Spontaneity, Entropy, and Free Energy

A ball rolls spontaneously down a hill but not up….

Spontaneous Processes

A reaction that will occur without outside intervention; product favored

Examples: ice melting at temps above 0°C; NaCl dissolving in water, iron rusting

Non-examples: water doe not boil at 75°C and 1 atm, water does not freeze at 15°C

****A reaction is spontaneous in one direction and non-spontaneous in the opposite direction****

Reversible Reactions

•A reversible process is one that can go back and forth between states along the same path. The reverse process restores the system to its original state.

• The path taken back to the original state is exactly the reverse of the forward process.

• There is no net change in the system or the surroundings when this cycle is complete

Example: At 0º C, water freezing and melting is reversible, but it is irreversible at other temperatures.

Quick Facts:

  • Chemical systems in equilibrium are reversible.
  • Completely reversible processes are too slow to be attained in practice.
  • In any spontaneous process, the path between reactants and products is irreversible.

Why are certain processes Spontaneous???

  • Early idea: exothermicity

HOWEVER

  • Current Conclusion: the characteristic common thread of all spontaneous reactions and the driving force is an increase in entropy

The BEST SCENARIO of a spontaneous reaction: the eaction moves to lower its energy(- ΔH) and increase its entropy (+ ΔS)

Recall the 1st law of Thermodyanics - Energy of the universe is constant: keeps track of how much energy is involved in the change

Entropy- (S)

  • is the measure of disorder or randomness in a system
  • natural tendency to go from order to disorder thus from a lower to higher entropy
  • this is a state function (independent of pathway)

2nd law- The entropy of the universe increases in a spontaneous process

  • Defined in terms of probability
  • Described by the number of arrangements available to a system
  • Substances take the arrangement that is most likely, which is the most RANDOM
  • See overhead for possible arrangements for a system

S solid < S liquid < S gas

  • More ways for molecules to be arranged as a liquid than a solid; gases have even more ways
  • Solutions form b/c there are many more possible arrangements of dissolved pieces than if they stay separate
  • Substances that are more complex(larger) have more entropy( vibrational and rotational motion)
  • In a gas, if the products have a smaller number of molecules, then the entropy has decreased

Examples

Predict whether the entropy of each scenario will increase or decrease and try to explain why that happens.

  1. A deck of playing cards is neatly tucked away in their box when Johnny decides to play 52-card pickup and scatters them all over the family room.
  1. A child’s room several days after you have asked him/her to make it spotless.
  1. A solid (ice) changing into a liquid (water) at 0ºC.
  1. A gas (steam) changing into a liquid (water) at 100º C.
  1. Dissolving a solute into a solvent. Example: Sugar is dissolved into coffee.
  1. Dissolving gas into a liquid. Example: Carbon dioxide gas is dissolved into Coke.
  1. A chemical reaction between gas molecules that results in a net decrease in the overall numbers of gas molecules.
  1. The volume of a container containing gas is increased from 5 liters to 10 liters.

9. For each pair, choose the substance with the higher entropy

a. Solid CO2 and gaseous CO2

b. N2 gas at 1 atm and N2 gas at .001 atm

10. Predict the sign of entropy change for each process

  1. solid sugar is added to water to form a solution
  2. iodine vapor condenses on a cold surface to form crystals
  3. The thermal decomposition of solid calcium carbonate

CaCO3(s)  CaO(s) + CO2(g)

  1. The oxidation of sulfur dioxide

2 SO2(g) + O2(g) 2 SO3(g)

Second Law of Thermodynamics- In detail

ΔSuniverse = ΔSsystem + ΔSsurroundings

  • If ΔSuniverse is positive(increase in entropy), the process is spontaneous(and irreversible) ∆ Suniv = ∆ Ssys + ∆ Ssurr > 0
  • If ΔSuniverse is negative(decrease in entropy), the process is nonspontaneous ( or spontaneous in the opposite direction)∆ Suniv = ∆ Ssys + ∆ Ssurr 0
  • If ΔSuniverse is zero, the process is at equilibrium ∆ Suniv = ∆ Ssys + ∆ Ssurr = 0

Entropy in the surroundings is primarily controlled by enthalpy

Recall: If ΔH is negative, heat is released into surroundings SO

ΔSsurroundings is positive ( more disorder due to faster movement)

If ΔH is positive, heat is absorbed from the surroundings SO

ΔSsurroundings is negative ( less disorder due to slower movement)

BUT WHAT HAPPENS WHEN..

H2O(l) H2O(g)

ΔSsys = positiveΔSsurr = negative

Which one controls it? ΔSuniv will be dependent on temperature

Temperature and Spontaneity

  • An exothermic process is favored because by giving up heat, the entropy of the surroundings increases
  • The sign of ΔSsurr depends on the direction of heat transfer and the magnitude of ΔSsurr depends on temperature : at constant temp/pressure
  • where : T is Kelvin temperature

ΔH is the enthalpy of system (negative if exothermic)

Example 11

Calculate the Ssurr for each reaction at 25 ºC and 1 atm

  1. Sb4O6 + 6C  4 Sb + 6COΔHº = 778 kJ/mol
  2. Sb2S3 + 3Fe  2 Sb + 3 FeSΔHº = -125 kJ/mol

Example 12

N2(g) + 3H2(g)  2NH3(g) ΔHº = -92.6 kJ/mol

What is the ΔSuniv for the reaction if the ΔSsys = -199 J/K·mol

Units: J/K

ΔSsys / ΔSsurr / ΔSuniv / Spontaneous?
+ / + / + / Yes
- / - / - / No, reverse
+ / - / ? / Yes if ΔSsys has a greater magnitude than ΔSsurr(at high temp)
- / + / ? / Yes, if ΔSsurr has a greater magnitude than ΔSsys(at low temp)

*****From now on no subscript next to the delta sign refers to the system*****

Third law of Thermodynamics

  • Entropy of a pure crystal at 0K is zero. All others must be > zero.

SO the standard enthalpies of elements in their standard states are

not zero(unlike standard enthalpies and standard free energies)

  • Standard entropy S˚ at 298 K and 1 atm of substances- see appendix for values
  • It is a state function
  • Can use

Example 13

What is the standard entropy change for the following reaction at 250C?

2CO (g) + O2 (g)  2CO2 (g)

S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol

Example 14

Calculate ºS at 25 C for the reaction 2 NiS(s) + 3 O2(g)  2 SO2(g) + 2 NiO(s)

substanceºS (J/K mol)

SO2(g) 248

NiO(s)38

O2(g) 205

NiS(s) 53

Gibb’s Free Energy

  • ΔSsurr can be hard to measure so we look at another thermodynamic function
  • Can be used to determine if a reaction is spontaneous at constant temp and pressure
  • Defined as the amount of energy available to do work at a given temp and pressure
  • If ΔG is negative, then the reaction is spontaneous in the forward direction
  • If ΔG is positive, then the reaction is not spontaneous, but in the reverse direction
  • If ΔG is zero, the reaction is at equilibrium

Factors Affecting the Sign of ΔG

ΔG° = ΔH° - TΔS°

ΔH / ΔS / Spontaneous in forward direction?
+ / + / Spontaneous at high temp “entropy driven”
(Exothermicity is unimportant)
+ / - / ΔG is always positive;
It will never be spontaneous at any temp; reverse is spontaneous
- / + / ΔG is always negative;
It will always be spontaneous at all temperatures
- / - / Spontaneous at low temp “enthalpy driven”
(Exothermicity is important)

Standard Free Energy- (ΔG˚ ) is the free energy change that occurs when reactants in their standard stated turn into products in their standard states

  • Is a state function
  • Can’t be measured directly but can be calculated from other measurements

3 Ways to Measure

1.

  1. Use Hess’s law with known reactions
  2. Use modified Hess’s Law: see tables in the appendix for standard free energy of formation ΔGf˚
  • The standard free energy of formation for any element in its standard state is zero.

Example 15

What is the standard free-energy change for the following reaction at 25 0C?

2C6H6 (l) + 15O2 (g)  12CO2 (g) + 6H2O (l)

ΔºGf CO2 = -394.4kJΔºGf H2O= -237.2 kJΔºGfC6H6 = 124.5 kJ

Example 16

Using the following data at 25ºC, calculate the ΔGº for the reaction

C (diamond)(s)  C (graphite)(s)

C (diamond)(s) + O 2(g) CO2(g) ΔºG = -397 kJ

C (graphite)(s) + O 2(g) CO2(g) ΔºG = -394 kJ

At Equilibrium, ΔºG = 0

So when you modify the equation, ΔG° = ΔH° - TΔS°

0 = ΔH° - TΔS°

Tequil = ΔH°

ΔS°

This will allow you to calculate the temperature at boiling point, freezing point, etc.

Example 17

At what temperature is the following process spontaneous at 1 atm? What is the

normal boiling point of liquid bromine? Br2(l)  Br2(g)

ΔSº = 93 J/K·mol and ΔHº = 31.0 kJ/mol

Dependence of Free Energy on Pressure (specific to gaseous systems)

  • So far we have seen that at constant temp and pressure, a reaction will be spontaneous in the direction that lowers its free energy

But what if we are not at standard conditions


- where : R = universal gas constant (8.3145J/K∙mol)

T = Kelvin temperature

Q = reaction quotient

This will show if a reaction will be spontaneous at varied pressures…..

Q= reaction quotient (pressure of the gaseous products) Do not look at solilds & liquids in a reaction when calculating this

For the reaction aA + bB  cC + dD

Q = [C]c [D] d

[A] a [B] b

ΔG˚= free energy change of the gas at 1 atm

ΔG = free energy change of gas at a specified pressure, non standard conditions

T= Kelvin temp

R= 8.3145 J/K mol

If ΔG is higher in value than ΔG˚, then it is more spontaneous at higher pressures

Example 18

Would the reaction be spontaneous at 25º C with the hydrogen gas pressure of

3.0 atm and the CO pressure of 5.0 atm? CO(g) + 2H 2(g) CH 3OH(l)

The Relationship between Free Energy & the Equilibrium Constant

ΔG˚= / K =
0 / 1
< 0 / > 0
> 0 / < 0

Example 19

Determine the standard free energy change ΔºG for the formation of 1 mole of ammonia from nitrogen and hydrogen gas and use this value to calculate the equilibrium constant for this reaction at 25ºC. N2(g) + 3H2(g) 2 NH3(g)