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Entropy and Free Energy. How to predict if a reaction can occur, given enough time? THERMODYNAMICS. How to predict if a reaction can occur at a reasonable rate? KINETICS. Thermodynamics.

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Entropy and free energy
Entropy and Free Energy

How to predict if a reaction can occur, given enough time?

THERMODYNAMICS

How to predict if a reaction can occur at a reasonable rate?

KINETICS


Thermodynamics
Thermodynamics

  • If the state of a chemical system is such that a rearrangement of its atoms and molecules would decrease the energy of the system---

  • AND the K is greater than 1,

  • then this is a product-favoredsystem.

  • Most product-favored reactions are exothermic—but this is not the only criterion


Thermodynamics1
Thermodynamics

  • Both product- and reactant-favored reactions can proceed to equilibrium in a spontaneousprocess.

    AgCl(s) e Ag+(aq) + Cl–(aq) K = 1.8 x 10-10

    Reaction is not product-favored, but it moves spontaneously toward equilibrium.

  • Spontaneous does not imply anything about time for reaction to occur.


Thermodynamics and kinetics
Thermodynamics and Kinetics

Diamond is thermodynamically favored to convert to graphite, but not kinetically favored.

Paper burns — a product-favored reaction. Also kinetically favored once reaction is begun.


Spontaneous reactions
Spontaneous Reactions

In general, spontaneous reactions are exothermic.

Fe2O3(s) + 2 Al(s) ---> 2 Fe(s) + Al2O3(s)

∆H = - 848 kJ


Spontaneous reactions1
Spontaneous Reactions

But many spontaneous reactions or processes are endothermic or even have ∆H = 0.

NH4NO3(s) + heat ---> NH4NO3(aq)


Entropy s
Entropy, S

One property common to spontaneous processes is that the final state is more DISORDERED or RANDOM than the original.

Spontaneity is related to an increase in randomness.

The thermodynamic property related to randomness is ENTROPY, S.

Reaction of K with water



Directionality of reactions
Directionality of Reactions solid water (ice) at 0˚ C.

How probable is it that reactant molecules will react?

PROBABILITY suggests that a spontaneous reaction will result in the dispersal

* ofenergy

* or ofmatter

* or of energy & matter.


Directionality of reactions1
Directionality of Reactions solid water (ice) at 0˚ C.

Probability suggests that a spontaneous reaction will result in the dispersal of energy or of matter or both.

Matter Dispersal


Directionality of reactions2
Directionality of Reactions solid water (ice) at 0˚ C.

Probability suggests that a spontaneous reaction will result in the dispersal of energy or of matter or both.

Energy Dispersal


Directionality of reactions energy dispersal
Directionality of Reactions solid water (ice) at 0˚ C.Energy Dispersal

Exothermic reactions involve a release of stored chemical potential energy to the surroundings.

The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules.

The final state—with energy dispersed—is more probable and makes a reaction spontaneous.


Entropy s1
Entropy, S solid water (ice) at 0˚ C.

So (J/K•mol)

H2O(liq) 69.95

H2O(gas) 188.8

S (gases) > S (liquids) > S (solids)


Entropy and states of matter
Entropy and States of Matter solid water (ice) at 0˚ C.

S˚(Br2 liq) < S˚(Br2 gas)

S˚(H2O sol) < S˚(H2O liq)


Entropy s2
Entropy, S solid water (ice) at 0˚ C.

Entropy of a substance increases with temperature.

Molecular motions of heptane at different temps.

Molecular motions of heptane, C7H16


Entropy s3
Entropy, S solid water (ice) at 0˚ C.

Increase in molecular complexity generally leads to increase in S.


Entropy s4
Entropy, S solid water (ice) at 0˚ C.

Entropies of ionic solids depend on coulombic attractions.

So (J/K•mol)

MgO 26.9

NaF 51.5

Mg2+ & O2-

Na+ & F-


Entropy s5
Entropy, S solid water (ice) at 0˚ C.

Entropy usually increases when a pure liquid or solid dissolves in a solvent.


Processes that increase entropy
Processes that Increase Entropy solid water (ice) at 0˚ C.

  • Changing from a more ordered state to a less ordered state.

    • Solidliquidgas

  • Decomposing (changing from one reactant to many products)

  • Going from a small molecule to a large and complex one (more freedom for atoms to move, mostly important in organic compounds)

  • When a pure solid or liquid dissolves in a solvent.

  • When a gas molecule escapes from a solvent.


Standard molar entropies
Standard Molar Entropies solid water (ice) at 0˚ C.


Entropy changes for phase changes
Entropy Changes for Phase Changes solid water (ice) at 0˚ C.

For a phase change, ∆S = q/T

where q = heat transferred in phase change

For H2O (liq) ---> H2O(g)

∆H = q = +40,700 J/mol


Entropy and temperature

S increases slightly with T solid water (ice) at 0˚ C.

S increases a large amount with phase changes

Entropy and Temperature


Calculating s for a reaction
Calculating ∆S for a Reaction solid water (ice) at 0˚ C.

∆So =  So (products) -  So (reactants)

Consider 2 H2(g) + O2(g) ---> 2 H2O(liq)

∆So = 2 So (H2O) - [2 So (H2) + So (O2)]

∆So = 2 mol (69.9 J/K•mol) - [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)]

∆So = -326.9 J/K

Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid.


2nd law of thermodynamics
2nd Law of Thermodynamics solid water (ice) at 0˚ C.

A reaction is spontaneous if ∆S for the universe is positive.

∆Suniverse = ∆Ssystem + ∆Ssurroundings

∆Suniverse > 0 for spontaneous process

First calc. entropy created by matter dispersal (∆Ssystem)

Next, calc. entropy created by energy dispersal (∆Ssurround)


2nd Law of Thermodynamics solid water (ice) at 0˚ C.

Dissolving NH4NO3 in water—an entropy driven process.

∆Suniverse =

∆Ssystem + ∆Ssurroundings


2nd Law of Thermodynamics solid water (ice) at 0˚ C.

2 H2(g) + O2(g)  2 H2O(liq)

∆Sosystem = -326.9 J/K

Can calc. that ∆Horxn = ∆Hosystem = -571.7 kJ

∆Sosurroundings = +1917 J/K


2 H solid water (ice) at 0˚ C.2(g) + O2(g) ---> 2 H2O(liq)

∆Sosystem = -326.9 J/K

∆Sosurroundings = +1917 J/K

∆Souniverse = +1590. J/K

The entropy of the universe is increasing, so the reaction is product-favored.

2nd Law of Thermodynamics


Spontaneous or not
Spontaneous or Not? solid water (ice) at 0˚ C.

Remember that –∆H˚sys is proportional to ∆S˚surr

An exothermic process has ∆S˚surr > 0.


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