Ch. 17 Spontaneity, Entropy and Free Energy

Spontaneous Processes and Entropy First Law “Energy can neither be created nor destroyed" The energy of the universe is constant Spontaneous Processes Processes that occur without outside intervention Spontaneous processes may be fast or slow Many forms of combustion are fast Conversion of diamond to graphite is slow

Entropy (S) A measure of the randomness or disorder The driving force for a spontaneous process is an increase in the entropy of the universe Entropy is a thermodynamic function describing the number of arrangements (position and/or energy levels) that are available to a system existing in a given state Nature proceeds toward the states that have the highest probabilities of existing (½)^n

p. 775

Figure 17.4 Arrangements & Microstates Also see Table 17.1

Positional Entropy The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved Ssolid < Sliquid << Sgas The tendency to mix is due to the increased volume available to the particles of each component of the mixture.

Second Law of Thermodynamics "In any spontaneous process there is always an increase in the entropy of the universe" "The entropy of the universe is increasing" For a given change to be spontaneous, Suniverse must be positive (∆S > 0) Suniv = Ssys + Ssurr

Relationship between ∆H (kJ) and ∆S (J/K) - ∆H => + ∆S of surroundings, -∆Ssys +∆H => - ∆S of surroundings, +∆Ssys The ∆S’s are opposing, so who controls the ∆Suniv? In other words, which ultimately says whether a process is spontaneous? Temperature The impact of energy transfer of a given quantity of energy as heat to or from the surroundings will be greater at lower temperatures.

What about the entropy in the universe?! ∆Ssurr = -∆H/T, at constant pressure Why is there a negative in the equation? The minus sign changes the point of view from the enthalpy in the system to the entropy of the surroundings. Table 17.3 to determine the sign ∆Suniv Classwork: p. 807 #13, 15, 22, 23, 28, 30

H, S, G and Spontaneity G = H - TS G is free energy, H is enthalpy, T is Kelvin temperature, S is entropy Value of H Value of TS Value of G Spontaneity Negative Positive Spontaneous Nonspontaneous ??? Spontaneous if the absolute value of H is greater than the absolute value of TS (low temp); |∆H| > |T∆S| Spontaneous if the absolute value of TS is greater than the absolute value of H (high temp); |T∆S| > |∆H|

Calculating Entropy Change in a Reaction Entropy is an extensive property (a function of the number of moles) Generally, the more complex the molecule, the higher the standard entropy value

Standard Free Energy Change G0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states G0 cannot be measured directly The free energy of a rxn system changes as the reaction proceeds The more negative the value for G0, the farther to the right the reaction will proceed in order to achieve equilibrium Equilibrium is the lowest possible free energy position for a reaction

For reactions at constant T and P: Calculating Free Energy Method #1 For reactions at constant T and P: G0 = H0 - TS0 < 0 to be spontaneous at a given T Ex: Is it a given reaction spontaneous at 400 K? H0 : 79.3 kJ/mol S0 : 162.1 J/K mol Plug into equation: =79.3 kJ/mol – (400K)(0.1621kJ/Kmol) G0 = 14.46 kJ/mol <- Both are positive… which one dominates? <- not very spontaneous at 400 K.

Calculating Free Energy: Method #2 An adaptation of Hess's Law: Cdiamond(s) + O2(g)  CO2(g) G0 = -397 kJ Cgraphite(s) + O2(g)  CO2(g) G0 = -394 kJ Cdiamond(s) + O2(g)  CO2(g) G0 = -397 kJ CO2(g)  Cgraphite(s) + O2(g) G0 = +394 kJ Cdiamond(s)  Cgraphite(s) G0 = -3 kJ

Calculating Free Energy Method #3 Using standard free energy of formation (Gf0): Gf0 of an element in its standard state is zero

Classwork Appendix 4 begins on p.A19 P.808 #37, 41, 46, 49 choose 2 of 4, 51, 56, 57

Sec. 17.7 The Dependence of Free Energy on Pressure Enthalpy, H, is not pressure dependent Entropy, S entropy depends on volume, so it also depends on pressure Slarge volume > Ssmall volume Slow pressure > Shigh pressure

Sec. 17.8 Free Energy and Equilibrium A given process will be spontaneous until the free energy is minimized and G =0 Equilibrium point occurs at the lowest value of free energy available to the reaction system Not necessarily to completion At equilibrium, G = 0 and Q = K (See figure 17.8)

G K Shift G = 0 K = 1 None G < 0 K > 1 Right G > 0 To find k: G = -RT ln(K) G K Shift G = 0 K = 1 None G < 0 K > 1 Right G > 0 K < 1 Left

Temperature Dependence of K Notice: Y=mx + b So, ln(K)  1/T

This agrees well with the low K at 400 K Remember this example? A given rxn has the following T-dependent values of K. Is it spontaneous at 400 K? H0 : 79.3 kJ/mol S0 : 162.1 J/K mol Plug into equation: =79.3 kJ/mol – (400K)(0.1621kJ/Kmol) G0 = 14.46 kJ/mol This agrees well with the low K at 400 K T 350 400 450 K 3.98 x 10^-4 1.41 x 10⁻² 1.86 x 10⁻¹

Sec. 17.9 Free Energy and Work The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy The amount of work obtained is always less than the maximum Henry Bent's First Two Laws of Thermodynamics First law: You can't win, you can only break even Second law: You can't break even