University:
Kentucky State University
Course:
CHE 101 | General Chemistry I
Academic year:

2021
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355

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37
Author:

aprivantuqy

Thermodynamics CHEM-102 1 Thermodynamics It deals with the relationships between heat and other forms of energy. Thermodynamics Functions: DH DS DG CHEM-102 : Enthalpy (-) Exothermic (+) Endothermic (-) : Entropy (+) (-) Spontaneous : Gibbs Energy (+) Non-Spontaneous 2 Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Examples: energy, enthalpy, pressure, volume, temperature DH0 / DS0 / DG0 CHEM-102 3 Spontaneous Physical and Chemical Processes • A waterfall runs downhill • A lump of sugar dissolves in a cup of coffee • At 1 atm, water freezes below 0 oC and ice melts above 0 oC • Heat flows from a hotter object to a colder object • A gas expands in an evacuated bulb • Iron exposed to oxygen and water forms rust spontaneous nonspontaneous CHEM-102 4 spontaneous nonspontaneous CHEM-102 5 Favorable Reaction → proceed to forward (favor the product side CHEM-102 DH0 → (-) Exothermic DS0 → (+) DG0 → (-) Spontaneous 6 Does a decrease in enthalpy mean a reaction proceeds spontaneously? Spontaneous reactions CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = -890.4 kJ/mol H+ (aq) + OH- (aq) H2O (l) DH0 = -56.2 kJ/mol H2O (s) NH4NO3 (s) CHEM-102 H2O (l) DH0 = 6.01 kJ/mol H2O NH4+(aq) + NO3- (aq) DH0 = 25 kJ/mol 7 Entropy (S) is a measure of the randomness or disorder of a system. order disorder S S DS = Sf - Si If the change from initial to final results in an increase in randomness Sf > Si DS > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state. Ssolid < Sliquid << Sgas H2O (s) CHEM-102 H2O (l) DS > 0 8 First Law of Thermodynamics Energy can be converted from one form to another, but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. CHEM-102 Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 9 Entropy Changes in the Surroundings (DSsurr) Exothermic Process DSsurr > 0 CHEM-102 Endothermic Process DSsurr < 0 10 Processes that lead to an increase in entropy (DS > 0) CHEM-102 11 Example: Br2(l) DS > 0 CHEM-102 Br2(g) Example: I2(s) I2(g) DS > 0 12 DG0 DH0 J kJ kJ/mol CHEM-102 13 Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. CHEM-102 14 Entropy Changes in the System (DSsys) When gases are produced (or consumed) CHEM-102 • If a reaction produces more gas molecules than it consumes, DS0 > 0. • If the total number of gas molecules diminishes, DS0 < 0. • If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number. 15 Example 1: Predict whether the entropy change is greater or less than zero for each of the following processes: (a) freezing ethanol ΔS < 0 (b) evaporating a beaker of liquid bromine at room temperature ΔS > 0 (c) dissolving glucose in water ΔS > 0 (d) cooling nitrogen gas from 80°C to 20°C ΔS < 0 CHEM-102 16 Example 2: Predict whether the entropy change of the system in each of the following reactions is positive or negative. (a) 2H2(g) + O2(g) (b) NH4Cl(s) (c) H2(g) + Br2(g) CHEM-102 2H2O(l) NH3(g) + HCl(g) ΔS < 0 ΔS > 0 2HBr(g) 17 Solution (a) Two reactant molecules combine to form one product molecule, the fact that there is a net decrease of one molecule and gases are converted to liquid ensures that the number of microstates will be diminished and hence ΔS° is negative. (b) A solid is converted to two gaseous products. Therefore, ΔS° is positive. (c) The same number of molecules is involved in the reactants as in the product. Furthermore, all molecules are diatomic and therefore of similar complexity. As a result, we cannot predict the sign of ΔS°, but we know that the change must be quite small in magnitude. CHEM-102 18 Entropy Changes in the System (DSsys) The standard entropy of reaction (DS0rxn ) is the entropy change for a reaction carried out at 1 atm and 250C. aA + bB DS0rxn = cC + dD [ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ] DS0rxn = S nS0(products) - S mS0(reactants) CHEM-102 19 The standard free-energy of reaction (DG0rxn) is the freeenergy change for a reaction when it occurs under standardstate conditions. aA + bB cC + dD 0 DGrxn = [cDG0f (C) + dDG0f (D) ] - [aDG0f (A) + bDG0f (B) ] 0 DGrxn = S nDG0f (products)- S mDGf0 (reactants) Standard free energy of formation (DG0f ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. DGf0 of any element in its stable form is zero. CHEM-102 20 aA + bB DS0rxn = cC + dD [ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ] DS0rxn = S nS0(products) - S mS0(reactants) 0 DGrxn = [cDG0f (C) + dDG0f (D) ] - [aDG0f (A) + bDG0f (B) ] 0 DGrxn = S nDG0f (products)- S mDGf0 (reactants) CHEM-102 21 CHEM-102 22 Example 3: From the standard entropy values provided, calculate the standard entropy changes for the following reactions at 25°C. (a) CaCO3(s) CaO(s) + CO2(g) (b) N2(g) + 3H2(g) 2NH3(g) (c) H2(g) + Cl2(g) 2HCl(g) Solution (a) ΔS°rxn = [S°(CaO) + S°(CO2)] - [S°(CaCO3)] = [(39.8 J/K·mol) + (213.6 J/K·mol)] - (92.9 /K·mol) = 160.5 J/K·mol (b) ΔS°rxn = [2S°(NH3)] - [S°(N2) + 3S°(H2)] = (2)(193 J/K·mol) - [(192 J/K·mol) + (3)(131 J/K·mol)] = -199 J/K·mol CHEM-102 23 Example 4: Calculate the standard free-energy changes for the following reactions at 25°C. (a) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) (b) 2MgO(s) 2Mg(s) + O2(g) Solution: (a) ΔG°rxn = [ΔG°f (CO2) + 2ΔG°f (H2O)] - [ΔG°f (CH4) + 2ΔG°f (O2)] ΔG°rxn =[(-394.4 kJ/mol) + (2)(-237.2 kJ/mol)] [(-50.8 kJ/mol) + (2) (0 kJ/mol)] = -818.0 kJ/mol (b) ΔG°rxn = [2ΔG°f (Mg) + ΔG°f (O2)] - [2ΔG°f (MgO)] ΔG°rxn = [(2)(0 kJ/mol) + (0 kJ/mol)] - [(2)(-569.6 kJ/mol)] = 1139 kJ/mol CHEM-102 24 Gibbs (Free) Energy For a constant temperature and constant pressure process: Gibbs Energy (G) DG = DHsys -TDSsys DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium. CHEM-102 25 DG = DH - TDS CHEM-102 26 DG = DH - TDS CHEM-102 27 Temperature and Spontaneity of Chemical Reactions CaCO3 (s) CaO (s) + CO2 (g) Equilibrium Pressure of CO2 DH0 = 177.8 kJ/mol DS0 = 160.5 J/K·mol DG0 = DH0 – TDS0 At 25 oC, DG0 = 130.0 kJ/mol At 835 oC → DG0 = 0 CHEM-102 28 Gibbs Free Energy and Phase Transitions DG0 =0= DH0 – H2O (l) DH DS = T TDS0 H2O (g) DH = 40.79 kJ/mol DS = 373 K T = 1.09 x 10-1 kJ/K·mol = 109 J/K·mol CHEM-102 29 Example 5: The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid → liquid and liquid → vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C. ΔSfus ΔH fus = Tf (10.9 kJ/mol)(1000 J/1 kJ) = (5.5 + 273)K = 39.1 J / K  mol CHEM-102 30 The Efficiency of Heat Engines Efficiency = CHEM-102 Th - Tc X 100% Th 31 Gibbs Free Energy and Chemical Equilibrium DG = DG0 + RT lnQ R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium DG = 0 Q=K 0 = DG0 + RT lnK DG0 = - RT lnK CHEM-102 32 DG0 = - RT lnK CHEM-102 33 Example 6: Using the solubility product of silver chloride at 25°C (1.6 x 10-10), calculate ΔG° for the process AgCl(s) Ag+(aq) + Cl-(aq) Solution The solubility equilibrium for AgCl is AgCl(s) Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] = 1.6 x 10-10 ΔG° = -(8.314 J/K·mol) (298 K) ln (1.6 x 10-10) = 5.6 x 104 J/mol = 56 kJ/mol CHEM-102 34 Example 7: Using provided data, calculate the equilibrium constant (KP) for the following reaction at 25°C: 2H2O(l) 2H2(g) + O2(g) Solution: ΔG°rxn = [2ΔG°f(H2) + ΔG°f(O2)] - [2ΔG°f(H2O)] = [(2)(0 kJ/mol) + (0 kJ/mol)] - [(2)(-237.2 kJ/mol)] = 474.4 kJ/mol o ΔGrxn = -RTlnK p 1000 J 474.4 kJ/mol × = -(8.314J/K  mol)(298 K)lnK p 1 kJ lnK p = -191.5 K p = e -191.5 = 7 × 10-84 CHEM-102 35 Example 8: The equilibrium constant (KP) for the reaction N2O4(g) 2NO2(g) is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.40 kJ/mol. In a certain experiment, the initial pressures are PNO2 = 0.122 atm and PN2O4 = 0.453 atm. Calculate ΔG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium. CHEM-102 36 Solution ΔG = ΔG o + RTlnQp = ΔG o + RTln 2 PNO 2 PN 2O4 2 (0.122) = 5.40 × 103 J/mol + (8.314J/K  mol)(298 K) × ln 0.453 = 5.40 × 103 J/mol - 8.46 × 10 3 J/mol = -3.06 × 103J/mol = - 3.06kJ / mol Because ΔG < 0, the net reaction proceeds from left to right to reach equilibrium. CHEM-102 37

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