Thermodynamics Chapter 19

19.1 Spontaneous Processes – process that occurs without any outside intervention, the internal energy alone determines if a reaction will occur

Spontaneous Processes example:

Spontaneous Processes example: when T>0 o C, ice melts spontanously; when T<0 o C water turns to ice spontaneously; when T= 0 o C, the system is at equilibrium and neither process is preferred

Reversible process - when a system and surroundings is changed in such a way that they can be restored by reversing the change exactly

Reversible process Whenever a chemical system is in equilibrium, reactants and products can interconvert reversibly

Irreversible Process - a change to a system and surroundings that can be restored by taking a new path to get back and changing the surroundings

Sum Up 1. Whenever a chemical system is in equilibrium, reactants and products can interconvert reversibly

Sum Up 2. In any spontaneous process, the path between reactants and products is irreversible

19.2 Entropy disorder of a system The more disorder of a system, the larger the entropy  S = S final -S initial

Entropy When  S > 0, the system is disordered When  S < 0, the system is more ordered or less random

Second Law of Thermodynamics states the entropy of the universe increases in any spontaneous process (entropy is not conserved)

19.3 The Molecular Interpretation of Entropy Entropy changes are related to the way the particles in a system can be arranged

19.3 The Molecular Interpretation of Entropy Particles change through translational motion, vibrational motion, and rotational motion

19.3 The Molecular Interpretation of Entropy Entropy and movement decrease with decreasing T

Third Law of Thermodynamics

entropy of a pure crystalline substance at absolute zero (0 Kelvin) is zero

Entropy Generalizations: Entropy is expected to increase for processes in which: 1. Liquids or solutions are formed from solids

Entropy Generalizations: 2. Gases are formed from either solids or liquids

Entropy Generalizations: 3. The number of molecules of gas increases during a chemical rxn

Sample Exercise 19.3 Predicting the Sign of ΔS Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature:

Sample Exercise 19.4 Predicting Which Sample of Matter Has the Higher Entropy Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a)1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C, (b)2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C, (c)1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.

19.4 Entropy Changes in Rxns Standard Molar Entropies (S o ) - the molar entropy values of substances in their standard states, pure substance at 1 atm; found on appendix C

Standard Molar Entropies (S o ) Characteristics: 1. Molar entropies of elements are not 0 2. S o (g) > S o (l) >S o (s) 3. S o increases w/ increasing MM 4. Increase w/ increasing atoms

Entropy Change  S o =  nS o (products) -  mS o (reactants), where m and n are coefficients of the balanced chemical eq

Sample Exercise 19.5 Calculating ΔS from Tabulated Entropies Calculate ΔSº for the synthesis of ammonia from N 2 (g) and H 2 (g) at 298 K: N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

19.5 Gibbs Free Energy G=H-TS  G =  H-T  S at constant temperature

Relationship between G and spontaneity 1. If  G is negative, the rxn is spontaneous in the forward direction

Relationship between G and spontaneity 2. If  G is zero, the rxn is at equilibrium

Relationship between G and spontaneity 3. If  G is positive, the rxn in the forward direction is nonspontaneous; work must be supplied from the surroundings to make it occur. The reverse rxn is spontaneous

Sample Exercise 19.6 Calculating Free-Energy Change from ΔH°, T, ΔS° Calculate the standard free energy change for the formation of NO(g) from N 2 (g) and O 2 (g) at 298 K: N 2 (g) + O 2 (g) → 2 NO(g) given that ΔH° = kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these circumstances?

Standard Free-Energy Changes  G o =  nG f o (products) -  mG f o (reactants)

Standard Free-Energy Changes

when  G>0 (nonspontaneous), it is the measure of the minimum amount of work that must be done to cause the process to occur

Sample Exercise 19.7 Calculating Standard Free-Energy Change from Free Energies of Formation (a) Use data from Appendix C to calculate the standard free-energy change for the following reaction at 298 K: P 4 (g) + 6 Cl 2 (g) → 4 PCl 3 (g)

C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(l) ΔH° = –2220 kJ (a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH°. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct? Sample Exercise 19.8 Estimating and Calculating ΔG°

19.6 Free Energy and Temperature The value of  H and  S do not change much with temperature The value of T plays a major role in determining  G

19.6 Free Energy and Temperature

Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity The Haber process for the production of ammonia involves the equilibrium Assume that ΔH°(-92.38kJ) and ΔS°(-198.3J/K) for this reaction do not change with temperature. (a) Predict the direction in which ΔG° for this reaction changes with increasing temperature. (b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C.

19.7 Free Energy and the Equilibrium Constant Calculating  G at nonstandard conditions  G=  G o + RTlnQ, where R is the ideal gas constant 8.314J/mol-K, T is abs T, and Q is the reaction quotient

Free Energy and the Equilibrium Constant  G is related to Keq, at equilibrium;  G=0 and Q = Keq  G o = - RT lnKeq Keq= e -  G/RT

Sample Exercise Calculating an Equilibrium Constant for ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25 °C for the reaction involved in the Haber process: (see 19.9)