1. Spontanaity, Entropy, Enthalpy & Free Energy
Chapter 16

2. Thermodynamics vs. Kinetics

3. Spontaneous Processes and Entropy
Thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process.
A spontaneous process is one that occurs without outside intervention.

4. Spontaneous Processes
A.    Spontaneous Processes - Occur without any outside intervention (ex. rusting of iron).
B.    Reversing spontaneous processes is possible with the expenditure of energy.

5. Chemical Reactions That Involve Heat TWO Trends in Nature
Order  Disorder
 
High energy  Low energy 

6. Enthalpy
A.    Enthalpy of reaction (ΔH= ΔHp- ΔHr) - Heat transferred in reaction.
a.  Exothermic reactions - Negative ΔH, release heat.
b. Endothermic reactions - Positive ΔH, absorb heat.
c.  Usually spontaneous processes are exothermic but
entropy must also be considered.

7. DH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
Hproducts > Hreactants
DH < 0
DH > 0
6.4

8. Heat and Enthalpy Changes
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
H2O (s) H2O (l)
DH = 6.01 kJ
6.4

9. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
Exothermic Reactions
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
DH = kJ
6.4

10. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s)
DH = kJ
If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
2H2O (s) H2O (l)
DH = 2 x 6.01 = 12.0 kJ
6.4

11. Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burns in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ
1 mol P4
123.9 g P4 x
-3013 kJ
1 mol P4 x
266 g P4
= kJ
6.4

12. 2H2 (g) + O2 (g) 2H2O (l) + energy
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
Heat is a product!
2H2 (g) + O2 (g) H2O (l) + energy
H2O (g) H2O (l) + energy
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
Heat is a reactant!
energy + H2O (s) H2O (l)
energy + 2HgO (s) Hg (l) + O2 (g)
6.2

13. Exothermic
Endothermic
6.2

14. Spontaneous Physical and Chemical Processes
A waterfall runs downhill
A lump of sugar dissolves in a cup of coffee
At 1 atm, water freezes below 0 0C and ice melts above 0 0C
Heat flows from a hotter object to a colder object
A gas expands in an evacuated bulb
Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
18.2

15. spontaneous
nonspontaneous
18.2

16. Does a decrease in enthalpy mean a reaction proceeds spontaneously?
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = kJ
H+ (aq) + OH- (aq) H2O (l) DH0 = kJ
H2O (s) H2O (l) DH0 = 6.01 kJ
NH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ
H2O
18.2

17. Entropy A. Entropy (ΔS) - A quantitative measure of the disorder or
randomness in the substances involved in a chemical reaction. 
B. ΔS = Sproducts-Sreactants
a.Gases formed from liquids or solids ΔS>0
b. Solutions formed from liquids and solids ΔS>0
c. More mole of gases as products than moles
of gases as reactants ΔS>0.
d. The temperature of Substance increases ΔS>0
C. 2nd law of Thermodynamics - In any spontaneous process the overall entropy of the universe increases.
ΔSuniverse= ΔSsurroundings + ΔSreaction

18. Entropy (S) is a measure of the randomness or disorder of a system.
DS = Sf - Si
If the change from initial to final results in an increase in randomness
Sf > Si
DS > 0
For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state
Ssolid < Sliquid << Sgas
H2O (s) H2O (l)
DS > 0
18.3

19. Thermodynamics
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
energy
, pressure, volume, temperature
DE = Efinal - Einitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.
6.3

20. Entropy Changes in the Surroundings (DSsurr)
Exothermic Process
DSsurr > 0
Endothermic Process
DSsurr < 0
18.4

21. Interplay of Ssys and Ssurr in Determining the Sign of Suniv

22. Calorimetry
First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.
DEsystem + DEsurroundings = 0 or
DEsystem = -DEsurroundings
C3H8 + 5O CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
system
surroundings
6.3

23. Heat (q) absorbed or released:
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
C = J/gº•C
Heat (q) absorbed or released:
q = mCDt
Dt = tfinal - tinitial
6.5

24. Dt = tfinal – tinitial = 50C – 940C = -890C
How much heat is given off when an 869 g iron bar cools from 940C to 50C?
C of Fe = J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = mCDt
= 869 g x J/g • 0C x –890C
= -34,000 J
6.5

25. Constant-Volume Calorimetry
qsur = -qrxn
qrxn = - qwater
qwater = mCDt
Reaction at Constant V
DH = qrxn
DH ~ qrxn
No heat enters or leaves!
6.5

26. Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = mCDt
qcal = CcalDt
Reaction at Constant P
DH = qrxn
No heat enters or leaves!
6.5

27. 6.5

28. Chemistry in Action: Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) DH = kJ/mol
1 cal = J
1 Cal = 1000 cal = 4184 J

29. Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f
Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f
The standard enthalpy of formation of any element in its most stable form is zero.
DH0 (O2) = 0 f
DH0 (C, graphite) = 0 f
DH0 (O3) = 142 kJ/mol f
DH0 (C, diamond) = 1.90 kJ/mol f
6.6

30. 6.6

31. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
6.6

32. Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ
rxn +
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ
rxn
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = (2x-296.1) = 86.3 kJ
rxn
6.6

33. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[ 12(–393.5) + 6(–285.8) ] – [ 2(49.04) ] = kJ
-6535 kJ
2 mol
= kJ/mol C6H6
6.6

34. DHsoln = Hsoln - Hcomponents
The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be used for melting ice?
Which substance(s) could be used for a cold pack?
6.7

35. The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
6.7

36. What is Heat? Caloric Theory
Heat is a fluid that transfers between objects.
This was widely accepted through the 1700’s
Count Rumford showed that heat produced by friction could not be a fluid.
James Joule demonstrated the mechanical equivalent of heat in 1850.
This proved heat is a form of energy and destroyed the caloric theory
Kinetic Theory suggested that heat results from the motion of particles in matter.

37. Kinetic Molecular Theory
The volume of individual particles is so small it can be assumed to be zero
Collision of particles with walls of the container are the cause of pressure
Particles exert no force on each other, neither attract nor repel
Average kinetic energy of a collection of gas particles is assumed to be directly proportional to Kelvin Temp. of the gas.
Heat is the transfer of kinetic energy from a hot object to a cooler one, or the total kinetic and potential energy of the particles in a substance.
Temperature is the average kinetic energy of the particles (how fast they move on average).

38. Gibbs Free Energy
A.     Gibbs Free Energy Equation - Incorporates enthalpy, entropy, and temperature to determine spontaneity of a given reaction.
a.  ΔG = ΔH – TΔS
b. If ΔG is negative the reaction is spontaneous and will occur on its own.
c.  If ΔG is positive the reaction is not spontaneous and requires input of energy.
d. If ΔG is zero the reaction is at equilibrium.
B.    Copy table 16-5 for summary of free energy change and reaction spontaneity
C.    ΔG˚ is the free energy change at standard conditions.
a.  If ΔG˚ is positive the reaction is not spontaneous and Keq << 1.
b. If ΔG˚ is negative the reaction is spontaneous and Keq >>1.
c. If ΔG˚ = 0 the reaction is at equilibrium and Keq= 1.
C.     Spontaneous reactions release free energy that is available to do work.
The work done cannot exceed the free energy available.

39. Spontaneous Reactions

40. Effect of H and S on Spontaneity

41. Change in Free Energy to Reach Equilibrium

42. Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction