CHEM 163 Chapter 20 Spring

3-minute exercise Is each of the following a spontaneous change? Water evaporates from a puddle A small amount of sugar dissolves in hot tea Methane burns in air A hamburger becomes uncooked 2

Thermodynamics First Law: Law of Conservation of Energy 3 Second Law: Systems change towards more disorder Internal E of a system heatwork Limitation: Explains change, but not direction

Spontaneous Change Change that occurs without continuous E input Change can only be spontaneous in one direction, under a given set of conditions 4 Enthalpy (∆H): heat gained or lost at constant P Sign of ∆H Exothermic or endothermic No information about spontaneity Entropy ( ∆ S): Freedom of particle motion (dispersed E of motion)

Energy Levels Each atom or molecule has quantized E levels –Electronic –Kinetic Vibrational, rotational, translational Microstate: combined E at any given point –each microstate is equally possible (equal E) for a given set of conditions 5 entropy Number of microstates Boltzmann constant = R/N A = 1.38 x J/K (J/K)

Entropy Change: Microstates 6 for 1 mol

Entropy Change: Heat Changes 7 Remove 1 grain of sand Gas does work on piston absorbs heat to maintain E Works for tiny changes (totally reversible)

Always Increasing Entropy All real processes occur spontaneously in direction that increases the entropy of the universe. 8 Perfect crystal at T = 0 K has S = 0 1 microstate Standard Molar Entropy (S°) S increase from 0 to standard state 1 atm (gases) 1 M (solutions) Pure substance, most stable form (liquids/solids)

What affects S°? 1.Temperature change ↑ T ↑ S 2.Phase change absorb heat ↑ S 9 As # of microstates (or kinetic E) increases, S increases

10 4.Atomic Size heavier atoms allotropes 5.Molecular Complexity more complex  closer E levels  more microstates  more types of movement  more microstates Only applies to molecules in same physical state 3.Dissolution Ions: increased S, except small, highly charged ions Molecules (solid or liquid): ∆ S ≈ 0 Gases: decreased S

3-minute practice What is the sign of ∆ S sys ? A pond freezes in winter Atmospheric CO 2 dissolves in the ocean 2 K (s) + F 2 (g)  2KF (s) 11

Standard Entropy of Reaction Increasing disorder: Predict ∆S rxn : Decreasing disorder: N 2 (g) + 3H 2 (g)  2 NH 3 (g) Change in # moles of gas Calculate ∆S rxn :

∆S universe Decrease in ∆S sys only if greater increase ∆S surr System acts as heat sink or drain 1.Exothermic 2.Endothermic at constant P Measure ∆H sys to determine ∆S surr

Spontaneous at 298 K? N 2 (g) + H 2 (g)  NH 3 (g) 32 1.Balance equation! 2.Calculate ∆S sys 3.Calculate ∆H sys 4.Calculate ∆S surr 5.Calculate ∆S univ

Entropy at Equilibrium Approaching equilibrium: At equilibrium: No net change

5-minute Practice Calculate ∆S sys for the combustion reaction of ammonia (producing nitrogen dioxide and water vapor).

Gibbs Free Energy Measure of spontaneity Combines enthalpy and entropy Spontaneous if… ∆S univ > 0 T∆S univ > 0 -T∆S univ < 0 ∆G sys < 0

∆G Spontaneous process Nonspontaneous process Process at equilibrium Standard Free Energy Change Standard Free Energy of Formation : E change when 1 mol of compound is made from its elements in standard states Element in standard state:

Nonspontaneous process: Process may occur if work is done to the system How much work is needed? ∆G and work (constant T & P) Spontaneous process: ∆G = maximum useful work done by the system w = ∆G w max only if process is totally reversible Actually does less w < w max Extra E lost as heat

Useful Work Excludes work done by or on atmosphere Some free energy is always lost to heat ∆ H sys ? ∆ S sys ? ∆ G sys ? < 0 > 0 < 0 ∆ G sys = w max that can be done by system ∆ G sys > w actually done by system In some multistep reactions, ∆G from one reaction can cause an otherwise nonspontaneous reaction to occur. “coupling of reactions”

What about T? Typically ∆H > T∆S For ∆G to be negative, need ∆H to be… What about at high T? negative T ∆S term can dominate Sign of ∆S becomes important 1.∆H 0 2.∆H > 0 & ∆S < 0 3.∆H > 0 & ∆S > 0 4.∆H < 0 & ∆S < 0 4 situations: ∆G > 0 at low T;∆G < 0 at high T ∆G < 0 at low T;∆G > 0 at high T ∆G < 0 ∆G > 0

High T v. Low T? Spontaneous “limit” at what temperature?

Equilibria and ∆G If Q < K, reaction… If Q > K, reaction… If Q = K, at equilibrium → ← ∆G < 0 ∆G > 0 ∆G = 0 Q/K < 1 > 1 = 1 proportional Make Q standard state (all values = 1) 0

Homework due MONDAY, May 11 th Chap 20: #19, 26, 30, 41, 50, 52, 58, 70, 78,