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Chapter 18 Chemical Thermodynamics – Entropy and Free Energy S 3 Laws of Thermodynamics 1st- Energy is Conserved (Chapter 5) 2nd-Spontaneous 3rd-Entropy Processes and Entropy at Absolute Zero 1st- Energy is Conserved (Chapter 5) System – portion of the universe that we are concerned with Surroundings- everything outside the system ΔE = q + w q = ΔH at constant pressure - ΔH = exothermic + ΔH = endothermic 1st- Energy is Conserved (Chapter 5) Energy can be transferred between a system and its surroundings Energy can be converted from one form to another ΔE = q + w “ΔE” is the change in the internal energy of a system “q” is the heat absorbed or released by the system or from the surroundings “w” is the work done on the system or by the system 1st- Energy is Conserved (Chapter 5) Helps us determine how the energy is transferred and if work is done on the system or by the system If q > 0 means the system is absorbing heat from the surroundings If w > 0 then the surroundings are doing work on the system 1st- Energy is Conserved (Chapter 5) State functions- properties that depend on the initial and final state, not on how the change was made. ΔH and temperature are state functions Work is not a state function Calorimetry – process used to measure heat changes, by using ΔT of water in a calorimeter 1st- Energy is Conserved (Chapter 5) Two ways a system can exchange energy with the surroundings: Change ΔE in heat(q) or work (w) =q+w W= fxd Doing work on a system increases the potential energy Work done by the system decreases the potential energy of the system 1st- Energy is Conserved (Chapter 5) Hess’s Law –total energy change is a sum of all the changes(steps) that occur during a reaction ΔHf- the energy required to form a mole of a substance from its elements ΔHf of an element = 0 ΔHreaction = ΔHf products –ΔHf reactants Chapter 18 Energy lost to universe (entropy- amount of disorder) During a reaction, energy is transferred in 3 directions Energy lost or gained (enthalpy) Energy that can be used to do work (Gibbs free energy) Spontaneous Processes A process that occurs on its own without any outside assistance Spontaneity is independent of time it takes for the process to occur (that is kinetics) Most reactions and processes have a directionality A spontaneous process occurs in one direction. The reverse reaction is nonspontaneous Some processes are spontaneous without being energy-driven; these processes are driven by an increase in disorder Spontaneous Processes Example 1: Predict whether each process is spontaneous as described, spontaneous in the reverse direction, or at equilibrium A. Water gets hotter when a piece of metal heated to 150° C is added B. Water at room temperature decomposes into hydrogen and oxygen gas Spontaneous – heat is always transferred from the hot object to the cool one Not spontaneous C. ice cube melts at room temperature Spontaneous- heat is transferred from the hot object to the cool one 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. © 2012 Pearson Education, Inc. © 2012 Pearson Education, Inc. Second Law of Thermodynamics 2nd Law of Thermodynamics- States that for any spontaneous process the total entropy of the universe must increase. (The universe is becoming more disorderly) Entropy (S) – degree of disorder. Maximum entropy, most disorder S > 0 Second Law of Thermodynamics If S > 0 disorder increases If S < 0 disorder decreases (more ordered) Minimal entropy, most order, S < 0 What causes a reaction to be spontaneous? 2 Laws of Nature drive spontaneous processes. Law of Nature #1 Systems tend to attain a state of minimum energy, (lose energy, cool off) large negative ∆H are spontaneous 2Na + 2H2O NaOH + H2 ∆H = -281.9 kJ (exothermic) Reaction is spontaneous forward 2Ag2S + 2H20 Ag + 2H2S + O2 ∆H = 595.6kJ (endothermic) not spontaneous, but the reverse reaction is. Some endothermic processes may be spontaneous ex. cloth drying, ice melting Law of Nature #2 –Systems tend to attain a state of maximum disorder (entropy) Entropy (S) -randomness, or the amount of disorder, of a system. State function ( doesn’t depend on how you got there) Entropy increases in spontaneous processes. Greater the disorder, the higher its entropy. Every substance has a characteristic entropy ∆S = Sfinal –S initial Calculate change in Entropy (ΔS) If a process is at constant temperature (isothermal) ΔS = qrev T Units of ΔS are J/K qrev = heat T = temp in Kelvin Calculate change in Entropy for phase changes ΔS for phase changes – remember that during a phase change temp remains constant so substitute ΔH of the phase change for qrev ΔSphase change = qrev = ΔHphase change T T Tabulated value Example 2: Elemental mercury is a silver liquid at room temperature. Its normal freezing point is -38.9 °C, and its molar enthalpy of fusion for mercury is ΔHfusion = 2.29 kJ /mol. What is the entropy change of the system when 50.0 g of Hg (liquid) freezes at the normal freezing point? Example 2: A few notes : fusion is melting, the question is asking for entropy change when mercury freezes, so reverse the sign of ΔH ΔH is given to us in kJ / mol, so use stoichiometry to get q into J Answer = -2.44 J/K ΔH = 50.0g Hg x 1 mole Hg 200.59g Hg x -2.29 kJ 1 mol Hg x 1000J 1kj = -571J Step 2 –convert -38.9oC to 234.1K Step 3 ΔSsys = ΔH = T -571J = -2.44J/K 234.1K ΔS is negative b/c heat is flowing from the system to the surroundings when it is freezing Total Entropy Reversible : ΔSuniverse = ΔSsystem + ΔSsurroundings = 0 Irreversible (spontaneous) = ΔSuniverse = ΔSsystem + ΔSsurroundings = 0 The entropy of the universe increases in any spontaneous process It is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases Molecular Entropy Microstate – single possible arrangement of the positions and kinetic energies of molecules in a particular thermodynamic state Each thermodynamic state has a characteristic number of microstates, W Boltzmann developed an equation S = k ln W k = Boltzmann constant, 1.38 x 10-23 J/K Molecular motion When a substance is heated, the motion of its molecules increases The higher the temperature the faster the molecules move Hotter systems have a larger distribution of molecular speeds 3 different types of motion Translational Vibrational rotational invisible motion Computational chemistry Entropy and Physical States Entropy increases with the freedom of motion of molecules. S(g) > S(l) > S(s) © 2012 Pearson Education, Inc. Entropy increases a. in a reaction that goes from a solid to a liquid or gas b. in a reaction from fewer moles to more moles c. simpler molecules to more complex molecules d. smaller molecules to longer molecules e. ionic solids with strong attractions to ionic solids with weaker attractions f. ionic salts dissolve in water (involves disorder of ionic solid but ordering of water molecules – overall increase in disorder) g. gas dissolved in water to escaped gas Example 3 Predict whether ΔS is positive or negative or 0 for each process, assuming constant temperature H2O (l) → H2O (g) Ag+ (aq) + Cl- (aq) → AgCl (s) Negative 4 Fe(s) + Positive 3 O2 (g) → 2 Fe2O3 (s) Negative N2 (g) + O2 (g) → 0 2 NO (g) Entropy changes in chemical reactions standard molar entropies (S°) 1. standard state – the state of a pure substance at 1atm 2. standard molar entropy of ELEMENTS at 298 K are NOT 0 like enthalpy 3. standard molar entropy of gases are greater than solids and liquids 4. standard molar entropy generally increase with increasing mass of compound The entropy change for a chemical reaction The entropy change for a chemical reaction ΔS° = 𝑛𝑆 ° 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝑛𝑆 ° (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) Example 4 Calculate the change in the standard entropy of the system, ΔS° , for the synthesis of ammonia from N2 (g) and H2 (g) at 298 K N2 (g) + H2 (g) → 2 NH3 (g) ΔS° = 2 S° (NH3) – [S° (N2) + 3 S°(H2) ] Look up S° values in table and plug in ΔS° = -198.3 J/K Entropy changes and the surroundings . ΔSsurroundings = - qsystem T recall that if a reaction is taking place at constant pressure that the q = ΔH so For a chemical reaction, ΔSsurroundings = - ΔH system T Can use to find ΔS °universe = ΔS °system + ΔS °surroundings Gibbs Free Energy (G) Another way to determine if a process is spontaneous or not is to look at the energy that’s used to do work…. Gibbs Free energy There are two equations to determine ΔG One involves reaction in standard conditions ( 298K 0r 25oC and 1 atm)- use Appendix C Other involves calculating values for ΔH, temperature and ΔS and using them to determine Gibbs free energy They may differ a little due to errors in experimentation Entropy Change in the Universe Since Ssurroundings = and qsystem T qsystem = Hsystem This becomes: Suniverse = Ssystem + Hsystem T Multiplying both sides by T, we get TSuniverse = Hsystem TSsystem © 2012 Pearson Education, Inc. Gibbs Free Energy TSuniverse is defined as the Gibbs free energy, G. When Suniverse is positive, G is negative. Therefore, when G is negative, a process is spontaneous. © 2012 Pearson Education, Inc. 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. © 2012 Pearson Education, Inc. Standard Free Energy Changes similar to standard enthalpies of formation are standard free energies of formation f G: AT STANDARD CONDITIONS G = SnGf (products) SmG f (reactants) where n and m are the stoichiometric coefficients. © 2012 Pearson Education, Inc. Free Energy Changes At temperatures other than 25 C, G = H TS How does G change with temperature? © 2012 Pearson Education, Inc. Determination of Gibbs Free Energy Changes ∆G = ∆H - T (∆S) Gibbs free energy Temperature in Kelvin Change in enthalpy (heat) in kJ Change in entropy (randomness) ΔS is change in entropy ∆S is given in joules in Apprendix C so convert to kJ T is in kelvin Look at Appendix C in textbook Use out Appendix C page 1059-1061 to find ΔH (kJ/mol) ΔGf S (kJ/mol (J/mol) * be careful to convert to kJ Now let’s try a problem….. Example 5: Calculate ΔG for the reaction below at 25oC (298K) N2(g) + 3H2(g) 2NH3(g) Appendix C ΔHf 0 0 -46.19 ΔGf 0 0 -16.66 .13058 .1925 Convert to kJ ΔS .19150 • ΔHf = 2mol(-46.19) – (0) +3(0) = - 92.38kJ • ΔS = 2 mol(.1925) – [1(.19150) + 3(.13058)] =-.19824 • ΔG = ΔH –TΔS • ΔG = -92.38- 298(-.19824) • ΔG = -33.33kJ Example 6: Using Appendix C, calculate ΔG at 298K for the following reaction using both equations. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Ans. -800.7kJ vs. (-853kJ) Why the difference? Experimental data can be off Interpreting ΔG if ∆G is negative, the reaction is spontaneous in the forward direction if ∆G is positive, the reaction is spontaneous in reverse If ∆G is zero, the reaction is at equilibrium The equilibrium constant is related to both ∆H and the ∆S in the following way: ln Keq = ∆H/(RT) + ∆S/R Or………. lnKeq = -∆H + ∆S RT R Since ∆G =∆H - T∆S then…… Simplified ∆G = -RT ln Keq Free Energy and Equilibrium At equilibrium, Q = K, and G = 0. The equation becomes 0 = G + RT ln K Rearranging, this becomes G = RT ln K or K=e G/RT Example 7: The standard free-energy change for the formation of NH3 from N2 + H2 is -33,000 J/mol. Calculate the equilibrium constant for the process at 25°C. Answer : K = 7 x 105
