Chemical Thermodynamics

Chapter 19

Chemical Reactions: Why ?

• Why are some reactions reversible and others are not?

• HC2H3O2 <--------> C2H3O2 + H

What does it mean that the Ka = 1.8 x 105 ?

• Not all reactions that you can write on paper will occur. CO2 (g) ---> C (s) + O2 (g) does not, but the reverse does.

• What determines if a reaction will occur?

• Most chemical reactions are exothermic. Why?

• Thermodynamics answers these questions.

Spontaneous Processes

• Certain processes will always occur.

ice melts on a table at 20 C

an iron nail left outside will rust

• Spontaneous Processes occur without outside intervention

• May be fast or slow (combustion vs. conversion of coal to a diamond)

• A spontaneous process has a definite direction, while energy is conserved in both forward and reverse directions.

• Temperature plays a role

• Exothermic processes tend to occur spontaneously.

• Spontaneous processes produce “Free Energy”.

Reversible vs. Irreversible Processes

• A reversible process:

the system can be restored to it’s exact state by simply reversing the process which led to the change

+Q

Example: ice ----> liquid water

Q ice <---- liquid water

Reversible vs. Irreversible Processes

• An irreversible process: the system cannot be restored by simply reversing the procedure. It must be restored to its original state by a

new pathway which changes the surroundings.

Example: Freezing water at 30C, it takes work to keep it from freezing between 0 and 30 C.

Reversible vs. Irreversible Processes

• When a chemical system is in equilibrium, reactants and products can move back and forth reversibly.

• For any spontaneous process, the path between reactants and products is irreversible.

Keep in Mind . . .• a spontaneous reaction always has a certain

direction, while energy is conserved in both forward and reverse directions.

• Some spontaneous reactions are VERY slow.

First Law of Thermodynamics

• Energy is neither created or destroyed• The energy of the universe is constant• Energy can be transferred between the

system and the surroundings

E = q + w• E = energy absorbed by the system• q = heat absorbed by system• w = work don on the system by the

surroundings

Second Law of Thermodynamics• The Entropy of the universe is

increasing.

• Entropy, S, is disorder or randomness.

• The more disorder, the more randomness

• Units: J/K

• ∆S = Sf Si

• +S = final state more disorderedS = initial state more disordered

Entropy (S)• What would drive spontaneous forces? An

increase or decrease in disorder?

• An increase in spontaneous forces

• Unlike energy, entropy is NOT conserved.

• The Suniv is continually increasing.

• No process that produces order within the system can proceed without producing an equal or greater disorder in its surroundings.

• Entropy is defined in terms of probability.• Substances take the arrangement that is most likely.• The most likely is the most random

• 1 molecule

• 2 possible arrangements

• 50 % chance of finding the left empty

Calculate the number of arrangements for a system.

2 molecules

4 possible arrangements

25% chance of finding the left empty

50 % chance of them being evenly dispersed

4 molecules12 possible

arrangements8% chance of

finding the left empty

50 % chance of them being evenly dispersed

Substances take

the most random

arrangement

Entropy• Which State of matter will have the most

entropy? The least? Explain.

•Ssolid < Sliquid <<< Sgas• there are many more ways for gas molecules

to be arranged than as a liquid than a solid.

• Gases have a huge number of positions possible.

• Solids are very orderly.

Entropy increases when:

• Temperature increases.• Gases are formed from solids or liquids.• The number of molecules of gas

increases during a reaction.• Solutions are formed from solids or

liquids.

Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate.

Entropy

Suniv = Ssys + Ssurr

• Reversible: Suniv = Ssys + Ssurr = 0

• Irreversible: Suniv = Ssys + Ssurr >0

EntropySuniv = Ssys + Ssurr

• IfSuniv is positive the process is spontaneous.

• IfSuniv is negative the process is spontaneous in the opposite direction.

• For exothermic processes: Ssurr is +. • For endothermic processes: Ssurr is .• Consider this process: H2O(l)H2O(g)

Ssys is positive

Ssurr is negative

Suniv depends on temperature.

Temperature and Spontaneity• Entropy changes in the surroundings are

determined by the heat flow.• An exothermic process is favored because

by giving up heat the entropy of the surroundings increases.

• The size of Ssurr depends on temperature

Ssurr = -H/T

Ssys Ssurr Suniv Spontaneous?

+ ++

+

+

- - -

-

-

?

?

Yes

No, Reverse

At High Temp

At Low Temp

Suniv = Ssys + Ssurr

Entropy Changes in Chemical Reactions

• Standard Molar Entropies, So: the molar entropy values of substances in their standard states.

• Where standard enthalpies of elements are 0, standard entropies are not 0.

• What state would have the highest standard molar entropies? The lowest?

• Gases = Highest, Solids = Lowest

• So increase or decrease with molar mass?

• Increases

• So increase or decrease with increasing number of molecules involved in the formula?

• Increases

Third Law of Thermodynamics

• 3rd Law- The entropy of a pure crystal at absolute zero ( 0 K) is 0.

• Gives us a starting point. All others must be > 0.

Entropy Changes in Chemical ReactionsSo = nSo (products) – mSo (reactants)• = “sum of”• n and m are the coefficients in the chemical equation.• Calculate the So for the synthesis of ammonia from

N2 (g) and H2 (g) at 298 K

N2 (g) + 3H2 (g) 2NH3 (g)

• So for N2 = 191.6 J/mol K

• So for H2 = 130.7 J/mol K

• So for NH3 = 192.5 J/mol K

• So = 2 (192.5) – [(191.6) + 3(130.7)• So = -198.3 J/K … spontaneous or not?

Gibb's Free Energy• Free Energy= energy available to do work,

(however, much is in the form of heat)

• All spontaneous processes produce free energy.

• Whether a process is spontaneous or not depends on 2 thermodynamic concepts:

1. Enthalpy, ∆H and 2. Entropy, ∆S

Gibbs Free Energy• Formula: G = H-TS• Never used this way. But rather as:• G = H -TS at constant temperature

(use Kelvin!)• Divide by -T -G/T = -H/T-S -G/T = Ssurr + S

-G/T = Suniv

• If G is negative at constant T and P, the process is spontaneous.

• If ∆G is positive, the process is nonspontaneous.• If ∆G is 0, the process is at equilibrium.

G = H -TS

HS Spontaneous?

+ - At all Temperatures

+ + At high temperatures,

“entropy driven”

- - At low temperatures,

“enthalpy driven”

+- Not at any temperature,

Reverse is spontaneous

∆G-

+

?

?

Is this process spontaneous?• For the reaction H2O(s) H2O(l)Sº = 22.1 J/K mol Hº =6030 J/mol• Calculate G at 10ºC and -10ºCG=H-TS∆G = 6030 J/mol - (283K)(22.1J/K mol)∆G = 6030 - 6254.3∆G = -224.3 yes, at 10C

G= 6030 J/mol - (263K)(22.1J/K mol)∆G = 6030 - 5812.3∆G = +217.7 no, at -10C

Free Energy and Work

• Free energy is that energy free to do work.

• The maximum amount of work possible at a given temperature and pressure.

• Never really achieved because some of the free energy is changed to heat during a change, so it can’t all be used to do work.

In Summary

• If ∆G = , reaction is spontaneous• If ∆G = +, reaction is Non-spontaneous;

work MUST be done to reverse

the reaction• If ∆G = 0, reaction is at equilibrium• If ∆G = very large (-) #, then a lot of energy

is released to surr. and is capable of doing much work. (burning of gasoline)

Entropy in Reactions

• Enthalpy, ∆H, is measured experimentally

• No easy way to measure Entropy, ∆S

• Use absolute Entropy, S, values from tables in back of textbook

• These values are based on difficult calculations using absolute zero data.

Free Energy in Reactions

• Gº = standard free energy change.

• Gº = Hº - TSº

• Can’t be measured directly, must be calculated from other measurements.

• Free energy change that will occur if reactants in their standard state turn to products in their standard state.

• ∆G = ∑ n Gf products ∑ m Gf reactants

Free energy in ReactionsGo = Ho –TSo

• Example:• The Haber process for the production of

ammonia involves the following equillibrium: N2 + 3H2 2NH3

• Assume Ho and So don’t change with temperature.

1. Predict the direction in which Go changes for this reaction with increasing temperature.

• Expect So to be (–) because number of moles of gas is smaller on product side, thus –TS term will be positive and get bigger as temperature increase.

• Calculate the values of Go for the reaction at 25oC and 500oC.

• First we need to Calculate the Ho and So.

• Ho values = N2 (0); H2 (0); NH3 (-46.19 kJ/mol)

• So values = N2 (191.6 J/mol K); H2 (130.7 J/mol K); NH3 (192.5 J/mol K)

• At 298K, Go = -33.3 kJ

• At 773K, Go = 61 kJ

Free Energy and Equilibrium• For a system at equilibrium, ∆G = 0.• we can find the ∆G for any conditions, not

only ∆G, with this formula:G = Gº +RT ln (Q) where Q is the reaction quotients

(pressure of the products / pressure of the reactants), and R is the ideal gas law constant, 8.31 J/mol K

• At equilibrium, Q = KeqG = Gº +RT ln (Q) 0 = Gº + RT ln (Keq)

Gº = RT ln (Keq)

Free Energy and Equilibrium

Gº = RT ln (Keq)

• If ∆Gis negative, Keq >1

• If ∆Gis zero, Keq =1

• If ∆Gis positive, Keq <1

How far will an equilibrium reaction continue to go?

• During the course of a reaction, the free energy always decreases.

• G tells us spontaneity at current conditions. When will it stop?

• It will go to the lowest possible free energy which may be an equilibrium.

• Analogy of boulder rolling down a hill

Equilibrium/Free Energy analogy

Free Energy

Course of ReactionPosition

PEEquilibrium position in valley

Reactants

Products

Equilibrium Position = Minimum free energy available to the system

Mixture at Equilibrium

Free Energy and Work

• “Free Energy” is the energy “free” to do “work”

• The amount of work obtained is always less than the max.

Henry Bent’s First two Laws of Thermodynamics:

1. “You can’t win, you can only break even”

2. “ You can’t even break even”

• The amount of free energy of our system (Earth) is decreasing