UNIT 5: THERMODYNAMICS II Text: Chapter 16 1 st Law of Thermodynamics: __________________________ q = heat help us w = work keep track  E = change in internal energy of the  H = change in enthalpy flow of energy q p (heat constant pressure) “SPONTANEOUS” describes a _______________________ that occurs without ______________________. (for chemical change, products___________________reactants. energy in the universe is constant. chemical or physical change constant input of energy are more stable than more natural

“NON-SPONTANEOUS” describes a change that requires a ____________________________________. (for chemical change, products___________________reactants. continuous input of energy in order to occur are less stable than reactants products heat products reactants needs constant energy input to push it uphill natural Thermodynamically favored Not thermodynamically favored

Spontaneous Processes Spontaneous processes are those that can proceed without any outside intervention. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B. © 2012 Pearson Education, Inc.

Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Above 0  C, it is spontaneous for ice to melt. Below 0  C, the reverse process is spontaneous. © 2012 Pearson Education, Inc.

“SPONTANEOUS” processes: 1) have a natural direction examples: 2) depend on the conditions (ie, what’s spontaneous at one temp may be non-spont. at another temp) examples: 3) we cannot predict “speed” of a process with thermodynamics only WHETHER the process will occur example: copper tarnishing ice melting at room temp dye mixing in water marker drying out without a cap C diamond ----> C graphite spont non-spont (spont. but very SLOW process) water 25 o C -10 o C

WHAT IS SPONTANEOUS IN ONE DIRECTION IS NON-SPONTANEOUS IN THE OPPOSITE DIRECTION! Processes that “FAVOR” spontaneity: 1)exothermic 2) an increase in the DISORDER of the system Spont: Cu + O 2 ---> CuO (tarnish) Non-Spont: CuO ---> Cu + O 2 PE release energy before after  H = (-)  S = (+)

2nd Law of Thermodynamics: ___________________________ ____________________________________________________ in any spontaneous process, there is always an increase in the entropy of the universe

Solutions Generally, when a solid is dissolved in a solvent, entropy increases. © 2012 Pearson Education, Inc.

Entropy Changes In general, entropy increases when – Gases are formed from liquids and solids; – Liquids or solutions are formed from solids; – The number of gas molecules increases; – The number of moles increases. © 2012 Pearson Education, Inc.

Standard Entropies Larger and more complex molecules have greater entropies. © 2012 Pearson Education, Inc.

Entropy Changes Entropy changes for a reaction can be estimated in a manner analogous to that by which  H is estimated:  S  =  n  S  (products) —  m  S  (reactants) where n and m are the coefficients in the balanced chemical equation. © 2012 Pearson Education, Inc.

So while _______ of the universe is CONSTANT ( ) ________ in universe is INCREASING ( )  S > 0 means “universe is tending toward greater disorder”  S universe =  S system +  S surroundings To predict whether a process is SPONTANEOUS, We must know the sign of  S universe If (+), entropy increases, process is SPONT in direction written If (-), entropy decreases, process is SPONT in opposite direction energy entropy  S =(+) (non-spont) 1 st law 2 nd law

same (+) (-)

 S surr = -  H (J) T (K) by gaining or losing heat

Entropy Entropy (S) is a term coined by Rudolph Clausius in the nineteenth century. Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered,. qTqT © 2012 Pearson Education, Inc.

Entropy Entropy can be thought of as a measure of the randomness of a system. It is related to the various modes of motion in molecules. © 2012 Pearson Education, Inc.

Entropy Like total energy, E, and enthalpy, H, entropy is a state function. Therefore,  S = S final  S initial © 2012 Pearson Education, Inc.

Second Law of Thermodynamics In other words: For reversible processes:  S univ =  S system +  S surroundings = 0 For irreversible processes:  S univ =  S system +  S surroundings > 0 © 2012 Pearson Education, Inc.

Second Law of Thermodynamics These last truths mean that as a result of all spontaneous processes, the entropy of the universe increases. © 2012 Pearson Education, Inc.

Xe (g) + 2F 2(g) --> XeF 4(s)  H = Room Temp 25 o 100 o C  S surr = -  H T K 298K  S surr = -  H T = - (-251kJ) = - (-251kJ) 298K 373K = 0.842kJ/K = 0.673kJ/K  S surr = 842J/K  S surr = 673J/K For cooler surroundings (lower temps) there is more impact on entropy of surr.

To predict SPONTANEITY of a process, which we have seen is TEMPERATURE dependent, we can use 2 different equations: 1)  S universe =  S system +  S surroundings for ALL processes (+) must!  G =  H - T  S for a constant T & P (-) must! “Going Home To Supper” When  G is (-) then  S universe is (+)!!

Free Energy and Temperature There are two parts to the free energy equation:  H  — the enthalpy term – T  S  — the entropy term The temperature dependence of free energy then comes from the entropy term. © 2012 Pearson Education, Inc.

G “Gibbs Free Energy” is defined as maximum energy free to do useful work.  G = (-) SPONTANEOUS (products more stable than reactants)  G = (+) NON- SPONTANEOUS (reactants more stable than products)  G = 0 EQUILIBRIUM (reactants <----> products) “product favored” “reactant favored” “equally favored”

T  S =  H We can calculate the TEMPERATURE boundary between SPONTANEOUS & NON-SPONTANEOUS, when  G = 0 :  G =  H - T  S +T  S 0 SS SS T =  H  S

SIGNS OF  H AND  S are IMPORTANT: 1)when  H = (-) and  S = (+) ex: “exothermic”process that has an entropy increase is ALWAYS spontaneous!  G =  H - T  S always ( - ) = ( - ) - ( + ) 2) when  H = (+) and  S = (-) ex: “endothermic”process that has an entropy decrease (becomes more ordered) is ALWAYS non-spontaneous!  G =  H - T  S always ( + ) = ( + ) - ( - ) Ex: combustion ( - ) HEAT out of SYSTEM: more disorder HEAT into SYSTEM: more order ( + )

3) when  H = (-) and  S = (-) (both negative) reaction is spontaneous at LOW temperatures.  G =  H  - T  S then ( - ) = ( - ) - lowT( - ) larger neg# smaller pos# 4) when  H = (+) and DS = (+) (both positive) reaction is spontaneous at HIGH temperatures.  G =  H - T  S then ( - ) = ( + ) - highT( + ) smaller pos# larger neg#

Which are “SPONTANEOUS”?  H = 25kJ  G =  H - T  S  S = 5.0J/K  T = 27 o C K --> J = (25,000J)-(300.K)(5.0J) K = (25000J)-(1,500J) GG = 23,500J = +23.5kJ Since is  G =(+), then non-spontaneous.

Which are “SPONTANEOUS”?  H = -3.0kJ  G =  H - T  S  S = 45.0J/K  T = 25 o C K --> J = (-3,000J)-(298K)(45.0J) K = (-3000J)-(13,400J) GG = -16,400J = -16.4kJ Since is  G =(-), then spontaneous. - -

Now see other handout with 4 different scenarios: At what Temperature will this be SPONTANEOUS?  H = -18kJ T =  H  S = -60.J/K  S = -18,000J -60.J/K --> J T = 300K - Test it: Pick an easy # like 1 to substitute for T in Going Home To Supper  G =  H - T  S = -18,000J (1K)(-60.J) K - -18,000J= +60.J  G = -17,940J Since SPONT at 1 K (which is below 300K), then Ans: SPONT below 300K SPONT

RUBBERBAND stretching isnon-spont  G =  H - T  S  S must be ____ + - ? (-) smaller(-)# greater(+)# (+) - T = feels warm so exo (-) (+) (-) so stretched rubber band has less disorder (less entropy)

contracting isspont “natural” RUBBERBAND  G =  H - T  S  S must be ____ + - ? (+) smaller(+)# greater(-)# (-) - T = feels cold so endo (+) (-) (+) so relaxed rubber band has more disorder (more entropy)

CHEMICAL REACTIONS WITH GASES: The relative # of moles of gaseous reactants vs. products dominates “positional entropy” 1) N 2(g) + 3H 2(g) ---> 2NH 3(g)  S = ___ meaning________________ 2) 4NH 3(g) + 5O 2(g) ---> 4NO (g) + 6H 2 O (g)  S = ___ meaning_________________ 2mol 4mol (-) less disorder 9mol10mol (+) more disorder

3) CaCO 3(s) ---> CaO (s) + CO 2(g)  S = _____ 4) C (s) + 2H 2(g) ---> CH 4(g)  S = _____ 5) F 2(g) ---> 2F (g)  S = _____ 0mol 1mol 2mol 1mol 2mol 1mol (-) (+)

3 rd LAW OF THERMODYNAMICS : _________________ ___________________________________________________ meaning that its internal arrangement is “absolutely regular” with “motionless” particles so NO “DISORDER”: S = 0 at zero Kelvin the entropy of a perfect crystal at zero Kelvin is zero

As TEMPERATURE increases from 0 Kelvin, random vibrational motions increase, thus disorder increases. (NO motion) rotational motion vibrational motion

S o = “standard molar entropy” or “absolute entropy”  units are _________  always______________________  for an element _________________________________  values _________________________________________  S o reaction =  n p S o (products) -  n r S o (reactants) J/molK (+) value (never zero) S o ≠ 0 as for  H o f &  G o f of S o are at Room Temp (298K) & 1atm (use chart)

Problem: Calculate  S o rxn at 25 o C for the reaction: 2NiS (s) + 3O 2(g) ---> 2SO 2 (g) + 2NiO (s) S o _____________ _______ _____  S o rxn = 53J/molK205J/molK248J/molK38J/molK 2mol(248J) mol∙K + 2mol(38J) mol∙K 2mol(53J) + mol∙K 3mol(205J) mol∙K = 496J/K + 76J/K 106J/K + 615J/K  S o rxn = -149J/K 572J/K 721J/K products - reactants

 G o “standard free energy change” is defined as the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.(298K and 1 atm) There are 3 ways to solve for  G o : 1) Given: “Heats of Formation”  H o f & Std. Entropy Values S o Find :  H o and  S o Then substitute answers:  G =  H - T  S

2SO 2(g) + O 2(g) -----> 2SO 3(g)  H o f :  H o rxn = -297kJ/mol 0kJ/mol-396kJ/mol 2mol(-396kJ) mol 2mol(-297kJ) mol = products - reactants -792kJ-594kJ  H o rxn = -198kJ to use later

2SO 2(g) + O 2(g) -----> 2SO 3(g) S o :  S o = 248J/mol ·K 205J/mol ·K 257J/mol ·K 2mol(257J) mol∙K 2mol(248J) + mol∙K 1mol(205J) mol∙K =514J/K 496J/K + 205J/K  S o rxn = -187J/K 514J/K 701J/K products - reactants = to use later

Nowsubstitute   G o =  H o - T  S o for  H o rxn and  S o T = 298K (std.state) = (-198kJ)-(298K)(-0.187kJ) K = -198kJ kJ GG = -142kJ

2) Given  G o for related reactions, we can use Hess’s Law procedures since free energy G is a state function like enthalpy & entropy: Find  G o for the following reaction: 2CO (g) + O 2(g) -----> 2CO 2(g) Given: 2CH 4(g) + 3O 2(g) ----> 2CO (g) + 4H 2 O (g)  G o = kJ CH 4(g) + 2O 2(g) ----> CO 2(g) + 2H 2 O (g)  G o = -801 kJ flip 2 2CO (g) + 4H 2 O (g) ---->2CH 4(g) + 3O 2(g)  G o = kJ 2CH 4(g) + 4O 2(g) ----> 2CO 2(g) + 4H 2 O (g) 2( )  G o = kJ 2CO (g) + O 2(g) ----> 2CO 2(g)  G o = -514 kJ

3)Given  G o f for reactants and products.  G o f “standard free energy of formation” is defined as the change in free energy that accompanies the formation of 1 mole of a substance from its constituent elements with all reactants and products in their standard states. (most stable 298K, 1 atm). Using formula:  G o rxn =  n p  G o f -  n r  G o f 2CH 3 OH (g) + 3O 2(g) ---> 2CO 2(g) + 4H 2 O (g)  G o f :  G o rxn = -163kJ/mol0kJ/mol-394kJ/mol-229kJ/mol products - reactants

2mol(-394kJ) mol + 4mol(-229kJ) mol 2mol(-163kJ) mol = -788kJ kJ -326kJ  G o rxn = -1378kJ -1704kJ + 326kJ  G o rxn = products - reactants =

chemical change physical change absorbed (+) or released(-) always (-) /mol J/g or J/mol

physical change /mol J/g or J/mol

chemical change /mol 1 mole dissociating absorbed (+) or released(-) physical change

State Function Name Meaning SignsFor Elements H enthalpy heat content (-) exo, heat released by system (+) endo, heat absorbed by system  H o f = 0 kJ/mol S entropy disorder(+) more disorder (-) less disorder S o > 0 J/Kmol G Gibbs free energy spontaneity (products favored over reactants) (-) spontaneous (o)at equilibrium (+) non- spontan  G o f = 0 kJ/mol

At Room Temp 25 o C “evaporating”

Ex: calculate “entropy change”  S o vap when H 2 O(l) ----> H 2 O(g) IF For H 2 O(l) S o = 70.J/mol K and For H 2 O(g) S o = 189J/mol K  S o vap = products - reactants -242kJ/mol -286kJ/mol  H o vap = +44kJ/mol products - reactants 189J/mol∙K 70J/mol∙K  S o vap = 119J/mol∙K

Calculating Gibbs Free Energy using enthalpy and entropy changes:  G =  H - T  S “going home to supper” kJ kJ K(J/K) careful with units!!! (-) = (-) exo - (+) SPONT at all temps! (+) = (+) endo - (-) NON-SPONT at all temps!

The greater the magnitude of the negative  G for a process, the more thermodynamically favorable it is. The only time that there is ”temperature dependence” is when you have “like signs” for  H and  S. Substitute and solve. When  G = 0 (both products and reactants are favored equally) then we can calculate the boundary temp between SPONT and NON-SPONT: T =  H  S Ex: Normal Melting Pt. S L Boiling Pt. L G (1 atm)

Processes that result in “ENTROPY” changes: 1)phase changes sol--->liq--->gas  S = (+) 2) temp changesT incr, motion incr  S = (+) 3) volume changes (due to  pressure) P decr, V incr  S = (+) 4) mixing (incr V)  S = (+) or separating (decr V)  S = (-) 5) # atoms/molecule # incr  S = (+) (more rotations & vibrations possible) 6) # moles of gases # moles incr, V incr  S = (+)

Which of the following is SPONTANEOUS at LOW temps only? a) NH 4 Br (s) + 188kJ ----> NH 3 (g) + Br 2 (l) b) NH 3 (g) + HCl (g) ----> NH 4 Cl (s) + 176kJ c) 2H 2 O 2 (l) ----> 2H 2 O (l) + O 2 (g) + 196kJ d) both a and b e) both a and c endo  H = (+) exo  H = (-)  S = (+)  S = (-)  S = (+) Decomposition Synthesis MnO 2 Decomposition

Now if B is spontaneous at all Temps below 347 o C, find  S o rxn We know  H = -176kJ  G =  H - T  S And 347 o C is the “boundary” Temp when  G = 0 then 0 =  H - T  S T  S =  H +T  S TT  S =  H T  S =  H T = -176kJ K  S = kJ/K  S = -284J/K

The “enthalpy of vaporization”  H vap is a function of temperature  H vap kJ/mol Temp o Cfor WATER: Note: as temp increases, the energy required to overcome the inter-molecular forces in the liquid decreases  H o vap = (Standard Heat of Vaporization) at Rm Temp!!!!  H vap = Heat of Vaporization at Normal Boiling Point

 H o vap called “standard heat of vaporization” what temp? ____ So it is NOT the same as  H vap which is at Normal BPt Problem: Find  H o vap when bromomethane vaporizes (evaporates at Rm Temp) CH 3 Br (l) > CH 3 Br (g) (at 25 o C!!)  H o f CH 3 Br (l) = -59.8kJ/mol  H o f CH 3 Br (g) = -35.4kJ/mol  H o vap = products - reactants -35.4kJ/mol -59.8kJ/mol  H o vap = +24.4kJ/mol endo 25 o C

2) Ethanol 78.4 o C (N. BPt) and its  H vap = 38.56kJ/mol while its  H o vap = 42.32kJ/mol Find the change in entropy during the boiling of 1 atm. Formula?? Hint: C 2 H 5 OH (l) <-----> C 2 H 5 OH (g) both favored so  G = 0 T =  H  S  S =  H T = 38.56kJ/mol 78.4 o C K  S = = 0.110kJ/mol ·K 110J/mol ·K  H vap L G

1)Estimate the BPt of SnCl 4 at 1 atm. If SnCl 4(l) :  H o f = kJ/mol S o = J/mol · K SnCl 4(g) :  H o f = kJ/mol S o = 366 J/mol · K Equation: SnCl 4(l) < > SnCl 4(g) Since reactants and products are equally favored at the N BPt (1atm) then  G = 0 (equilibrium) so we can use T =  H  S Normal BPt

 H rxn =  S rxn = products - reactants 1mol ( kJ ) mol  H o rxn = +39.8kJ endo 1mol ( kJ ) mol products - reactants 1mol (366J) mol·K 1mol (258.6J) mol·K  S rxn = T =  H  S 107J/K T = 39.8kJ 0.107kJ/K … T = 372K -273 T = 99 o C (BPt)