔT

• ΔT = 4.78 ºC

______ ______ 209 J 209 J

Thermochemical Equations and Stoichiometry

• The combustion of 5.00 grams of methane releases 250. kJ of heat.– Now we’ll calculate how hot the water in the container will get if it

absorbs all of the heat.– First convert 25.0 kJ to J

– q = m x c x T– 2.50x104 J = (100. g) (4.18 J/gºC) T– 2.50x104 J = (418 J/ºC) T T = 59.8ºC– The water will get 59.8ºC warmer.– The final temperature will be 20.0ºC + 59.8ºC = 79.8ºC.

25.0 kJkJ 1

J 1000x = 2.50x104 J