Thermodynamics of Reactions

Spontaneity, Entropy, and Free Energy

Chapter 16

1st Law of Thermodynamics

• The energy of the universe is constant

• (conservation of energy)

Spontaneous Processes

• Can be fast OR slow. • Occurs without outside

intervention

True, or Not True?

• Ball rolls downhill spontaneously.• Ball rolls uphill spontaneously.• Wood burns spontaneously in O2.• CO2 and H2O spontaneously form

wood.• Steel rusts spontaneously.• Rust spontaneously turns into iron

and water.

Its All About Entropy

• Entropy is…• All about the ability of energy to spread

out• Related to probability.• Over-simplified to be about disorder.

Positional Entropy

• When a substance has more opportunities to exist, (ie, particles can have more than one arrangement) it is also said to have more positional entropy.

• Increasing moles of gas• Increasing volume• Decreasing pressure• Changing state of matter ( s l g )• Exothermic rxns **

Try Me Conceptual

• Determine which of the following pairs has the most positional entropy:

− 1 mol H2 at STP or 1 Mol H2 at 100oC, 0.5 atm

− 1 mol N2 at STP or 1 mol N2 at 100 K 2atm

− 1 mol H2O(s) at 0oC or 1 mole H2O(l) at 20oC

2nd Law of Thermodynamics

• Spontaneous processes increase the entropy of the universe.

Suniverse = Ssystem + Ssurroundings

The Value of S

• Sign is determined by enthalpy:− exothermic− endothermic

• Magnitude is determined by temperature

T

HS gssurroundin

Try Me Calculation

• The melting point of tungsten (W) is the second highest among the elements, at 3680 K. The enthalpy of fusion for this metal is 35.2 kJ/mol. What is the entropy of fusion?

3rd Law of Thermodynamics

The entropy of a perfect crystal at 0 Kelvin is zero.

In PhaseOut of Phase

Angle BendingRotation

Translation

Entropy for a chemical reaction

Sorxn = nSo

p - nSor

Try Me

• Calculate the change in entropy at 25oC for the reaction:

2 NiS + 3 O2 2 SO2 + 2 NiO

Given Entropy Values:SO2 = 248 J/KmolNiO = 38 J/KmolO2 = 205 J/KmolNiS = 53 J/Kmol

Free Energy

• Gibbs Free Energy- another method for determining spontaneity.

• Also indicates the amount of available energy that is capable of doing work.

• As an energy source is used, the energy is not destroyed, only converted to a non-usable form.

G = H - TS

• Defines Gibbs energy in terms of enthalpy and entropy.

• All three factors will contribute to reaction spontaneity.

• When G is negative, the reaction is spontaneous.

Rearrange that formula

• Lets start with the following 3 formulae:

G = H –TSSsurr = -H/T

Suniv = Ssurr + Ssyst

Suniverse = - G /T

Fancy Pants Charts

HH SS GG Spontaneous?

__ + __ Always

__ __ +/- When temp is low

+ + +/- When temp is high

+ __ + Never

Free Energy in Chemical Reactions

• Standard Free energy is used so that we can compare the relative tendency to occur.

• Go = Ho – TSo

2SO2(g) + O2(g) 2SO3(g)

• The above rxn. Occurs at 25oC and 1 atm. Calculate Ho, So, and Go using the following data:

substance Hof So

kJ/mol J/Kmol

SO2(g) -297 248

SO3(g) -396 257

O2(g) 0 205

It also works like Hess’s Law problems:

What is the Go for the reaction if the mechanism is:

Cdi(s) + O2(g) CO2(g) Go = -397kJ

Cgr(s) + O2(g) CO2(g) Go = -394 kJ

Dependence on Pressure

• Enthalpy does not depend on pressure.

• Entropy does depend on temperature.

S low pressure > S high pressure

G = Go + RT ln(P)

• This can be adapted to reflect partial pressures for the reaction it describes. When that occurs, the formula can be re-written:

Go = -RT ln(Q)

@ Equilibrium

• K = Q• The free energy is the lowest

possible it will ever be for the system.

Try Me Out

• The overall rxn for rusting iron by oxygen is

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

@25oC, find the equilibrium constant given:

Substance Ho So

Fe2O3(s) -826 90Fe(s) 0 27O2(g) 0 205

Free Energy and Work

• Achieving the maximum amount of work from a process is highly unlikely because of transfers of energy.

• G represents the maximum possible quantity of work a system is capable of doing.