AP Chemistry Chapter 17

Spontaneity of Reaction

Spontaneous reactions

What does that mean?

Some occur without any “help” Others require some “help” No help – ice cube melting Help – wood burning

• If a reaction is spontaneous under a If a reaction is spontaneous under a certain set of conditions, the reverse certain set of conditions, the reverse reaction must be nonspontaneousreaction must be nonspontaneous

• In any spontaneous change, the amount In any spontaneous change, the amount of free energy available decreases toward of free energy available decreases toward zero as the process proceeds towards zero as the process proceeds towards equilibrium.equilibrium.

Reactions tend to be spontaneous when:

• it leads to lower energy = -it leads to lower energy = -∆H∆H• But not always!!!!But not always!!!!• Also tend to be spontaneous if the reaction Also tend to be spontaneous if the reaction

results in an increase in randomnessresults in an increase in randomness• Entropy SEntropy S• Greater entropy – more random the system is. Greater entropy – more random the system is.

+∆S increase in entropy ∆S>0+∆S increase in entropy ∆S>0• - ∆S decrease in entropy ∆S<0- ∆S decrease in entropy ∆S<0

• Page 448 Example 17.1 Predict sign of Page 448 Example 17.1 Predict sign of ∆S∆S

• In general, nature tends to move In general, nature tends to move spontaneously from more ordered to more spontaneously from more ordered to more random state (less ordered)random state (less ordered)

• Entropy increases in the order:Entropy increases in the order:

• s <l < gs <l < g

• g>l>sg>l>s

• Increasing temperature of a substance Increasing temperature of a substance increases its entropyincreases its entropy

Third Law of Thermodynamics

• A completely ordered pure crystalline A completely ordered pure crystalline solid has an entropy of zero at 0Ksolid has an entropy of zero at 0K

∆S for reactions

• Pg. 450 Table of Standard EntropiesPg. 450 Table of Standard Entropies

• Used to calculate the standard entropy Used to calculate the standard entropy change, change, ∆S∆Soo, for reactions., for reactions.

• ∆∆SSoo = ∑ S = ∑ Soo products – ∑S products – ∑Soo reactants reactants

• Must remember to multiply by the Must remember to multiply by the number of moles from balanced equationnumber of moles from balanced equation

• Note that SNote that Soo is a positive quantity for both is a positive quantity for both compounds and elements; can be negative compounds and elements; can be negative for ions in solutionsfor ions in solutions

• Reactions which SReactions which Soo is positive tend to be is positive tend to be spontaneous, at least at high temperatures.spontaneous, at least at high temperatures.

• HH22OO(s)(s) H H22OO(l)(l) ( ( ∆S > 0)∆S > 0)

• HH22OO(l)(l) H H22OO(g)(g) (∆S > 0) (∆S > 0)

• FeFe22OO3(s)3(s) + 3H + 3H2(g) 2(g) 2Fe 2Fe(s)(s) + 3H + 3H22OO(g)(g) (∆S > 0) (∆S > 0)

• All of these reactions are endothermic (∆H>0)All of these reactions are endothermic (∆H>0)• They become spontaneous at high temperaturesThey become spontaneous at high temperatures

A reaction that results in an increase in the number of moles of gas is accompanied by an increase in entropy.

If the number of moles of gas decreases, ∆S is a negative quantity

Elements have nonzero standard entropies

Standard molar entropies of pure substances are always positive quantities

Aqueous ions may have negative So values

Among substances of similar structure and physical state, entropy usually increases with molar mass

Molecule becomes more complex, more ways for the atoms to move about with respect to one another (higher entropy)

• Pg. 449-451 samplesPg. 449-451 samples

• Example 17.2Example 17.2

Second Law of Thermodynamics

• In a spontaneous process, there is a net increase in entropy, taking into account both system and surroundings.

• ∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0

• spontaneous

Gibbs Free Energy G

• Two quantities affect reaction Two quantities affect reaction spontaneity;spontaneity;

• enthalpy, H and entropy, Senthalpy, H and entropy, S• Put them together in a way that the Put them together in a way that the

signs will give us a cluesigns will give us a clue• G = H – TSG = H – TS• T = kelvin tempT = kelvin temp

• ∆∆G – for a reaction at constant temp and G – for a reaction at constant temp and pressure, represents that portion of the total pressure, represents that portion of the total energy change that is available to do useful energy change that is available to do useful work – is a state propertywork – is a state property

• Depends only on the nature of products and Depends only on the nature of products and reactants and the conditions reactants and the conditions (temp/pressure/concentration), not on the (temp/pressure/concentration), not on the path by which the reaction is carried outpath by which the reaction is carried out

- ∆G = spontaneous+ ∆G = not spontaneous (reverse is

spontaneous

∆G = 0 system is at equilibrium (no tendency for reaction to occur in either direction)

∆G measure of the driving force of a reaction

• Reaction, at constant pressure and Reaction, at constant pressure and temperature, go in such a direction as to temperature, go in such a direction as to decrease the free energy of the systemdecrease the free energy of the system

• Products have lower free energy, reaction Products have lower free energy, reaction will go in that directionwill go in that direction

• Reactants have lower free energy, reaction Reactants have lower free energy, reaction will go in that direction (means the reverse will go in that direction (means the reverse rxn spontaneous)rxn spontaneous)

Gibbs-Helmholtz equation ∆G = ∆H - T∆S To make ∆G negative; Negative value for ∆H (exothermic) Positive value for ∆S (less

ordered)

Gibbs-Helmholtz equation

• Valid under all conditions but we will apply it only under “standard conditions”

• Meaning: gases are at one atmosphere partial pressure

• Ions or molecules in solution are at one molar concentration

• ∆G = standard free energy change

• ∆Go = ∆Ho - T∆So

• now we can use the tables in the book

• If ∆Go is negative = spontaneous at standard conditions

• If ∆Go is positive = nonspontaneous at standard conditions

• ∆G = 0 system is at equilibrium at standard conditions

Calculation of ∆G at 25oCFree Energies of Formation!!

Make sure units are correctUse ∆H is kJ, convert ∆S for J/K to kJ/KPg. 455 Example 17.3Pg. 456 Example 17.4, 17.5Pg. 458 IMPORTANT TABLE!!