Spontaneity Entropy and Free Energy

Spontaneity Entropy and Free Energy WHAT DRIVES A REACTION TO BE SPONTANEOUS?

ENTHALPY (H) heat content (exothermic reactions are generally favored)

1st Law of Thermodynamics The total energy in the universe is constant. DE = q (heat) + w (work) -heat out -work on surroundings + heat in + work on system Calculating DH: • stoichiometry—using info given with equation • H rxn = ∑ n DHf° products -- ∑ n DHf° reactants • Hess’s Law—adding up equations and DH’s • calorimetry q = mCDT or q = CDT • bond energies ∑ bonds broken -- ∑ bonds formed • Note: pure elements have DHf° values of 0

ENTROPY ‘S’The Second Law of Thermodynamics The universe is constantly increasing disorder. DSuniv = DSsystem + DSsurroundings ΔSsurroundingsbased on heat flow exothermic + DSsurr endothermic -DSsurr

ENTROPY (S) Disorder of a system (more disorder is favored) Nature tends toward chaos! Think about your room at the end of the week! Factors that can indicate entropy changes: • Phase changes • Temperature changes • Volume changes • Mixing substances • Change in number of particles • Change in moles of gas

Spontaneous reactions are those that occur without outside intervention. They may occur fast OR slow (that is kinetics). Some reactions are very fast (like combustion of hydrogen) other reactions are very slow (like graphite turning to diamond)

Predicting the entropy of a system based on physical evidence: • The greater the disorder or randomness in a system, the larger the entropy. • The entropy of a substance always increases as it changes from solid to liquid to gas. • When a pure solid or liquid dissolves in a solvent, the entropy of the substance increases (carbonates are an exception! --they interact with water and actually bring MORE order to the system)

Predicting the entropy of a system based on physical evidence: • When a gas molecule escapes from a solvent, the entropy increases. • Entropy generally increases with increasing molecular complexity (crystal structure: KCl vs CaCl2) since there are more MOVING electrons! • Reactions increasing the number of moles of particles often increase entropy.

Exercise 2 Predicting Entropy Changes Predict the sign of the entropy change for each of the following processes. A: Solid sugar is added to water to form a solution. B: Iodine vapor condenses on a cold surface to form crystals.

Solution: A: +∆S Sugar molecules have been randomly dispersed in the water. Greater entropy + DS B: -∆S Gaseous iodine is forming a solid. Less entropy - DS

Sample Problem A Which of the following has the largest increase in entropy? a) CO2(s)  CO2(g) b) H2(g) + Cl2(g)  2 HCl(g) c) KNO3(s)  KNO3(l) d) C(diamond) C(graphite)

Answer: • the substance changes from a highly organized state to a more disorganized state.

Choose the substance expected to have the greater absolute entropy. • Pb(s) or Cgraphite • He(g) at 1 atmosphere or He(g) at 0.5 atmosphere • H2O(l) or CH3CH2OH(l) at the same temperature • Mg(s) at 0°C or Mg(s) at 150 °C both at the same pressure

Pb has greater molar entropy. Pb, with metallic bonding, forms soft crystals with high amplitudes of vibration; graphite has stronger (covalent) bonds is more rigid and thus more ordered. • He(g) at 0.05 has greater molar entropy. At lower pressure (greater volume) He atoms have more space in which to move so are more random. • CH3CH2OH has greater molar entropy. Ethanol molecules have more atoms and thus more vibrations; water exhibits stronger hydrogen bonding • Mg(s) at 150 °C has greater molar entropy. At the higher temperature the atoms have more kinetic energy and vibrate faster and, thus, show greater randomness.

ENTROPYThe Third Law of Thermodynamics • The entropy of a perfect crystal at 0 K is zero. • not a lot of perfect crystals out there so, entropy values are RARELY ever zero—even elements

This means the absolute entropy of a substance can then be determined at any temperature higher than 0 K. (Handy to know if you ever need to defend why G & H for elements = 0. . . BUT S does not!) So what?

BIG MAMMA, verse 2 S°rxn = S°(products) - S°(reactants) S is + when disorder increases (favored) S is – when disorder decreases Units are usually J/K mol (not kJ ---tricky!)

Sample Problem B Calculate the entropy change at 25C, in J/K for: 2 SO2(g) + O2(g)  2 SO3(g) Given the following data: SO2(g) 248.1 J/K mol O2(g) 205.3 J/Kmol SO3(g) 256.6 J/Kmol Entropy change = -188.3 J/K

ENTROPY CHANGES FOR REVERSIBLE PHASE CHANGES (that’s a phase change at constant temperature) ΔS = heat transferred = q temperature at which change occurs T **where the heat supplied (endothermic) (q > 0) or evolved (exothermic) (q < 0) is divided by the temperature in Kelvins

EX: water (l @ 100) → water (g @ 100) the entropy will increase. Taking favored conditions into consideration, the equation above rearranges into: S surr = - H T Give signs to ΔH following exo/endo guidelines! (If reaction is exo.; entropy of surroundings increases—makes sense!)

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

Solution: ∆Ssurr= 419 J/K ∆Ssurr = -2.61 × 103 J/K

SUMMARY ENTROPY: S = + MORE DISORDER (FAVORED CONDITION) S = - MORE ORDER

Whether a reaction will occur spontaneously may be determined by looking at the S of the universe. ΔS system + ΔS surroundings = ΔS universe • IF ΔS universe is +, then reaction is spontaneous • IF ΔS universe is -, then reaction is NONspontaneous

Consider 2 H2 (g) + O2 (g) → H2O (g) ignite & rxn is fast! ΔSsystem = -88.9J/K (this was found with S°rxn = S°(products) - S°(reactants)) Entropy declines (due mainly to 3 → 2 moles of gas!)

First law of thermodynamics demands that this energy is transferred from the system to the surroundings so... -ΔHsystem = ΔHsurroundings OR - (- 483.6 kJ) = + 483.6 kJ

ΔS°surroundings = ΔH°surroundings = + 483.6 kJ T 298 K = 1620 J/K

Now we can find ΔS°universe ΔS system + ΔS surroundings = ΔS universe (-88.9 J/K) + (1620 J/K) = 1530 J/K Even though the entropy of the system declines, the entropy change for the surroundings is SOOO large that the overall change for the universe is positive.

Bottom line: A process is spontaneous in spite of a negative entropy change as long as it is extremely exothermic. Sufficient exothermicity offsets system ordering.

Spontaneity Entropy and Free Energy