Chapter 19: Tricks and Tips

Chapter 19: Tricks, Tips and Common Mistakes

1. Forgetting to watch units.
   • Two VERY common unit errors are:
     1. Entropies tend to be in joules/K^.mol whereas heats of formation tend to be listed in kilojoules/mol. If you are doing a calculation based on G = H - T(S) and you mismatch units between the H and T(S) term your answer will be wrong.

     2. Remember that temperature is in Kelvins. The units on entropy are J/K^.mol, so the temperature you use must be in Kelvins for the units to cancel properly.

     To avoid problems like this (which can easily lose you points on an exam), be sure to ALWAYS write units when doing even "trivial" calculations. If you run into problems you'll know that your units are working correctly and checking for math errors is much easier.

2. Appendix II of Chang lists G⁰_f, H⁰_f and S⁰ for compounds. But the relationship G⁰_f = H⁰_f + T(S⁰) doesn't hold when I check the numbers!
   • That's correct. Because the values listed in the table are not delta S⁰, but simply S⁰. Remember, everything has an entropy associated with it except at absolute zero.

     Let's look at the example of H₂S (g). Here are some values of interest:

     	H^0f kJ/mol	G^0f kJ/mol	S^0 J/K^.mol
     H2S(g)	-20.15	-33.0	205.64
     H2(g)	0	0	131.0
     S(s)	0	0	31.88

     To determine the G⁰_f of H₂S we need to write a balanced equation for the formation of one mole of hydrogen sulfide from the elements.

     S(s) + H₂(g) H₂S(g)

     We know then that the S⁰ for this reaction is the net entropy change, i.e. S⁰ = S⁰(products) - S⁰(reactants):

     S⁰ = S⁰(H₂S(g)) - S⁰(S(s)) - S⁰(H₂(g))

     S⁰ = 205.64 - 31.88 - 131.0 = 42.76 J/Kmol

     We can now determine G⁰_f = H⁰_f - T S⁰:

     G⁰_f = -20.15 kJ/mol - (298 K)(42.76 J/molK)(1 kJ/1000J) = -32.9 kJ/mol

     Notice that this matches the value in our table.

3. Confusing the G of a reaction with the G of formation
   • This is a pretty common mistake. Here's an example. Given that the G⁰_f of CO is -137.3 kJ/mol and that

     2 H₂(g) + CO(g) CH₃OH (l) G⁰ = -29 kJ/mol

     what is the G⁰_f of methanol (CH₃OH)?

     Do NOT make the mistake of saying that the G written after the reaction is our answer. It's not. It's the G of the reaction. It's not a G of formation because methanol is not being formed from the elements in this reaction.

     However, we can determine the G⁰_f of methanol. Notice that the G of the reaction is given by:

     G⁰_rxn = G⁰_f(products) - G⁰_f(reactants)

     G⁰_rxn = G⁰_f(methanol) - 2G⁰_f(H₂) - G⁰_f(CO)

     We can look up the G⁰_f of the reactants in Appendix 2 of Chang. We also know G⁰_rxn because we are given this quantity in the problem:

     -29 kJ/mol = G⁰_f(methanol) - 2(0) - (-137.3 kJ/mol)

     Solving, we get:

     G⁰_f(methanol) = -166.3 kJ/mol

4. Energies of Formation
   • People seem to have problems remembering these definitions. The standard enthalphy of formation of a substance, H⁰_f, is defined in Chapter 6.6 of Chang as:

     "the heat change that results when ONE MOLE of a compound is formed from its ELEMENTS in their STANDARD STATES."

     Likewise, G⁰_f is the free energy change that results when one mole of a compound is formed from its elements in their standard states.

     So, how do we recognize when the H of a reaction is the H⁰_f for a substance? It's easy. Make sure you are making one mole of product and that the reactants are elements in their standard states.

     For example, in this reaction,

     2 H₂(g) + CO₂(g) CH₃OOH(l)

     the H_rxn is not the heat of formation because CO₂ is not an element.

     However, in this reaction,

     2 C(graphite) + 3 H₂(g) C₂H₆(g)

     we make one mole of product and the reactants are all elements in their standard state. Therefore the H of the reaction is the heat of formation.

     Finally, note that if we doubled the coefficients of the previous reaction:

     4 C(graphite) + 6 H₂(g) 2 C₂H₆(g)

     that the H of this reaction would be twice as large. And it would not be the heat of formation of ethane because (look at our definition) we are not making one mole of the product.

5. More to come....