thermodynamics

Indicated by a degree symbol Gases at 1 atm Liquids are pure. Solids are pure. Solutions are at 1 M. Temperature is 25o C (298 K).

Conservation of Energy—Energy can be neither created nor destroyed.

The amount of energy in the universe is constant.

Energy can change forms, but the amount does not change.

Chemical (potential) energy is stored in the bonds of methane gas.

Combustion releases the energy stored in the bonds as heat.

Energy exchange between system & environment = enthalpy

Spontaneous processes need no outside cause; they happen naturally.

Spontaneous = Fast (Thermodynamics is not

concerned with how a reaction happens—just whether it will happen.)

Exothermicity?? (Tendency to reach lowest potential energy)

Partial explanation, but does not always hold true

Probability—how likely are arrangements of atoms (randomness or entropy)

The amount of disorder or randomness in a system

Higher entropy is favored because random arrangements are more likely.

Ex.: Gas molecules are more likely to be evenly dispersed than clumped up in one place.

1 mole of gas or 1 mole of liquid?

Pure water or a 0.5 M salt solution?

Gas at 25o C or gas at 100o C? 1 mole of gas or 5 moles of gas?

In state changes--solid to liquid, liquid to gas or solid to gas

When solids dissolve When temperature increases When volume of a gas increases

(or pressure decreases) When a reaction occurs that

produces more moles of gases

In any spontaneous process, there is an increase in entropy of the universe.

Think of the universe as being inclusive of a system and its surroundings—combined entropy must increase (Cell)

Sign depends on direction of heat flow—heat added to surroundings increases entropy & heat absorbed from surroundings decreases entropy

Magnitude—depends on temperature—greater impact of exothermic system if temperature is low

Analogy—giving $50 to a millionaire or to you—to whom does it mean more?

Equal to –H/T negative—because the

surroundings are opposite from the system

H –change in heat in the system (from system’s perspective)

T—temperature in Kelvin

ssurr is larger at lower temperatures because the denominator of the fraction is smaller .

The effect of heat flow for a process is more significant when temperature is lower.

G Uses enthalpy, entropy and

temperature to predict the spontaneity of a reaction

If G is negative, then a process is spontaneous.

G = H – TS Gis free energy is change in enthalpy Sis change in entropy All terms are from the point

of view of the system.

Mathematical relationships: If H is negative and S is

postive—always spontaneous If H is positive and S is

negative—never spontaneous If both H and S have the

same sign—temperature dependent

At what temperature is the following process spontaneous?

Br2 (l) Br2 (g)

H = 31.0 kJ/mol & S = 93.0 J/K mol

Calculate temperature when G is zero. (G = H – TS)

At the exact boiling point, G is zero.

Set known values of H, S and/or T equal to zero and solve for unknowns.

The enthalpy of vaporization of mercury is 58.51 kJ /mol and the entropy of vaporization is 92.92 J/K mol. What is mercury’s normal boiling point?

The entropy of a perfect crystal at 0 K is zero.

Standard entropy values have been determined—Appendix 4

To determine entropy change for a reaction, subtract the entropy values

S = npSoproducts – nrSo

reactants

n = number of moles (State functions depend on amount of substance present.)

Calculate So at 25o C for the reaction:

2NiS(s) + 3O2 (g) 2SO2 (g) + 2NiO (s) So in J/K mol:

› SO2 248 J/K mol

› NiO 38 J/K mol

› O2 205 J/K mol

› NiS 53 J/K mol

Calculate So for the reduction of aluminum oxide by hydrogen gas:

Al2O3(s) + 3H2(g) 2 Al(s) + 3 H2O(g)

Note: equal number of gas molAns: 179 /K (large & positive

because of water’s asymmetry)

Cannot be directly measured Can be calculated using

enthalpy and entropy changes (G = H – TS)

Can be calculated from already determined standard free energy values: G = npGo

products – nrGoreactants

Calculate H, S, and G using the following information for the reaction 2SO2 (g) + O2 (g) 2SO3(g).

SO2: H = -297 kJ/mol; So = 248 J/K mol

SO3: H = -396 kJ/mol; So = 257 J/K mol

O2: H = 0 kJ/mol; So = 205 J/K mol

Using the following data, calculate G for the reaction Cdiamond (s) C graphite (s)

Cdiamond (s) + O2 (g) CO2 (g) G = -397 kJ

Cgraphite (s) + O2 (g) CO2 (g) G = -394 kJ

Given the following free energies of formation, calculate the G for the reaction

2CH3OH (g) + 3O2 (g) 2 CO2 (g) + 4 H2O (g)

CH3OH (g): G = -163 kJ/mol

O2 (g): G = 0 kJ/mol

CO2 (g): G = -394 kJ/mol

H2O (g): G = -229 kJ/mol

Entropy depends on pressure because changes in pressure affect volume.

G = Go + RTln(P) G = free energy at new pressure Go = free energy at 1 atm R = gas constant (8.3145 J/K*mol) T = Kelvin temperature P = new pressure

From the previous equation, a relationship between G and Q or K can be derived.

(See p. 807) G = Go + RT ln (Q)

CO (g) + 2H2 (g) CH3OH (l) Calculate G for this process

where CO is at 5.0 atm and H2 is at 3.0 atm.

Calculate Go using standard values for reactants & products.

Then use G = Go + RT ln (Q)

At equilibrium, a system has used all of the free energy available.

G equals zero because forward & reverse reactions have the same free energy.

G = Go + RT ln (Q) = Go + RT ln (K) Go = - RT ln (K)

For the synthesis of ammonia, Go = -33.33 kJ/mol of N2 consumed at 25o C.

Predict how each system below will shift in order to reach equilibrium:› NH3 = 1.00 atm; N2 = 1.47 atm; H2 = 1.00 x 10-2

atm

› NH3 = 1.00 atm; N2 = 1.00 atm; H2 = 1.00 atm

NH3 = 1.00 atm; N2 = 1.47 atm; H2 = 1.00 x 10-2 atm

G = Go + RT ln (Q)

G = Go + RT ln (Q)

4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s) Use the data to calculate K for this

reaction at 25o C. Fe2O3: H = -826 kJ/mol; So = 90

J/K mol Fe: H = 0 kJ/mol; So = 27 J/K mol O2: H = 0 kJ/mol; So = 205 J/K mol