Chemistry 2000 (Spring 2006)
Problem Set #9: Chapter 19
Answers to Practice Problems
1.
(a)
(b)
There are two isomers of C3H6. Both are gases at room temperature.
Draw the Lewis structure of each isomer.
H
H
H
C
H
C
H
H
H
C
C
C C
H
H
H
H
H
propene
cyclopropane
Which isomer would you expect to have a higher entropy at room temperature? Why?
Propene should (and does) have a higher entropy at room temperature. It is possible to
rotate about the C-C single bond as well as all of the C-H bonds. In cyclopropane, the three
carbon atoms are locked into a triangle (though the C-H bonds can still rotate).
2.
Hydrogen gas has ∆H˚f = 0 J·mol-1 and ∆G˚f = 0 J·mol-1, but its S˚ = 130.7 J·mol-1·K-1. Why
are the enthalpy and free energy values zero, but the entropy is not?
Enthalpy of formation (and free energy of formation) is the enthalpy change (and free
energy change) involved in producing a compound from its elements in their standard states.
Since hydrogen gas is an element in its standard state, its enthalpy of formation (and free
energy of formation) is zero.
Entropy is derived from the number of ways to distribute energy in a molecule, and is
measured in comparison to a perfect crystal of the substance at 0 K. There are many more
ways to distribute energy in a sample of hydrogen gas under standard conditions (298 K)
than there are in a sample of crystalline hydrogen at 0 K. As such, S˚ for hydrogen has a
positive value – not zero.
3.
One of the most common titration reactions is between aqueous sodium hydroxide and
aqueous hydrochloric acid.
HCl(aq)
+
NaOH(aq)
H2O(l)
+
NaCl(aq)
or
H3O+
(a)
+
OH-
2 H2O
Based on your knowledge of acid-base chemistry, what should the equilibrium constant be
for this reaction at 25 ˚C?
Kw-1 = 1.0 × 1014
(b)
(c)
Use thermodynamic data from Appendix L of your text to calculate the enthalpy change for
this reaction.
∆H˚rxn = [∆H˚f(H2O(l)) + ∆H˚f(NaCl(aq))] - [∆H˚f(HCl(aq)) + ∆H˚f(NaOH(aq))]
= [(-285.83 kJ/mol) + (-407.27 kJ/mol)] - [(-167.159 kJ/mol) + (-469.15 kJ/mol)]
∆H˚rxn = -56.79 kJ/mol
Use your answers to parts (a) and (b) to calculate the equilibrium constant for this titration
reaction at 60 ˚C.
K1 = 1.0 × 1014
T1 = 298.15 K
K1
∆H o ⎛ 1 1 ⎞
⎜⎜ − ⎟⎟
ln
=−
K2 = ???
T2 = 333.15 K
K2
R ⎝ T1 T2 ⎠
∆H˚ = -56.79 kJ/mol = -56 791 J/mol
J
− 56791
K
1
1
⎞
mol ⎛⎜
ln 1 = −
−
⎟
J ⎝ 298.15 K 333.15 K ⎠
K2
8.3145
mol ⋅ K
K
ln 1 = 2.41
K2
K1
= e 2.41
K2
K1 1.0 ×1014
= 9.0 ×1012
K 2 = 2.41 =
2.41
e
e
(d)
What is pKw at 60 ˚C?
pK w = − log K w
where Kw = Krxn-1 for this titration
⎛ 1 ⎞
1
⎞
⎟⎟ = − log⎛⎜
= 12.95
pK w = − log⎜⎜
12 ⎟
⎝ 9.0 × 10 ⎠
⎝ K rxn ⎠
(e)
Sometimes, you have to heat a solution to get all of your solid (e.g. KHP, oxalic acid)
dissolved. Explain why it is essential to cool the solution back to room temperature before
beginning the titration. In other words, what will happen if you don’t?
If the temperature is too high (or too low, for that matter), pKw will not be 14. This means
that pH + pOH is no longer 14. As such, the pH of the warm solution at the equivalence
point will not be the same as it would be if the solution was at room temperature.
The main problem this will cause is due to the fact that we normally use an indicator when
titrating. The indicator will change colour at a particular pH – and the indicator is chosen
such that the endpoint pH and the equivalence point pH will be as close as possible. If we
alter the equivalence point pH by using a warm solution, it is possible that the endpoint pH
and equivalence point pH will no longer be as close. In other words, the indicator may
change colour at a volume other than the equivalence point. This would mean that the
indicator would no longer be doing its job properly (through no fault of its own!), and the
accuracy of the titration would be poor.