General Physical Chemistry I Lecture 11 Aleksey Kocherzhenko March 12, 2015"
Last time "
W Entropy" Let be the number of microscopic configurations that correspond to the same macroscopic state" Ø Entropy is a measure of how disordered a system is" Define entropy as:" S = k B ln W Change in entropy on going from state 1 to state 2:" S = S 2 S 1 = k B ln W 2 k B ln W 1 = k B (ln W 2 ln W 1 ) or:" S = k B ln W 2 W 1 > 0 if" W 2 >W 1 < 0 if" W 2 <W 1 Rudolph Clausius" 2 nd law of thermodynamics" In spontaneous processes, the entropy increases" Alternatively:" S = q rev he entropy difference between two states is equal to the heat that must be transferred to the system in a reversible transition between these states at " his is the original (thermodynamic) definition of entropy given by R. Clausius" he statistical and the thermodynamic definitions of entropy are equivalent!
H Heater" U Maximum efficiency of a heat engine" q in he laws of thermodynamics:" q out System" C 1 st law:" dq =du +da 2 nd law: for a spontaneous process," S>0 A Cooler" Heat engine operation must be spontaneous and must use a cyclic process" S U sys =0 S sys =0 and are state functions à at the end of each cycle: and à Entropy change for the heat engine = entropy change for the heater and the cooler:" S = S {z H } + S C 0 {z } = q in = q Heat loss is minimum out H when equality holds" H C ) q out q in C à he difference between the heat that flows into and out of the system in one cycle is completely converted into work:"q in q out = A ) A max = q in q min out C = q in q in H ) max =1 We define the efficiency of a heat engine as:" = A q in Maximum efficiency of a heat engine:" Ø he maximum efficiency for a heat engine is realized in the Carnot cycle" C H
he third law of thermodynamics"
he third law of thermodynamics" =0 Only a single microscopic configuration for a perfect crystal at = 0 (no vibrations)" W = 1 ) S = kb ln 1 = 0 he third law of thermodynamics:" CO" At absolute zero, the entropies of all perfect crystals are equal to zero " Walther Nernst" Many possible microscopic configurations for a crystal at > 0 (vibrations present)" W > 1 ) S = kb ln W > 0 In reality, at = 0 the entropy of a crystal may be 0 due to orientational disorder:" Ø for CO:" S (0) = kb ln 2N = N kb ln 2 Residual entropy" H2O" wo possible for each of" orientations" N molecules" Ø for H2O: freedom in the position of H-atom that does not participate in an H-bond" >0
Calculating the absolute entropy" By definition:"ds = dq he entropy is a function of temperature" ) S ( )= Z 0 Entropy relative to C ( )d S (0), where we can express"dq = C ( )d minimum value" What if there is a phase transition?" Entropy of a phase transition:" Solids more ordered than liquids, liquids more ordered than gasses" à here must be an entropy change associated with a phase transition" S pt = q pt pt = pt H pt pt à positive for transitions from a more ordered to less ordered phase (e.g., or ) and negative otherwise" S fus S vap
Calculating the absolute entropy" he heat capacities of different phases often differ significantly" he absolute entropy of a substance can be calculated by adding:" =0 Ø he residual entropy at ; " Ø All entropies associated with heating the substance from =0to the temperature of interest;" Ø All entropies associated with phase transitions" Entropy increase between melting and boiling temperatures" Entropy of vaporization" S ( )=S (0) + Residual entropy" Z fus 0 C (s) p + fus H fus + Entropy increase between 0 K and the melting temperature" Z vap fus Entropy of fusion" (l) Cp + vap H + vap Z vap (g) Cp Entropy increase between boiling temperature and temperature of interest"
Standard molar entropies" he entropies calculated as described on the previous slide are absolute values, and are sometimes referred to as third law entropies" Ø he entropy of a substance depends on its state (pressure, temperature);" Ø Define standard molar entropies (under standard pressure: 1 bar)" Ø he table reports standard molar entropies of some substances at 298.15 K" Ø For a given substance, molar entropy for gas phase is greater than molar entropy for liquid phase is greater than molar entropy for solid phase"
he Gibbs free energy"
Standard reaction entropies" Ø If a gas is produced in a chemical reaction, the entropy likely increases" Ø If a gas is consumed in a chemical reaction, the entropy likely decreases" he standard reaction entropy is the difference between the standard entropies of reactants and products:" Standard molar entropies for products and reactants " rs = X j Sum over all products" b j S m,j i Sum over all reactants" X a i S m,i Stoichiometric coefficients for products and reactants " Example: calculate the standard entropy for the reaction of water formation from H and O gasses at 25 C if S m (H 2 O, l) = 70 J K 1 1 mol,"," S m (H 2, g) = 131 J K 1 mol Write reaction: 2H 2 + O 2 à 2H 2 O" 1 S m (O 2, g) = 205 J K 1 mol rs =2S m (H 2 O, l) [2S m (H 2, g) + S m (O 2, g)] = 327 J K 1 mol 1 1
Spontaneity of chemical reactions" Wait! here is something wrong!" For the reaction 2H 2 + O 2 à 2H 2 O" rs =2S m (H 2 O, l) [2S m (H 2, g) + S m (O 2, g)] = 327 J K 1 mol he standard reaction is negative the reaction should not be spontaneous, but it is! " How come?" 1 < 0 he reaction mixture is not an isolated system à to apply the second law correctly, " " " " " " " " " " "we need to consider the surroundings" rs sur = q r = H r Reaction is at constant pressure" he reaction 2H 2 + O 2 à 2H 2 O is exothermic, the reaction enthalpy is " H r = 527 kj mol he total change in entropy for the system and the surroundings is thus:" rs total = r S + r S sur = = 327 J K 1 mol 1 + =+1.59 10 3 J K 1 mol " 527 10 3 J mol 1 298 K he total change in entropy is positive à the reaction is spontaneous" 1 # = 1
Gibbs free energy" In order to analyze the spontaneity of chemical reactions, we need to calculate the entropy changes in both the system and the surroundings" Notice, however, that the change in the entropy of the surroundings is expressed in terms of a property of the system the reaction enthalpy:" rs sur = H r hus, the condition for spontaneity of a chemical reaction," ds universe =ds sys +ds sur > 0 can be expressed in terms of the properties of the system only:" ds sys, dh sys ds sys < 0 J. Willard Gibbs" dh sys > 0 Introduce the Gibbs free energy:" G = H S At constant temperature:" dg =dh ds he Gibbs free energy only depends on state functions, so it is a state function itself" Criterion for spontaneity of chemical reactions (or other processes):" dg <0
Spontaneity and equilibrium conditions" ds universe =ds sys +ds sur ) ds sys =ds universe ds sur he change in the Gibbs free energy can then be written as follows:" dg =dh sys ds sys = dh sur =dh sys ds universe {z } =ds sys =dh sys ds universe + dh sys {z }!! because"dh sys = dh sur = = ds universe At constant temperature and pressure, the criterion for spontaneity of a process is:" ds universe > 0, dg = ds universe < 0 Equilibrium is achieved when"dg =0
Gibbs free energy and nonexpansion work" he work performed by a system may include expansion work and nonexpansion work (i.e., electrical work in an electrochemical cell)" Expansion Nonexpansion he first law of thermodynamics:" dq = da +du work" work" For reversible heat transfer:" ds = dq thermodynamic definition of entropy" At constant temperature and pressure:" dg =du + pdv {z } =dh ds ) dq = ds =dq pdv +da 0 +pdv {z } du ) du =dq pdv +da 0 {z } =da ds At constant temperature and pressure, the Gibbs free energy is equal to the maximum nonexpansion work! G<0: exergonic (work-producing) process" G>0: endergonic (work-consuming) process"