Advanced Thermochemistry Mrs. Stoops Chemistry

Chapter Problems Ch 19 p742: 16, 20, 28, 34, 38, 40, 46, 52, 56, 58, 75, 93

Important For Thermo.. Hess’s Law =  H for reaction will equal the sum of the  H for the steps  H rxn =  n  H products -  m  H reactants Q=mc  t Endothermic  H >0; exothermic  H <0 kJ; cal1 cal = 4.184J Heat capacity  E = q + w Potential and kinetic energy

1 st Law Energy is conserved – can transfer Spontaneous – happens on its own without outside help Spontaneous reactions – no extra energy is needed. –No reversible, has a definite direction –On a warm day, ice will only go in one direction, melt.

Homework Day 1 NO assigned problems, but LOOK over the odds as they are fair game for the chapter test.

2 nd Law of Thermodynamics Gases will expand spontaneously Examples - Febreeze, perfume, smelly stuff Molecules will spread out to become more disordered Processes where disorder increases are spontaneous which are less organized? –Solid or liquid –Liquid or gas –3 mol or 10 mol

Entropy Disorder S Higher disorder, the higher the entropy value (number)  S = S f – S i  S positive = the final is more disordered, spontaneous  S negative = less disorder, nonspontaneous  S can be related to temperature and heat

Example Hg is a liquid at room temperature. Normal freezing point is o C. Its molar heat of fusion is  H fus = kJ/mol. What is the entropy change when 50.0 g Hg (l) freezes at the normal freezing point? (fusion = melting) Since its freezing = kJ/mol q = kJ/mol x 0.25 mol = kJ q = kJ = -583 J Not spontaneous

Calculating Entropy changes Standard molar entropy = S o (like H o) Appendix C & page 729 Units = J/mol K  S o =  n S o products -  m S o reactants Example: Find  S o for: N 2 + 3H 2  2NH 3  S o = [(2x192.5)] – [ (3x130.6)] = J/mol K

Observations 1.S o for elements is NOT zero! (  H are zero) 2.S o for gases are greater than liquids which are greater than solids 3.S o tends to increase with molar mass 4.S o tends to increase with the # of atoms in the formula

Homework Day 2 Book: 23, 25, 31, 33, 35, 39, 41

Molecular Interpretation of Entropy Entropy is disorder = S (least) Solid  Liquid  Gas (most) # of moles will increase S More mass = More disorder Molecular level 2 NO (g) + O 2 (g)  2NO 2(g) Negative entropy = products have less mole “tying up more atoms” Reducing degrees of freedom (more freedom, more entropy)

3 Modes of Freedom 1.Translational motion = moving physically from point to point in space 2.Vibrational motion = shortening and lengthening of bond = vibrate like a spring 3.Rotational motion = spinning around an axis Energy is required for all 3 types more energy = more entropy (think 3yr old on a sugar high) Increase temperature = more entropy

Gibbs Free Energy S = entropyH = enthalpy Decide if the reaction is spontaneous G = H – TST in KELVIN!!!!  G =  H - T  ST is held constant, but used in eqn

What it tells us: If  G: Is negative = forward reaction is spontaneous Is zero = reaction is at equilibrium Is positive = the forward reaction is NOT spontaneous –Supply energy “to do” the rxn –Reverse will be spontaneous

Finding  G o  G o =  n  G f o products -  m  G f o reactants can find these values in appendix C  G o =  H o - T  S o At standard conditions

Homework Day 3 P 45, 49, 53, 55a,b, 57

Influence of Temperature  G o =  H o - T  S o HoHo - T  S o GoGo ---spontaneous +++Nonspontaneous -+- (low T)Spontaneous -++ (high T)Nonspontaneous +-+(low T)Nonspontaneous +-- (high T)spontaneous

Free Energy and Equilibrium  G = 0 is at equilibrium  G o can be related to K Most reactions occur at nonstandard conditions –Standard conditions: solid = pure, liquid = pure, gas = 1 atm, solutions = 1 M, element = 0 in std state  G =  G o + RT ln Q –R = J/mol K, T = Kelvin, Q = reaction quotient, –If Q = 1 at std conditions,  G =  G o

Continued Eqm  G = 0so...  G o = -RT ln KK = e -  G/RT  G o negative then, K>1  G o 0 then, K = 1  G o positive then, K<1

Homework Day 4 Book: 77, 81