Thermodynamics ❮ 135❯ Free-Response Question
You have 10 minutes to answer the following question. You may use a calculator and the
tables in the back of the book.Question
Xe(g) + 3 F 2 (g) XeF 6 (g)
Under standard conditions, the enthalpy change for the above reaction going from left to
right (forward reaction) is DH° = - 294 kJ.
(a) In the above reaction, is the value of DS° positive or negative? Justify your conclusion.
(b) The above reaction is spontaneous under standard conditions. Predict what will
happen to DG for this reaction as the temperature is increased. Justify your prediction.
(c) Will the value of K remain the same, increase, or decrease as the temperature increases?
Justify your prediction.
(d) Show how the temperature at which the reaction changes from spontaneous to
nonspontaneous can be predicted. What additional information is necessary?
(e) Draw the Lewis structure of XeF 6.- A—Heat is required to boil a substance; there-
fore, ΔH is greater than 0. A transformation
from a liquid to a gas gives an increase in
entropy (ΔS > 0). - C—The value is negative because there is a
decrease in the number of moles of gas during
the reaction. - B—The key relationship is ΔG = ΔH – TΔS.
From this relationship, it is apparent that the
value of ΔG will increase (become less negative)
as the temperature increases. In general, both
ΔH and ΔS are relatively constant with respect
to small temperature changes. As the tempera-
ture increases, the value of the entropy term,
TΔS, becomes more negative. The negative sign
in front of this term leads to a positive contri-
bution. The value of ΔG will first become less
negative (more positive) and eventually become
positive (no longer spontaneous).
22. A—Recall that ΔG = –RT ln K. As the value of
ΔG increases, the value of K will decrease.
23. D—Recalling ΔG = ΔH – TΔS, it is possible to
determine ΔG, ΔH, and ΔS from the standard
thermodynamic values. The change from sponta-
neous to nonspontaneous occurs when ΔG = 0.
Rearranging this equation and setting ΔG = 0
gives T = ΔH/ΔS, which will allow the tempera-
ture to be estimated.
24. A—Carbon monoxide exhibits two molecular
orientations, CO and OC, while the orienta-
tions of N 2 both appear the same. For this
reason, there are more states (disorder and
higher entropy) present in solid CO.