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“Why do we want to burn propanol?” C 3 H 8 +  H 2 O + CO 2 C 3 H 8 +  H 2 O + CO 2 There is another non chemical product…. There is another non.

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Presentation on theme: "“Why do we want to burn propanol?” C 3 H 8 +  H 2 O + CO 2 C 3 H 8 +  H 2 O + CO 2 There is another non chemical product…. There is another non."— Presentation transcript:

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4 “Why do we want to burn propanol?” C 3 H 8 +  H 2 O + CO 2 C 3 H 8 +  H 2 O + CO 2 There is another non chemical product…. There is another non chemical product…. HEAT! »The science of chemical reactions that involve HEAT is called THERMODYNAMICS

5 Thermochemistry The study of energy changes that accompany physical or chemical changes in matter The study of energy changes that accompany physical or chemical changes in matter All thermodynamics depends on the LAW OF CONSERVATION OF ENERGY: the total energy of a system and its surroundings remains the same All thermodynamics depends on the LAW OF CONSERVATION OF ENERGY: the total energy of a system and its surroundings remains the same Changes that occur in matter may be classified as physical, chemical or nuclear Changes that occur in matter may be classified as physical, chemical or nuclear

6 Heat and Energy Changes Physical, chemical and nuclear changes are all accompanied by a change in energy –Physical: a change in the form of a substance. No chemical bonds are broken. Ex. hydrogen boils at -252°C H 2 (l) + heat  H 2 (g) –Chemical: a change in the chemical bonds between atoms resulting in new substances Ex. hydrogen is burned as fuel in the space shuttle’s engines. 2H 2 (g) + O 2 (g)  2H 2 O (l) + heat –Nuclear: a change in the protons or neutrons in an atom resulting in a new atom. Ex. hydrogen undergoes nuclear fusion in the Sun, producing helium.

7 Heat and Energy Changes cont’d Chemical system – the substances undergoing a change Surroundings – the system’s environment Thermal Energy – energy available from a substance as a result of the motion of its molecules Temperature (T) – the average kinetic energy of the molecules in a sample, measured in °C or K Heat (q) – amount of energy transferred between substances, measured in Joules (J) Energy flows between substances because of their difference in temperature.

8 Heat and Energy Changes cont’d A. Exothermic: releasing thermal energy, heat (q) flows from the system to the surroundings, usually causing an increase in the temperature of the surroundings. q has a negative value q has a negative value B. Endothermic: absorbing thermal energy, heat (q) flows into the system from the surroundings, usually causing a decrease in the temperature of the surroundings. q has a positive value q has a positive value A B A B

9 Measuring Energy Changes Calorimetry: Calorimetry: experimental technique used to measure energy changes/transfers (the quantity of heat (q) in chemical systems different substances vary in their ability to absorb amounts of heat Substances used in a system have a specific heat capacity: amount of energy required to raise the temperature of 1 g of a substance by 1 °C or 1K amount of energy required to raise the temperature of 1 g of a substance by 1 °C or 1K (see Table 1, pg.301)

10 Equation used to determine the value of q: The amount of heat transferred (q) depends on »measured in grams »measured in J/g°C or J/gK – –measured in °C or K Measuring Energy Changes

11 Sample Problem 1 Many water heaters use the combustion of natural gas (assume methane) to heat the water in the tank. When 150.0 L of water at 10.0°C is heated to 65.0°C, how much heat flows into the water? m = 150L x 1kg/L = 150 kg = 150 000 g c = 4.18 J/g°C T = (65-10) = 55°C q = mcΔT = 150 000g x 4.18 x 55.0 = 34485000 q = 3.45 x 10 7 J

12 Sample Problem 2 If 25.0 g of aluminum cools from 310°C to 37°C, how many joules of heat energy are lost by the sample? m = 25.0 g c = 0.902 T = 37 – 310 = -273 q = 25.0 x 0.902 x –273 = -6156.15 q = -6.2 x 103 J * Negative sign indicates loss of energy

13 Sample Problem 3 Calculate the molar heat capacity (C) of water (J/mol°C). c = 4.18 J/g °C M = 18.02 g/mol C p = c x M = 4.18 J/g °C x 18.02 g/mol C p = 75.3 J/mol °C

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