© University of South Carolina Board of Trustees Chapt. 17 Thermodynamics Sec. 4 2 nd Law - Quantitative

© University of South Carolina Board of Trustees  H,  S,  G from Tables Hess’ Law for  H ( Chapt , Eq. 5.13) Same for  S Same for  G (at 25 ° C = 298 K)

© University of South Carolina Board of Trustees Example:  S from Tables Calculate the standard entropy change for the reaction 2C 2 H 6(g) + 7O 2(g)  4CO 2(g) + 6H 2 O (liq)

© University of South Carolina Board of Trustees Example:  G from Tables Standard Conditions What is  G rxn for the reaction CO 2(g) + C (s)  2CO (g) at 25 °C?

© University of South Carolina Board of Trustees  G at a Non-Standard T  G =  H - T  S

© University of South Carolina Board of Trustees  G at a Non-Standard T  GT  G =  H - T  S strong temperature dependence

© University of South Carolina Board of Trustees  G at a Non-Standard T  H  S  G =  H - T  S weak temperature dependence strong temperature dependence

© University of South Carolina Board of Trustees  G at a Non-Standard T  H  S  G =  H - T  S use values from table (25 °C)

© University of South Carolina Board of Trustees  G at a Non-Standard T T  G =  H - T  S use values from table (25 °C) use non-standard T here

© University of South Carolina Board of Trustees  G at a Non-Standard T  G  G =  H - T  S use values from table (25 °C) use non-standard T here get  G at non- standard T

© University of South Carolina Board of Trustees Example:  G from Tables Non-Standard Temperature What is  G rxn for the reaction CO 2(g) + C (s)  2CO (g) at 1500 K? 25 °C

© University of South Carolina Board of Trustees 2 nd Law of Thermodynamics In a system at constant pressure and temperature: ● the Gibbs Free Energy must decrease in a spontaneous reaction  G < 0 ● at equilibrium, the Gibbs free energy does not decrease in either direction  G = 0

© University of South Carolina Board of Trustees 2 nd Law of Thermodynamics In a system at constant pressure and temperature: spontaneous reaction  G < 0 ● the Gibbs Free Energy must decrease in a spontaneous reaction  G < 0 ● at equilibrium ● at equilibrium, the Gibbs free energy does not decrease in either direction  G = 0

© University of South Carolina Board of Trustees Example: Find Equilibrium T Calculate the boiling point for CS 2 given:  H ° = kJ/mol,  S ° = J/K-mol. ● Boiling point: Liquid and gas in equilibrium CS 2(liq)  CS 2(g) ● Equilibrium:  G = 0

© University of South Carolina Board of Trustees Chapt. 17 Sec. 5 K eq from  G° (and  H° )

© University of South Carolina Board of Trustees  G at Non-Standard Concentrations

© University of South Carolina Board of Trustees Overview Energy (  E = w + q ) ● work ( w ) ● heat ( q ) Enthalpy (  H =  E + P  V) Entropy ( S ) Gibb Free Energy ( G )  G =  H - T  S K eq = exp(-  G / RT )

© University of South Carolina Board of Trustees Example: K eq from  G o Evaluate the equilibrium constant at 298 K for 2NO (g) + Br 2(g)  2NOBr (g) using the standard free energies of formation at 298 K given below.

© University of South Carolina Board of Trustees Example: K eq at a New T Evaluate the equilibrium constant at 500 K for 2NO (g) + Br 2(g)  2NOBr (g) (Recall: K eq = 298 K)

© University of South Carolina Board of Trustees K eq at a Non-Standard T 25 °CNew T Wrong !

© University of South Carolina Board of Trustees K eq at a Non-Standard T 25 °C T ≠ 25 °C Wrong ! Right!  G°  K 1 at T 1 = 25 ° C  H°  K 2 at T 2

© University of South Carolina Board of Trustees Example: K eq at a New T Evaluate the equilibrium constant at 500 K for 2NO (g) + Br 2(g)  2NOBr (g) (Recall: K eq = 298 K)

© University of South Carolina Board of Trustees Student Example What is the equilibrium constant for N 2 O 4(g)  2NO 2(g) at 25 °C?

© University of South Carolina Board of Trustees Student Example What is the equilibrium constant for N 2 O 4(g)  2NO 2(g) at 0 °C?