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Thermochemistry

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Thermochemistry

Chemical reactions are accompanied by changes in energy.

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Definitions:

Energy: The ability to do work, w = f x d

K.E. =1/2mv2 energy in motion

What are some forms of kinetic energy?

P.E. potential energy

What are some forms of potential energy?

Chemical

Positional

Mechanical, electrical, thermal, sound, etc (Motion)

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Energy Units:

Joule = J 1 J = kg ·m/s2

Since a J is very small we usually use kJ’s

calorie = cal 1 cal = 4.184 J (exact by definition)

1 cal is the amount of heat necessary to heat 1 g of water 1oC (from 14.5 to 15.5oC)

1 cal 4.184 J

A conversion factor!

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Food Calories:

1 Cal (note capital for food) = 1000 cal = 1 kcal

So... That donut with only 500 Calories is really....500,000 calories... (eat 2 to make it an even million).

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1st Law of Thermodynamics:

Energy cannot be created nor destroyed.

System SurroundingsSystem: whatever we are interested in.

Surroundings: the rest of the universe

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Internal Energy (E): The sum of all K.E. and P.E. of a system.

Internal energy is too complex to be measured exactly,so we look at the change in internal energy:

Esystem = Efinal - Einitial

Sign of E: (+) means: system has gained energy (-) means: system has lost energy

From the 1st Law of Thermodynamics: Esystem = -Esurroundings

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Sign of E: (+) means: system has gained energy (-) means: system has lost energy

2H2O(g) 2H2(g) + O2(g)2H2(g) + O2(g) 2H2O(g)

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Sign of E: (+) means: system has gained energy (-) means: system has lost energy

2H2O(g) 2H2(g) + O2(g)2H2(g) + O2(g) 2H2O(g) + energy(given off)

+ energy(added to system)

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Heat and Work: energy exchange

E = q + w

Work done on (+) or by (-) the sys. Heat gained (+) or lost (-) by the sys.

Internal energy of the sys. (both K.E. and P.E.)

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Reactions are Energy Driven:

Energy given off Exothermic (-) sign “Exit”

Energy absorbed Endothermic (+) sign “Enter”

Remember: 1st Law of Thermo.:

Esystem = -Esurroundings

So for every endothermic process there is an exothermic process.

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Decide if each of the following are Endo or Exothermicprocesses:

Ice melting?

Water freezing?

A candle burning?

Dynamite exploding?

A plant growing?

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Misconception: Energy vs Temperature

Both containersare at the same temperature, whichone contains the mostenergy?

Temperature is an intensive property (amount independent)Energy is an extensive property (amount dependent)

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Enthalpy (H): heat content per amount

Extensive property

Enthalpy Change (H)

In a reaction: H = HProducts - Hreactants

H (+) endoH (-) exo

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Enthalpy Change (H)

In a reaction: H = HProducts - Hreactants

H (+) endoH (-) exo

Enthalpy is a state function.

“Net Change” “Path indepedent”

A

B

Both accomplished the same net

result!

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Problem:Given the reaction, which occurs in the presence ofthe catalyst MnO2):

2KClO3(s) 2KCl(s) + 3O2(g) H = -89.7 kJ

a. What would be the value of H for the reverse RXN?

+89.7 kJ

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Problem:Given the reaction (which occurs in the presence ofthe catalyst MnO2):

2KClO3(s) 2KCl(s) + 3O2(g) H = -89.7 kJ

The H value listed is really a conversion factor!

-89.7 kJ -89.7 kJ -89.7 kJ 2 mol KClO3 2 mol KCl 3 mol O2

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Problem:Given the MnO2 catalyzed reaction:

2KClO3(s) 2KCl(s) + 3O2(g) H = -89.7 kJ

-89.7 kJ -89.7 kJ -89.7 kJ 2 mol KClO3 2 mol KCl 3 mol O2

Calculate the value of H for the decomposition 9 moles of KClO3 ?

2KClO3(s) 2KCl(s) + 3O2(s) H = -89.7 kJ

9 moles KClO3 -------------------------------------------------- = kJ-403.73KClO 2mol

89.7kJ

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Problem:Given the MnO2 catalyzed reaction:

2KClO3(s) 2KCl(s) + 3O2(g) H = -89.7 kJ

-89.7 kJ -89.7 kJ -89.7 kJ 2 mol KClO3 2 mol KCl 3 mol O2

Calculate the value of H for the formation 25.0 grams of KCl .

2KClO3(s) 2KCl(s) + 3O2(s) H = -89.7 kJ

25.0 g KCl -------------------------------------------------- = kJ2 mol KCl-89.7 kJ -15.0

74.6 g KClmol KCl

23Look at overhead

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CS2(l) + 3O2(g) CO2(g) + 2SO2(g)

From tables: 87.9kJ/mol 0 kJ/mol -393.5 kJ/mol -296.8 kJ/mol

H = H(products) - H(reactants)

Each value must bemultiplied by the numberof moles in the equation/

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CS2(l) + 3O2(g) CO2(g) + 2SO2(g)

From tables: 87.9kJ/mol 0 kJ/mol -393.5 kJ/mol -296.8k J/mol

H = H(products) - H(reactants)

H = H(products) - H(reactants) = [-393.5 + 2(-296.8)] - [87.9 + 3(0)]H = -1073.4 kJ

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