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Thermochemist ry

Thermochemistry

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Thermochemistry. When energy is changed from one form to another, all of the energy can be accounted for. This is called the Law of Conservation of Energy. Heat is energy. If heat is released by a chemical system, an equal amount of heat will be absorbed by the surroundings. - PowerPoint PPT Presentation

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Page 1: Thermochemistry

Thermochemistry

Page 7: Thermochemistry

• In an exothermic process, the system gives off heat to the surroundings.

Page 9: Thermochemistry

• In an exothermic reaction, the energy stored in the chemical bonds of the reactants is greater than the energy stored in the bonds of the products.

Page 10: Thermochemistry

• As perspiration evaporates from your skin, your body is cooled.

• With respect to your body, this process is said to be exothermic.

Page 11: Thermochemistry

• A calorie is the quantity of heat that raises the temperature of 1 gram of pure water by 1C.

• 1 Calorie = 4.18 kJ of heat

Page 12: Thermochemistry

• How many kJ of energy can be released by a banana containing 150 Cal?

(1 Calorie = 4.18 kJ)

kJCal

kJCal 630

1

18.4150

Page 13: Thermochemistry

• An object's heat capacity is the amount of heat required to change the temperature of an object by exactly 1°C.

Page 14: Thermochemistry

• The specific heat capacity or specific heat is the heat divided by the mass and temperature change.

• Where c = specific heat Q = heat m = mass and T = change of temperature

• The temperature of a 10. g sample of metal changed from 25°C to 50°C when it absorbed 500. J of heat. What is the specific heat of this sample?

• c = ?

• m = 10. g T = 50°C-25°C = 25°C

• Q = 500. J Tm

Qc

Page 15: Thermochemistry

Cg

J

Cg

J

Tm

Qc

0.2

25.10

.500

Page 16: Thermochemistry

• Determine the specific heat of a material if an 18 g sample absorbed 75 J as it was heated from 15°C to 40.°C.

• c = ?

• m = 18 g T = 40.°C-15°C = 25°C

• Q = 75 J

Page 17: Thermochemistry

Cg

J

Cg

J

Tm

Qc

17.0

2518

75

Page 18: Thermochemistry

• What is the specific heat of a substance if 2000. cal are required to raise the temperature of a 300. g sample by 20.C?

• c = ?

• Q = 2000. cal

• m = 300. g T = 20.C

Page 19: Thermochemistry

Cg

cal

Cg

cal

Tm

Qc

33.0

.20.300

.2000

Page 20: Thermochemistry

• What is the amount of heat required to raise the temperature of 200.0 g of aluminum by 10.C?

• (specific heat of aluminum = 0.21 cal/gC)

• Q = ? • m = 200. g • c = 0.21 cal/gC T = 10.C

Page 21: Thermochemistry

• Q = mcT

• Q = (200. g)(0.21 cal/gC)( 10.C) = 420 cal

Page 22: Thermochemistry

• Water has the highest specific heat of all common substances.

C°g

cal00.1=c

Page 23: Thermochemistry

• The symbol H stands for the heat of reaction for a chemical reaction.

• This is also called the change in enthalpy. H = Q

• The heat content of a system is equal to the enthalpy only for a system that is at constant pressure.

• Calorimetry depends on the law of conservation of energy.

Page 24: Thermochemistry

• If 27.0 mL of water containing HCl is mixed with 28.0 mL of water containing NaOH in a calorimeter such that the initial temperature of each solution was 24.0°C and the final temperature of the mixture is 33.0°C, how much heat (in kJ) is released in the reaction? Assume that the densities of the solutions are 1.00 g/mL.

Page 25: Thermochemistry

H = ?

• m = 27.0 g + 28.0 g = 55.0 g T = 33.0°C - 24.0°C = 9.0°C

• c = 0.00418 kJ/g°C

Page 26: Thermochemistry

kJ1.2C0.9Cg

kJ00418.0g0.55TmcH

Page 27: Thermochemistry

• A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperature of 20.0°C. How many joules of heat were released by the lead?

Page 28: Thermochemistry

H = ?

• m = 40.0 g T = 20.0°C - 17.0°C = 3.0°C• c = 4.18 J/g°C

Page 29: Thermochemistry

J502C0.3Cg

J18.4g0.40TmcH

Page 30: Thermochemistry

• Fusion = solid liquid

• Solidification = liquid solid

• Molar heat of fusion –

–heat required to melt a mole = Hfusion

• Molar heat of solidification –

–heat required to freeze a mole =Hsolid

• Hfusion = -Hsolid

Page 31: Thermochemistry

• Vaporization is liquid vapor

• Condensation is vapor liquid

• Molar heat of vaporization – – heat required to vaporize a mole = Hvap

• Molar heat of condensation – – heat required to condense a mole = Hcond

• Hvap = -Hcond

Page 32: Thermochemistry

• Given the equation

I2(s) + 62.4 kJ I2(g) H = +62.4 kJ

Page 33: Thermochemistry

• How much heat, in kJ, is released when 108 g of water at O°C freezes to ice at O°C if Hsolid

for water = -6.01 kJ/mol?

Page 34: Thermochemistry

kJ1.36OHmol1

kJ01.6

OHg18

OHmol1g108

22

2

Page 35: Thermochemistry

• How much heat is released in the condensation of 27.0 g of steam at 100°C to water at 100°C if

Hcond for water = -40.7 kJ/mol?

Page 36: Thermochemistry

kJ0.61OHmol1

kJ7.40

OHg18

OHmol1g0.27

22

2

Page 37: Thermochemistry

• How many grams of ice at 0°C can be melted into water at 0°C by the addition of 75.0 kJ of heat?

Hfus for water = 6.01 kJ/mol

Page 38: Thermochemistry

g225kJ01.6

OHmol1

OHmol1

OHg18kJ0.75 2

2

2

Page 39: Thermochemistry

• The heat of solution is the amount of heat absorbed or released when a solid dissolves.

• If the molar heat of solution of NaOH is -445.1 kJ/mol, how much heat (in kJ) will be released if 80.00 g of NaOH are dissolved in water?

Page 40: Thermochemistry

kJ2.890NaOHmol1

kJ1.445

NaOHg.40

NaOHmol1NaOHg00.80

Page 41: Thermochemistry

• Hess's law makes it possible to calculate H for complicated chemical reactions.

• Hess’s Law – If you add 2 or more thermochemical equations to give a final equation, then you also ADD the heats of reactions to give the final heat of reaction.

Page 42: Thermochemistry

• Given the equation: • C2H4 (g) + 3O2(g) + 2H2O(l) + 1411 kJ

• How much heat is released when 8.00 g of O2 react?

Page 43: Thermochemistry

kJ118NaOHmol3

kJ1411

Og.32

Omol1Og00.8

2

22

Page 44: Thermochemistry

• What is the heat of reaction (H) for the combustion (with O2) of benzene, C6H6 to form carbon dioxide and water? Write the balanced equation for the reaction.

Page 45: Thermochemistry

• Standard heats of formation:

• C6H6 = +48.50 kJ

• O2 = 0.0 kJ

• CO2 = -393.5 kJ

• H20 = -285.8 kJ

Page 46: Thermochemistry

• 2 C6H6 + 15 O2 12 CO2 + 6 H2O

H + (2) (48.50 kJ) + (15)(0.0 kJ)

= (12) (-393.5 kJ) + (6) (-285.8 kJ)

H = -3266.9 kJ