Spontaneity, Entropy, and Free Energy

Transcription

1 Spontaneity, Entropy, and Free Energy A ball rolls spontaneously down a hill but not up. Spontaneous Processes A reaction that will occur without outside intervention; product favored Most reactants are converted to products Says nothing about the rate of the reaction (KINETICS) Thermodynamics (tells us the direction in which a reaction will occur) Kinetics describes the pathway Examples: ice melting at temps below 0 C; NaCl dissolving in water, Fe rusting Non-examples: water does not boil at 75 C and 1 atm, water does not freeze at 15 C **** A reaction is spontaneous in one direction and non-spontaneous in the opposite direction.***** Example: H 2 O (l) H 2 O (s) at -10 C spontaneous H 2 O (l) H 2 O (s) at -10 C non-spontaneous Thermodynamics 1 st law- Energy of the universe is constant: keeps track of how much energy is involved in the change Why are certain processes spontaneous??? Early idea: exothermicity HOWEVER Current Conclusion: the characteristic common to all spontaneous reactions and the driving force is an increase in entropy Entropy (S) is the measure of disorder or randomness in a system this is a state function (independent of pathway) 2 nd law- The entropy of the universe increases there is a natural tendency to go from order to disorder thus from a lower to higher entropy need to look at it in terms of the number of possible ways the molecules can be arranged Substances take the arrangement that is most likely, which is the most RANDOM S solid < S liquid << S gas More ways for molecules to be arranged as a liquid than a solid; gases have even more ways Solutions form b/c there are many more possible arrangements of dissolved pieces than if they stay separate

2 Substances that are more complex (larger) have more entropy(vibrational and rotational motion) In a gas, if the products have a smaller number of molecules, then the entropy has decreased Example 1 a. A deck of playing cards is neatly tucked away in their box when Johnny decides to play 52-card pickup and scatters them all over the family room. b. Your child s room several days after you have asked him/her to make it spotless. c. A solid (ice) changing into a liquid (water) at 0ºC. d. A gas (steam) changing into a liquid (water) at 100º C. e. Dissolving a solute into a solvent. Example: Sugar is dissolved into coffee. f. Dissolving gas into a liquid. Example: Carbon dioxide gas is dissolved into Coke. g. A chemical reaction between gas molecules that results in a net decrease in the overall numbers of gas molecules. h. The volume of a container containing gas is increased from 5 liters to 10 liters. Example 2 For each pair, choose the substance with the higher entropy a. Solid CO 2 and gaseous CO 2 b. N 2 gas at 1 atm and N 2 gas at.001 atm Example 3 Predict the sign of entropy change for each process a. solid sugar is added to water to form a solution b. iodine vapor condenses on a cold surface to form crystals c. The thermal decomposition of solid calcium carbonate CaCO 3(s) CaO (s) + CO 2(g) d. The oxidation of sulfur dioxide 2 SO 2(g) + O 2(g) 2 SO 3(g) Second Law of Thermodynamics- In detail ΔS universe = ΔS system + ΔS surroundings If ΔS universe is positive(increase in entropy), the process is spontaneous(and irreversible) If ΔS universe is negative(decrease in entropy), the process is nonspontaneous ( or spontaneous in the opposite direction)

3 If ΔS universe is zero, the process is at equilibrium Entropy in the surroundings is primarily controlled by enthalpy Recall: If ΔH is negative, heat is released into surroundings SO ΔS surroundings is positive ( more disorder due to faster movement) If ΔH is positive, heat is absorbed from the surroundings SO ΔS surroundings is negative ( less disorder due to slower movement) BUT WHAT HAPPENS WHEN.. H 2 O (l) H 2 O (g) ΔS sys = positive ΔS surr = negative Which one controls it? ΔS univ will be dependent on temperature Temperature & Spontaneity An exothermic process is favored because by giving up heat, the entropy of the surroundings increases The sign of ΔS surr depends on the direction of heat transfer and the magnitude of ΔS surr depends on temperature : at constant temp/pressure where : T is Kelvin temperature ΔH is the enthalpy of system (negative if exothermic) Units: J/K ΔS sys ΔS surr ΔS univ Spontaneous? Yes No, reverse + -? Yes if ΔS sys has a greater magnitude than ΔS surr (at high temp) - +? Yes, if ΔS surr has a greater magnitude than ΔS sys (at low temp) *****From now on no subscript next to the delta sign refers to the system***** Example 4 Calculate the ΔS surr for each reaction at 25 ºC and 1 atm a. Sb 4 O 6 + 6C 4 Sb + 6CO ΔHº = 778 kj/mol

4 b. Sb 2 S 3 + 3Fe 2 Sb + 3 FeS ΔHº = -125 kj/mol Example 5 What is the ΔS univ for the reaction if the ΔS sys = -199 J/K mol? N 2(g) + 3H 2(g) 2NH 3(g) ΔHº = kj/mol Third law of Thermodynamics Entropy of a pure crystal at 0K is zero. All others must be > zero. SO the standard enthalpies of elements in their standard states are not zero(unlike standard enthalpies and standard free energies) Standard entropy S at 298 K and 1 atm of substances- see appendix for values It is a state function Can use Example 6 Calculate ΔºS at 25 C for the reaction 2 NiS (s) + 3 O 2(g) 2 SO 2(g) + 2 NiO (s) substance ΔºS (J/K mol) SO 2(g) 248 NiO (s) 38 O 2(g) 205 NiS (s) 53 Example 7 Is the reaction of hydrogen and chlorine to give hydrogen chloride gas predicted to be spontaneous? H 2(g) + Cl 2(g) 2HCl (g) Gibb s Free Energy ΔS surr can be hard to measure so we look at another thermodynamic function Can be used to determine if a reaction is spontaneous at constant temp and pressure Defined as the amount of energy available to do work at a given temp and pressure

5 Never really achieved because some of the free energy is changed to heat during a change so it can t be used to do work. At constant temp and pressure ΔG = ΔH TΔS If ΔG is negative, then the reaction is spontaneous If ΔG is positive, then the reaction is not spontaneous, but in the other direction If ΔG is zero, the reaction is at equilibrium Example 8 For the reaction H 2 O (s) H 2 O (l) ΔSº = 22.1 J/K mol and ΔHº = 6030 J/mol, calculate ΔG at 10º C, 0º C and 10º C. ΔG = ΔH - T ΔS ΔH ΔS Spontaneous? - + At all temperatures + + At high temp entropy driven (exothermicity is unimportant) - - At low temp enthalpy driven Exothermicity is important) + - Not at any temp; reverse is spontaneous Standard Free Energy- (ΔG ) is the free energy change that occurs when reactants in their standard stated turn into products in their standard states Is a state function Can t be measured directly 3 Ways to Measure Use Hess s law with known reactions The standard free energy of formation for any element in its standard state is zero. See tables in the appendix for standard free energy of formation ΔG f

6 Example 9 Calculate the standard free energy change for the formation of methane at 298K using ΔH and ΔS C (graphite) + 2H 2(g) CH 4(g) Example 10 Using the following data at 25ºC, calculate the ΔGº for the reaction C (diamond)(s) C (graphite)(s) C (diamond)(s) + O 2(g) CO 2(g) ΔºG = -397 kj C (graphite)(s) + O 2(g) CO 2(g) ΔºG = -394 kj Example 11 Calculate ΔGº for the following equation: 2CH 3 OH(g) + 3O 2(g) 2CO 2(g) + 4 H 2 O (g) ΔG f º CH 3 OH(g) = -163 kj/mol ΔG f º O 2 (g) = 0 ΔG f º CO 2 (g) = -394 kj/mol ΔG f º H 2 O(g) = -229 kj/mol Dependence of Free Energy on Pressure So far we have seen that at constant temp and pressure, a reaction will be spontaneous in the direction that lowers its free energy The equilibrium position of a reaction system represents the lowest free energy value available. - reason why some reactions will not go to completion b. For an ideal gas : - enthalpy is independent of pressure - entropy is pressure-dependent (due to the inverse relationship of pressure and volume S larger volume > S small volume S low pressure > S high pressure - where : R = universal gas constant (8.3145J/K mol) T = Kelvin temperature Q = reaction quotient

7 This will show if a reaction will be spontaneous at varied pressures.. Q= reaction quotient (pressure of the gaseous products) For the reaction aa + bb cc + dd Q = [C] c [D] d /[A] a [B] b ΔG = free energy change of the gas at 1 atm ΔG = free energy change of gas at a specified pressure, non standard conditions T= Kelvin temp R= J/K mol If ΔG is higher in value than ΔG it is more spontaneous at higher pressures Example 12 Would the reaction be spontaneous at 25º C with the hydrogen gas pressures of 3.0 atm and the CO pressure of 5.0 atm? CO (g) + 2H 2(g) CH 3OH (l) ΔG tells us spontaneity at current conditions. When will it stop? It will go to the lowest possible free energy which may be at equilibrium G prod = G react, or ΔG = G prod - G react = 0 At equilibrium ΔG = 0 and Q = K SO ΔG = K = 0 1 < 0 > 0 > 0 < 0 Example 13 Determine the standard free energy change ΔºG for the formation of 1 mole of ammonia from nitrogen and hydrogen gas and use this value to calculate the equilibrium constant for this reaction at 25ºC. N 2(g) + 3H 2(g) 2 NH 3(g)