Topic: Thermodynamics

Do Now: packet p.1

Every physical or chemical change is accompanied by

energy change

Energy released = _________________

Energy absorbed = _______________

exothermic

endothermic

G

L

S

• Thermodynamics is the study of entropy and enthalpy changes that accompany chemical reactions

Thermodynamics Tells us if a reaction will occur

The total amount of energy a substance contains depends on many factors, some of which are not totally understood…it’s impossible to know the total heat content of a substance.

So scientists measure ΔH

H = enthalpy = the heat content of a system at constant pressure. It describes chemical PE stored in matter

ΔH = enthalpy (heat) of reactionΔH = Hproducts – Hreactants

--the difference between the enthalpy(heat) of the substances that exist at the end of the reaction and the enthalpy(heat) of the substances that exist at the beginning of the reaction

Thermochemical Equations

• balanced chemical equation• shows physical state of all reactants &

products

• gives energy change – It can be written 2 ways

• energy term can be written as reactant or product

OR• H is given right after equation

4Fe(s) + 3O2(g) 2Fe2O3(s) + 1625 kJ

OR

4Fe(s) + 3O2(g) 2Fe2O3(s)

H = -1625 kJ

NH4NO3(s) + 27 kJ NH4+(aq) + NO3

-(aq)

OR

NH4NO3(s) NH4+(aq) + NO3

-(aq)

H = 27 kJ

If ΔH is negative…- ΔH = exothermic

- Hproducts < Hreactants

- PE of Products < PE of Reactant- Example: 4Fe(s) + 3O2(g) 2Fe2O3(s) + 1625 kJ

- ΔH = -1625kJ

If ΔH is positive…+ ΔH = endothermic

- Hproducts > Hreactants

- PE of Products > PE of Reactant

Example: NH4NO3(s) + 27 kJ NH4+(aq) + NO3

-(aq)- ΔH = +27kJ

Universe

Environment

System A

B

Since energy is conversed…the system changes in one direction and the surrounding have to change in the opposite direction

A. Reaction is Exothermic, environment gets _________________

B. Reaction is Endothermic, environment gets _________________

warmer

colder

For any reaction occurring at constant pressure ΔH = Q

• Q = mCT– Q = Energy change

– m = mass of water

– C = specific heat of water

– T = temperature change = Tf – Ti

Since there are different types of reactions, you have various ΔH’s

• ΔHcomb = enthalpy (heat) of combustion – the enthalpy change for the complete bunring of one mole of the substance

• ΔHformation = enthalpy (heat) of formation – the enthalpy change for the formation of a compound from its constituent elements

• ΔHsolution = enthalpy (heat) of solution – the enthalpy change when 1 mole of an ionic substance is dissolved in water.

Look at Table I: Heats of ReactionLets label the various types from above

• Rxns 1-6: combustion rxns

-H = heat of combustion

• Rxns 7-18: formation (synthesis) rxns

– Substance is formed from its elements

– H = heat of formation

• Rxns 19-24: dissolving equations

– H = heat of solution

Table I

Many other processes other than chemical reactions absorb or release energy

like, Changes of state

Hvaporization = molar heat of vaporization

= amount of heat required to vaporize one mole of a liquid

Hfusion = molar heat of fusion

= amount of heat required to melt one moles of a solid

Energy depends on amount

• Remember – it takes more energy to heat up water in the ocean than to make a cup of tea

CH4(g) + 2O2(g) CO2(g) + 2H2O (l) H = -890.4 kJ

1 mole of methane + 2 mole of oxygen →

1 mole of carbon dioxide gas & 2 moles of liquid water

reaction is ____________ (negative sign for ΔH)

890.4 kJ energy released per mole of CH4(g) burned

exothermic

What would happen if we had 2 moles of methane?

Twice as much energy would be released

2 x 890.4 kJ = 1780.8 kJ will be released

Reactions: Energy depends on direction too!

• N2(g) + 3H2(g) 2NH3(g) H = -91.8 kJ

• 2NH3(g) N2(g) + 3H2(g) H = _______

If reverse equation, reverse sign of H

91.8 kJ

If we can’t calc. H for one individual substance how are we

able to calc. ΔH???!

FYI: Hess’s Law(not on regents)

Can add 2 or more equations by adding the H’s

Enables you to calculate H for # of rxns

Say you’re interested in

2S(s) + 3O2(g) 2SO3(g)

2S(s) + 3O2(g) 2SO3(g)Have H’s for the following:

a) S(s) + O2(g) SO2(g) H = -297 kJ

b) 2SO3(g) 2SO2(g) + O2(g) H = 198 kJ

2 2 2

2SO2(g) + O2(g) 2SO3(g) H = -198 kJ

H = -594KJ + (- 198kJ) H = -792 kJ

x (2)