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 that are available to a system Nature proceeds toward the states that have the highest probabilities of existing

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

Example Which has higher positional S? Solid CO2 or gaseous CO2? N2 gas at 1 atm or N2 gas at 0.01 atm? Predict the sign of DS Solid sugar is added to water + because larger volume Iodine vapor condenses on cold surface to form crystals - because gs, less volume

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 Suniv = Ssys + Ssurr

2nd Law of Thermodynamics In any spontaneous reactions, there is always an increase in entropy of the universe ∆Suniv = ∆Ssys + ∆Ssurr If ∆Ssys is negative, it can still be spont. as long as the ∆Ssurr is larger and positive ∆Suniv > 0 : spontaneous ∆Suniv = 0 : no tendency, at equilibrium ∆Suniv < 0 : not spontaneous

Ch. 16: Spontaneity, Entropy, and Free Energy 16.3 The Effect of Temperature on Spontaneity

H2O(l)  H2O(g) ∆Ssys has + sign b/c of the increase in # of positions ∆Ssurr is determined mostly by heat flow Vaporization is endothermic so it removes heat from the surroundings So decreases random motion of surroundings Negative ∆Ssurr

Temperature’s Effects If the ∆Ssys and ∆Ssurr have different signs, the temperature determines the ∆Suniv For vaporization of water Above 100°C, ∆Suniv is positive Below 100°C, ∆Suniv is negative Impact of the transfer of heat will be greater at lower temperatures

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

Gibb’s Free Energy D Suniv= D Ssystem + D Ssur. Entropy of something will increase as it gets more heat. The particles will move more. The effect this heat has on the particles is dependant upon the temp. A small amount of heat at a high temp will not make much of an impact, while a small amount of heat at a small temp. will.

This leads to this relationship. S= D H T

Thus the D Ssurr. Can be expressed as - DHsys T The sign should be opposite that of the systems, so it is negative because from the systems perspective, the surrounding have an opposite sign.

Can be changed to D Suniv= D Ssys + - DHsys So D Suniv= D Ssys + D Ssur Can be changed to D Suniv= D Ssys + - DHsys T

T D Suniv= T D Ssys + - DHsys Multiply the whole thing be T and this gives: T D Suniv= T D Ssys + - DHsys Multiple by –1 and you get :

T D Suniv= -T D Ssys + DHsys Rearranging this gives: T D Suniv= DHsys -T D Ssys

-T D Suniv So: D G= DHsys - T D Ssys DG is what we call the relationship -T D Suniv So: D G= DHsys - T D Ssys

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 more negative the value for G0, the farther to the right the reaction will proceed

G0 = H0 - TS0 Calculating Free Energy Method #1 For reactions at constant temperature: G0 = H0 - TS0

H, S, G and Spontaneity H is enthalpy, T is Kelvin temperature G = H - TS H is enthalpy, T is Kelvin temperature 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 temperature) Spontaneous if the absolute value of TS is greater than the absolute value of H (high temperature)

An adaptation of Hess's Law: 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 CO2(g)  Cgraphite(s) + O2(g) G0 = +394 kJ Cdiamond(s)  Cgraphite(s) G0 = -3 kJ

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

Sample problem using free energy to find the boiling point of a substance. At what temperature is the following process spontaneous at 1 atm? Br2 (l)  Br2(g) DH= 31.0 Kj/mol & DSo = 93.0 j/k.mol G0 = H0 - TS0 make G0 = 0, so 0 = H0 - TS0 0 = 31000 j/mol - T (93.0 j/k.mol), solve for T…. T = 333 K is boiling point

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