 # Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved.  E = q + w.

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Thermodynamics Chapter 18

1 st Law of Thermodynamics Energy is conserved.  E = q + w

SPONTANEOUS: occur without any outside intervention Example: drop an egg The REVERSE is not spontaneous!!

REVERSIBLE PROCESS: change can be restored to its’ original state by exactly reversing the change. Example: ice water at 0 o C

IRREVERSIBLE PROCESS: cannot simply be reversed to original state. Example: gas expanding

Processes in which the disorder of the system increases tend to occur spontaneously. Ex: gas expanding, ice melting, salt dissolving

ENTROPY: (S) the change in disorder. (Change in randomness) The more disorder, the larger the entropy.  S = S final - S initial

 S = > 0 when the final state is in more disorder  S = < 0 when the final state is more ordered than original

H 2 O (l)  H 2 O (s) Ag + (aq) + Cl - (aq)  AgCl (s)

- a solid melts - a liquid vaporizes - a solid dissolves in water - a gas liquefies

For a process at constant temperature, the entropy change is the value of q rev divided by the absolute temperature.  S = q rev /T

Example: Calculate the entropy change when 1 mol of water is converted into 1 mol of steam at 1 atm pressure. (  H vap = 40.67 kJ/mol)

(1 mole)(40.67 kJ/ mol)(1000 J/1 kJ) 373 K  S = 109 J/K

The normal freezing point of mercury is -38.9 o C, an its molar enthalpy of fusion is  H fus = 2.331 kJ/mol. What is the entropy change when 50.0 g of Hg (l) freezes at the normal freezing point?

-2.48 J/K The answer is negative because the process brings more order

The normal boiling point of ethanol, is 78.3 o C and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C 2 H 5 OH (g) condenses to liquid at the normal boiling point?

-61.4 J/K

The element gallium, Ga, freezes at 29.8 o C, and its enthalpy of fusion is 5.59 kJ/mol. Calculate the value of  S for the freezing of 90.0 g of Ga (l).

 S = -23.8 J/K

2 nd Law of Thermodynamics: In any reversible process,  S universe = 0. In any irreversible (spontaneous) process,  S universe > 0.  S universe =  S system +  S surroundings

On a Molecular Level TRANSLATIONAL MOTION: movement of molecules VIBRATIONAL MOTION: the movement of atoms within the molecule. ROTATIONAL MOTION: the molecules spinning

Increasing Temperature Increases Entropy

3 rd Law of Thermodynamics: the entropy of a pure crystalline substance at absolute zero is zero. S(0K) = 0

In general, the entropy increases when: Liquids or solutions are formed from solids Gases are formed from either solids or liquids The number of molecules of gas increases during a chemical reaction.

CaCO 3(s)  CaO (s) + CO 2(g) N 2(g) + 3H 2(g)  2NH 3(g)

Standard molar entropies: (S o ) absolute entropies for substances in their standard state. (J/mol-K) 1. Unlike enthalpies of formation, the standard molar entropies of elements are not zero. 2. The S o of gases are greater than those of liquids and solids. 3. The S o generally increases with increasing molar mass. 4. The S o generally increase with the number of atoms in the formula.

 S o =  nS o (products) -  mS o (reactants) Calculate  S o for the synthesis of ammonia: N 2(g) + 3H 2(g)  2NH 3(g)

 S o = (2 mol)(192.5 J/mol-K) - [(1 mol)(191.5 J/mol-K) + (3 mol)(130.6 J/mol-K)] = -198.3 J/K

Using Appendix C, calculate the standard entropy change, for the following reaction: Al 2 O 3(s) + 3H 2(g)  2Al (s) + 3H 2 O (g)

180.4 J/K

C 2 H 4(g) + H 2(g)  C 2 H 6(g) NH 3(g) + HCl (g)  NH 4 Cl (s)

-120.5 J/K -284.6 J/K

GIBBS FREE ENERGY The spontaneity of a reaction involves both enthalpy and entropy. The relationship is known as free energy.  G =  H - T  S

1. If  G is negative, the reaction is spontaneous in the forward direction. 2. If  G is zero, the reaction is at equilibrium. 3. If  G is positive, the reaction in the forward direction is nonspontaneous; work must be supplied from the surroundings to make it occur. However, the reverse reaction will be spontaneous.

GG

 G o =  n  G f o (products) -  m  G f o (reactants) N 2(g) + 3H 2(g)  2NH 3(g)

-33.32 kJ

CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g)

-800.7 kJ

Assuming no change for  H o and  S o, what happens to  G o with an increase in temperature? N 2(g) + 3H 2(g)  2NH 3(g)

Calculate  G at 298K for a reaction mixture that consists of 1.0 atm N 2, 3.0 atm H 2 and 0.50 atm NH 3. N 2(g) + 3H 2(g)  2NH 3(g)

Use standard free energies of formation to calculate the equilibrium constant K at 25 o C for the reaction involved in the Haber process.

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