Chapter 17 Review:  Thermodynamics 

First we must review material from chapter 8.  You are responsible for knowing material from first semester, specifically sections 2, 5, 7, 9 and 10 except where "work" is mentioned.  

  1. First Law of Thermodynamics - Energy cannot be created nor destroyed, only converted and transferred from one form to another.
  2. Enthalpy = DH  = change in heat energy for a process or chemical reaction.  Recall how to calculate  DHorxn  (pages 317-319)
    1. DHorxn =  S DHof products  -  S DHof reactants (get heat of formation values, DHof, from appendix B)
      1. The o symbol is "knot" and refers to standard conditions (1.00 atm pressure, 25.0oC and 1.00M for solutions) not standard state.  We can get values for bromine liquid (standard state) but also bromine gas (not standard state) also since the liquid sets up equilibrium with it's vapor in a closed container under standard conditions. 
      2. Subscripts rxn refers to a chemical reaction and  f  refers to formation of that compound - a formation reaction is element + element g compound. (example:  C(s) + O2(g)  g  CO2(g))
      3. So DHof refers to the heat energy or enthalpy to form one mole of a compound from elements
      4. DHof  for elements is zero!  They aren't "formed" from elements, they ARE elements, so we set their heat of formation change value to zero. 
      5. Units for DHof  are kJ/mol and for DHrxn are kJ.  
      6. We can't get H values - we can only get changes in H values!  That's why we always find DH and not plain H.  It's like someone saying they lost 5 pounds.  We don't know how much they actually weigh, we only know their change in weight is - 5 pounds.
    2. Enthalpy refers to heat flowing.  Endothermic is when heat flows into a reaction, heat is a reactant and DHorxn is +.  Exothermic is when heat flows out of a reaction, heat is a product and DHorxn is -. 

Now on to the chapter. 

  1. Spontaneous (section 17.1)
    1. A process is spontaneous when it occurs naturally under the given conditions without assistance once started.  Notice spontaneous does not mean fast.  Metals rust spontaneously, but not quickly. 
    2. Examples:  fire, combustion reactions, decomposition of dead matter, gas dispersing through a room, salt dissolving in water, rusting, ice melting at room temperature
    3. The reverse of a spontaneous process is not natural and will not occur unless the energy needed is provided - it takes interference/effort/energy to make a non spontaneous process occur.  Try to stop ice from melting at room temperature! Try to stop decomposition of a dead body.
    4. Nature tends towards lower energy!  That is natural usually.  This makes sense as most chemical reactions are breaking bonds and forming new bonds with the purpose of obtaining lower energy to be stable like noble gases.  So usually exothermic reactions are spontaneous as they are losing energy in the form of heat.
      1. combustion reactions are spontaneous (fire) and they all have DHorxn  (-), exothermic.
      2. Acid base reactions are spontaneous (making water which is very stable) and  DHorxn is (-), exothermic.
      3. Ice melting is spontaneous at room temperature BUT DHorxn is (+), endothermic.  
      4. NaCl dissolves in water spontaneously BUT DHorxn is (+), endothermic.
      5. CRAP  - it seems there is more to being spontaneous than being exothermic.  Many exothermic reactions are spontaneous, but a few endothermic reactions are also spontaneous.  I guess DHorxn is not the only determination to being spontaneous.
    5. Something else besides enthalpy is important in determining if a rxn is spontaneous, what is it?    Try problem 17.1
  2. Entropy S (section 17.2) Entropy = disorder
    1. Go back to Section 8.13    Nature also tends towards disorder (entropy).  Entropy (S) increases if things get more disordered.  Entropy measures disorder / randomness.  A process is usually natural (spontaneous) if it increases the disorder so DS is +.  If the process creates order, then the entropy (disorder) decreases and DS is -. The units are Joules per Kelvin or J/K.
      1. States of Matter.  Figure 8.13
        1. Solids are very ordered, the atoms have 1 arrangement possible in their "pattern" called the crystal lattice structure.  Their entropy S is low.
        2. Liquids are more disordered, the atoms and molecules have many possible arrangements in a liquid, it is more random.
        3. Gases are the most disordered, the atoms and molecules have lots of space to spread out so there are many, many possible arrangements.  
        4. S (s) < S (l) << S (g)
        5. A solid turning into a liquid increases in entropy, DS is +.  A liquid turning into a solid decreases entropy (it becomes ordered) so DS is negative. 
        6. A liquid turning into a gas increases in entropy, DS is +.  A gas turning into a liquid decreases entropy (it becomes ordered) so DS is -. 
      2. Dissolving - usually increases the entropy because the solid (order) is now dissolved and spread throughout the liquid ( thus more arrangements and more disorder) so DS is +. There are only a few examples where dissolving decreases entropy, DS is -, and that is due to the water becoming very ordered around the ions. 
      3. Heating increases the entropy because the atoms and molecules are moving faster, rotating faster and vibrating faster (thus more arrangements and more disorder) so DS is +.  Check this out. 
      4. Try example 8.11 and problems 8.21-22.
      5. So nature tends to lower energy and disorder.  So if a process is exothermic (DHorxn is - )AND increases disorder (DS is +) then it is always spontaneous. 
    2. Go back to Section 10.4 Phase Changes
      1. Melting, boiling and sublimation increase entropy, DS is +, but they are endothermic as heat is needed to break IMF, DH is +.  Whether or not these process will occur spontaneously depends on temperature.  Below the mp melting will not occur.  Above the mp melting will occur and so on. 
      2. Condensation, deposition and freezing decrease entropy, DS is -, and they are exothermic as heat is released as a product as stabilization occurs when IMF form, DH is -.  These also depend on temperature.  Freezing will not occur above the mp and so forth. 
      3. See figure 10.9 and figure 17.3
      4. Perspiration - as water evaporates from your body and vaporization process is endothermic and heat is removed from our body so we feel cooler.  It is our natural cooling process. 
      5. Freezing releases heat as the substance forms rigid IMF in the solid lattice pattern. The surrounding gain this heat.  This is why you spray plants with water when you think it will freeze.  That water will freeze and release heat to the plant protecting it. 
      6. We will return to this section later to do calculations.  Fun, fun, fun!!! 
    3. Go back to Section 11.2  Salts Dissolving
      1. Watch it!   
      2. A solid salt has ion-ion forces holding it together in the solid pattern.  Water has H bond forces.  For the solid to dissolve those forces must be broken which requires energy, DH is +.  For the solid to dissolve and mix with the solvent ion-dipole forces must be formed which releases energy, DH is -. The total DHsoln is those added together and may be positive or negative. Athletes take advantage of this all the time - hot and cold packs. 
      3. Entropy almost always increases when salts dissolve - there is less order and DSsoln is +.  
      4. Enthalpy can increase or decrease.  DHsoln is + or -.  
      5. See figure 11.2 and figure 17.5
      6. See figure 11.4 and try example 11.1 and problem 11.1.
      7. So even if DHsoln is positive (endothermic and not usually spontaneous) the salt can still dissolve because the entropy is increasing, DSsoln is +. Basically you can't predict spontaneity on either enthalpy or entropy alone.  Play with this  
      8. Examples 17.1-2 and problems 17.2-3.
  3. skip section 17.3
  4. The Third Law - section 17.4
    1. Third Law of Thermodynamics - the entropy (S) of a perfectly ordered crystal at 0.0 Kelvin is zero.  In other words a perfect solid crystal with no flaws at the coldest temperature possible has absolute order and zero disorder. Because of this law we can calculate entropy (S) values as well as changes in entropy (DS).  Entropy has a "zero" point on the scale, enthalpy does not which is why we can only calculate changes in enthalpy (DH) and never just H. 
    2. Figures 17.7 and 17.8
  5. section 17.5:  Standard Entropy So  - units are J/molK (standard conditions are 1 atm, 1M and 25oC)
    1. Entropy, unlike enthalpy, can be measured directly (not just changes) and these values are in Table 17.1 and Appendix B. Remember only changes in enthalpy can be measured.
    2. Compare S values:  C diamond = 2.4 J/molK, C graphite = 5.7 J/molK, Br2 (l) = 152 J/molK, Br2 (g) = 245 J/molK, I2 (g) = 261 J/molK.  What trends so we see?  Solids have less entropy than liquids which have less entropy than gases.  The more ordered the solid, the less entropy it has. If the state is the same (both gases) the larger atoms have more entropy due to more arrangements of electrons, protons and neutrons.  
    3. An important game:  Predict if  DS is positive or negative! 
      1. CO2 (g) g  CO2 (s)    DS is - since the reaction became more ordered so entropy went down (solid is more ordered than gas)
      2. NH4Cl (s) g  NH4Cl (aq)      DS is + since the reaction became more disordered so entropy went up (dissolved is more disordered than solid)
      3. He (g) at 100oC  g  He (g) at 0oC      DS is - since the reaction became more ordered so entropy went down (colder is more ordered than hotter)
      4. Freezing?   well that is liquid to solid so DS is -
      5. Sublimation?  well that is solid to gas so DS is +
    4. We calculate DSorxn just like we calculate DHorxn   DSorxn =  S Sof products  -  S Sof reactants
      1. Calculate DSorxn for 2Na(s) + Cl2(g) g  2NaCl(s).  Get the Sof values from the appendix.  DSorxn = [ 2 mol (72.38 J/molK)] - [ 2 mol (51.05 J/molK) + 1 mol (223.0 J/molK)] = -180.34 J/K.  Notice this is NOT spontaneous w/respect to entropy which makes sense as the reactants have a gas (disorder) but the product is solid (order).  The reaction creates order, not disorder.  So the entropy goes down.  
      2. Example 17.3 and problem 17.5
      3. In general when predicting if a reaction has DSorxn + or - consider the following
        1. the more gas molecules, the higher the entropy
        2. when there is no change in the number of gases the entropy change will be small (positive or negative)
        3. when products and reactants are similar and same state, the more molecules and larger molecules have the larger entropy since there are more possible arrangements with more particles
      4. Let's play some more:  Predict if  DSorxn  is positive or negative! 
        1. 2NH3 (g) D 3H2(g) + N2(g)      DSorxn is positive (rxn made more gases with more entropy)
        2. 3O2(g)  D 2O3(g)       DSorxn is negative (rxn made less gas with less entropy)
        3. PCl5(s)  D  PCl3(s) + Cl2(g)      DSorxn is positive (rxn made a gas with more entropy)
  6. Second Law of Thermodynamics (section 17.6)
    1. States that entropy is increasing (spontaneous processes have increasing entropy, at equilibrium there is no change in entropy)
    2. For spontaneous processes DStot is +, at equilibrium DStot = 0, note that  DStot = DSsys + DSsurr  (remember in chemistry the system is a reaction)
    3. So  DStot being positive will tell us that a process is spontaneous!!!  Finally! 
    4. Now that we know how to calculate DSsys we need to know how to calculate DSsurr and then we can calculate DStot 
      1. Figure 17.9  Consider an exothermic system - heat is exiting the system and entering the surroundings.  So the surroundings are being heated which gives them more disorder so DSsurr is positive when DHrxn is negative. And for an endothermic system heat is entering the system and leaving the surroundings thus the surroundings are cooling giving them less disorder so DSsurr is negative when DHrxn is positive.  We get this equation:  DSsurr = - DHrxn / T
      2. Calculate DStot at 25oC for this reaction:  2 CuO(s) D  2Cu(s) + O2(g)
        1. First calculate DSorxn using standard entropy values from the appendix and make sure you get 184.6 J/K
        2. Next calculate DHorxn using standard enthalpy values from the appendix and make sure you get 310.4 kJ
        3. Now find DSosurr = - 310.4 kJ / 298K = - 1.042 kJ/K then convert to Joules = -1042 J/K
        4. Finally DStot = 184.6 J/K + -1042 J/K = - 857 J/K (not spontaneous)
      3. Wait a minute!  Why isn't this reaction spontaneous?  Afterall we made a gas product!  Entropy should be positive - well it is for the reaction, but still overall the process is NOT spontaneous.  So although the entropy says this reaction should be spontaneous, notice the enthalpy is very, very endothermic which is not spontaneous.   
      4. Neither entropy nor enthalpy individually can decide if a reaction is spontaneous.  With respect to entropy (DSorxn was +) so spontaneous but with respect to enthalpy (DHorxn was + )so not spontaneous and it seems the enthalpy won.  
      5. Example 17.4 and problem 17.6
      6. To summarize, considering enthalpy only DHorxn being (-) is spontaneous.  Considering entropy only DSorxn being (+) is spontaneous.  But you must consider both factors and sometimes one outweighs the other.  So neither of these alone can tell you for sure if a reaction is spontaneous.  You must calculate DStot which contains both of them for the final definitive answer. Or you can calculate Gibb's Free Energy - coming up next. 
  7. Gibb's Free Energy (DG)  (section 17.7)
    1. Go back to section 8.14 also. 
    2. We want to know if a reaction is spontaneous or not and we've already learned that considering DSorxn and DHorxn is not enough to tell us.  We can calculate DStot but getting surrounding data is hard.  So we derived this equation in class:  DG = DH - TDS 
    3. DG is Gibb's Free Energy and it is similar to enthalpy in that we can only measure changes (there is no zero on the Gibb's scale), the units are kJ/mol like DH, and the DGof for elements is zero like DHof. 
    4. The definitive answer to the spontaneous questions is this:   DG positive is not spontaneous, DG = 0 is at equilibrium, and DG negative is spontaneous, no exceptions!  
    5. Consider all the possible sign combinations for DH and DS.  

    6. DG     =

      		 DH  -TDS Spontaneous?
      - - + exothermic (natural) and disorder (natural) so reaction is always spontaneous
      + + - endothermic (not natural) and order (not natural) so reaction is always NOT spontaneous
      ? - - exothermic (natural) but order (not natural) so can't tell - depends on T:  at low temperature the enthalpy term is more important and the reaction is spontaneous
      ? + + endothermic (not natural) but disorder (natural) so can't tell - depends on T:  at high temperature the entropy term is more important and the reaction is spontaneous

      See table 17.2 and example 17.5

    7. For the combinations above that depend on temperature, how can we find that temperature where the reaction will be spontaneous?  Remember that DG is zero at equilibrium - so that is where a reaction switches from spontaneous to non spontaneous.  So solve DG = DH - TDS   for T using zero for DG.  Calculate the temperature at which 2 CuO(s) D  2Cu(s) + O2(g) will become spontaneous.  
      1. First calculate DSorxn using standard entropy values from the appendix and make sure you get 184.6 J/K
      2. Next calculate DHorxn using standard enthalpy values from the appendix and make sure you get 310.4 kJ = 310,400 J
      3. Now plug in:  0 = 310,400 J - T(184.6 J/K) and solve for T getting 1682 K.  Since entropy is + and enthalpy is also +, the reaction is spontaneous ABOVE 1682 Kelvin. It is not spontaneous below 1682 Kelvin.
      4. Problems like this are very important!  Examples 8.12-13, problems 8.23-25, example 17.5, problems 17.7-9.
    8. Calculating entropy changes during Phase Changes (section 10.4 again for calculations)
      1. During phase changes ( s D  l  D g) we are at equilibrium.  For example during freezing we have liquid and solid in equilibrium together as the liquid is freezing into solid.  Since DG = zero at equilibrium, 0 = DH - TDS.  So we can solve for the temperature at which phase changes occur - the melting and boiling points.  
      2. Another type of problem would be solving for DS =  DH / T.
      3. Remember for the solid D liquid change the heat of fusion applies DHfus and the temperature is the melting point.
      4. Remember for the liquid D gas change the heat of vaporization applies DHvap and the temperature is the boiling point.
      5. Calculate the entropy change for ethanol boiling.  Given:  bp for ethanol is 78.3oC and DHvap = 39.3 kJ/mol.  So just plug in changing kJ to J and changing Celsius to Kelvin.  DSvap for ethanol (l)g (g) = (39,300 J/mol)  /  351.3 K = 112 J/molK.  It is a positive change - is that correct?  Yes, since we made a disordered gas.  
      6. Example 10.4 and problems 10.7-8.   These are important for lab. 
  8. Calculating DGorxn (section 17.8)
    1. What about standard Gibb's free energy DGorxn? Well it tells us whether or not a reaction will favor products are reactants (like K does).  If DGorxn is + the reactants are favored.  If DGorxn is - the products are favored.  Think of the standard as just that - the standard.  It is a reference point at standard conditions and it is a constant.  How to calculate DGorxn?   DG = DH - TDS  For this equation they are either all standard or not.
    2. Calculate DGorxn for O2(g) + 2 Cu(s)  D   2 CuO(s).   
      1. First find DHorxn using the appendix and make sure you get -310.4 kJ
      2. Now find DSorxn using the appendix and make sure you get -184.6 J/K
      3. Plug in DGorxn = (-310.4 kJ) - 298K (-0.1846 kJ/K) = -255.4 kJ.  The products are favored.  Equil constant K will be large.
    3. Example 17.6 and problem 17.10
  9. Calculating DGorxn using DGof (section 17.9)
    1.  Similar to  DHof,  DGof  is the free energy to FORM 1 mole of a substance.  Values are in Table 17.3 and appendix B. 
    2.   DGorxn =  S DGof products  -  S DGof reactants
    3. Calculate DGorxn for 2 CuO(s) D  2Cu(s) + O2(g).  Use appendix B and find the DGof values for the products and reactants.  Remember that the free energy to form an element from itself is of course zero.  DGorxn = [0] - [2 mol (-127.2 kJ/mol)] = 254.4 kJ.  The reactants are favored - this reaction barely goes forward till equilibrium is reached, few products are formed.  
    4. Example 17.7 and problem 17.12. 
  10. Relating DGo  to DG and K (section 17.10 and 11)
    1. DG is not standard and it changes during a reaction as concentrations change and/or as temperature changes.  DGo is standard and is a constant - it does not change - it is more of a reference point to which non standard conditions can be compared.  How do they compare?  DG = DGo + RT lnQ.  You can think of RT lnQ as a correction term for the non standard conditions.  R is the gas constant = 8.314 J/molK, and Q is calculated like K but we may not be at equilibrium - it is still [products] / [reactants] and don't forget to raise them to coefficients from the balanced equation. There is a graph in your book that is super important with DG on the y axis and rxn progress on the x axis.  Look at Figure 17.10. 
    2. What happens when we are at equilibrium?  Well DG = zero and Q = K.  So the equation becomes DGo = -RT lnK
    3. For reversible reaction DG tells us spontaneity and rxn direction:  When DG is - (spontaneous), the rxn goes forward until equilibrium is reached at which point DG = 0.  So DG changes from a negative number to zero as the reaction proceeds forwards to reach equilibrium.  When DG is + (not spontaneous), the rxn goes backwards until equilibrium is reached at which point DG = 0.  So DG changes from a positive number to zero as the reaction proceeds backwards to reach equilibrium.
    4. For a reversible reaction DGo tells us about the equilibrium constant.  If K is a large number, DGo is a negative number and the products are favored.  As K gets larger, DGo gets more negative.  If K is a tiny number, DGo is a positive number and the reactants are favored.  As K gets smaller, DGo gets more positive.  
    5. Calculations 
      1. Calculate K for 2 CuO(s) D  2Cu(s) + O2(g) at 25oC.  Earlier we found that DGo is 254.4 kJ.  So 254.4 kJ = (-8.314 J/molK)(298 K)(ln K).  Changing kJ to J we get -102.68 = ln K.  Solving for K = 2.55 x 10-45 which is a tiny number, so reactants favored.
      2. Calculate DGo for CH3COOH  D   H+(aq) + CH3COO-(aq).  Recognize that this is a weak acid equation and goes with Ka.  We find from the book that Ka for acetic acid is 1.8 x 10-5.  Thus DGo = (-8.314 J/molK)(298 K)(ln 1.8 x 10-5)(kJ / 1000 J) = 27.1 kJ and the reactants are favored which makes sense because acetic acid is weak and few ions are formed - barely dissociates into ions.
      3. Calculate DG at 25.0oC for S(s) + O2(g)  D SO2(g) when PO2 = 0.140 atm and PSO2 = 1.24 atm.  First we need to find DGorxn so we look up the DGof values for the products and reactants.  DGorxn = [ 1 mol (-300.4 kJ/mol)] - [0] = -300.4 kJ.  Now we can find DG  = -300,400 J + (8.314 J/molK)(298 K)(ln 1.24 / 0.14) = -295 kJ.  So this reaction is spontaneous and will proceed in the forward direction until equilibrium is reached.
      4. Look at Table 17.4, examples 17.8-11 and problems 17.13-17.
  11. Gibb's Free Energy and Biochemistry (just for fun) not in book
    1. Consider this reaction:  ADP +  H3PO4  D ATP + H2O.    DGo = 31 kJ/ mol
    2. ATP stores energy for cells, when needed the reverse reaction occurs and releases 31 kJ per one mol. 
    3. This reverse reaction can "drive" other reactions like peptide synthesis which is the joining of amino acids
    4. Consider this reaction: 
    5. ATP + H2O + 2 amino acids  D ADP +  H3PO4  +  dipeptide.     DGo = -2 kJ.  This works because the dipeptide synthesis is 29 kJ which is not spontaneous, BUT the ATP releases 31 kJ, so the overall DGo is -2 kJ which is spontaneous. Our body is fabulous!

Final Notes: