Chapter 16 Spontaneity, Entropy and Free Energy DE Chemistry Dr. Walker

Laws Of Thermodynamics The first law of thermodynamics is a statement of the law of conservation of energy: Energy can be neither created nor destroyed. In other words, the energy of the universe is constant.

Energy in Chemical Reactions A + B C + D + Energy The potential energy is broken in chemical bonds in compounds A and B The potential energy is the chemical bonds of C and D is lower The excess has been given off (energy) as thermal energy, or heat, which is kinetic energy transferred to the surroundings

Spontaneity Spontaneous Processes Processes that occur without outside intervention Spontaneous processes may be fast or slow Many forms of combustion are fast Conversion of graphite to diamond is slow Kinetics is concerned with speed, thermodynamics with the initial and final state

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 (one of the laws of thermodynamics!) The universe favors chaos!! Entropy is a thermodynamic function describing the number of arrangements that are available to a system Nature proceeds towards the states that have the highest probabilities of existing

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

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

Entropy and Temperature Suniv = Ssys + Ssurr Entropy changes in the surrounding are primarily determined by heat flow The magnitude of entropy is dependent on the temperature The lower the temperature, the higher the impact to the surroundings of the transfer of energy Remember the books’ $50 example

Entropy and Exothermic Processes Entropy and a process being exothermic are related, but a process doesn’t have to be both Exothermic process DSsurr = positive, exothermic = more disorder in surroundings Endothermic process DSsurr = negative, endothermic = less disorder in surroudings Dssystem must increease to obey 2nd Law of Thermodynamics

More on Entropy Why is the sign different? The enthalpy (H) concerns the system Our entropy here concerns the surroundings As usual, temperatures must be in Kelvin

Example

Example 1st reaction DH = -125 kJ, 25 oC + 273 = 298 K DSsurr = -125 kJ/298 K = 419 J/K Remember 125 kJ = 125000 J Dssurr is positive, calculation refers to system!

Example 2nd reaction DH = 778 kJ, 25 oC + 273 = 298 K DS = 778 kJ/298 K = -2.61 kJ/K Remember 778 kJ = 778000 J

Free Energy, G G = H - TS Typically defined as the energy available to perform work G determines whether a process is spontaneous or not (DG = negative = spontaneous) Takes into account enthalpy, entropy, and temperature Symbol honors Josiah Gibbs (sometimes called Gibbs Free Energy or Gibbs energy), a physics professor at Yale during the late 1800’s who was important in developing much of modern thermodynamics

Energy Diagrams Left – A spontaneous reaction The products have a lower free energy (G) than the reactants (DG < 0) Right – A nonspontaneous reaction The reactants have a higher free energy than the products (DG > 0)

What If? DG = 0? As a result The reaction is at equilibrium DG < 0 The reaction is spontaneous. DG > 0 The reaction is nonspontaneous. DG = 0 The reaction mixture is at equilibrium. -It’s not moving forward or reverse overall – multiple processes, same speed

Dependence on Temperature DG = DH - TDS Notice there are two terms in the equation Spontaneity can change when the temperature changes. Temperature and spontaneity are not necessarily correlated. A reaction with a negative entropy (loss of randomness) would be LESS spontaneous at higher temperatures – it doesn’t want to happen, but is pushed by the extra heat.

Reaction Rates Free energy (G) values tell us if reactions will occur They do not tell us how fast reactions will occur. During a reaction Reactant particles must physically collide They must collide with enough energy to break the bonds in the reactant Some reactions require the addition of heat energy. This gives the reactants the extra energy needed (more collisions, harder collisions) for this process to occur.

H, S, G and Spontaneity 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)

Example of Gibbs Energy For the reaction at 298 K, the values of DH and DS are 58.03 kJ and 176.6 J/K, respectively. What is the value of DG at 298 K?

Example of Gibbs Energy For the reaction at 298 K, the values of DH and DS are 58.03 kJ and 176.6 J/K, respectively. What is the value of DG at 298 K? G = H - TS G = 58.03 kJ – (298 K)(176.6 J/K) G = 5.40 kJ Not spontaneous at this temperature

Second Example Calculate the standard free-energy change at 25 oC for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes: 2NH3(g) N2(g) + 3H2(g) DHo = 92.2 kJ DSo = 198.7 J/K

Second Example DG = DH  TDS Calculate the standard free-energy change at 25 oC for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes: 2NH3(g) N2(g) + 3H2(g) DHo = 92.2 kJ DSo = 198.7 J/K DG = DH  TDS DG = 92.2 kJ  (298 K)(-198.7 J/K)(1 kJ/1000 J) = -33.0 kJ

Third Example Iron metal can be produced by reducing iron(III) oxide with hydrogen: Fe2O3(s) + 3 H2(g)  2 Fe(s) + 3 H2O(g) H = +98.8 kJ; S = +141.5 J/K (a) Is this reaction spontaneous under standard-state conditions at 25 °C? (b) At what temperature will the reaction become spontaneous?

Third Example To determine whether the reaction is spontaneous at 25 °C, we need to determine the sign of G = H  TS. At 25 C (298 K), G for the reaction is G = H  TS = (98.8 kJ)  (298 K)(0.1415 kJ/K) = (98.8 kJ)  (42.2 kJ) = 56.6 k Reaction is NOT spontaneous!

Third Example At what temperature does this become spontaenous? At temperatures above 698 K, the TS term becomes larger than H, making the Gibbs energy negative and the process spontaneous Why is this postive this time? Remember, here we’re dealing with the system, not the surroundings!

Entropy Changes in Chemical Reactions Constant Temperature and Pressure Reactions involving gaseous molecules The change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products Typically, more moles of gas, more entropy

Third Law of Thermodynamics "The entropy of a perfect crystal at O K is zero" (NO disorder, since everything is in perfect position) No movement = 0 K No disorder = no entropy (DS = 0)

Standard State Conditions We’ve seen that quantities such as entropy (S), enthalpy (H), and free energy (G) are dependent upon the conditions present Standard State One set of conditions at which quantities can be compared Pure solids/liquids/gases at 1 atm pressure Solutes at 1 M concentration Typically @ 25 oC (298 K) Designated by o sign (similar to degree sign)

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

Calculating Standard Entropy - Example

Calculating Standard Entropy - Example 2 (28 J/Kmol) + 3 (189 J/Kmol) – 51 J/K mole – 3 (131 J/Kmol) = 179 J/K System gained entropy – water more complex than hydrogen!

Second Example Evaluate the entropy change for the reaction: CO + 3 H2 -> CH4 + H2O in which all reactants and products are gaseous. So values: CO 198 J/Kmol H2 131 J/Kmol CH4 186 J/Kmol H2O 189 J/Kmol

Second Example Evaluate the entropy change for the reaction: CO + 3 H2 -> CH4 + H2O in which all reactants and products are gaseous. So values: CO 198 J/Kmol H2 131 J/Kmol CH4 186 J/Kmol H2O 189 J/Kmol Dsoreaction = [186 + 189] – [198 + 3(131)] = 216 J/Kmol

Standard Free Energy Change DG0 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 DG0 cannot be measured directly The more negative the value for DG0, 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

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

Calculating Free Energy of Formation - Example

Calculating Free Energy of Formation - Example 2 (-394kJ/mol) + 4 (-229 kJ/mol) – 2 (-163 kJ/mol) – 3 (0) = - 1378 kJ

Second Example Calculate the standard free energy for the reaction below and determine whether it is spontaneous at standard conditions. Fe2O3(s) + 3 CO (g) 2 Fe (s) + 3 CO2(g) DGo = -742.2 -137.2 0 -394.4

Second Example Calculate the standard free energy for the reaction below and determine whether it is spontaneous at standard conditions. Fe2O3(s) + 3 CO (g) 2 Fe (s) + 3 CO2(g) DGo (kJ/mol)= -742.2 -137.2 0 -394.4 DGo = [0 + (3 x -394.4)] – [-742.2 + (3 x -137.2)] = -29.4 kJ/mol

Additional Note on Energies of Formation Examples on enthalpy are not shown, but could be calculated the same way. There are numerous other possibilities for calculating these Another type of problem could be where DHo and DSo values are given --- you find DHoreaction and DSoreaction, then plug in to get DGoreaction

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

Free Energy and Equilibrium DG = DGo + RTln(Q) DGo = -RTln(K) at equilibrium R = Universal gas constant = 8.314 J/K mol T = Temperature in Kelvin K = Equilibrium constant = [Pproducts]/[Preactants] System is at equilbrium (no net movement) when DGo = 0 (K = 1) If the system is NOT at equilibrium, use Q instead of K

Example For the synthesis of ammonia where DGo = -33.3 kJ/mole, determine whether the reaction is spontaneous at 25 C with the following pressures: PN2 = 1.0 atm, PH2 = 3.0 atm, PNH3 = 0.020 atm

Example For the synthesis of ammonia where DGo = -33.3 kJ/mole, determine whether the reaction is spontaneous at 25 C with the following pressures: PN2 = 1.0 atm, PH2 = 3.0 atm, PNH3 = 0.020 atm DG = DGo + RTln(Q) DG = -33.3 kJ/mole + (8.314 J/K mol)(298 K) ln( [0.02 atm]2 ) ___________________ [1.0 atm][3.0 atm]3 DG = -33.3 kJ/mole + (8.314 J/K mol)(298 K) ln(1.5 x 10-5) DG = -60.5 kJ/mol

Example The synthesis of methanol from carbon dioxide and hydrogen has a DGo = -25.1 kJ/mol. What is the value of the equilibrium constant at 25 oC?

Example The synthesis of methanol from carbon dioxide and hydrogen has a DGo = -25.1 kJ/mol. What is the value of the equilibrium constant at 25 oC? DG = DGo + RTln(K) DGo = - RTln(K) -25.1 kJ/mol = -(8.314 J/molK)(298)lnK Solve for ln K = 10.1, K = 2 x 104

Example At 298 K, DGo = -5.40 kJ/mole. Give the equilibrium K for this process.

Example At 298 K, DGo = -5.40 kJ/mole. Give the equilibrium constant K for this process. DGo = -RTln(K) -5.40 kJ/mole = -(8.31 J/K mol)(298 K)ln K 2.18 = ln K 8.85 = K

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 the maximum Work is changed to heat in surroudings – You lose efficiency, but the DSsurr increases (favorable!) Henry’s Bent’s First Two Laws of Thermodynamics 1st Law: You can’t win, you can only break even 2nd Law: You can’t break even