Presentation on theme: "Chapter 16 Spontaneity, entropy and free energy. Spontaneous l A reaction that will occur without outside intervention. l We cant determine how fast."— Presentation transcript:
Chapter 16 Spontaneity, entropy and free energy
Spontaneous l A reaction that will occur without outside intervention. l We cant 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 1st Law- the energy of the universe is constant. l Keeps track of thermodynamics doesnt correctly predict spontaneity. l Entropy (S) – Number of ways things can be arranged – Looks like disorder or randomness l 2nd 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 S solid
Entropy l Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate. l 2nd Law S univ = S sys + S surr If S univ is positive the process is spontaneous. If S univ is negative the process is spontaneous in the opposite direction.
For exothermic processes S surr is positive. For endothermic processes S surr is negative. Consider this process H 2 O(l) H 2 O(g) S sys is positive S surr is negative S univ depends on temperature.
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. The size of S surr depends on temperature S surr = - H/T
S sys - H/T S surr S univ Spontaneous? ? +-? Yes No, Reverse At Low temp. At High temp.
Gibb's Free Energy l G=H-TS l Never used this way. G= H-T S at constant temperature l Divide by -T - G/T = - H/T- S - G/T = S surr + S - G/T = S univ If G is negative at constant T and P, the Process is spontaneous.
Lets Check For the reaction H 2 O(s) H 2 O(l) Sº = 22.1 J/K mol Hº =6030 J/mol Calculate G at 10ºC and -10ºC l When does it become spontaneous? Look at the equation G= H-T S Spontaneity can be predicted from the sign of H and S.
G= H-T S H S Spontaneous? +- 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. Products - reactants to find Sº (a state function). l More complex molecules higher Sº.
Free Energy in Reactions 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 Cant be measured directly, can be calculated from other measurements. Gº= Hº-T Sº l Use Hesss Law with known reactions.
Free Energy in Reactions There are tables of Gº f. 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.
Free energy and Pressure G = Gº +RTln(Q) where Q is the reaction quotients (P of the products /P of the reactants). CO(g) + 2H 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? Gº f CH 3 OH(l) = -166 kJ Gº f CO(g) = -137 kJ Gº f H 2 (g) = 0 kJ
How far? 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. At equilibrium G = 0, Q = K Gº = -RTlnK
Gº K =0=1 1 >0<1 Gº = -RTlnK
At 1500°C for the reaction CO(g) + 2H 2 (g) CH 3 OH(g) the equilibrium constant is K p = 1.4 x Is H° at this temperature: A. positive B. negative C. zero D. can not be determined
The standard free energy ( G rxn 0 for the reaction N 2(g) + 3H 2(g) 2NH 3(g) is kJ. Calculate the equilibrium constant for this reaction at 25 o C. A B. 5.8 x 10 5 C. 2.5 D. 4.0 x E. 9.1 x 10 8
Temperature dependence of K Gº= -RTlnK = Hº - T Sº l A straight line of lnK vs 1/T 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 cant be used to do work. l Cant be 100% efficient