Chapter 19. Chemical Thermodynamics

Transcription

1 Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 19 John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc.

2 First Law of You will recall from Chapter 5 that energy cannot be created nor destroyed. Therefore, the total energy of the universe is a constant. Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.

3 Thermodynamically favorable Processes Thermodynamically favorable processes are those that can proceed without any outside intervention. The gas in vessel B will move into vessel A on its own (or spontaneously ), but once the gas is in both vessels, it will not spontaneously all go back into B

4 Thermodynamically Favorable Processes that are thermodynamically favorable in one direction are not favorable in the reverse direction (at a particular temp!) Processes that are favorable at one temperature may be not favorable at other temperatures. Ø Above 0 C it is favorable for ice to melt, but below 0 C the reverse process is favorable (water freezes) Ø Under what conditions is boiling of water favorable? When is the reverse favorable?

5 Thermodynamically Favorable What determines if a process is thermodynamically favorable? Three factors affect this. Ø ENTHALPY, ΔH (endo or exo) exo processes are more likely to favorable Ø TEMPERATURE Ø ENTROPY CHANGE, ΔS What is entropy? The randomness or disorder in a system. The universe favors disorder, so processes that are disordering are favored. Disordering processes have positive values of ΔS

6 Entropy Entropy (S) can be thought of as a measure of the randomness of a system or how disorganized it is. On the particle level, it is related to the various modes of motion in molecules basically how many ways they can move or be arranged. The units of S are J/K Like total energy, E, and enthalpy, H, entropy is a state function. Therefore, ΔS = S final - S initial

7 Entropy on the Molecular Scale Ludwig Boltzmann described the concept of entropy on the molecular level. Temperature is a measure of the average kinetic energy of the molecules in a sample. The higher the temp. of a given sample, the more entropy it has.

8 Entropy on the Molecular Scale Molecules exhibit several types of motion: Ø Translational, Vibrational, Rotational A microstate is a possible way the molecules in a sample could be arranged at a particular moment. More microstates = more randomness = higher S Faster movement of molecules (higher T) = more microstates More molecules = more microstates The more atoms and single bonds a given molecule has the more microstates there are. What about volume?

9 Entropy and Physical States Entropy increases with the freedom of motion of molecules. Therefore, S(g) > S(l) > S(s)

10 Solutions Generally, when a solid is dissolved in a solvent, entropy increases.

11 Entropy Changes in Reactions (19.3) In general, entropy increases when Ø Gases are formed from liquids and solids. Ø Liquids or solutions are formed from solids. Ø The number of gas molecules increases. Ø The number of moles increases.

12 Third Law of The entropy of a pure crystalline substance at absolute zero is 0. In reality nothing is ever at 0K, so all substances have nonzero (positive) entropy values. Molar entropy values under standard conditions can be found in tables.

13 Standard Entropies These are molar entropy values of substances in their standard states. What are the enthalpy values of three hydrocarbons? Look up methane, CH 4, ethane (C 2 H 6 ) and propane (C 3 H 8 )in the appendix Based on the above, how do standard enthalpy changes change with increasing molar mass?

14 Standard Entropies 19.4 Larger and more complex molecules have greater entropies. WHY?

15 Entropy Changes Entropy changes for a reaction can be estimated just like the way ΔH is estimated: ΔS = ΣnΔS (products) - ΣmΔS (reactants) where n and m are the coefficients in the balanced chemical equation. Calculate ΔS for the equation below using the thermodynamic data in the appendix: N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

16 Entropy Changes Entropy changes for a reaction can be estimated just like the way ΔH is estimated: ΔS = ΣnΔS (products) - ΣmΔS (reactants) where n and m are the coefficients in the balanced chemical equation. Calculate ΔS for the equation below using the thermodynamic data in the appendix: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) S values (J/mol-K) are NH3 = 192.5, N2 = 191.5, H2 = Delta S = [ (2*192.5) ((3* 130.6) ))] J/K =-192.3J/K

17 Calculate ΔS for the equation below using the thermodynamic data: N2(g) + 3 H2(g) 2 NH3(g)