1. Chapter 16
Thermodynamics

2. Objectives Define key terms and concepts
Explain the three laws of thermodynamics.
Calculate the entropy for a system.
Determine if a reaction is spontaneous or nonspontaneous.
Calculate the Gibbs Free Energy for a reaction.

3. The First Law of Thermodynamics
Energy cannot be created or destroyed, only converted from one form to another.
ΔH = ΔU + PΔV
Δ H°rxn = ΣΔH°f(products) - ΣΔH°f(reactants)

4. Chemistry is Spontaneous
Spontaneous processes/reactions are favored over non-spontaneous processes.
Entropy
How dispersed the energy of a system is
The more dispersed, the greater the entropy

5. Chemistry is Spontaneous

6. Entropy Solids have less entropy than liquids
Liquids have less entropy than gases
As a solute dissolves in solution, entropy increases.
Where k = 1.38x10-23 J/K
W = the number of microstates
ΔS = Sf - Si
S = k ln W
ΔS = k ln Wf - k ln Wi

7. This system has 16 microstates

8. Would entropy increase or decrease in each of the following system?
Melting aluminum
Iodine crystals subliming
A car accident
Water freezing
Dissolving sodium chloride in water

9. The Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.
For a spontaneous process, ΔSuniv > 0
For an equilibrium process, ΔSuniv = 0
ΔSuniv = ΔSsys + ΔSsurr

10. The Second Law of Thermodynamics
For the general reaction:
aA + bB  cC + dD
ΔS°rxn =[cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]
If the entropy for a system increases, ΔS° is positive.
If the entropy for a system decreases, ΔS° is negative.
ΔS°rxn = ΣnS°(products) - ΣnS°(reactants)

11. The Second Law of Thermodynamics

12. Predicting the Sign of ∆S°
If a reaction produces more gas molecules than it consumes, ∆S° is positive.
If the total number of gas molecules diminishes, ∆S° is negative.
If there is no net change in the total number of gas molecules, then ∆S° may be negative or positive, but the numerical value will be small.

13. Predict the sign of ∆S° in the following reactions.
C6H12O6(s)  C2H5OH(l) + CO2(g) NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l) CO(g) + H2O(g)  CO2(g) + H2(g)

14. Calculate the standard entropy of reaction for the following reaction at 25°C. Standard molar enthalpy of glucose = J/(mol*K) C6H12O6(s)  C2H5OH(l) + CO2(g)

15. Calculate the standard entropy of reaction for the following reaction at 25°C. Standard Entropy of Urea is 174 J/(mol*K) NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l)

16. Calculate the standard entropy of reaction for the following reaction at 25°C. CO(g) + H2O(g)  CO2(g) + H2(g)

17. What are your questions?

18. Entropy and Temperature
Heat
Heat
Surroundings
Entropy
System 1
System 2
Entropy

19. Entropy and Temperature
If the temperature of the surroundings is low, the energy released by a system will increase the entropy.
The Third Law of Thermodynamics
The Entropy of a perfect crystalline substance is zero at absolute zero temperature.
∆Hsys T
∆Ssurr =

20. The heat of vaporization, ∆Hvap, of carbon tetrachloride at 25°C is 43
The heat of vaporization, ∆Hvap, of carbon tetrachloride at 25°C is 43.0 kJ/mole. If 1 mole of liquid carbon tetrachloride at 25°C has an entropy of 214 J/K, what is the entropy of 1 mole of the vapor in equilibrium with the liquid at this temperature?

21. Liquid ethanol at 25°C has an entropy of 161 J/mole•K
Liquid ethanol at 25°C has an entropy of 161 J/mole•K. If the heat of vaporization, ∆Hvap, at 25°C is 42.3kJ.mole, what is the entropy of the vapor in equilibrium with the liquid at 25°C?

22. Gibbs Free Energy The energy available to do work.
If ∆G<0, the reaction is spontaneous in the forward direction.
If ∆G >0, the reaction is nonspontaneous in the forward direction.
If ∆G=0, the system is at equilibrium and no net change in energy will be observed.
∆G= ∆H-T∆S

23. Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following reaction at 25°C.
N2(g) + H2(g)  NH3(g)

24. CaCO3(s) ↔ CaO(s) + CO2(g)
Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following reaction at 25°C.
CaCO3(s) ↔ CaO(s) + CO2(g)

25. KClO3(s) ↔ KCl(aq) + O2(g)
Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following reaction at 25°C.
KClO3(s) ↔ KCl(aq) + O2(g)

26. Factors Affecting the Sign of ∆G+
Reaction is spontaneous in the forward direction at high temperatures, but is spontaneous in the reverse direction at low temperatures. -
∆G is always positive. Non-spontaneous reaction.
∆G is always negative. Spontaneous reaction.
Reaction is spontaneous in the forward direction at low temperatures, but is spontaneous in the reverse direction at high temperatures.

28. ΔG°rxn = ΣnG°f(products) - ΣnG°f(reactants)
Gibbs Free Energy
Standard Free Energy of a Reaction (∆G°rxn)
The change in free energy when 1 mole of a compound is synthesized from its elements in their standard state.
For the general reaction:
aA + bB  cC + dD
ΔS°rxn =[cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]
ΔG°rxn = ΣnG°f(products) - ΣnG°f(reactants)

29. Gibbs Free Energy

30. Determine if the following reaction is spontaneous using Gibbs Free Energy if the reaction is conducted at 25°C. CaCO3(s)  CaO(s) + CO2(g)

31. Determine if the following reaction is spontaneous using Gibbs Free Energy if the reaction is conducted at 25°C. KClO3(s)  KCl(aq) + O2(g)

32. Determine if the following reaction is spontaneous using Gibbs Free Energy if the reaction is conducted at 25°C. NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l)

33. Free Energy and Chemical Equilibrium
ΔG° = -RTlnK
For calculating reactions where not all reactants and products are in their standard states.
Where R= J/K•mole
T is temperature in Kelvin
K is the equilibrium constant.

34. Free Energy and Chemical EquilibriumK
InK
ΔG°
Affect at Equilibrium
> 1 + -
Products are favored
= 1
Neither side of the reaction is favored
< 1
Reactants are favored

35. What is the value of the equilibrium constant K at 25°C for the following reaction? Predict whether the products or reactants are favored at equilibrium. CaCO3(s) ↔ CaO(s) + CO2(g)

36. What is the value of the equilibrium constant K at 25°C for the following reaction? Predict whether the products or reactants are favored at equilibrium. KClO3(s) ↔ KCl(aq) + O2(g)

37. What is the value of ΔG° at 25°C for the following reaction
What is the value of ΔG° at 25°C for the following reaction? The Ksp of magnesium hydroxide is 1.8x Predict whether the products or reactants are favored at equilibrium. Mg(OH)2(s) ↔ Mg2+(aq) + OH-(aq)

38. What is the value of the equilibrium constant K at 25°C for the following reaction? The Ksp of lead (II) iodide is 6.5x10-9. Predict whether the products or reactants are favored at equilibrium. PbI2(s) ↔ Pb2+(aq) + I2-(aq)

39. What are your questions?