2. Wednesday, Nov. 10 th : “A” Day Agenda  Section 10.4: Order and Spontaneity Entropy, Standard Entropy, Gibbs energy  Homework: Sec. 10.4 review, pg. 367, #3-5, 7-11 Sec 10.4 concept review *Quiz next time over section 10.4*

3. Entropy  Entropy: a measure of the randomness or disorder of a system  Symbol: S  Units: J/K  A process is more likely to occur if it is accompanied by an increase in entropy ( ΔS is positive)

4. Factors that Affect Entropy  Entropy increases as molecules or ions become dispersed. (Diffusion)  Entropy increases as solutions become more dilute or when the pressure of a gas is reduced.  Mixtures of gases have more entropy than a single gas.  Entropy increases when total # moles product > total # moles reactant  Entropy increases when a reaction produces more gas particles, because gases are more disordered than liquid or solids.

5. Hess’s Law Also Applies to Entropy  Standard Entropy, S o : the entropy of 1 mole of a substance at a standard temperature, 298.15 K. The entropy change of a reaction can be calculated by: ΔS reaction = ΔS products - ΔS reactants  Elements can have standard entropies of formation that have values other than zero.

6. Sample Problem F (pg. 361) Use Table 4 to calculate the entropy change that accompanies the following reaction: ½ H 2(g) + ½ CO 2 (g) ½ H 2 O (g) + ½ CO (g) ΔS reaction = ΔS product - ΔS reactants ΔS product =[(.5mol)(188.7J/K·mol)+(.5mol)(197.6 J/K·mol)] = 193.2 J/K ΔS reactant =[(.5mol)(130.7J/K·mol)+(.5mol)(213.8 J/Kmol)] = 172.3 J/K ΔS reaction = 193.2 J/K – 172.3 J/K = 20.9 J/K

7. Practice #1 Pg. 361 Find the change in entropy for the reaction below by using Table 4 and that S˚ for CH 3 OH (l) is 126.8 J/K·mol. CO (g) + 2 H 2( g ) CH 3 OH (l) ΔS reaction = ΔS product - ΔS reactants ΔS product = (1mol)(126.8 J/K·mol) = 126.8 J/K ΔS reactants =[(1mol)(197.6 J/K·mol)+(2mol)(130.7J/K·mol)] = 459 J/K ΔS reaction = 126.8 J/K – 459 J/K = -332.2 J/K

8. Gibbs Energy  If ΔH is negative and ΔS is positive for a reaction, the reaction will likely occur.  If ΔH is positive and ΔS is negative for a reaction, the reaction will NOT occur.  How can you predict what will happen if ΔH and ΔS are both positive or negative?

9. Gibbs Energy  Gibbs Energy: the energy in a system that is available for work. (free energy)  Symbol: G G = H – TS OR ΔG = ΔH – TΔS H = enthalpy (kJ or J) S = entropy (J/K) T = temperature (K)

10. Gibbs Energy Determines Spontaneity  Spontaneous reaction: a reaction that does occur or is likely to occur without continuous outside assistance, such as the input of energy.  Non-spontaneous reaction: a reaction that will never occur without assistance.

11. Spontaneous vs. non- spontaneous  On a snow-covered mountain in winter, an avalanche is a spontaneous process because it may or may not occur, but it always CAN occur.  The return of snow from the bottom of the mountain to the mountaintop is a non- spontaneous process, because it can NEVER happen without assistance.

12. Gibbs Energy Determines Spontaneity If ΔG is negative, reaction is spontaneous If ΔG is greater than 0, reaction is non- spontaneous If ΔG is exactly 0, reaction is at equilibrium

13. Entropy and Enthalpy Determine Gibbs Energy  Standard Gibbs energy of formation: the change in energy that accompanies the formation of 1 mole of the substance from its elements at 298.15 K.  Symbol: ΔG f o.  Unit: kJ or J ΔG reaction = ΔG products – ΔG reactants

14. Sample Problem G Pg. 364 Given that the change in enthalpy and entropy are - 139 kJ and 277 J/K respectively for the reaction given below, calculate the change in Gibbs energy. Then, state whether the reaction is spontaneous at 25˚C. C 6 H 12 O 6 (aq) 2 C 2 H 5 OH (aq) + 2 CO 2 (g) ΔG = ΔH – TΔS ΔH = -139 kJ ΔS = 277 J/K T = 25°C + 273 = 298K ΔG = (-139 kJ) – [(298K) (0.277 kJ/K)] = -222 kJ Spontaneous because ΔG is negative.

15. Sample Problem H Pg. 365 Use Table 5 to calculate ΔG for the following water- gas reaction with graphite. C (s) + H 2 O (g) CO (g) + H 2(g) ΔG reaction = ΔG products - ΔG reactants ΔG products = [(1mol)(-137.2 kJ/mol) + (1mol)(0)] = -137.2 kJ ΔG reactants = [(1mol)(0) + (1mol)(-228.6 kJ/mol)] = -228.6 kJ ΔG reaction = -137.2 kJ – (- 228.6 kJ) = 91.4 kJ

16. Predicting Spontaneity ΔHΔHΔSΔSΔGΔGSpontaneous ? NegativePositiveNegativeYes, at all Temps Negative Either Positive or Negative Only if T < ΔH/ΔS Positive Either Positive or Negative Only if T > ΔH/ΔS PositiveNegativePositiveNever

17. Predicting Spontaneity  Since ΔG = ΔH – TΔS, temperature may greatly affect ΔG.  Increasing the temperature of a reaction can make a non-spontaneous reaction spontaneous.

18. Homework  Sec 10.4 review Pg. 367, #3-5, 7-11  Concept Review: “Order and Spontaneity” Next time:  Quiz over section 10.4  Chapter 10 review Chapter 10 test/concept review due: Friday, Nov. 19th