chemical thermodynamics chapter 19 (except 19.7!)

Chemical Thermodynamics

Chapter 19

(except 19.7!)

THERMODYNAMICS

The BIG question!!!

How do we know whether a reaction will happen or not?

THERMODYNAMICSSample Problem 1Which of the following processes are spontaneous (happen naturally)?

(a) a ball rolling up a hill (b) the freezing of water at -3oC and 1 atm pressure(c) the freezing of water at 3oC and 1 atm pessure

(d) the decomposition of carbon dioxide into carbon and oxygen at 25oC and 1 atm pressure(e) the dissolving of NaCl in water

(a) not spontaneous(b) spontaneous(c) not spontaneous(d) not spontaneous(e) spontaneous

Processes that occur spontaneously in one direction are NOT spontaneous in the opposite direction.

THERMODYNAMICS• The two driving forces for reactions are

(1) the release of energy (gain of stability)

(2) increasing entropy• In nature, reactions move toward

lowest energy and highest entropy

THERMODYNAMICS• Sample Problem 2

At 25oC, the ΔH value for the reaction 2 H2(g) + O2(g) 2 H2O(l) is - 572 kJ Can you conclude from the sign of ΔH that the reaction is spontaneous?

• NO! Many exothermic reactions are spontaneous, but not all!

THERMODYNAMICS• Entropy is a thermodynamic state function

represented by S.• Entropy is a measure of the number of ways

energy is distributed in a system. The more ways energy is distributed in a system, the more random or the more disordered the system is said to be. The greater the entropy of a system, the greater is its randomness or disorder.

• Entropies are usually positive for both elements and compounds.

• The units of entropy are J/mole•K

THERMODYNAMICSWe usually look at the change in entropy during a process:• ΔS = Sfinal - Sinitial (state function)

• ΔS > 0 (is positive) if the system changes to a state of greater disorder.

• ΔS < 0 (is negative) if the system changes to a state of greater order

THERMODYNAMICS2nd Law of Thermodynamics:• In any spontaneous process, the entropy of

the universe always increases.

ΔSuniverse = ΔSsystem + ΔSsurroundings > 0

• A process that involves an increase in order (ΔSsystem < 0) CAN be spontaneous provided it causes an even greater positive entropy changes in the surroundings.

• An isolated system (no exchange of energy between the system and surroundings) always increases its entropy when it undergoes a spontaneous change. For such a change, ΔSsurroundings = 0

THERMODYNAMICS• Standard entropies (S°) have been

determined for many substances, and these values are based on the 3rd law of thermodynamics.

• 3rd Law of Thermodynamics

The entropy of a perfect, pure crystalline substance at 0 K is zero.

THERMODYNAMICSQualitative "Rules" About Entropy:1) Entropy increases as one goes from a solid to a liquid,

or even more dramatically, a liquid to a gas2) Entropy increases as the number of moles of gas

increases3) The Entropy of any material increases with increasing

temperature4) Entropy increases if a solid or liquid is dissolved in a

solvent.5) Entropy is higher for weakly bonded compounds than

for compounds with very strong covalent bonds6) Entropy increases as the number of particles in a

system increases7) Entropy increases as the number of electrons increases:

Entropy increases as the mass, # of atoms, # of heavier atoms, etc. of a molecule increases

THERMODYNAMICS

Entropy changes dramatically at a phase change.

THERMODYNAMICSSample Problem 3• For each of the following pairs, which

has the greater entropy: (a) CO2(s) or CO2(g) (b) NH3(l) or NH3(g) (c) a crystal of pure Mg at 0 K or a crystal at 200 K

(a) CO2(g)

THERMODYNAMICS• Sample Problem 3• For each of the following pairs, which


(a) CO2(g)

(b) NH3(g)

THERMODYNAMICS• Sample Problem 3• For each of the following pairs, which


(a) CO2(g)

(b) NH3(g)

(c) crystal at 200 K

THERMODYNAMICS• Sample Problem 4• Predict whether ΔS is > 0 or < 0 for

each of the following: (a) sugar + water sugar dissolved in water (b) Na2SO4(s) 2 Na+(aq) + SO4

-2(aq) (c) 2 H2(g) + O2(g) 2 H2O(g)



-2(aq) (c) 2 H2(g) + O2(g) 2 H2O(g)

(a) > 0 (positive)



-2(aq) (c) 2 H2(g) + O2(g) 2 H2O(g)

(a) > 0 (positive)

(b) > 0 (positive)



-2(aq) (c) 2 H2(g) + O2(g) 2 H2O(g)

(a) > 0 (positive)

(b) > 0 (positive)

(c) < 0 (negative)

THERMODYNAMICS• Entropy is a state function and can be

calculated for a reaction:• ΔS°rxn = Σ n S°products - Σ m S°reactants

• values are tabulated in appendix C in your textbook

On AP Equation Sheet!

THERMODYNAMICS• Sample Problem 5

Calculate ΔS°rxn for the following at 25oC 2 SO2(g) + O2(g) 2 SO3(g) standard entropy values at 25oC (from book)SO2(g) 248.5 J/K mole O2(g)

205.0 J/K mole SO3(g)256.2 J/K mole

• ΔS°rxn = Σ n S°products - Σ m S°reactants

• = 2 S° SO3 – (2 S° SO2 + S° O2)• = 2(256.2) – [2(248.5) + 205.0]• = - 189.6 J/K• negative sign indicates a decrease in entropy• (decrease in # of gas molecules also indicates a

decrease in entropy)

THERMODYNAMICS• Exothermic processes (-ΔH) are more

likely to be spontaneous than endothermic processes (+ΔH).

• Processes that involve an increase in entropy (ΔS > 0) are more probable than those showing a decrease in entropy (ΔS < 0).

THERMODYNAMICS• Enthalpy and entropy are connected together

by the thermodynamic state function called free energy (G). The change in free energy for any process or reaction at constant temperature and pressure is

• ΔG = ΔH - T ΔS• if ΔG < 0 (is negative), a reaction

spontaneously proceeds in the forward direction

• if ΔG = 0, a system is at equilibrium• if ΔG > 0 (is positive), a reaction is not

spontaneous as written, but the reverse direction is spontaneous


THERMODYNAMICS• ΔG is a state function and tabulated

values exist for standard free energy of formation for substances - in appendix C

• ΔG°f for any element in its most stable state is zero.

• ΔG°rxn = Σ n ΔG°f (products) - Σ m ΔG°f (reactants)


THERMODYNAMICS• Sample Problem 6• Given the ΔG value for the following phase

changes at 1 atm, predict whether each change is spontaneous:

• (a) at 283 K, ΔG = -250 kJ/mole for H2O(s) H2O(l)• (b) at 273 K, ΔG = 0 kJ/mole for H2O(s) H2O(l)• (c) at 263 K, ΔG = 210 kJ/mole for H2O(s) H2O(l)




(a) spontaneous since ΔG < 0




(a) spontaneous since ΔG < 0(b) at equilibrium – no net reaction




(a) spontaneous since ΔG < 0(b) at equilibrium – no net reaction(c) non spontaneous since ΔG > 0

THERMODYNAMICS• Sample Problem 7 Given:

ΔG°f for C6H12O6(s) = -907.9 kJ/mole, ΔG°f for CO2 (g) = -394.6 kJ/mole ΔG°f for H2O(l) = -237.2 kJ/mole calculate ΔG° for the oxidation of glucose

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) ΔG°rxn = Σ n ΔG°f (products) - Σ m ΔG°f (reactants) = [6(ΔG°f CO2) + 6(ΔG°f H2O)] – [ΔG°f C6H12O6 + 6(ΔG°f O2)]

= [6(-394.6) + 6(-237.2)] – [1(-907.9) + 6(0)] = - 2882.9 kJ (spontaneous)

THERMODYNAMICSEffect of temperature on Reaction Spontaneity

ΔG = ΔH - T ΔSΔH ΔS ΔG Rxn Characteristics

- + always - always spontaneous as written

+ - always + always nonspontaneous as written reverse rxn is spontaneous

- - - at low T spontaneous at low T+ at high T nonspontaneous at high T

+ + + at low T nonspontaneous at low T- at high T spontaneous at high T

THERMODYNAMICS• Sample Problem 8 What is

the temperature at which sodium chloride reversibly melts (solid and liquid states are in equilibrium)? The enthalpy of melting is 30.3 kJ/mole and the entropy change upon melting is 28.2 J/mole K.

ΔG = ΔH - T ΔS at equilibrium ΔG = 0 = ΔH - T ΔSso ΔH = T ΔS and T = ΔH / ΔS

30300 J/moleT = -------------------- = 1070 K (797oC) 28.2 J/mole K

above this temp (1070 K), melting is spontaneous

below this temp, melting would be nonspontaneous

THERMODYNAMICS• Sample Problem 9 At

298 K, ΔG° = -190.5 kJ and ΔH° = -184.6 kJ for the reaction

H2(g) + Cl2(g) 2 HCl(g) calculate the standard entropy ΔS° for the reaction

ΔG° = ΔH° - T ΔS° Δ S° = (ΔH° - ΔG° ) / T

- 184.6 kJ – (-190.5 kJ) ΔS° = ------------------------------- = .020 kJ/K = 20 J/K

298 K

Typically, S values or S values are in J/K not kJ/K

THERMODYNAMICS• Sample problem 10Given the following data:

Substance Hof (kJ/mol) So (J/K mol)

Fe2O3(s) -826 90.0

Fe(s) 0 27.0

O2(g) 0 205.0

Calculate ΔGo for the reaction: 4 Fe(s) + 3O2(g) 2 Fe2O3(s)

Go = Ho - TSo

Ho = prod - react = [2(-826)] – [0 + 0] = - 1650 kJ

So = prod - react = [2(90.0)] – [3(205.0) + 4(27.0)] = -543 J/K

Notice that H is kJ and S is J/K -543 J/K = -0.543 kJ/K

Go = Ho - TSo

= - 1650 kJ - (298 K)(-0.543 kJ/K)

= - 1490 kJ

chemical thermodynamics chapter 19 (except 19.7!)

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