Thermodynamics. Chem 36 Spring The study of energy changes which accompany physical and chemical processes

Transcription

1 Thermodynamics Chem 36 Spring 2002 Thermodynamics The study of energy changes which accompany physical and chemical processes Why do we care? -will a reaction proceed spontaneously? -if so, to what extent? It won t tell us: -how fast the reaction will occur -the mechanism by which the reaction will occur 2 1

2 What is Energy? (review!) Energy is the capacity to do work or to transfer heat -Kinetic Energy: energy associated with mass in motion (recall: E k = ½mv 2 ) -Potential Energy: energy associated with the position of an object relative to other objects (energy that is stored - can be converted to kinetic energy) 3 The System (review) We must define what we are studying: System «Surroundings System: portion of the universe under study Surroundings: everything else If exchange is possible, system is open 4 2

3 First Law of Thermodynamics (review) The total energy of the universe is constant. Energy is neither created or destroyed in a process, only converted to another form. -Conservation of Energy You can t win... you can only break even. DE = q + w Change in energy of the system (state function) Heat Flow: + is into system - is out of system (path function) Work: + is done on system - is done by system (path function) 5 Enthalpy (review) Chemistry is commonly performed at constant pressure, so: -it is easy to measure heat flow (q p ) -work (P V) is small (but finite) and hard to measure Define a new term: Enthalpy (H) H = E + PV 6 3

4 Recall: Relating Enthalpy and Heat (review) DE = q + w At constant pressure: DE = q p - PDV Rearranging: q p = DE + DPV q p = (E + PV) Substituting: q p = DH So, if we measure q p, then we can obtain the enthalpy change ( H) directly (Calorimetry!) 7 How are E and H related? (review) From the definition of Enthalpy: DH = DE + PDV -for an ideal gas: PDV = RTDn So: DH = DE + RTDn q p q v PV work 8 4

5 H and Spontaneity Shouldn t H be a good indicator of reaction spontaneity? If H < 0, products at a lower energy than reactants Examples: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) H = -802 kj 4Fe (s) + 3O 2 (g) 2Fe 2 O 3 (s) H = kj 9 But what about... These are spontaneous processes too: NaCl (s) Na + (aq) + Cl - (aq) H = + H 2 O (s) H 2 O (l) H = + Both are nearly enthalpy neutral Both result in increased disorder We need another LAW of thermodynamics! 10 5

6 The 2nd Law of Thermodynamics Spontaneous processes are accompanied by an increase in Entropy. The Entropy of the Universe is constantly increasing. So, what is Entropy? Entropy (S): A measure of the degree of disorder or randomness of a system Processes will tend towards conditions with the highest probability of occuring 11 A Little Experiment Consider the following: A B # molecules relative probability 1 1/2 2 1/(2) 2 (= 1/4) 5 1/(2) 5 (= 1/32) 10 1/(2) 10 (=1/1024) Two bulbs of equal volume Equal number of molecules in each bulb What is the probability of finding ALL of the molecules in Bulb A? For 1 mol gas, changing orientations 100 times/sec: Event would occur once every years 12 6

7 Entropy of Physical States S solid < S liquid < S gas increasing number of possible orientations So, for a phase change: Why spontaneous? Solid Liquid (spontaneous at T>T f ) Liquid Gas (spontaneous at T>T b ) But, also: Gas Liquid (spontaneous at T<T b ) Liquid Solid (spontaneous at T<T f ) S > 0 S < 0 13 How is Entropy Quantified? On the microscopic level: S = k B lnw Boltzmann s constant (= R/N o ) # of microstates Entropy increases with an increasing number of microstates (increased disorder ) Units: k b = 1.38 x J/K (so units of S are: J/K) 14 7

8 Macroscopic Entropy For a reversible process at constant pressure: S sys = q p, rev = H And don t forget: T S surr = - H T sys T surr Now, let s revisit the freezing process for water Freezing Water: Revisited H 2 O (l) H 2 O (s) H = kj/mol Freezing is an exothermic process Energy flows from the system to the surroundings: Entropy of the surroundings increases Entropy of the system decreases Suppose: T surr < T sys such that T surr is less than FP S surr is more positive than S sys is negative S univ = S sys + S surr = positive (freezing is spontaneous) 16 8

9 Quantifying Spontaneity Making some substitutions, we find: S univ = S sys - H/T Re-arranging gives: T S univ = T S sys - H Reversing signs: -T S univ = H - T S sys G (Gibb s Free Energy) 17 The Gibbs-Hemholtz Equation So, at constant temperature and pressure: DG = DH - TDS Has units of Joules (energy): Free Energy Tells us how much energy is available to do work All values are for the system What happens at T=0? We break even ; all energy is available to do work. 18 9

10 Balancing H and S Remember: DG = DH - TDS Two considerations: 1. Reactions seek a minimum in Enthalpy ( H) 2. Reactions seek a maximum in Entropy ( S) - + spontaneous + - nonspontaneous - - spontaneous at low temps + + spontaneous at high temps 19 Calculating G for a reaction G is a state function Treat like we do H: DG o = S DG o f (products) - S DGo f (reactants) At 1 atm and 25 o C Standard Molar Free Energy of Formation: for 1 mol of compound formed from elemental constituents in their standard states 20 10

11 The 3rd Law of Thermodynamics We can obtain absolute values for Entropy because we have a set reference point: If W = 1 (only 1 microstate possible), then S = 0 So, the 3rd Law says: At the absolute zero of temperature, the entropy of a perfect crystalline solid is zero. So, we can tabulate S o (standard molar entropy) values (NOT S o ) - unlike H and E and G 21 Calculating S for a reaction Entropy is also a state function: DS o = S S o (products) - S S o (reactants) NOTE: There are no S o f values S o 0 for pure elements in their standard states 22 11

12 Let s Get Quantitative! Consider the following process: CH 3 OH (l) CH 3 OH (g) We know: H o vap S o vap = kj/mol = J/K For most liquids: 88 J/K (Trouten s Rule) S o vap At what temperature does methanol boil? 23 The BP of Methanol G o = H o - T S o At its BP (T b ), system is at equilibrium so G o =0, and we can write: 0 = H o vap - T b So vap Rearranging: T b = H o vap / So vap Substituting: T b = x 10 3 J = K = 63.4 o C J/K Actual BP of Methanol is o C 24 12

13 There are many paths to G o rxn If G o f values are available for products and reactants: DG o = S DG o f (products) - S DGo f (reactants) If H o f and So values are available for products and reactants: Calculate H o rxn and So rxn values DG o rxn = DHo rxn - TDSo rxn Or, use a combination of both methods! 25 Limits to the utility of G o rxn G o rxn predicts reaction spontaneity only when products and reactants are in their standard states How does G vary with: Pressure (for gases) Concentration (for solutions) Temperature 26 13

14 Effect of Pressure Change What happens to free energy of a gas when we change its pressure? G = H - T S G - G o = 0 - [-nrt Ln(P/P o )] Independent of pressure It can be derived! G = G o + nrt Ln (P/P o ) 27 Now, Apply to a Reaction Let s make some ammonia: N 2 (g) + 3H 2 (g) 2NH 3 (g) For 1 mol of each compound, we can write: G NH3 = G o NH3 + RTLn(P NH3 /Po ) G N2 = G o N2 + RTLn(P N2 /Po ) G H2 = G o H2 + RTLn(P H2 /Po ) 28 14

15 So, G for the Reaction Since we know: G rxn = Σ G products - Σ G reactants Substituting: Finally giving: G = 2G NH3 - (G N2 + 3G H2 ) G = G o + RTLn (P NH3 /P o ) 2 (P N2 /P o ) (P H2 /P o ) 3 29 Some Tweaking What is P o? P o = 1 atm (pressure of a gas in its standard state) Numerically: P NH3 = P NH3 /P o Defined as the ACTIVITY (= a NH3 ) So, now we can write: G = G o + RTLnQ Where: Q = (a NH3 ) 2 (a N2 ) (a H2 ) 3 Reaction Quotient 30 15

16 The Reaction Quotient: In General For any reaction: aa + bb cc We can write: Q = (a C ) c (a A ) a (a B ) b Activity (a) is UNITLESS! Where, a X = activity of species X For gases: a X = P x /P o For solutions: a X = [X]/[X] o For pure cmpds: a X = 1 P o = 1 atm [X] o = 1 mol/l 31 An Example Calculate G for the following reaction: CO (g) + 2H 2 (g) CH 3 OH (l) Reaction Conditions: T = 25. o C P CO = 5.0 atm P H2 = 3.0 atm First: Find DG o G o rxn = Go f (CH 3OH) - [ G o f (CO) + 2 Go f (H 2)] G o rxn = ( kj) - [( kj) + 2(0)] G o rxn = kj (= x 104 J) 32 16

17 On to G! Next: Find Q Q = a CH3OH = 1 = x 10-2 (a CO ) (a H2 ) 2 (5.0) (3.0) 2 Finally: Calculate G G = G o + RTLnQ G = x 10 4 J + ( J/mol-K)( K) Ln(2.222 x 10-2 ) G = x 10 4 J + ( x 10 3 J) G = x 10 4 J = -39. kj Spontaneous! 33 How Does G Change as the Reaction Proceeds? Let s make it simple: A (g) B (g) G < 0 (spontaneous) So, we can write: G A = G o A + RTLn(a A ) G B = G o B + RTLn(a B ) As the reaction proceeds: So, we start at: G A > G B a A decreases G A DECREASES a B increases G B INCREASES DG fi

18 Equilibrium Reaction will proceed until G = 0 System is then at Equilibrium: A (g) B (g) Equilibrium is dynamic: Rate of forward rxn = rate of reverse rxn Relationship with G: Equilibrium Constant G = G o + RT LnQ = 0 DG o = -RTLnK 35 G and K DG o = -RTLnK G o = 0 then K = 1 (rxn at equilibrium when all in std states) G o < 0 then K > 1 (fwd rxn spontaneous from std states) G o > 0 then K < 1 (rev rxn spontaneous from std states) K quantifies the relative balance between products and reactants at equilibrium

19 Calculating K from G o Rearrange equation to solve for K: LnK = G o -RT For our CH 3 OH formation example: LnK = x 10 4 J = (8.3145)(298.15K) only 1 sig fig! K = e = x 10 5 = 1. x 10 5 So: at equilibrium, product (CH 3 OH) should predominate: K = 1. x 10 5 = 1/[(a CO )(a H2 ) 2 ] 37 How about an example? What is the vapor pressure of benzene at 25 o C, if DG o = 5.16 kj/mol for: C 6 H 6 (l) C 6 H 6 (g) What is K? K = a gas /a liquid = P gas (vapor pressure of benzene!) Knowing that: G o = -RTLnK LnK = (5.16 x 10 3 J/mol)/[( J/mol-K)( K)] LnK = K = = P gas P o benzene = atm = 95. torr 38 19

20 Effect of Temperature on K G is temperature dependent, so: At T 1 : G o 1 = Ho - T 1 S o = -RT 1 LnK 1 Solve for S o : S o = RLnK 1 + H o /T 1 At T 2 : S o = RLnK 2 + H o /T 2 Combine, collect terms, rearrange: Ln(K 2 /K 1 ) = -(DH o /R)(1/T 2-1/T 1 ) Van t Hoff Equation Endothermic: K increases with increasing temp Exothermic: K decreases with increasing temp 39 Clausius-Clapeyron Revisited Recall, at any temperature T: H o - T S o = -RTLnK Now, solve for LnK: LnK = (-DH o /R)(1/T) + DS o For the vaporization of a liquid: A (l) A (g) K = a A(g) /a A(l) = P A /1 = P A Just substitute and, voila, we have C-C: LnP A = (-DH o vap /R)(1/T) + DSo 40 20

21 Fun with Van t Hoff Use to measure H o Measure K at different temperatures Plot LnK versus 1/T Straight line plot with slope = -DH o /R Can calculate G o from K Can calculate S o (from H o & G o ) 41 Big Example Problem Let s look at this process to make ammonia: ½N 2 (g) + 3/2H 2 (g) NH 3 (g) Some thermodynamic constants: DG o f : kj/mol DH o f : kj/mol 42 21

22 Spontaneous at 25 o C? Calculate DG o rxn : G o = Σ G o f (products) - Σ Go f (reactants) = G o f (NH 3 ) - [½ Go f (N 2 ) + 3/2 Go f (H 2 )] = kj - 0 G o = kj Yes, reaction is spontaneous 43 What is the value of K? From: DG o = -RTLnK LnK = G o = x 10 3 J = RT -( J/mol-K)( K) Solving for K: K = e = x 10 2 = 7.71 x

23 What is S o? We need H o first: H o = Σ H o f (products) - Σ Ho f (reactants) = H o f (NH 3 ) - [½ Ho f (N 2 ) + 3/2 Ho f (H 2 )] = kj - 0 H o = kj 45 Now, on to S o! Rearranging Gibbs-Helmholtz: S o = ( H o - G o )/T Substituting: S o = [(-46,110 J - (-16,480 J))]/ K S o = J/K S o = J/K Why negative? 46 23

24 At What Temperature Will NH 3 Decompose? G o = H o - T S o negative negative At a high enough temperature, G o will be positive and the reverse rxn will be spontaneous: T > H o / S o = -46,110 J/ J/K T > K 47 What is K at 1000 K? Here s what we know: K 1 = 7.71 x 10 2 K 2 =? T 1 = K T 2 = K Invoking van t Hoff: Ln(K 2 /K 1 ) = -(DH o /R)(1/T 2-1/T 1 ) Ln(K 2 /K 1 ) = -(-46,110 J/ J/mol -K)[1/1000-1/298.15] Ln(K 2 /K 1 ) = ( x 10 3 )[ x 10-3 ] = K 2 /K 1 = e = x 10-6 K 2 = ( x 10-6 )(7.71 x 10 2 ) = x 10-3 = 1.7 x