Chapter 16 Spontaneity entropy and free energy Spontaneous

Chapter 16 Spontaneity, entropy and free energy

Spontaneous l. A reaction that will occur without outside intervention. l We can’t determine how fast. l We need both thermodynamics and kinetics to describe a reaction completely. l Thermodynamics compares initial and final states. l Kinetics describes pathway between.

Thermodynamics l 1 st Law- the energy of the universe is constant. l Keeps track of thermodynamics doesn’t correctly predict spontaneity. l Entropy (S) – Number of ways things can be arranged – Looks like disorder or randomness l 2 nd Law the entropy of the universe increases in any change

Entropy l Defined in terms of probability. l Substances take the arrangement that is most likely. l The most likely is the most random. l Calculate the number of arrangements for a system.

l 2 possible arrangements l 50 % chance of finding the left empty

l 4 possible arrangements l 25% chance of finding the left empty l 50 % chance of them being evenly dispersed

l 4 atoms l 8% chance of finding the left empty l 50 % chance of them being evenly dispersed

Gases l Gases completely fill their chamber because there are many more ways to do that than to leave half empty. l. Ssolid <Sliquid <<Sgas l there are many more ways for the molecules to be arranged as a liquid than a solid. l Gases have a huge number of positions possible.

l Solutions Entropy form because there are many more possible arrangements of dissolved pieces than if they stay separate. l 2 nd Law l Suniv = Ssys + Ssurr l If Suniv is positive the process is spontaneous. l If Suniv is negative the process is spontaneous in the opposite direction.

exothermic processes Ssurr is positive. l For endothermic processes Ssurr is negative. l Consider this process H 2 O(l)® H 2 O(g) l Ssys is positive l Ssurr is negative l Suniv depends on temperature. l For

Temperature and Spontaneity l Entropy changes in the surroundings are determined by the heat flow. l An exothermic process is favored because by giving up heat the entropy of the surroundings increases. l The size of Ssurr depends on temperature l Ssurr = - H/T

Ssys - H/T Ssurr Suniv + + + No, Reverse + - ? At High temp. - + ? At Low temp. Spontaneous? Yes

Gibb's Free Energy l G=H-TS l Never used this way. l G= H-T S at constant temperature l Divide by -T l - G/T = - H/T- S l - G/T = Ssurr + S l - G/T = Suniv l If G is negative at constant T and P, the Process is spontaneous.

Let’s Check the reaction H 2 O(s) ® H 2 O(l) l Sº = 22. 1 J/K mol Hº =6030 J/mol l Calculate G at 10ºC and -10ºC l When does it become spontaneous? l Look at the equation G= H-T S l Spontaneity can be predicted from the sign of H and S. l For

G= H-T S Spontaneous? S H + - At all Temperatures + At high temperatures, “entropy driven” - At low temperatures, “enthalpy driven” + Not at any temperature, Reverse is spontaneous + -

Third Law of Thermo l The entropy of a pure crystal at 0 K is 0. l Gives us a starting point. l All others must be>0. l Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. l Products - reactants to find Sº (a state function). l More complex molecules higher Sº.

Free Energy in Reactions l Gº = standard free energy change. l Free energy change that will occur if reactants in their standard state turn to products in their standard state. l Can’t be measured directly, can be calculated from other measurements. l Gº= Hº-T Sº l Use Hess’s Law with known reactions.

Free Energy in Reactions are tables of Gºf. l Products-reactants because it is a state function. l The standard free energy of formation for any element in its standard state is 0. l Remember- Spontaneity tells us nothing about rate. l There

Free energy and Pressure l G = Gº +RTln(Q) where Q is the reaction quotients (P of the products /P of the reactants). l CO(g) + 2 H 2(g) ® CH 3 OH(l) l Would the reaction be spontaneous at 25ºC with the H 2 pressure of 5. 0 atm and the CO pressure of 3. 0 atm? l Gºf CH 3 OH(l) = -166 k. J l Gºf CO(g) = -137 k. J Gºf H 2(g) = 0 k. J

How far? l G tells us spontaneity at current conditions. When will it stop? l It will go to the lowest possible free energy which may be an equilibrium. l At equilibrium G = 0, Q = K l Gº = -RTln. K

Gº =0 <0 >0 K =1 >1 <1 Gº = -RTln. K

At 1500°C for the reaction CO(g) + 2 H 2(g) → CH 3 OH(g) the equilibrium constant is Kp = 1. 4 x 10 -7. Is H° at this temperature: A. positive B. negative C. zero D. can not be determined

The standard free energy ( Grxn 0 for the reaction N 2(g) + 3 H 2(g) → 2 NH 3(g) is -32. 9 k. J. Calculate the equilibrium constant for this reaction at 25 o. C. A. 13. 3 B. 5. 8 x 105 C. 2. 5 D. 4. 0 x 10 -6 E. 9. 1 x 108

Temperature dependence of K l Gº= l. A -RTln. K = Hº - T Sº straight line of ln. K vs 1/T l With slope - Hº/R

Free energy And Work l Free energy is that energy free to do work. l The maximum amount of work possible at a given temperature and pressure. l E = q + w l 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. l Can’t be 100% efficient