Chapter 19 Chemical Thermodynamics Entropy and free energy Learning goals and key skills: Understand the meaning of spontaneous process, reversible process, irreversible process, and isothermal process. State the second law of thermodynamics. Describe the kinds of molecular motion that a molecule can possess. Explain how the entropy of a system is related to the number of accessible microstates. Predict the sign of S for physical and chemical processes. State the third law of thermodynamics. Calculate standard entropy changes for a system from standard molar entropies. Calculate entropy changes in the surroundings for isothermal processes. Calculate the Gibbs free energy from the enthalpy change and entropy change at a given temperature. Use free energy changes to predict whether reactions are spontaneous. Calculate standard free energy changes using standard free energies of formation. Predict the effect of temperature on spontaneity given H and S. Calculate G under nonstandard conditions. Relate G and equilibrium constant. Energetics of solutions An endothermic process is not favored based on the change in enthalpy, yet they occur. Why? We must also look at the disorder (entropy, S) of the system. 0 th Law of Thermodynamics If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then C is also in thermal equilibrium with A. 1 st Law of Thermodynamics Energy of the universe is constant (conserved). U = q + w q = heat absorbed by the system, w = work done on the system 2 nd Law of Thermodynamics In a spontaneous process, the entropy of the universe increases. S universe = S sys + S surr > 0 (if spontaneous) 3 rd Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is zero: S(0 K) = 0. 1
Review Chapter 5: energy, enthalpy, 1 st law of thermo Thermodynamics: the science of heat and work Thermochemistry: the relationship between chemical reactions and energy changes Energy (E) The capacity to do work or to transfer heat. Work (w) The energy expended to move an object against an opposing force. w = F d Heat (q) Derived from the movements of atoms and molecules (including vibrations and rotations). +q, heat absorbed by system (endothermic) -q, heat evolved by the system (exothermic) Surroundings +w, work done onto the system System E = q + w -w, work done by the system 1 st law of thermodynamics: the law of conservation of energy Energy is neither created nor destroyed Internal Energy = heat + work E = q + w 2
Enthalpy, H the heat content of a substance at constant pressure. a state function an extensive property reversible The enthalpy change, H, is defined as the heat gained or lost by the system under constant pressure; it depends upon the identity and states of the reactants and products. H = q p Standard Enthalpy Values Most H values are labeled H o Measured under standard conditions P = 1 atm T = usually 298.15 K (25 o C) (Concentration = 1 M) H fo = standard molar enthalpy of formation (Appendix C) H o rxn = Σ H fo (products) - Σ H fo (reactants) Spontaneous Reactions: reactions that occur without outside intervention In general, spontaneous reactions are exothermic. Thermite reaction: Fe 2 O 3 (s) + 2 Al(s) 2 Fe(s) + Al 2 O 3 (s) H rxn = -848 kj But many spontaneous reactions or processes are endothermic or even have H = 0. 3
Spontaneous Processes Spontaneous processes are those that can proceed without any outside intervention. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B. 4
S = S final S initial Which has a positive value of S? Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction. Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Above 0 C it is spontaneous for ice to melt. Below 0 C the reverse process is spontaneous. 5
Reversible Processes In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process. Irreversible Processes Irreversible processes cannot be undone by exactly reversing the change to the system. Spontaneous processes are irreversible. Entropy For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the absolute temperature: S sys = q rev T S surr = -q = - H sys sys T T (at constant P,T) Like energy and enthalpy, entropy is a state function and S = S final S initial For any spontaneous process, the entropy of the universe increases: ( S univ >0) 6
2 nd Law of Thermodynamics For reversible processes: S univ = S system + S surroundings = 0 For irreversible processes: S univ = S system + S surroundings > 0 Entropy on the molecular scale Molecules exhibit several types of motion: Translational: Movement of the entire molecule from one place to another. Vibrational: Periodic motion of atoms within a molecule. Rotational: Rotation of the molecule on about an axis or rotation about σ bonds. Statistical treatment of entropy open the stopcock no work (w = 0) no heat (q = 0) spontaneous, nonetheless. Reverse is unimaginable. 7
Entropy: matter dispersal # particles probability they are all in A 2 1 / 4 = 0.25 3 1 / 8 = 0.125 4 1 / 16 = 0.0625 10 1 / 1024 = 0.0009765625 N = 6.022 10 23 2 -N 0 Entropy: energy dispersal A key contribution is the dispersal of energy over many different energy states (each state is called a microstate). (This is similar to matter dispersal.) Let W = number of microstates of a system S = k ln W where k = 1.38 10-23 J/K (Boltzmann constant) Entropy on the Molecular Scale The change in entropy for a process: S = k lnw final k lnw initial S = k ln W final W initial Entropy increases with the number of microstates in the system. 8
Entropy and microstates S = k ln W In general, the number of microstates available to a system increases with an increase in volume, an increase in temperature, or an increase in the number of molecules because any of these changes increases the possible positions and energies of the molecules of the system. Solutions Generally, when a solid is dissolved in a solvent, entropy increases. Entropy, S In general, S (solids) < S (liquids) << S (gases) S o (J/K mol) H 2 O (liq) 69.91 H 2 O (gas) 188.8 S (small molecules) < S (large molecules) S (simple molecules) < S (complex molecules) S increases as the temperature is raised 9
Entropy Changes In general, entropy increases when Gases are formed from liquids and solids; Liquids or solutions are formed from solids; The number of gas molecules increases; The number of moles increases. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is zero: S(0 K) = 0. S = k ln W Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass. See Appendix C. 10
Standard Entropies Larger and more complex molecules have greater entropies. Entropy equations Boltzmann Equation: S = k ln W k = 1.38 10-23 J/K S = q rev / T S sys = Σ ns (products) Σ ms (reactants) S univ = Σ S sys + Σ S surr S universe = S sys + S surr > 0 (for a spontaneous reaction) Entropy Change in the Universe S universe = S system + S surroundings Since S surroundings = and q system = H system This becomes: q system T S universe = S system + H system T Multiplying both sides by T, we get T S universe = H system T S system 11
Gibbs Free Energy T S universe is defined as the Gibbs free energy, G. When S universe is positive, G is negative. Therefore, when G is negative, a process is spontaneous. G = H T S Gibbs Free Energy 1. If G is negative, the forward reaction is spontaneous. 2. If G is 0, the system is at equilibrium. 3. If G is positive, the reaction is spontaneous in the reverse direction. Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation, G. f G = Σn G f (products) Σm G f(reactants) where n and m are the stoichiometric coefficients. 12
Free Energy Changes At temperatures other than 25 C, G = H T S How does G change with temperature? There are two parts to the free energy equation: H the enthalpy term T S the entropy term The temperature dependence of free energy comes from the entropy term. Free Energy and Temperature Gibbs free energy change = total energy change for system - energy lost in disordering the system If G o is negative, the reaction is spontaneous (and product-favored). If G o is positive, the reaction is not spontaneous (and reactant-favored). Calculating G a) Determine H o and S o and use Gibbs equation. G o = H o - T S o b) Use tabulated values of free energies of formation, G fo. G o rxn = Σ G fo (products) - Σ G fo (reactants) 13
Example: Gibbs free energy Determine G rxn for the following combustion reaction. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) H f (kj/mol) -74.87 0-393.509-285.83 S (J/mol K) 186.26 205.07 213.74 69.95 G f (kj/mol) -50.8 0-394.359-237.15 H rxn = -890.30 kj S rxn = -242.76 J/K G rxn = H o rxn - T S o rxn = -817.92 kj exothermic more ordered spontaneous G rxn = -817.92 kj spontaneous When G = 0, equilibrium A + B C + D irreversible (easy Gen Chem I) A + B C + D reversible (most reactions) forward reaction = reverse reaction No drive to make all products (spontaneous reaction) or all reactants (nonspontaneous) Examples: melting point boiling point A (s) A (l) A (l) A (g) Example: melting point of water Given ice or H 2 O (s) H f = -292.70 kj/mol S = 44.8 J/mol K water or H 2 O (l) H f = -285.83 kj/mol S = 69.95 J/mol K Find the melting point of water. It better be 0 C! 14
Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way: G = G + RT lnq (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnq = 0; the last term drops out.) Free Energy and Equilibrium At equilibrium, Q = K, and G = 0. The equation becomes 0 = G + RT lnk Rearranging, this becomes G = RT lnk or, - G RT K = e 15
Example: Gibbs free energy and equilibrium constant Use tabulated * free energies of formation to calculated the equilibrium constant for the following reaction at 298 K. N 2 O 4 (g) 2NO 2 (g) *See Appendix C, textbook pages 1059-1061 16