Spontaneity, Entropy, and Free Energy Chapter 17 Spontaneity, Entropy, and Free Energy

Fist Law of Thermodynamics Energy is not created nor destroyed Total energy in the universe is always held constant Energy is transformed from one form to another

Spontaneity Spontaneous means without intervention Can be fast OR slow Chapter 12 = rate laws (talked about rxn rates - activation energy, temperature, concentration, catalysts, etc.) Chapter 17 = thermodynamics show reaction direction and only initial (reactants) and final (products) - don’t need information about the pathway

Examples of Spontaneity A ball rolling down a hill Steel rusting exposed to air and moisture Gas uniformly fills its container Heat flows from hot to cold Water freezes <0°C and melts >0°C All of these processes happen spontaneously in one direction and never in reverse Gravity? Exothermicity? WHY?

Entropy (not to be confused with Enthalpy) Characteristic common to all spontaneous processes Represented by “S” Defined as a measure of molecular randomness or disorder It is natural for disorder to increase

Claudia Witt: Sock Drawer Examples Closely associated with probability (nature more likely to proceed towards states that have the highest probability of existing - called positional probability)

Positional Probability Largest entropy Increases as a gas expands Increases as solids change to liquids and gases When solutions are made, volume, positional probability, and entropy increase

Examples Choose the substance with the higher positional entropy (per mole) at a given temperature Solid CO2 and gaseous CO2 N2 gas at 1 atm and N2 gas at 1.0 X 10-2 atm Predict the sign of the entropy change for each of the following processes: Solid sugar is added to water to form a solution Iodine vapor condenses on a cold surface to form crystals

Second Law of Thermodynamics In any spontaneous process there is always an increase in the entropy of the universe (first = energy of universe is constant) ∆Suniv = ∆Ssys + ∆Ssurr Must consider both the system and surroundings to determine if a given process is spontaneous (∆Suniv must be +)

Sign of ∆Ssurr Depends on the direction of the heat flow Exothermic - heat flows into the surroundings, increasing random motions and entropy of the surroundings --> ∆Ssurr is positive Endothermic - heat flows into the system (away from the surroundings), decreasing random motions and entropy of the surroundings --> ∆Ssurr is negative

Magnitude of ∆Ssurr Depends on temperature At a high temperature, energy transferred as heat won’t really make a percent change in randomness (entropy) At a low temperature, energy transferred as heat will make a percent change in randomness (entropy)

Entropy vs. Enthalpy ∆Ssurr can be expressed in terms of ∆H (needs a sign - shows direction - and a number - shows quantity of energy) at constant pressure: ∆Ssurr = - ∆H/T Temperature in Kelvins

Example Calculate ∆Ssurr for each of these reactions at 25°C and 1 atm: Sb2S3(s) + 3Fe(s) -> 2Sb(s) + 3FeS(s) ∆H = -125 kJ Sb4O6(s) + 6C(s) -> 4Sb(s) + 6CO(g) ∆H = 778 kJ

MEMORIZE

Free Energy (G) G = H -TS H = enthalpy T = Kelvin temperature S = entropy At a constant temperature, the change in free energy is ∆G = ∆H -T∆S. ∆Suniv = - ∆G/T only at constant temperature and pressure! **All quantities refer to the system If you see a degree (°) sign after a substance, it represents that it is in its standard state (pg. 255)

∆H° = 31.0 kJ/mol and ∆S° = 93.0 J/Kmol Example At what temperatures is the following process spontaneous at 1 atm? Br2(l) -> Br2(g) ∆H° = 31.0 kJ/mol and ∆S° = 93.0 J/Kmol What is the normal boiling point of liquid Br2?

MEMORIZE! FYI: if a reaction is not spontaneous as written, it is spontaneous in reverse

Entropy in Chemical Reactions Compare coefficients of products and reactants in balanced chemical reactions Fewer molecules means fewer possible configurations (means entropy decreases) STATES of matter must be taken into account (gases have more entropy than solids)

Example Predict the sign of ∆S° for each of the following reactions: (NH4)2Cr2O7(s) -> Cr2O3(s) + 4H2O(l) + N2(g) Mg(OH)2(s) -> MgO(s) + H2O(g) PCl5(g) -> PCl3(g) + Cl2(g)

Third Law of Thermodynamics The entropy of a perfect crystal at 0K is zero (“perfect crystal” is an unattainable ideal - taken as standard, but never actually observed) Standard entropy values S° in Appendix 4 in your book ∆S°reaction = ∑npS°products - ∑nrS°reactants Generally, the more complex the molecule, the higher the standard entropy value (∆H° is calculated the same way)

Example Calculate ∆S° for each of the following reactions using Appendix 4 in your book N2O4(g) -> 2NO2(g) Fe2O3(s) + 2Al(s) -> 2Fe(s) + Al2O3(s) 4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)

Free Energy and Chemical Reactions Standard free energy change (∆G°): the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states N2(g) + 3H2(g) <-> 2NH3(g) ∆G° = -33.3 kJ Can not be directly measured, but it can be calculated from other measured quantities The more negative ∆G°, the further to the RIGHT the reaction will go to reach equilibrium Pressure and concentration must be held constant Three ways to calculate ∆G°

Calculating ∆G° ∆G° = ∆H° - T∆S° - Requires constant temperature 2. Similar to Hess’s Law - Flip equation = negates, multiply by constants… 3. ∆G°rxn = ∑npG°f(prod) - ∑nrG°f(reactants) - Uses standard free energy of formations (appendix 4)

Example: ∆G° = ∆H° - T∆S° Using data for ∆H° and ∆S°, calculate ∆G° for the following reactions at 25°C and 1 atm. Cr2O3(s) + 2Al(s) -> Al2O3(s) + 2Cr(s) Answer: -537 kJ; reaction is spontaneous C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g) Answer: -2075 kJ; highly spontaneous

Example: Like Hess’s Law Given the following data, calculate ∆G° for S(s) + O2(g) -> SO2(g): Eq 1: S(s) + 3/2O2(g) -> SO3(g) ∆G° = -371 kJ Eq 2: 2SO2(g) + O2(g) -> 2SO3(g) ∆G° = -142 kJ Answer: -300 kJ

Example: ∆G°rxn = ∑npG°f(prod) - ∑nrG°f(reactants) Calculate ∆G° for the reaction: C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g) using ∆G° data from Appendix 4. Compare the answer with the answer we calculated previously for this reaction Answer: -2074 kJ

Pressure vs. Free Energy Entropy depends on volume, which inversely relates to pressure via gas laws Slarge volume > Ssmall volume Slow pressure > Shigh pressure Equations used in this chapter can be manipulated (pg. 795) to give: G = G° + RT lnQ Q = reaction quotient using initial pressures of gases, R = 8.3145 J/K mol, T = Kelvins

Example Calculate ∆G at 700K for the following reaction: C(s, graphite) + H2O(g) <-> CO(g) + H2(g) Initial Pressures: PH2O = 0.85 atm, PCO = 1.0 X 10-4 atm, PH2 = 2.0 X 10-4 atm Answer: +92kJ = not spontaneous under standard conditions…when temperature and pressure are taken into account, ∆G = -10kJ and is spontaneous

Free Energy and Equilibrium Equilibrium Point occurs at the lowest value of free energy available to the reaction system Gforward rxn = Greverse rxn If ∆G < 0, it means that ∆Greactants > ∆Gproducts and rxn will go to the RIGHT until Greactants = Gproducts If ∆G > 0, it means that Greactants < Gproducts and rxn will go to the LEFT until Greactants = Gproducts

Relationship Between ∆G° and K ∆G° = -RT ln K K = equilibrium constant Example: C(s, graphite) + H2O(g) <-> CO(g) + H2(g) where T = 700 K and ∆G° = 92 kJ, determine the direction of the reaction with the following initial pressure of each gas is PH2O = 0.67 atm, PCO = 0.23 atm, PH2 = 0.51 atm. Answer: ∆G = 82kJ and reaction will go to the left

Cr2O3(s) + 2Al(s) <-> Al2O3(s) + 2Cr(s) Example #2 Given ∆G° = -537 kJ, calculate K for the following reaction at 25°C Cr2O3(s) + 2Al(s) <-> Al2O3(s) + 2Cr(s) Answer: K = 1.3 X 1094, reaction is spontaneous