Chapter 19: Chemical Thermodynamics Thermodynamics-the study of energy in chemical reactions. Three Laws. First Law-The total energy of the universe is fixed. CHEM 103 (CHAP 5) ∆E= q + w ∆H (for constant Pressure)

Is ∆H alone an accurate indicator if a process will occur? NOOOOOOO! Spontaneous a process

that occurs without intervention. It has a certain order (sequence). Spontaneous in one direction non spontaneous in the reverse.

Second Law-The total entropy of the universe is increasing Entropy, So , S,- a measure of the disorder (randomness) of the system. State function. ∆S = Sfinal-S initial ∆S= q(rev)/T Entropy explains certain observations regarding spontanaeity

-mixing -decay Third Law-the entropy of a perfect crystalline solid is zero at absolute zero, 0K. The third Law allows the absolute entropies of substances to be evaluated.

Ssolid < Sliquid < Sgas In general S increases for: -liquids/solutions from

solids -gases from solids/liquids - # gas molecules increase -temp. subs increases Calculate ∆SRx

o of the following reactions using Appendix C ∆SRx

o= ΣΣΣΣSprod.o - ΣΣΣΣSreact.

o N2(g) + 3H2(g)→2NH3(g) 2KClO3(s)→2KCl(s)+3O2(g) Which wins in predicting

spon.? ∆Hsys or ∆Ssys? neither A Rx is SPON if: ∆Suniverse>0 ∆Suniverse=∆Ssurroundings +∆Ssystem ∆Suniverse= -∆Hsystem/T +∆Ssystem -T∆Suniverse= ∆Hsystem-T∆Ssystem by defin. -T∆Suniverse= ∆G

∆G=∆H-T∆S

∆Go=∆Ho-T∆So

Gibbs Free Energy, ∆G-the thermodynamic state function defining ∆G as, ∆Hsys-T∆Ssys. The sign of ∆G indicates Rx. spontaniety. (Free???) ∆G < 0 Rx. Spon. ∆G >0 Rx. Non-Spon. ∆G=0 Rx. @ Equil.

Effect of sign of ∆H and ∆S on spontainiety of a reaction. (Table 19.4)

∆H ∆S result <0 >0 spon. at all T >0 <0 non-spon. at all T <0 <0 spon. at low T T<∆H/∆S >0 >0 spon at high T T>∆H/∆S

Calculation of ∆GRx can be done by 1) the above approach or 2) from ∆Gf

o data analagous to ∆H and ∆S calc. (Appendix C)

∆GRx

o=ΣΣΣΣ∆Gof, prod-ΣΣΣΣ∆Go

f,react Calculate ∆Go for the following reaction using the ∆Gf

o data of App. C. 2KClO3(s)→2KCl(s)+3O2(g) The standard molar free energy of formation, ∆Gf

o and enthalpy of formation, ∆Hf

o values of elements (stable form) are defined as ZERO!!!

Calculate ∆Go for the foll. Rx using the Gibbs Eq and from ∆Gf

o data of App. C. ∆Ho= -92.38 kJ ∆So= -198.3J/K N2(g) + 3H2(g) 2NH3(g) Calculate the temperature at which the production of ammonia becomes spontaneous. At equilibrium ∆GRx=0 (∆GRx

o ≠0) ∆G = 0= ∆H-T∆S

T= ∆H/∆S T=-92.38kJ/-.1983kJ/K T= 465.9 K SPON below, non SPON above 466 K Under non-standard conditions ∆G=∆G0 + RTlnQ Calculate ∆G when P of N2,

H2, and NH3 are 1.00, 3.00, and 1.00 ATM respectively. (non STD. conditions) This result is consistent with LeChat. Prin. Relationship between ∆G and the equilibrium constant, K. @equilibrium, Q=K, ∆G=0 ∆G0 = -RTlnK= -2.30RTlogK ∆Go < 0 K>1 Spon. ∆Go >0 K<1 Non-Spon.

Calculate the equilibrium constant for the following Rx at 298K. N2(g) + 3H2(g) 2NH3(g) Energy is released on the

way to equilibrium not at equilibrium How do you get unfavored reactions to occur? -temperature (in some cases) -couple to another favored reaction(many bio Rxs) ∆G0(kJ) Cu2S →→→→ 2Cu +S 86.2kJ S + O2 →→→→ SO2 -300.4 Cu2S+ O2 →→→→ 2Cu+SO2

-214.2

Review of methods of calculating ∆HRx

o, ∆GRxo,

and ∆SRxo.

1) From data of App C ∆HRx

o= ΣΣΣΣ∆Hof, prod. - ΣΣΣΣ∆Ho

f,react

∆GRx

o= ΣΣΣΣ∆Gof, prod. - ΣΣΣΣ∆Go

f,react.

Remember that ∆Gfo and

∆Hfo values of elements

are ZERO ∆SRx

o= ΣΣΣΣSprod.o - ΣΣΣΣSreact.

o S terms ARE NOT ZERO except at 0K. 2) From Gibbs equation ∆Go=∆Ho-T∆So