Errors in Ch 19 - sample and practice exercises, Go Figure, Self-Assessment, etc.

[Zellmer, Robert]

The following also has corrections for sample exercises, Go Figure questions and
Self-Assessment exercises.  Some are simple formatting errors in the text.  Some
are errors in sig. fig.

Section 19.3:

Sample Exercise 19.3:

In part (a) it's supposed to be liquid water but the state symbol is a capital "I".
change the capital "i" to a lower case "L".


Section 19.4:

Self-Assessment 19.11:


Section 19.5:

Sample Ex. 19.8:

Answer given is -2108 kJ.  It should be to -2107.6 kJ, to 1 decimal place since
all the values involved are to at least 1 decimal place.

In the balanced eqn it's supposed to be liquid water but the state symbol is a
capital "I".  change the capital "i" to a lower case "L".

Self-Assessment 19.15:

All numbers are to 2 decimal places but the answer is to 1 decimal place
The answer should be to 2 decimal places, -139.66 kJ


Section 19.6:

Sample Ex. 19.9:

Both parts (a) and (b) ask for delta(G).  Since you're given delta(H^o) and
delta(S^o) you're determining delta(G^o) not delta(G).  If you're assuming the
del(H^o) and del(S^o) don't change with temp and the reactants and products
are still in their standard states at the new temp (which they will be even at the
higher temp) then you're getting del(G^o) at this new temp.   Changing the
temp from the reference temp of 25 C to another temp doesn't necessarily
mean del(G^o) turns into del(G).  It would still be del(G^o) but at 500 C.

Practice Ex. 19.9:

Part (b) gives del(G^o) of -120.7 kJ.
I get -140.1 kJ.


Section 19.7:

Self-Assessment 19.18:

Has answer as -91.4 kJ

I'm getting delta(G^o) = -103.68 kJ and Q=400.  This gives RT*ln(Q) of 14.844
and a delta(G) of -88.835 kJ, -88.8 kJ to 3 s.f.


Exam Prep Questions:

EPQ 19.13:

correct answer given is -192.7 kJ
I'm getting -192.2 kJ

EPQ 19.15:

answer given is 193 C but that's not a choice.
The correct answer is 191 C, which is answer (d)


Dr. Zellmer




Dr. Zellmer
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