Dr. Namphol Sinkaset Chem 201: General Chemistry II Ch. 12: Thermodynamics Dr. Namphol Sinkaset Chem 201: General Chemistry II

I. Chapter Outline Introduction Spontaneity Entropy The 2nd and 3rd Laws of Thermodynamics The Gibbs Free Energy Gibbs Free Energy and Equilibrium

I. Introduction In this chapter, we look more closely at what causes chemical change. We will see that everything comes back to entropy, and we seek to understand what it is. Understanding thermodynamics allows us to predict chemical or physical changes under specified conditions.

II. Spontaneity spontaneous process: a process that occurs naturally under certain conditions e.g. objects falling towards the center of a gravitational field, like water flowing downhill nonspontaneous process: a process that only occurs if dirven by the continual input of energy

II. Kinetics vs. Thermodynamics A spontaneous process is NOT equivalent to a fast process. Thermodynamics tells us…* Kinetics tells us…*

II. All Nuclear Decays are Spontaneous, but…

II. Spontaneous, But Slow

II. Nonspontaneity Note that a nonspontaneous process is not impossible! e.g. Iron metal in an iron ore won’t just come out on its own. A nonspontaneous process can be made to occur by using external energy or coupling it to a spontaneous process.

II. What Causes Something to Happen? Up until this point, what have you been taught to say why something occurs? Consider the following expansion of an ideal gas into a vacuum.

II. Ideal Gas Expansion What is q for this expansion? What is w for this expansion (w = -PΔV)? What is ΔU? So why is this happening?

II. Heat Flow When objects at two different T’s come into contact, heat flows spontaneously. Note that in this system, no net gain/loss of energy, just a redistribution.

II. Dispersal of Matter and Energy Based on these two examples, spontaneity is tied to changes in the dispersal or distribution of matter and/or energy. Specifically, spontaneous processes result in a more uniform distribution of matter or energy.

III. Carnot and Clausius In 1824, Nicolas Carnot published the results of his extensive studies on the efficiency of steam engines. In a later review of Carnot’s work, Rudolf Clausius introduced a new thermodynamic property that related spontaneous heat flow of a process to the T at which it took place.

III. Entropy Clausius named the new thermodynamic property entropy. Entropy was defined as the ratio of the reversible heat and the kelvin temperature. Note that entropy is a state function.

III. Equation for Entropy Ludwig Boltzmann developed a molecular-scale statistical model to describe entropy. k is Boltzmann’s constant, (k = R/NA = 1.38 x 10-23 J/K). W is the # of energetically equivalent ways to arrange the components of the system (a microstate).

III. Understanding W A microstate is a specific configuration of locations and energies of particles in a system. To understand microstates, we’ll look at a “simple” type of system: arrangement of N particles distributed among n boxes. Mathematically, the number of microstates for this system is given by nN.

III. Expansion of 4 Particles If we allow expansion to occur for this system, how many microstates are possible? It’s equal to # of “boxes” raised to # of particles…

III. 16 Microstates

III. Microstates and Probability Equivalent microstates are grouped together in distributions (sometimes called macrostates). The distribution w/ the highest # of microstates is the most probable. Since entropy increases with the # of microstates, the most probable distribution has the greatest entropy.

III. Calculating ΔS Like other state functions, we are most often interested in changes in entropy.

III. Sample Problem Calculate ΔS for a process that converts the system between the two distributions.

III. Predicting/Quantifying S Now that we have an idea about what entropy is, we can try to analyze systems from an entropy point of view. Entropy differences between states of matter. Entropy changes associated with phase changes. Entropy differences between substances. Entropy changes associated with reactions.

III. Entropy and State The structure of a substance influences its entropy. Consider how many energetically equivalent ways you can arrange each phase.

III. Entropy and Phase Changes To change phase, a substance must absorb more energy. The greater the distribution of KE, the more entropic…

III. Entropy of Substances The entropy of a substance is influenced by the structure and the number of its particles. Thus, the more “places” the compound can put the energy, the more “entropic” that compound will be. # of “places” depends on structure. Specifically, look at molar mass and molecular complexity.

III. Entropy and Molar Mass As the molar mass increases, the entropy increases. The “why” is outside the scope of this course, but it has to do with spacing of energy levels.

III. S and Molecular Complexity The more complex the compound, the more places it can put energy due to its different modes of motion. e.g. Ar(g) vs. NO(g).

III. S and Molecular Complexity Similar molar mass, but more complex (bonds): Increasing molar mass and increasing complexity:

III. Entropy in Mixtures Dissolved ionic solids have more entropy than the nondissolved solid. Energy that was concentrated in the crystalline solid becomes dispersed when dissolved in a solution and you have more interactions and orientations between different particles. e.g. KClO3(s) with S° = 143.1 J/mole.K vs. KClO3(aq) with S° = 265.7 J/mole.K.

IV. Entropy and Spontaneity It appears that entropy is related to whether or not something is spontaneous. Processes that have systems w/ increasing entropy are spontaneous, but there are counterexamples – e.g. water freezing. We extend our look at entropy to the surroundings to see if we can find a universal answer to spontaneity.

IV. Entropy of the Universe As we know, the universe can be split into system and surroundings. Thus, ΔSuniverse = ΔSsys + ΔSsurr. We will consider the process of heat flow between two objects – one designated as system and the other as surroundings.

IV. Hot Flows to Cold Let’s say that 5 J of heat flows from 5 K object (system) to a 1 K object (surroundings) – a process known to be spontaneous. What’s ΔSuniv?

IV. Cold Flows to Hot? Let’s see what happens if 5 J of heat were to flow from 1 K object (system) to a 5 K object (surroundings) – something that would never be observed to be spontaneous. What’s ΔSuniv?

IV. 2nd Law of Thermodynamics 2nd Law of Thermodynamics: for any spontaneous process, the entropy of the universe increases (ΔSuniverse > 0). The criterion for spontaneity is thus the entropy of the universe. A system, chemical or otherwise, proceeds in the direction that will increase the entropy of the universe.

IV. Expression of the 2nd Law Because of the large scale of the surroundings, qrev is practically the same as qsurr. Therefore, we derive the following equation:

IV. Sample Problem Consider the reaction 2N2(g) + O2(g)  2N2O(g) with a ΔHrxn = 163.2 kJ. Calculate the entropy change in the surroundings associated w/ this reaction at 25 °C. What is the sign of the entropy change for the system? Will the reaction be spontaneous?

IV. 3rd Law of Thermodynamics 3rd Law of Thermodynamics: the entropy of a perfect crystal at absolute zero (0 K) is zero. Standard molar entropies can thus be tabulated based on this zero point.

IV. Standard Molar Entropies

IV. Calculating ΔS° Recall that the “not” symbol means standard conditions of 1 bar pressure and: Standard state would be the state the substance exists as at these conditions. For a solution, [ ] must be exactly 1 M. The standard entropy change for a reaction (ΔS°) is the change in entropy for a process in which all reactants and products are in their standard states.

V. Determining Spontaneity It would be easier if we could determine spontaneity by just considering changes in the system. We introduce a new thermodynamic quantity, the Gibbs free energy, and relate that to the entropy of the universe.

V. The Gibbs Free Energy Josiah Gibbs introduced the Gibbs free energy change, G, which is defined using the system’s enthalpy and entropy: G = H – TS. As usual, we are most concerned with changes, so: ΔG = ΔH – TΔS. We see how this relates to ΔSuniv...

V. Relationship Between ΔG and ΔSuniv Thus, ΔG = -TΔSuniv

V. Using ΔG The change in free energy can be calculated w/ respect to the system, and the result can be used to determine spontaneity. If ΔG is negative, the process is spontaneous. If ΔG is positive, the process is nonspontaneous.

V. Possible Combos in ΔG There are 3 factors that determine the outcome of the sign on ΔG. Sometimes ΔH and ΔS work together; sometimes they don’t.

V. Sample Problem The reaction C2H4(g) + H2(g)  C2H6(g) has ΔH = -137.5 kJ and ΔS = -120.5 J/K. Calculate ΔG at 25 °C and determine whether the reaction is spontaneous at this temperature. Does ΔG become more negative or more positive as temperature increases?

V. Rxn Free Energy Changes So, to determine whether or not a process is spontaneous, we can calculate ΔG°. Of course, this is the standard free energy change of a reaction. There are three ways to calculate ΔG°, depending on what information is given.

V. Calculating ΔG° with ΔH° and ΔS° The first method involves using the Gibbs free energy equation. ΔG° = ΔH° - TΔS° To use this method, typically need to do Hess’s Law type calculations for enthalpy and entropy.

V. Sample Problem Determine whether the reaction NO(g) + ½ O2(g)  NO2(g) is spontaneous at 25 °C. Note that the enthalpies of formation for NO(g) and NO2(g) are 91.3 kJ/mole and 33.2 kJ/mole, respectively and the standard entropies for NO(g), O2(g), and NO2(g) are 210.8 J/mole.K, 205.2 J/mole.K, and 240.1 J/mole.K, respectively.

V. Calculating ΔG° Via Hess’s Law standard free energy of formation, ΔG°f: change in free energy when 1 mole of a compound forms from its constituent elements in their standard states. If a table of ΔG°f’s is available, ΔG° can be calculated using a Hess’s Law type of calculation.

V. Sample Problem Calculate ΔG° for the reaction 2CO(g) + 2NO(g)  2CO2(g) + N2(g) given that the standard free energies of formation for CO(g), NO(g), and CO2(g) are -137.2 kJ/mole, 87.6 kJ/mole, and -394.4 kJ/mole, respectively.

V. Calculating ΔG° Stepwise Just like thermochemical equations, reactions w/ associated free energies can be manipulated with the same rules applying. If the equation is multiplied by a factor, then ΔG is multiplied by the same factor. If an equation is reversed, then ΔG changes sign. If a series of reactions adds up to an overall reaction, the ΔG for the overall process is the sum of ΔG for each step.

V. Sample Problem Using the equations below, find the ΔG° for the reaction N2O(g) + NO2(g)  3NO(g). 2NO(g) + O2(g)  2NO2(g) ΔG° = -71.2 kJ N2(g) + O2(g)  2NO(g) ΔG° = 175.2 kJ 2N2O(g)  2N2(g) + O2(g) ΔG° = -207.4 kJ

VI. “To Lower Energy” We’ve always explained why things happen by explaining how they lower energy. To be technically correct, things happen to lower the Gibbs free energy (which is equal to increasing Suniverse). The Gibbs free energy is the driving force behind a process.

VI. Nonstandard Free Energies Of course, most of the time, a reaction will not be under standard conditions. We need a way to calculate spontaneity under a variety of conditions.

VI. Sample Problem The reaction 2H2S(g) + SO2(g)  3S(s, rhombic) + 2H2O(g) has a standard free energy change of -102 kJ. Calculate ΔG when the partial pressures of H2S, SO2, and H2O are 2.00 atm, 1.50 atm, and 0.0100 atm. Is the reaction more or less spontaneous under these conditions?

VI. Relationship Between ΔG° and K For a system at equilibrium, we know the Q = K. What about the value of ΔG at equilibrium? No “driving force” to go either direction at equilibrium, so logically, ΔG = ?

VI. Relationship Between ΔG° and K

VI. ΔG° and K