Ppt04 (PS8), Thermodynamics energy changes (heat flows) Very global (“the universe”) Applies to all fields of science Led to creation of heat engines/refrigerators/air conditioners In chemistry, focus is on physical and chemical changes How can we use “spontaneous processes” to “do work”? We can use tabulated data to make predictions (and even calculate K’s for possible processes!) Ppt04_Thermo

“Spontaneous”  “fast” !! Spontaneity refers to “directionality” Does it occur (on its own) in the forward [or stated] direction, or does it not? Thermodynamics will answer / address this question. Does not depend on how the process is carried out. Fast refers to rate at which it occurs (in the spontaneous direction) How fast will it occur? Kinetics will answer / address this question Does depend on how the process occurs (“pathway”; transition state, activation energy, etc.) Recall that we should be spontaneously combusting right now!! Ppt04_Thermo

Figure 18.4 Thermodynamics domain vs. Kinetics domain Initial and final states, spontaneity “Path independent” “Pathway dependent” Ppt04_Thermo

Reminder: 1st Law of Thermodynamics The total amount of energy in the universe never changes (it’s constant) Euniv= 0  Esys = -Esurr “If energy leaves the system, it must go to the surroundings, and vice versa” Has nothing to do with “spontaneity” 1st Law is consistent with a casserole dish coming out of the oven colder than when it was put in, as long as the oven would get hotter! Ppt04_Thermo

Spontaneity and the 2nd Law A spontaneous process is one that occurs (at a specified set of conditions) “without external intervention”. Gases expand into the volume of their containers Ice will melt at 10°C Water will freeze at -10°C A chemical reaction will occur in the forward direction if Q < K Ppt04_Thermo

Spontaneity and the 2nd Law (continued) At the same conditions, the reverse process of a spontaneous one is not spontaneous Gases won’t spontaneously “contract” into a smaller volume (perfume molecules going back into the bottle after opened?) Water will not freeze at 10°C Ice will not melt at -10°C A chemical reaction will not occur in the reverse direction if Q < K Ppt04_Thermo

Spontaneity and the 2nd Law (continued) What determines whether a given process is spontaneous or not? This question is not addressed by the 1st Law; it is addressed by the 2nd Law. 2nd Law: For a spontaneous process to occur, the entropy (S) of the universe must increase. Suniv > 0 for any spontaneous process Ppt04_Thermo

**Suniv  Ssys!** Be careful! The universe is made up of two “parts”—system and surroundings: ΔSuniv = ΔSsys + ΔSsurr It doesn’t ultimately matter if ΔSsys is positive, or if ΔSsurr is positive. The key is whether ΔSuniv is positive! (next slide). Ppt04_Thermo

Interplay of ΔSsys and ΔSsurr in Determining the Sign of ΔSuniv (Table 16.3, Zumdahl) ΔSsys + ΔSsurr = ΔSuniv This turns out to be very temperature dependent (See Section 18.5 in Tro). We will come back to this issue later. plus equals See next slide Ppt04_Thermo

Bar plots to visualize idea on prior slide (system and surroundings both contribute to DSuniv!) ΔSsys + ΔSsurr = ΔSuniv |ΔSsurr| > |ΔSsys| |ΔSsys| > |ΔSsurr| NOTE: These two examples both have DSsys < 0 and DSsurr > 0, but all possible variations are possible! Ppt04_Thermo

What is entropy? Hard thing to define conceptually! Something like “energy dispersal”--not necessarily “spatial” dispersal, but energy dispersed over the various motions of atoms & molecules. Reflects quality or usefulness of energy. Let’s start with individual substances Look at what affects the amount of entropy in a sample of a single substance (without necessarily understanding “what” it is”). Then we’ll look at mixtures of substances (in a reactive system, e.g.) When a change occurs in the system, Ssys changes because there are new substances present (or new physical states) We’ll consider the surroundings last Just a “bunch of substances not doing anything” (Prof. Mines’ view) Ppt04_Thermo

Patterns for Entropies of Substances (see handout/outline for details; Tro does this later; Section 18.7) Entropy of a sample of a substance increases with: Energy added (T increase or phase change) Solid < Liquid << Gases Volume (gas or solution species) Complexity of basic unit of substance Number of moles of the substance Entropy, like energy, is a “per mole” kind of quantity Recall that we’ll typically only specify substances that undergo some change to be part of the system. Ppt04_Thermo

Differences in entropy of physical states (assume a given T) Ppt04_Thermo

Fig. 18.5 Entropy of a substance increases with T, and depends (significantly) on state (s, l, g) Ppt04_Thermo

Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E System A (4 J) System B (4 J) E = 5 J One “way” to arrange the energy (one [micro]state) Two “ways” to arrange the energy (two [micro]states) Both systems have 4 J of energy, but the entropy of System B is greater because it has two ways to “arrange” the energy. A greater # of “accessible” (i.e., “low”) energy levels leads to greater entropy (see Slide 17)! Chapter 17, Unnumbered Figure 2, Page 775 What if an energy state at 5 J was present? Would the # of microstates change in either system? No, because there isn’t enough energy (in either system) to access that 5 J state. But… Ppt04_Thermo

Two “ways” to arrange the energy (vs. one before); two[micro]states Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E System A (6-7 J) System B (6-7 J) E = 5 J E = 5 J E Two “ways” to arrange the energy (vs. one before); two[micro]states Three “ways” to arrange the energy (vs two before); three [micro]states Both systems have 6-7 J of energy, but the entropy of System B is (still) greater because it has three ways to “arrange” the energy. A greater # of “accessible” (i.e., “low”) energy levels leads to greater entropy (see next slide)! Chapter 17, Unnumbered Figure 2, Page 775 What if 2-3 more units of energy were added (to make a total of 6-7 J)? In both cases, adding energy (raising T) leads to increased S! Ppt04_Thermo

More accessible energy levels in gas! Chapter 17, Figure 17.6 ”Places” for Energy Ppt04_Thermo

Adding more accessible energy states Patterns for Entropies of Substances (revisit earlier slide, with nanoscopic “explanation”) Entropy of a sample of a substance increases with: Energy added (T increase or phase change) Solid < Liquid << Gases Adding units of energy Adding more accessible energy states Volume (gas or solution species) Complexity of basic unit of substance Number of moles of the substance Entropy, like energy, is a “per mole” kind of quantity (at a given T) Ppt04_Thermo

(NOTE: Appendix has lots more data. Use appendix for PS8!) Ppt04_Thermo

(From another text [McMurry]) Ppt04_Thermo

Predicting whether processes are “entropy increasing” or “decreasing” (in the system) (the most dominant effect in most cases is the number of moles of gases made and lost [Dngas], not the complexity of the substance[s]): C(s) + 2 H2(g)  CH4(g) N2(g) + 3 H2 (g)  2 NH3(g) [see next slide] 2 CO(g) + O2(g)  2 CO2(g) CaCO3(s)  CaO(s) + CO2(g) 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g) Ppt04_Thermo

Can quantify the change in S (at a given T) using standard entropies of substances C(s) + 2 H2(g)  CH4(g) DSrxn° = 186.2 – (2.4 + 2 x 130.6) 2.4 2 x 130.6 186.2 = -77.4 J/K (per mol of C reacted) (consistent with prediction, S decreases) Although CH4 is more complex than H2, it doesn’t have twice the entropy (per mole) as H2, so the dominant effect here (as usual) is the change in the number of moles of gases on reaction, Dngas) N2(g) + 3 H2 (g)  2 NH3(g) DSrxn° = 2 x 192.3 – (191.5 + 3 x 130.6) 191.5 3 x 130.6 2 x192.3 = -198.7 J/K (per mol of N2) Again, consistent with prediction (S decreases). Though NH3 more complex, only 2 moles of it form compared to 4 moles of gas “lost”). Ppt04_Thermo

Recall (1st semester): Now: DS S° S° Ppt04 Thermo

The entropy of a perfect crystal at 0 K is zero (3rd Law). W = “the number of microstates” (ways to arrange energy) Tro, Fig. 17.8 Zumdahl, Fig 16.5 Ppt04_Thermo

Entropy increases for a substance as energy is added to it (either T increases, or a phase change occurs) Curve differs for different substances. This is how standard entropies of substances are determined (at, say, 298 K) Ppt04_Thermo

See Section IV of Handout Outline Relating Entropy of the Surroundings to a Property of the System--What is “free energy” (G)? See Section IV of Handout Outline Up through the derivation of the expression: DG = DH –TDS Ppt04_Thermo

Consider only (a) for now. EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from H and S Consider the reaction for the decomposition of carbon tetrachloride gas: (a) Calculate G at 25 C and determine whether the reaction is spontaneous. (b) If the reaction is not spontaneous at 25 C, determine at what temperature (if any) the reaction becomes spontaneous. Consider only (a) for now. SOLUTION (a) Use Equation 17.9 to calculate G from the given values of H and S. The temperature must be in kelvins. Be sure to express both H and S in the same units (usually joules). The reaction is not spontaneous. At the specific conditions / concentrations of the reaction system! © 2011 Pearson Education, Inc. Ppt04_Thermo

Back to Outline Conceptual connections… Now, return to discuss T-dependence of spontaneity (specifically, DSuniv and DGsys) Ppt04_Thermo

Now begin to consider (b)! EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from H and S Consider the reaction for the decomposition of carbon tetrachloride gas: (a) Calculate G at 25 C and determine whether the reaction is spontaneous. (b) If the reaction is not spontaneous at 25 C, determine at what temperature (if any) the reaction becomes spontaneous. Earlier did (a) Now begin to consider (b)! SOLUTION (a) Use Equation 17.9 to calculate G from the given values of H and S. The temperature must be in kelvins. Be sure to express both H and S in the same units (usually joules). The reaction is not spontaneous. At the specific conditions / concentrations of the reaction system! © 2011 Pearson Education, Inc. Ppt04_Thermo

RECALL: Interplay of ΔSsys and ΔSsurr in Determining the Sign of ΔSuniv (Table 16.3, Zumdahl) ΔSsys + ΔSsurr = ΔSuniv This turns out to be very temperature dependent (See Section 18.5 in Tro). We will come back to this issue later. plus equals See next slide Ppt04_Thermo

Revisiting Earlier Example If both of these plots represent the same exact process under the same exact conditions EXCEPT THAT ONE IS AT A HIGHER TEMPERATURE, then which plot represents the higher T and which the lower T? How do you know? What would the bar plot look like if the T went to infinity? To zero? (see next slides →) Ppt04_Thermo

As T → , |DSsurr| → 0! (so DSuniv → DSsys) T increasing Ppt04_Thermo

As T → 0, |DSsurr| →  ! (so DSuniv → DSsurr) (Y-axis scale much bigger (zoomed out); DSsys same as on right.) T decreasing T very small Ppt04_Thermo

Summary, T-dependence of DSuniv At very high T, DSuniv approaches DSsys Because DSsurr becomes tiny (and DSsys remains essentially unchanged) At very low T, DSuniv approaches DSsurr Because DSsurr becomes huge, while DSsys remains essentially unchanged **A more detailed summary table of this will be discussed a bit later. NOTE: Many people (most, actually) end up taking a different approach to this “T-dependence” issue: They look at DGsys rather than DSuniv. We’ll explore this next! Ppt04_Thermo

-TDS (DSsys dominates) DG = DH –TDS Thus: As T goes to… …infinity, DG → -TDS (DSsys dominates) …zero, DG → DH (DSsurr dominates) Ppt04_Thermo

Tro’s “version” of this DG = DH –TDS Thus: As T goes to… …infinity, DG → -TDS …zero, DG → DH DGsys “view” DH -TDS opposites NOTE: You could also look at each of these from a DSuniv “view”! Ppt04_Thermo

Add the linear “version” of DG vs T. Slope related to sign of DS Add the linear “version” of DG vs T. Slope related to sign of DS. Intercept is DH Ppt04 Thermo

“Global” (entropy) view (Review of earlier slide) DSuniv = DSsys + DSsurr; Thus: As T goes to… …infinity, DSuniv → DSsys …zero, DSuniv → DSsurr DSuniv “view” DSsys DSsurr DSsurr DSsys + + (DSsurr > 0) (DSsys > 0) - - (DSsurr < 0) (DSsys < 0) - + (DSsurr > 0) (DSsys < 0) + - (DSsurr < 0) (DSsys > 0) Ppt04_Thermo

T-Dependence Explored Further Table 16.4 (Zumdahl) Results of the Calculation of ΔSuniv and ΔG° for the Process H2O(s)  H2O(l) at -10°C, 0°C, and 10°C Ppt04_Thermo

C2H4(g) + H2(g)  C2H6(g) ∆H = –137.5 kJ; ∆S = –120.5 J/K EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from H and S Now consider part (b) Consider the reaction for the decomposition of carbon tetrachloride gas: (a) Calculate G at 25 C and determine whether the reaction is spontaneous. (b) If the reaction is not spontaneous at 25 C, determine at what temperature (if any) the reaction becomes spontaneous. (a) Use Equation 17.9 to calculate G from the given values of H and S. The temperature must be in kelvins. Be sure to express both H and S in the same units (usually joules). SOLUTION The reaction is not spontaneous. At the specific conditions / concentrations of the reaction system! (b) Since S is positive, G will become more negative with increasing temperature. To determine the temperature at which the reaction becomes spontaneous, use Equation 17.9 to find the temperature at which G changes from positive to negative (set G = 0 and solve for T). The reaction is spontaneous above this temperature. FOR PRACTICE 17.3 Consider the reaction: C2H4(g) + H2(g)  C2H6(g) ∆H = –137.5 kJ; ∆S = –120.5 J/K Calculate G at 25 C and determine whether the reaction is spontaneous. Does G become more negative or more positive as the temperature increases? © 2011 Pearson Education, Inc. Ppt04_Thermo

Relationship of DG to Q (and DG) See Outline (and board) ΔG = ΔG° + RT ln Q Ppt04_Thermo

Relationship of DG to K See Outline—Derive from prior slide’s relationship! Ppt04_Thermo

Table 16.6(Zumdahl) Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction DG = -RT lnK Ppt04_Thermo

Ppt04_Thermo Chapter 17, Figure 17.14 Free Energy and the Equilibrium Constant Ppt04_Thermo

From another text These data provide a more direct way to calculate the DG for a chemical reaction equation (i.e., you don’t need to calculate DH and DS if you know DGf values! Ppt04_Thermo

Fig. 17.10 (Tro). Why “free” energy? DH = -74.6 kJ DS = -80.8 J/K DG = -50.5 kJ Ppt04_Thermo

Table 16.1 The Microstates That Give a Particular Arrangement (State) Ppt04_Thermo

Table 16.2 Probability of Finding All the Molecules in the Left Bulb as a Function of the Total Number of Molecules Ppt04_Thermo