1. Free energy and Thermodynamics suroviec Spring 2014
Chapter 17
Free energy and Thermodynamics
suroviec
Spring 2014

2. I. Spontaneous Change and Equilibrium
Chemical changes – reactions
Physical changes – expansions, heat, precipitation
Spontaneous change – occurs without outside intervention
Does not tell us about rate of change
Does indicate that we are headed to equilibrium

3. II. Heat and Spontaneity
HCl (aq) + NaOH (aq)  H2O (l) + NaCl (aq)
DH = ?
CH4 (g) + O2 (g)  CO2 (g) + H2O (g)
NH4NO3 (s)  NH4+ (aq) + NO3- (aq)
gas expanding : ?
Melting of ice : DH = ?

4. III. Dispersal of Energy
Entropy – S
Second law of thermodynamics
In a spontaneous process the entropy of the universe increases

5. A. Dispersal of Energy
The dispersal of energy over as many different energy states is the KEY to entropy
Thermal energy has been transferred

8. B. Dispersal of Matter
Since it is difficult to calculate to exact energy levels and dispersal of total energy among them lets initially look at the dispersal of matter among a system.

9. B. Dispersal of Matter
Looking at the flask again, the entropy of a state increases with the number of energetically equivalent ways to arrange the components of the system.
2nd law states that for any spontaneous process, the entropy of the universe increases.

10. C. Energy change and change in state
Entropy of a sample of matter changes as it changes state.

11. IV. Heat transfer
There can also be spontaneous processes that occur even when entropy seems to decrease.

12. B. Temperature dependence of DSsurr
If freezing increases entropy of surroundings, why would water not freeze at all temps?

13. C. Quantifying entropy changes in the surrounding
When a system exchanges heat with the surroundings, it changes the entropy of the surroundings
At constant pressure we can use qsys to quantify the change in entropy for the surroundings.

14. C. Quantifying entropy changes in the surrounding
We also know that the higher the temperature the lower the magnitude of DSsurr

15. C3H8 (g) + 5O2(g)  3CO2(g) + 4H2O(l) ΔHrxn = -2044 kJ
Example
Consider the combustion reaction below:
C3H8 (g) + 5O2(g)  3CO2(g) + 4H2O(l) ΔHrxn = kJ
Calculate DSsurr at 25oC
Determine the sign of DSsys
Determine the sign of DSuniv

16. Example
Given DH°rxn = -125kJ and ΔS°rxn = 253J/K at 25°C, what is the ΔSuniv?

17. V. Gibbs Free Energy Knowing that:
We can use that to rearrange and solve for Gibbs free energy.

18. V. Gibbs Free Energy
The change in Gibbs Free Energy for a process occurring at a constant temperature and pressure is proportional to the negative of ΔSuniv

19. A. Effect of ΔH, ΔS and T
These terms are all related, but how?

20. VI. So how do we know if a reaction is going to be spontaneous?

21. Example Given: C2H4(g) + H2(g)  C2H6(g)
Where ΔH = kJ and ΔS = J/K
Calculate ΔG at 25°C
How does T affect ΔG

22. Example Given: 2Ca(s) + o2(g)  2CaO(s)
Where ΔH°rxn = kJ and ΔS°rxn = J/K
Calculate ΔG°rxn at 25°C
Is it spontaneous?

23. VI. Entropy changes in chemical reactions
As a reminder: ΔH°rxn is change in enthalpy for a process in which reactants and products are in their std. states.

24. A. S° and the 3rd Law of Thermodynamics
Remember that we said that ΔH°f for an element in its std. state is zero.
However since S is temperature dependent, we can better define the zero of entropy.

25. B. Relative standard entropies
As we have seen, the entropy of a substance depends on its state.

26. C. Calculating ΔS°rxn
To calculate ΔS°rxn we use the following equation:

27. Example
Compute ΔS°rxn for the following reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
J/molK
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
188.8

28. Example
Compute ΔS°rxn for the following reaction: 2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g)
J/molK
H2S(g)
205.8
O2(g)
205.2
SO2(g)
248.2
H2O(g)
188.8

29. VI. Calculating ΔG°rxn Free energy changes
We can use ΔH°rxn and ΔS°rxn

30. Example Calculate the ΔG°rxn for the following reaction at 25°C
2SO2(g) + O2(g)  2SO3(g)
ΔH°rxn
kJ/mol
ΔS°rxn
J/molK
SO2(g)
-296.8
248.2
O2(g)
205.2
SO3(g)
-395.7
256.8

31. Example
What is the previous reaction was not at 25°C but instead was at 125°C?

32. B. ΔG°rxn using free energies
Since ΔG°rxn is the change in free energy and fee energy is a state function, we can solve ΔG°rxn as we did for ΔH°rxn and ΔS°rxn

33. CH4(g) + 8O2(g)  CO2(g) + 2H2O(g) + 4O3(g)
Example
Use standard free energies of formation to determine ΔG°rxn at 25°C
CH4(g) + 8O2(g)  CO2(g) + 2H2O(g) + 4O3(g)
kJ/mol
CH4(g)
-50.5
O2(g)
CO2(g)
-394.4
H2O(g)
-228.6
O3(g)
163.2

34. C. Determining ΔG°rxn for a series of reactions
To determine ΔG°rxn for a series of reactions, we use the same rules that we did with ΔG°rxn

35. Example Determine the ΔG°rxn for the following reaction:
2NO(g) + O2(g)  2NO2(g) ΔG°rxn = kJ
N2(g) + O2(g)  2NO(g) ΔG°rxn = kJ
2N2O(g)  2N2(g) + O2(g) ΔG°rxn = kJ

36. VII. DGo, K and product favorability
K = equilibrium constant
How does K relate to ΔG°rxn ?
ΔG°rxn is the increase or decrease in free energy as the reactants are completely converted to products
Reactants are not always converted completely to products.

37. VII. DGo, K and product favorability
When reactants are mixed, they proceed to position of lowest ree energy and then reach equilibrium.

38. Fig , p.929

39. Combining Everything that we have learned
The free energy at equilibrium is lower than the free energy of reactants or products
DGorxn gives the position of the equilibrium
DGorxn describes the direction of the reaction

40. 17_13.JPG
Figure 17-13
Title:
FIGURE Free energy and the equilibrium constant
Caption:
(a) Free energy curve for a reaction with a small equilibrium constant. (b) Free energy curve for a reaction with a large equilibrium constant. (c) Free energy curve for a reaction in which K = 1.