Wednesday, Oct. 31 st : “A” Day Thursday, Nov. 1 st : “B” Day Agenda  Homework questions/quick review  Sec quiz: “Changes in Enthalpy During Chemical Reactions”  Section 10.4: “Order and Spontaneity” Entropy, Standard Entropy, Gibbs energy  Homework: Sec review, pg. 367: #3-5, 7-11 Sec 10.4 concept review *Quiz next time over section 10.4*

Homework  Pg. 15 practice worksheet  Sec review, pg. 357: #1-5

Calculating a Reaction’s Change in Enthalpy Sample Prob. E, pg.356  Calculate the change in enthalpy for the reaction below using data from Table 2 on pg H 2 (g) + 2 CO 2 (g) 2 H 2 O(g) + 2 CO(g) State whether the reaction is exothermic or endothermic. ΔH reaction = ΔH f 0 products - ΔH f 0 reactants ΔH f 0 prod = [(2 mol)( kJ/mol) + (2 mol)( kJ/mol)] = kJ ΔH f 0 reactants = [(2 mol)(0 kJ/mol) + (2 mol)( kJ/mol)] = -787 kJ ΔH reaction = ( kJ) – (-787 kJ) = 82.4 kJ *Reaction is endothermic because ΔH is positive.*

Sec Quiz: “Changes in Enthalpy During Chemical Reactions”  You can use your notes and your book to complete the quiz with a partner of your choice… Good Luck!

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

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.

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

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 J/K·mol CO (g) + 2 H 2( g ) CH 3 OH (l) ΔS reaction = S˚ product - S˚ reactants S˚ product = (1mol)(126.8 J/K·mol) = J/K S˚ reactant =[(1mol)(197.6 J/K·mol)+(2mol)(130.7J/K·mol)] = 459 J/K ΔS reaction = J/K – 459 J/K = J/K

Example  Calculate the change in entropy for the following reaction using Table A-11 on pg Na(s) + 2 HCl(g) 2 NaCl(s) + H 2 (g) ΔS reaction = S˚ product - S˚ reactants S˚ product = [(2 mol)(72.1 J/mol∙K)+(1 mol)(130.7 J/mol∙K)] = J/K S˚ reactants = [(2 mol)(51.5 J/mol∙K)+(2 mol)(186.8 J/mol∙K)] = J/K Δs reaction = J/K – J/K = J/K

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?

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

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.

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.

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

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 K.  Symbol: ΔG f o  Unit: kJ/mol ΔG reaction = ΔG f ˚ products – ΔG f ˚ reactants

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 (Change to kJ kJ) T = 25°C = 298 K ΔG = (-139 kJ) – [(298K) (0.277 kJ/K)] = -222 kJ Reaction is spontaneous because ΔG is negative.

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 f ˚ products - ΔG f ˚ reactants ΔG f ˚ products = [(1mol)( kJ/mol) + (1mol)(0)] = kJ ΔG f ˚ reactants = [(1mol)(0) + (1mol)( kJ/mol)] = kJ ΔG reaction = kJ – ( kJ) = 91.4 kJ

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

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

Homework  Sec 10.4 review, pg. 367: #3-5, 7-11  Concept Review: “Order and Spontaneity” Next Time:  Quiz over section 10.4/Lab Write-up  Tues/Wed: Calorimetry Lab Chapter 10 test/concept review due: Friday, 11-16: “A” Day Monday, 11-19: “B”: Day