Advanced chemistry entropy, Cheat Sheet of Chemistry

Download Advanced chemistry entropy and more Cheat Sheet Chemistry in PDF only on Docsity! GCh18-1 Chapter 18. Entropy, Free Energy, and Equilibrium What we will learn: • Three laws of theormodynamic • Spontaneous processes • Entropy • Second law of thermodynamics • Gibbs free energy • Free energy and chemical reactions GCh18-2 Thermodynamics The scientific discipline that deals with the interconversion of heat and other forms of energy First law of thermodynamics Energy may be converted from one form to another, but cannot be created or destroyed. (one way to measure: DH) Second law of thermodynamics The law explains why chemical processes tend to favor one direction Third law of thermodynamics The law is an extension of the second law and only briefly examined in this course GCh18-5 Macrostate (distribution) and microstates Example Four molecules distributed between two containers Macrostate (I) S is small (four molecules in left container) Macrostate (II) S is big (three molecules in left container) Macrostate (III) S is very big (two molecules in left container) GCh18-6 Entropy • Analogies: deck of cards, socks in drawer, molecular motions • Change in entropy: DS = Sfinal - Sinitial • For a reaction: DS = Sproducts - Sreactants • Positive DS means: • An increase in disorder as reaction proceeds • Products are more disordered (random) than reactants Sgas >> Sliquid > Ssolid Ssolution >> Ssolvent or solute • Temperature increase – more molecular motions GCh18-7 S = k ln W S - entropy k - constant (Boltzman constant = 1.38 x 10-23 J/K) W - disorder (how many individual microstates is involved in a macrostate) DS = Sf - Si DS = k ln(Wf) - k ln(Wi) DS = k ln (Wf / Wi) GCh18-10 A. Entropy change (DS) in the system aA + bB g cC + dD DS0 rxn = [ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ] DS0 rxn = S nDS0 products - S mDS0 reactants DS0 rxn - standard entropy of reaction n - mole number of products m - mole number of reactants S0 products - standard entropy of prodcuts S0 reactants - standard entropy of reactants GCh18-11 Problem Calculate the standard entropy change for the reaction N2 (g) + 3 H2 (g) g 2 NH3 (g) DS0(N2) = 191.5 J/K mol DS0(H2) = 131.0 J/K mol DS0(NH3 ) = 193.0 J/K mol Solution DS0 rxn = 2 S0(NH3) - [ S0(N2) + 3 S 0(H2) ] = -199 J/K mol GCh18-12 Problem Calculate the standard entropy change for the reaction CaCO3 (s) g CaO (s) + CO2 (g) DS0(CaCO3) = 92.9 J/K mol DS0(CaO) = 39.8 J/K mol DS0(CO2) = 213.6 J/K mol Solution DS0 rxn = S0(CaO) + S0(CO2) - S 0(CaCO3) = 160.5 J/K mol GCh18-15 Problem Predict the entropy change of the system in each of the following reactions is positive or negative 2 H2 (g) + O2 (g) g 2 H2O (l) DS < 0 Two molecules combine to form a product. There is a net decrease of one molecule and gases are converted into liquid. The number of microstates of product that will be smaller, and entropy will be negative NH4Cl (s) g NH3 (g) + HCl (g) DS > 0 A solid is converted into two gaseous products H2 (g) + Br2 (g) g 2HBr (g) DS = 0 Two diatomic molecules are involved in reactants and products. All moleculeas have similar complexity, and therefore we expect similar entropy of reactants and products GCh18-16 Third law of thermodynamics and absolute entropy • The entropy of a pure crystalline substance equals zero at absolute zero S = 0 at T = 0 K any substance above 0 K has entropy > 0 any impure crystal has entropy > 0 • Standard Entropy (at 25 C) = S0 entropy at 25 C which is relative to 0 (zero) entropy at absolute zero At absolute zero all molecular moves are frozen, and therefore there is only one microstate which forms a particular macrostate. S = k ln (W) W = 1 ln(1) = 0 S = 0 GCh18-17 Entropy changes in surroundings For exothermic processes the heat is tranfered from a system to surroundings increasing the number of microstates in the surroundings, therefore the entropy of surroundings increases If the temperature of the surroundings is high, the transfered heat does not change the number of microstates much, however if the temperature is low, the transfered heat changes the number of microstates more strongly DSsurr = -DHsys / T Example N2 (g) + 3 H2 (g) g 2 NH3 (g) -DH0 rxn = -92.6 kJ / mol DSsys = -199 J / K mol DSsurr = -(-92.6 x 1000 J/mol) / 298 K = 311 J/ K mol DSuniv = DSsys + DSsurr GCh18-20 -TDSuniv = DHsys - TDSsys < 0 -TDSuniv = DG DG = DH - TDS < 0 • For any spontaneous change, DG is negative (the free energy decreases) • G is a state function (like H and S) - independent of pathway Summary • DG < 0 reaction is spontaneous in forward direction • DG > 0 reaction is nonspontaneous or it is spontaneous in the opposite direction • DG = 0 at equilibrium GCh18-21 Standard Free Energy Changes DG0 • Free energy change for a reaction when it occurs under standard-state conditions (reactants and products are in their standard states) Standard Temperature and Pressure (STP) 1 atmosphere, solutions at 1 M, DGf = 0 for the elements • DG0 (at STP) is generally used to decide if a reaction is spontaneous. It is spontaneous if DG0 rxn is negative • Two ways to obtain DG0 rxn for a reaction: (1) From DH0 f and DS0 requires having DH0 f and DS0 data for all reactants and products DG0 = DH0 - TDS0 DH0 = S DH0 f(products) - S DH0 f(reactants) DS0 = S DS0(products) - S DS0(reactants) GCh18-22 (2) From standard “Free Energies of Formation” DG0 f DG0 rxn = S nDG0 f(products) - S mDG0 f(reactants) where DG0 f is the free energy change for the formation of one mole of the compound from its elements Standard free energy of formation The free energy change that occurs when 1 mole of the compound is synthesized from its elements in their standard states GCh18-25 Temperature and chemical reactions “Ball park” estimates of temperature at which a reaction becomes spontaneous Procedure: • Find DH° and DS° at 25 C • Set DH - TDS = 0 • Solve for T, estimated temperature at which reaction becomes spontaneous GCh18-26 Problem Calcium oxide (CaO) is produced according to the reaction CaCO3 (s) D CaO (s) + CO2 (g) What is the temperature at which the reaction favors the products First we calculate DH0 and DS0 using the standard data of the reactants and products DH0 = DH0 f(CaO) + DH 0 f(CO2) - DH 0 f(CaCO3) DH0 = 177.8 kJ / mol DS0 = S0(CaO) + S0(CO2) - S 0(CaCO3) DS0 = 160.5 J / K mol GCh18-27 DG0 = DH0 - TDS0 DG0 = 130.0 kJ / mol at 298 K Because DG0 is a large positive value, we conclude that the reaction is not favored for products at 25 C (298 K). In order to make DG negative, first we find the temperature at which DG is zero 0 = DH0 - TDS0 T = DH0 / DS0 T = 1108 K = 835 C At temperature higher that 835 C DG0 becomes negative indicating that the reaction now favors the product formation GCh18-30 Problem The molar heat of fusion and vaporization of benzene are 10.9 kJ / mol and 31 kJ / mol. Calculate the entropy changes for the solid g liquid and liquid g vapor transitions. At 1 atm, benzene melts at 5.5 C and boils at 80.1 C. The entropy change for melting 1 mole of benzene at 5.5 C is DSfus = DHfus / Tf = (10.9 kJ/mol) (1000 J/kJ) / (5.5 + 273)K = 39.1 J/K mol The entropy change for boiling 1 mole of benzene at 80.1 C is DSvap = DHvap / Tbp = (31 kJ/mol) (1000 J/kJ) / (80.1 + 273)K = 87.8 J/K mol GCh18-31 Free energy and chemical equilibrium During a chemical reactions not all reactants and products will be at their standard states. Under this condition, there is a relationship between DG and DG0 • For any chemical system: DG = DG0 + (RT) ln Q Q - concentration quotient • If DG is not zero, then the system is not at equilibrium and it will spontaneously shift toward the equilibrium state • But, at equilibrium: DG = 0 and Q = K DG0 = - (RT) ln K GCh18-32 • Relates equilibrium constant to standard free energy DG0 • Allows the calculation of K if DG0 is known and vice versa, e.g., when K is very small or large, may be difficult to measure the concentration of a reactant or product, so it is easier to find DG0 and calculate K • For gaseous reactions: K = Kp • For solution reactions: K = Kc • Units of DG must match those of RT value GCh18-35 At equilibrium DG = 0 DH - TDS = 0 T = DH / DS T = (60.7 x 103 J / mole) / [(175 - 76.1) J / mole K] T = (60.7 x 103 J / mole) / (273 K) = 614 K GCh18-36 Problem Calculate the equilibrium constant, KP, for the reaction shown below at 25 C 2 H2O (l) D 2 H2 (g) + O2 (g) Solution First we calculate DG0 DG0 = 2DG0 f(H2) + DG0 f(O2) - 2DG0 f(H2O) = 474.4 kJ/mol GCh18-37 DG0 = - (RT) ln Kp 474.4 kJ/mol (1000 J/ kJ) = - (8.314 J/K mol)(298 K) ln(Kp) ln(Kp) = -191.5 Kp = e-191.5 = 7 x 10-84 GCh18-40 DG = 5.4 x 103 J/mol - 8.46 x 103 J/mol = -3.06 x 103 J/mol Because DG < 0, the net reaction proceeds from left to right (from reactants to products) to reach equilibrium GCh18-41 Problem The DG 0 for the reaction H2 (g) + I2 (g) D 2 HI(g) is 2.6 kJ/mol at 25 K. The initial pressures are PH2 = 4.26 atm and PI2 = 0.024 atm and PHI = 0.23. Calculate DG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium Solution DG = DG0 + (RT) ln Qp Qp = P2 HI / (PH2 PI2 ) DG = DG0 + (RT) ln [P2 HI / (PH2 PI2)] = 2.6 x 103 J/mol + (8.314 J/K mol) (298 K) ln [(0.23)2 / (4.26 x 0.024)] = 0.97 x 103 J/mol = 0.97 kJ/mol