Enthalpy ·△Ho=heat released or absorbed during a chemical reaction at standard conditions(1atm,25 degrees C). ·Exothermic(-△)：Heat is released. Endothermic (+AH):Heat is absorbed. Reactions favor products with the lowest enthalpy(strongest bonds). 2013 Pearson Education,Inc. Chapter4 16
© 2013 Pearson Education, Inc. Enthalpy • DHo = heat released or absorbed during a chemical reaction at standard conditions (1atm, 25 degrees C). • Exothermic (-DH): Heat is released. • Endothermic (+DH): Heat is absorbed. • Reactions favor products with the lowest enthalpy (strongest bonds). Chapter 4 16
Entropy ·△So=change in randomness,disorder,. or freedom of movement. Increasing heat,volume,or number of particles increases entropy. Spontaneous reactions maximize disorder and minimize enthalpy. ·In the equation△Go=△Ho-T△S,the entropy value is often small. 2013 Pearson Education,Inc. Chapter 4 17
© 2013 Pearson Education, Inc. Entropy • DSo = change in randomness, disorder, or freedom of movement. • Increasing heat, volume, or number of particles increases entropy. • Spontaneous reactions maximize disorder and minimize enthalpy. • In the equation DGo = DHo - TDSo , the entropy value is often small. Chapter 4 17
Solved Problem 1 Calculate the value of AG for the chlorination of methane. Solution △G°=-2.303 RT(log Keg) Keg for the chlorination is 1.1 x 1019,and logK=19.04 At 25 C(about 298 K),the value of RTis RT=(8.314 J/kelvin-mol)(298 kelvins)=2478 J/mol,or 2.48 kJ/mol Substituting,we have △G°=(-2.303)2.478kJ/mol)19.04)=-108.7 kJ/mol(-25.9kcal/mol) This is a large negative value for AG,showing that this chlorination has a large driving force that pushes it toward completion. 2013 Pearson Education,Inc. Chapter4 18
© 2013 Pearson Education, Inc. Calculate the value of DG° for the chlorination of methane. DG° = –2.303RT(log Keq) Keq for the chlorination is 1.1 x 1019, and log Keq = 19.04 At 25 °C (about 298 K), the value of RT is RT = (8.314 J/kelvin-mol)(298 kelvins) = 2478 J/mol, or 2.48 kJ/mol Substituting, we have DG° = (–2.303)(2.478 kJ/mol)(19.04) = –108.7 kJ/mol (–25.9 kcal/mol) This is a large negative value for DG°, showing that this chlorination has a large driving force that pushes it toward completion. Solved Problem 1 Solution Chapter 4 18
Bond-Dissociation Enthalpies (BDEs) Bond dissociation requires energy (+BDE). Bond formation releases energy (-BDE). ·BDE can be used to estimate△Hfor a reaction. BDE for homolytic cleavage of bonds in a gaseous molecule. Homolytic cleavage:When the bond breaks,each atom gets one electron. Heterolytic cleavage:When the bond breaks,the most electronegative atom gets both electrons. 2013 Pearson Education,Inc. Chapter4 19
© 2013 Pearson Education, Inc. Bond-Dissociation Enthalpies (BDEs) • Bond dissociation requires energy (+BDE). • Bond formation releases energy (-BDE). • BDE can be used to estimate DH for a reaction. • BDE for homolytic cleavage of bonds in a gaseous molecule. ▪ Homolytic cleavage: When the bond breaks, each atom gets one electron. ▪ Heterolytic cleavage: When the bond breaks, the most electronegative atom gets both electrons. Chapter 4 19
Homolytic and Heterolytic Cleavages Homolytic cleavage (free radicals result) A·+·B△H°=bond-dissociation enthalpy 2:C AH=242 kJ/mol (58 kcal/mol) Heterolytic cleavage (ions result) A:B → A+ :B (CH)C-CI: (CH),C++:C: (AH varies with solvent) 2013PenEacafonn 2013 Pearson Education,Inc. Chapter 4 20
© 2013 Pearson Education, Inc. Homolytic and Heterolytic Cleavages Chapter 4 20