Chemical Thermodynamics BLB 12 th Chapter 19

Chemical Reactions 1. Will the reaction occur, i.e. is it spontaneous? Ch. 5, How fast will the reaction occur? Ch How far will the reaction proceed? Ch. 15

Review terms energy heat work pathway state function system surroundings

Review terms, cont. exothermic endothermic enthalpy enthalpy change standard state std. enthalpy of formation 1 st Law of Thermodynamics

19.1 Spontaneous Processes Spontaneous – proceeds on its own without any outside assistance  product-favored; K > 1  not necessarily fast  the direction a process will take if left alone and given enough time. Nonspontaneous  opposite direction of spontaneous  reactant-favored; K < 1  not necessarily slow

Spontaneity and Energy Examples of spontaneous systems:  Brick falling  Ball rolling downhill  Hot objects cooling  Combustion reactions Are all spontaneous processes accompanied by a loss of heat, that is, exothermic? Spontaneity is temperature-dependent.

Reversible & Irreversible Systems Reversible – a change in a system for which the system can restored by exactly reversing the change – a system at equilibrium ex. melting ice at 0°C Irreversible – a process that cannot be reversed to restore the system and surroundings to their original states – a spontaneous process ex. melting ice at 25°C See p (last paragraph of section)

19.2 Entropy and the 2 nd Law of Thermodynamics Entropy, S – measure of randomness  State function  Temperature-dependent A random (or dispersed) system is favored due to probability. “Entropy Is Simple – If We Avoid the Briar Patches” Frank Lambert, Occidental College, ret.

Entropy Change ΔS = S final − S initial (a state function) (isothermal) as for phase changes. ΔS > 0 is favorable

Calculating ΔS for Changes of State

Problem 26 The freezing point of Ga is 29.8°C and the enthalpy of fusion is 5.59 kJ/mol. a. Is ΔS + or − for Ga(l) → Ga(s) at the freezing point? b. Calculate the value of ΔS when 60.0 g of Ga(l) solidifies.

System & Surroundings Dividing the universe: 1.System – dispersal of matter by reaction: reactants → products 2.Surroundings – dispersal of energy as heat

2 nd Law of Thermodynamics The entropy of the universe increases for any spontaneous process.  ΔS univ > 0  ΔS univ = ΔS sys + ΔS surr For a “reversible” process: ΔS univ = 0. For an irreversible process:  Net entropy increase ►spontaneous  Net entropy decrease ► nonspontaneous

19.3 The Molecular Interpretation of Entropy Molecules have degrees of freedom based upon their motion  Translational  Vibrational  Rotational Motion of water (Figure 19.8, p. 796) Motion of water Lowering the temperature decreases the entropy.

Boltzman & Microstates S = k ln W (W = # of microstates) If # microstates ↑, then entropy ↑. Increasing volume, temperature, # of molecules increases the # of microstates.

Examples of systems that have increased entropy Entropy increases for: Changes of state: solid → liquid → gas (T) Expansion of a gas (V) Dissolution: solid → solution (V) Production of more moles in a chemical reaction (# of particles) Ionic solids: lower ionic charge S° (J/mol·K) Na 2 CO MgCO 3 66

Changes of State H 2 O stateS° (J/mol·K) l69.91 g188.83

# microstates ↑, system entropy ↑

Dissolution

Expansion of a Gas

2 NO(g) + O 2 (g) → 2 NO 2 (g) S° + or −?

Problem 22 TNT (trinitrotoluene) Detonation 4 C 3 H 5 N 3 O 9 (l) → 6 N 2 (g) + 12 CO 2 (g) + 10 H 2 O(g) + O 2 (g) a)Spontaneous? b)Sign of q? c)Can the sign of w be determined? d)Can the sign of ΔE be determined?

3 rd Law of Thermodynamics The entropy, S, of a pure crystalline substance at absolute zero (0 K) is zero.

19.4 Entropy Changes in Chemical Reactions Standard molar entropy values, S° (J/mol·K): increase in value as temperature increases from 0 K have been determined for common substances (Appendix C, pp ) increase with molar mass increase with # of atoms in molecule

Calculating ΔS° sys ΔS° sys = ∑nS°(products) - ∑mS°(reactants) (where n and m are coefficients in the chemical equation)

2 NO(g) + O 2 (g) → 2 NO 2 (g) S° =

Problem 54(c)

Problem

Entropy Changes in the Surroundings Heat flow affects surroundings. As T increases, ΔH becomes less important. As T decreases, ΔH becomes more important.

Calculating ΔS° univ ΔS univ = ΔS sys + ΔS surr by obtaining ΔS sys and ΔS surr If ΔS univ > 0, the reaction is spontaneous. But there is a better way – one in which only the system is involved.

19.5 Gibbs Free Energy The spontaneity of a reaction involves both enthalpy (energy) and entropy (matter). Gibbs Free Energy, ΔG makes use of ΔH sys and ΔS sys to predict spontaneity. ΔG sys represents the total energy change for a system. G = H – TS or ΔG = ΔH – TΔS or, under standard conditions: ΔG° = ΔH° – TΔS°

Gibbs Free Energy If:  ΔG < 0, forward reaction is spontaneous  ΔG = 0, reaction is at equilibrium  ΔG > 0, forward reaction is nonspontaneous In any spontaneous process at constant temperature and pressure, the free energy always decreases. ΔG is a state function. ΔG f ° of elements in their standard state is zero.

Calculating ΔG° sys ΔG° sys = ΔH° sys − TΔS° sys or ΔG° sys = ∑nΔG° f (products) - ∑mΔG° f (reactants) (where n and m are coefficients in the chemical equation)

19.6 Free Energy and Temperature ΔHΔHΔSΔS−TΔS−TΔSΔGΔGReaction −+−always − spontaneous at all T K > 1 +−+always + nonspontaneous at all T K < 1 −−+ low T T spontaneous at low T nonspontaneous at high T ++− low T T nonspontaneous at low T spontaneous at high T

Problem 60(b) & 83

Driving force of a reaction For a reaction where ΔG < 0: Enthalpy-driven – if ΔH < 0 and ΔS < 0; at low temp. Entropy-driven – if ΔH > 0 and ΔS > 0; at high temp. “cross-over point” is where ΔG = 0

Problem

19.7 Free Energy and K If conditions are non-standard: ΔG = ΔG° + RT lnQR = J/mol·K If at equilibrium: ΔG = ΔG° + RT lnQ = 0 ΔG° = −RT lnK

Problem

Problem 81