1. Chemical Thermodynamics
Chapter 19
Chemical Thermodynamics

2. Spontaneous Processes Section 19.1
A spontaneous process is one that occurs without the influence of an outside force
Occurs in a single direction and the opposite direction is nonspontaneous

3. Spontaneity and the Relationship Between Thermodynamics and Kinetics
A chemical or physical process may be highly spontaneous, but at the same time may take place slowly
Thermodynamics offers information about the direction and extent that a process occurs but yields no information as to the rate of reaction

4. The Link Between Spontaneity and Enthalpy
The vast majority of spontaneous chemical reactions are exothermic in nature; however, there are examples of spontaneous processes that are endothermic
For an endothermic process to be spontaneous there must be some type of “compensation” to offset the increase in energy between initial and final states
This “compensation” is referred to as entropy

5. Making Qualitative Predictions About S
While entropy itself is an abstract concept, making predictions about the change in entropy for a chemical process is simple
Some simple guidelines:
Temperature
Increase in T leads to higher entropy
Change of phase
Increasing entropy: solid<liquid<gas
Number of particles
More particles in solution results in higher entropy
Dissolution
Particles in solution have more entropy than when undissolved

6. Predicting the Sign of S
Predict whether S is positive or negative for the following processes:
CO2(s)  CO2(g)
CaO(s) + CO2(g)  CaCO3(s)
HCl(g) + NH3(g)  NH4Cl(s)
2 SO2(g) + O2(g)  2 SO3(g)
See Sample Exercise 19.3 (Pg. 814)

7. Third Law of Thermodynamics
The Third Law states that at 0 K, all motion completely stops (S = 0)

8. Entropy Changes in Chemical Reactions Section 19.4
Just like enthalpy, entropy change for a reaction can be calculated using tabulated standard molar enthalpies:
S = S(products) - S(reactants)
Example:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

9. Gibbs Free Energy Section 19.5
Because there are endothermic processes that take place spontaneously and exothermic processes that take place with a decrease in entropy, there must be a dependence upon both these quantities on reaction spontaneity
These terms have been combined along with a new term, Gibbs free energy:
G = H – TS or
G = H - T S
Std. Conditions: G = H - T S

10. Gibbs Free Energy and Spontaneity
Since Gibbs free energy combines both enthalpy and entropy, this is what ultimately determines reaction spontaneity:
If G is negative, the reaction is spontaneous
If G is positive, the reaction is nonspontaneous
If G = 0, the reaction is at equilibrium (both directions equally spontaneous)

11. Calculating Grxn
Calculate Grxn and determine whether the following reaction will take place spontaneously under standard state conditions
H2S(g) + 2H2O(l) → SO2(g) + 3H2(g)
See Sample Exercise 19.7 (Pg. 823)

12. Estimating G
Predict whether G for this reaction is more or less negative than H:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) Hrxn = kJ
See Sample Exercise 19.8 (Pg. 823)

13. Free Energy and Temperature Section 19.6
G = H - T S
G = H + (-T S)
Enthalpy term
Entropy Term
Ex:
H2O(s)  H2O(l) H > 0, S > 0

14. Effect of Temperature on Spontaneity

15. Determining Effect of Temperature on Spontaneity
Consider the production of carbon dioxide and methane from carbon monoxide and hydrogen.
2CO(g) + 2H2(g)  CO2(g) + CH4(g)
a.) Calculate G, S, and H for the reaction at 298 K
b.) Calculate G for the same reaction at 1000 K. Assume H and S do not change much with temperature
See Sample Exercise 19.9 (Pg. 826)

16. Relating G to a Phase Change at Equilibrium
a.) Write the chemical equation that defines the normal boiling point of liquid methanol (CH3OH).
b.) Determine the value of G for the equilibrium in part a.)
c.) Use thermodynamic data from Appendix C to estimate the normal boiling point of CH3OH.
See Sample Exercise (Pg. 827)

17. Free Energy and the Equilibrium Constant Section 19.7
G determines spontaneity of a process, but only under standard conditions:
[ ] = 1 M; P = 1 atm; T = 25 C (298 K)
If conditions are other than standard:
G = G + RT ln(Q)

18. Calculating the Free Energy Change Under Nonstandard Conditions
Calculate the Gibbs free energy change for the reaction shown below at 298 K. PNO = 0.1 atm and PNOBr = 2.0 atm. G(NOBr) = 82.4 kJ/mol
2NO(g) + Br2(l)  2NOBr(g)
See Sample Exercise (Pg. 828)

19. Calculating Equilibrium Constants Using Gibbs Free Energy
At equilibrium G = 0; therefore:
G = -RT ln(K)
or rearranging:

20. Calculating an Equilibrium Constant from G
The Gibbs free energy change for the following reaction is kJ.
AgCl(s)  Ag+(aq) + Cl-(aq)
Calculate the equilibrium constant for this reaction at 350 K.
See Sample Exercise (Pg. 829)