AP CHEMISTRYNOTES Ch 8.8 Covalent bond energies Ch. 16 Spontaneity, Entropy, and Free Energy

8.8 Covalent Bond Energies and Chemical Reactions Bond energy is determined by average of bond dissociation energies(given). Given bond energy  one pair of e- shared (single) –Double? Triple? – # of shared electrons increased  bond energy is increased

8.8 Covalent Bond Energies and Chemical Reactions Bond energy value can be used to calculate approximate energy for reactions Broken bonds (reactants)  energy must be added (endo) Formation bonds (products)  energy must be released (exo) D=the bond E per mol of bonds, +only

Copyright © Houghton Mifflin Company. All rights reserved. 8–48–4 Table 8.4 Average Bond Energies (kj/mol)

Copyright © Houghton Mifflin Company. All rights reserved. 8–58–5 Table 8.5 Bond Lengths for Selected Bonds

8.8 Covalent Bond Energies and Chemical Reactions Ex.H 2 +F 2  2HF ∆H (H 2 ) = 432kJ ∆H (F 2 ) = 154kJ ∆H (HF) = 565kJ ∆H = 1molx432kJ+1mol x154kJ-2mol x 565kJ= - 544kJ Refer to Appendix 4

Ex 8.5 CH 4 (g) +2Cl 2 (g)+F 2 (g)  CF 2 Cl 2 (g)+2HF(g) +2HCl(g) C-H : 413kJ Cl-Cl : 239 kJ F-F : 154kJ C-Cl : 339kJ C-F :485kJ H-F : 565kJ H-Cl : 427kJ Refer to Appendix 4

Ex 8.5 CH 4 (g) +2Cl 2 (g)+F 2 (g)  CF 2 Cl 2 (g)+2HF(g) +2HCl(g) CH 4 = 4mol x 413kJ= 1652kJ Cl-Cl : 2 mol x 239 kJ=478kJ F-F : 1 mol x154kJ=154kJ C-Cl : 2 mol x 339kJ=678kJ (products) C-F : 2 mol x 485kJ=970kJ H-F : 2 mol x 565kJ=1130kJ H-Cl : 2 mol x 427 kJ=854kJ ∆H = 2284kJ –3632kJ= kJ/mol Refer to Appendix kJ required 3632kJ released

WS problem Refer to Appendix 4

16.1 Spontaneous Process and Entropy Spontaneous processes – without outside intervention may be fast or slow

16.1 Spontaneous Process and Entropy Examples: A ball rolls down a hill but never spontaneously rolls back up the hill If exposed to air and moisture, steel rusts spontaneously. However, the iron oxide in rust does not spontaneously change back to iron metal and oxygen gas. Wood burns spontaneously in an exothermic reaction to form carbon dioxide and water, but wood is not formed when carbon dioxide and water are heated together.

16.1 Spontaneous Process and Entropy Entropy, S (increase in S, spontaneous) The driving force for a spontaneous process is an increase in the entropy of the universe Viewed as a measure of molecular randomness or disorder –order to disorder –Lower entropy to higher entropy Your room?

16.1 Spontaneous Process and Entropy Entropy associated probability –To higher probability

Possible arrangements (states) of four molecules in a two-bulbed flask. Which figure shows high probability? Most likely to occur

Positional entropy increases from S  l  g Few positionMany position

Which is with the higher positional entropy? 1.Solid CO 2 and gaseous CO 2 2.N 2 gas at 1 atm and N 2 gas at 1.0x 1 -2 atm (Think of volume)

Making solution mixing solution Higher volume liquid  more position  increase of entropy Increase S  ∆S is positive Decrease S  ∆S is negative ∆S= S final - S initial

16.2 Entropy and the Second Law of Thermodynamics In any spontaneous process there is always an increase in the entropy of the universe. ∆S univ = ∆S sys + ∆S surr ∆S univ = positive  spontaneous ∆S univ = negative  spontaneous in opposite direciton ∆S unit = 0  no tendency to occur or equilibrium

16.3 The effect of Temperature on Spontaneity Exothermic process Heat transfer from system to surrounding KE increase, V increase,  S increase Endothermic process Opposite of exo  S decrease

16.3 The effect of Temperature on Spontaneity H 2 O (l)  H 2 O (g) ∆S sys is + ∆S surr is  Negative (Randomness decreased) ∆S sys and ∆S surr are opposite sign

The entropy changes that occur in the surroundings 1.Sign of ∆S surr depends on the direction of the heat Nature tends to seek the lowest possible energy flow 2.Magnitude of ∆S surr depends on the temperature.

System for exo Higher temperature  less impact Lower temperature  more impact S inversely related on temperature

Magnitude of the ∆S of surr = - Quantity of heat (J) Temperature (K) Exo: ∆S surr is + (∆H is -) Endo: ∆S surr is – (∆H is +) necessary

Ex 16.4 Calculate ∆S surr for each reaction at 25 o C and 1 atm kJ/K  change to J/K = 419J/K x10 3 J/K Case 1 is spontaneous, opposite of case 2 is spontaneous

Table 16.3 Interplay of ΔSsys and ΔSsurr in Determining the Sign of ΔSuniv

16.4 Free Energy G = H-TS  ∆G =∆H - T∆S T) ∆S univ = - ∆G T If ∆G is T, P, spontaneous Gibbs free energy

Ex. H 2 O (s)  H 2 O (l) ∆H o = 6.03 x 10 3 J/mol ∆S o = 22.1 J/ K∙mol T) ∆G o C=+2.2 x 10 2 (J/mol) ∆G o C=-2.2 x 10 2 (J/mol) Gibbs free energy

Table 16.4 Results of the Calculation of ΔS univ and ΔG° for the Process H 2 O  H 2 O(l) at -10°C, 0°C, and 10°C

Table 16.5 Various Possible Combinations of ΔH and Δs for a Process and the Resulting Dependence of Spontaneity on Temperature

Low temp  ∆H dominates High temp  ∆S dominates

Ex 16.5 Br 2 (l)  Br 2 (g) ∆H o = 31.0 kJ/mol ∆S o = 93.0 J/mol What is the normal bp?

Ex 16.5 What is the normal bp? Br 2 (l)  Br 2 (g) ∆H o = 31.0 kJ/mol ∆S o = 93.0 J/mol Assume ∆G o = 0 ∆H o = T∆S o T=333K Be careful to unit. You need the same unit

16.5 Entropy Change in Chemical Reactions Gaseous molecules ∆S is dominated by relative number of molecule of gas Less S More S ∆S is -

16.5 Entropy Change in Chemical Reactions 2NH 3 (g)  N 2 (g) +3H 2 (g) 2 mol  4 mol Focus on the volume of gases ∆S is ∆S is +

Ex CaCO 3 (s)  CaO(s) +CO 2 (g) o mol(g)  1mol 2. 2SO 2 (g) +O 2 (g)  2SO 3 (g) 3 mol (g)2 mol (g) ∆S o is + ∆S o is -

Third Law of Thermodynamics The entropy of a perfect crystal at 0K is zero. zero)

Entropy is extensive prop. (depends on amount) Ex (Calculate ∆S o C) 2NiS(s) + 3O 2 (g)  2SO 2 (g) + 2NiO(s) SubstanceS o (J/K mol) SO 2 (g)248 NiO (s)38 O 2 (g)205 NiS(s)53

-149J/K

Ex (Calculate ∆S o C) Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g) 3mol H 2 and 3molH 2 O  Can you predict ∆S o ? SubstanceS o (J/K mol) Al 2 O 3 (a)51 H 2 (g)131 Al(s)28 H 2 O (g)189

179J/K 3mol H 2 and 3molH 2 O  Can you predict ∆S o is positive? Depends on the shape Water is bent shape, more rotational, vibrational motions than H 2 as linear shape. More complex shape, higher ∆S o

16.6 Free Energy and Chemical Reactions ∆G o (Standard free energy change) There is no direct measurement Compare the relative tendency of reaction to occur at constant P, T More negative ∆G o, reaction occur

16.6 Free Energy and Chemical Reactions (First method) ∆G o = ∆H o – T∆S o (Constant T, 298K) Ex SO 2 (g) +O 2 (g)  2SO 3 (g) Calculate ∆H o, ∆S o, and ∆G o Substance∆H o f (kJ/mol)S o (J/K mol) SO 2 (g) O 2 (g)0205 SO 3 (g)

∆H o  -198kJ ∆S o,  -187J/K ∆G o  -142 kJ

16.6 Free Energy and Chemical Reactions (Second method) Use Hess’s law method with ∆G o Pg 768

16.6 Free Energy and Chemical Reactions (Second method) Use Hess’s law method with ∆G o Ex 16.10

16.6 Free Energy and Chemical Reactions (Third method) Use Standard free energies of formation ∆G o for elements is zero Ex 16.11

2CH 3 OH(g) +3O 2 (g)  2CO 2 (g) + 4H 2 O(g) SubstanceG o f (kJ/mol) CH 3 OH (g)-163 O 2 (g)0 CO 2 (g)-394 H 2 O (g)-229

-1378kJ (very favorable) Ex C 2 H 4 (g)+H 2 O(l)  C 2 H 5 OH (l) SubstanceG o f (kJ/mol) C 2 H 4 (g)68 H 2 O (g)-237 C 2 H 5 OH (l)-175

-6kJ (spontaneous)