Thermodynamics II Thermodynamics & Kinetic Connection

Thermodynamics II Thermodynamics & Kinetic Connection • The conversion of graphite to diamond is a spontaneous process at ambient pressure. However, at low to moderate temperatures (i.e., room temperatures) the rate of formation, kinetically is not favored. Reference: slideplayer.com Physical Thermal Properties of Matter • Previously we studied the physical thermal property of Enthalpy, DH, in this chapter we will introduce two more thermal physical properties, Entropy, DS and Gibbs Free Energy, DG. – These properties are extensive in nature. – They are state functions, meaning they are independent of pathway. • Example: ∆H = Hfinal – Hinitial – Thermal properties have directionality. • Example: Exothermic process is flow of energy from system to surroundings (e.g., –DH). Spontaneity is Directional • Thermodynamics predicts whether a process will proceed towards product formation under given conditions. – If conditions are favorable, then it said to be spontaneous. • Spontaneity DOES NOT imply a fast or slow rate of reaction or a length of time. – If extra energy is required for the process to proceed forward that can not be gotten from the surroundings, then it is said to be nonspontaneous in that direction. Reference: www.m2c3.com Entropy: The 2nd Law of Thermodynamics • The 2nd law describes relationship between entropy and spontaneity. – All spontaneous changes, whether involving a phase change or a chemical reaction, results in an increase in the entropy of the universe. DSuniv > 0 Spontaneous DSuniv < 0 Nonspontaneous (spontaneous if in opposite direction) Reversible: system is at equilibrium. DSuniv = 0 Entropy & The 2nd Law of Thermodynamics • The total entropy change of the universe (DSuniverse) must be positive for a process to be spontaneous. – For irreversible (spontaneous) process, DSuniverse > 0. • The universe’s entropy increases for a spontaneous irreversible process. – For reversible process, DSuniverse = 0 • The entropy of the universe remains the same (i.e., does not change) for a nonspontaneous reversible process. DSuniverse = DSsystem + DSsurroundings • When the entropy of the system decreases, the entropy of the surroundings (DSsurroundings ) must increase by a larger amount. – DSsystem is negative, then DSsurroundings is positive Factors that Effect Entropy(DSosystem) • Size of the container increases: – the number of microstates accessible to the system and the density of states. • Phase changes effect the number of macro states possibilities. – Solids have fewer macro states than liquids, which have fewer macro states than gases. • Molecular complexity & Ion size – Small more compact; more ions per volume greater the density. • Example: NaF (52 J/mol) vs. MgO (27 J.mol) • Temperature elevation/ energy flow. – Hot flows to cold; always spontaneous. • Dissolution of solid into liquid. – Liquids or solids dissolve in a solvent in a spontaneous process owing to the increase in entropy. • Matter and energy are more dispersed. Problem: Predict whether the entropy of the system increases (+), decrease (-) or stays the same. 1. Snowflake melting. 2. Dry ice subliming. 3. Table salt dissolving in boiling water. 4. Magma cooling after a volcanic eruption. 5. Formation of stalagmites in caves. 6. Ag+(aq) + Cl−(aq) → AgCl(s) 7. CaCO3(s) ⟶ CaO(s)+CO2(g) 8. 2 NH3(g) → N2(g) + 3 H2(g) D So system −(DHosys ) DSsurroundings = T = S(S°product) − S(S°reactant) • When the entropy of the system decreases, the entropy of the surroundings increases. – DSsys is negative, then DSsurr is positive. • In an exothermic process: energy is released from the system into the surroundings. – The overall entropy of the system goes down, but the additional energy to the surroundings causes its entropy to increase. • If the entropy of the system increases, then the entropy of the surroundings must decrease. – DSsys is positive, then DSsurr is negative. • With an endothermic process the system takes in energy from the surroundings, cause the surrounding entropy to decrease while the system entropy increases. • The overall effect on the surroundings entropy is dependent on its initial temperature. – The higher the original temperature of the surroundings the less effect addition or removal of energy will have on the surroundings overall entropy. Examples of Standard Entropies Substance DSo (J/K mol) C (s) as graphite 5.740 C (s) as diamond 2.38 CO (g) 197.7 CO2 (g) 213.8 H2O (g) 188.71 H2O (l) 69.1 O2 (g) 205.03 • Standard conditions designated as DSo (subscript o). • Standard entropy conditions of reactions or phase changes were carried out under standard conditions (i.e., 1 mol, 25.0 oC and 1 atm). • Values are obtained from Thermodynamic Tables. • The units for entropy of chemical substances are J/K mole. • Entropy is an extensive thermochemical property of matter. • Entropy is a state function of matter. 1 DHo system DSo system Exothermic <0 Positive >0 Spontaneous or Not? • • 2 Exothermic <0 Negative <0 • • 3 Endothermic >0 Positive >0 • • 4 Endothermic >0 Negative < 0 • • Spontaneous at all temperatures. DSouniverse >0 Dependent on the relative magnitude of DHo & DSo. Spontaneous at low temperatures. Dependent on the relative magnitude of DHo & DSo. Spontaneous at high temperatures. Not spontaneous at all temperatures. DSouniverse < 0 Problems: 1. Calculate the standard entropy change for the following process: H2O(g) ⟶ H2O(l). Substance DSosystem = S(S°products) − S(S°reactants) DSo (J/K mol) C (s) as graphite 5.740 C (s) as diamond 2.38 CO (g) 197.7 CO2 (g) 213.8 H2O (g) 188.71 H2O (l) 69.1 O2 (g) 205.03 Problems: 1. For this reaction: Ca(OH)2(s) ⟶ CaO(s) + H2O(l) a. Determine the standard entropy change. b. Determine the DSsurr. c. Calculate the DSuniv. d. Predict if the reaction would be spontaneous or not. Justify answer. Substance DHosys ( kJ/mol) DSosys (J/K mol ) at 298 K CaO (s) -634.9 38.1 Ca(OH)2 (s) -985.2 83.4 H2O (l) -285.8 69.1 3rd Law of Thermodynamics • As temperatures approach absolute zero (i.e., zero K), theoretically matter should stop all modes of motion (e.g., translation, rotation, vibration, etc.). – According to Boltzmann – only one microstate when entropy =0 at absolute zero K. • Entropy (S) = k ln W where k is Boltzmann constant & W = number of microstates – S = k ln 1 then S = 0 • The 3rd Law of Thermodynamics describes this phenomenon; – At absolute zero the entropy of pure crystalline solid is zero.* (* not achievable). Gibbs Free Energy is the energy that a chemical reaction uses to do work. Example: Reaction: C(s, graphite) + 2 H2 (g) → CH4 (g) Where: DHorxn = -74.6 kJ DSorxn = - 80.8 J/k DGorxn = -50.5 kJ • The above reaction is exothermic, releasing 74.6 kJ energy into the surroundings. • The maximum about of energy available to do useful work, DGorxn is 50.5 kJ not 74.6 kJ (DHorxn). • Why the discrepancy of 24.1 kJ of energy? • DSorxn is - 80.8 J/K (- 0.080 kJ/K). • In reality, the amount of energy to do useful work is less than DGorxn because additional energy is lost to the surroundings as heat. More on Gibbs Free Energy (DGo) DGo = DHo − TDSo • DSuniv determines whether a process is spontaneous. • DGo also determines spontaneity. • When DSuniv is (+) positive, the process is spontaneous, so from the equation about, DG must be (−) negative. • DGo value predicts the direction of a chemical reaction – Positive (+)DGo the reaction is nonspontaneous (i.e.. reactant-favored). – Negative (-)DGo the reaction is spontaneous (i.e.. product-favored). – DGo = 0, the reaction is at equilibria. Relation Between DGo, DHo & DSo Conditions where DGo will have a negative value (spontaneous): 1. DHo has a negative (-) value & DSo is has a positive (+) value at any temperature. • Exothermic system: more energy available & more randomness. 2. DHo has a large negative (-) value & DSo has a small negative (-) value. 3. DHo has a small positive (+) value & DSo has a large positive (+) value. 4. DHo has a small positive (+) value & DSo is a large positive (+) value at high temperatures. Conditions when DGo will have a positive value (nonspontaneous): 1. DHo has a positive (+) value & DSo has a negative (−) value at any temperature. • Endothermic system and a very ordered system. At equilibrium DGo = 0 DG Calculations 1. ∆Go = ∆Ho - T∆So – Have known values for temperature, ∆Ho & ∆So 2. ∆Go = S∆Go (Products) - S∆Go (Reactants) – Use tabulated values of free energies of formation, ∆Go. • ∆Go formation values for elements is ZERO kJ/mol. • DGo is an extensive property of matter. 3. Indirect method – Given several chemical reactions with DG values to obtain the DG value for given reaction. Problems: 1. Calculate DGo and determine whether the reaction is spontaneous. Given the following information: C (s) + 2 Cl2(g) → CCl4(g) DHo = - 95.7 kJ DSo = - 142.2 J/K T = 25.00 oC Problem#2: The following chemical reaction: 2 SO2(g) + O2(g) → 2 SO3 (g) was carried out under standard conditions. Determine the following thermo properties for this reaction: DHo, DSo, DSurr , DSuniv & DGo. DHfo (kJ/mol) DSo (J/K mol) SO2 (g) -297 248 SO3 (g) -396 257 O2 (g) 0 205 Substance Problem #3. Using the following data taken at 25.0oC, determine the DG for the conversion of diamond to graphite. C(diamond) → C(graphite) DG C(diamond) + O2 (g) → CO2 (g) -397 kJ C (graphite) + O2 (g) → CO2 (g) -394 kJ Equilibrium (K) & DG Connection ▪ ▪ ▪ The driving force of process can be viewed as the change in free energy, ∆G. ▪ Negative DG means driving force is towards completion (product-favored). ▪ Positive DG says driving force does not favor completion (reactant-favored). The equilibrium constant, Keq is related to reaction favorability. ▪ Product-favored processes have Keq > 1. ▪ Reactant-favored processes have Keq < 1 So, both ∆G˚ & Keq are about reaction favorability. ∆Go = - RT lnKeq R = 8.31 J/K•mol – The larger the value of Keq the more negative the value of ∆Go – The smaller the value of Keq the more positive the value of ∆Go Spontaneity, DG & K Relations Under Standard Conditions Keq DGo >1 < 0 Spontaneous; product favored =1 <1 0 Comments Reaction is at equilibrium > 0 Not spontaneous; reactantfavored Calculating DG & Keq under Non-Standard Conditions • To calculate the change in free energy the following relationship is required: DG = DG + RTlnQ • • Q is the reaction quotient. DG = DG only when the reactants and products are in their standard states. When Q = K, the reaction is at equilibrium & ∆G = 0. – The above relationship then becomes DG = - RTlnK Problem: What is the free energy change for the decomposition of ammonia gas to its elements (i.e., hydrogen and nitrogen gas) given the following data: T = 25.0 oC PNH3 = 12.9 atm PH2 = 0.250 atm PN2 = 0.870 atm DGo = 33.0 kJ R = 8.314 x 10-3 kJ/K mol