[Disclaimer : If you disagree with the contents of this post
(which makes no claim whatsoever to be accurate or correct, or the
final say on this question), then please disregard the
contents herein and proceed as you would with the question. Perhaps
you could interpret the question in the (strange) way the
setter had originally intended, in which case, good for you.]
The question is surely flawed in its iteration, or at the very
least, misleading. As such, the only offer of (limited,
if at all) assistance will be of brief comments and/or
hyperlinks (somewhat) relevant to the topic being discussed
(by the question). Check with your teacher (or whomever gave you
this dubious question) what he, or the question, was actually
trying to get you to calculate or work out.
1) The term 'ozonized oxygen' means to (partially or usually
fully) convert oxygen from it's most common allotrope of dioxygen
O2, to it's ozone allotrope of O3. Hence, by definition, the
formula of ozone is O3 (see http://en.wikipedia.org/wiki/Ozone ). The
question is therefore not asking for "what is the formula of
ozone", but is asking for something else (thus the iteration of the
question is flawed and misleading).
2) Students of H2/H1 'A' level Chemistry are familiar with the
oxidative cleavage of an alkene double bond by hot,
concentrated KMnO4, but many students may not be aware that a
similar (but not identical) oxidative cleavage
occurs with ozone, known as ozonolysis. Note, however, with
ozonolysis you require a follow-up (reductive
work-up) step to reduce the cyclic intermediate
(molozonide rearranged to ozonide) to the carbonyl compounds, using
dimethyl sulphide (to produce aldehydes and ketones) or sodium
borohydride (to produce alcohols) or hydrogen peroxide (to produce
carboxylic acids).
http://en.wikipedia.org/wiki/Ozonolysis
Thus, the volume of ozone decreases as it is used up to oxidize
alkenes to carbonyl compounds (ie. aldehydes and/or ketones),
alcohols or carboxylic acids. Notice, in the (flawed) question,
you're not given data on what the alkene is, it's physical
state or any solvent used, or how much alkene is used, or
which is the excess/limiting reactant, etc. A 'contraction' in
volume seems to imply a physical change rather than a chemical
change, which is unreasonable. What actually happens, is
that the ozone is used up to oxidize the alkene.
3) What happens when you heat ozone? Let's consider the
conversion of ozone (O3) to dioxygen (O2). Note that bond
breaking of 2O3 to 6O is endothermic, and bond forming
of 6O to 3O2 is exothermic. Overall, conversion of 2O3 to
3O2 is exothermic. However, note that the forward reaction results
in an increase in entropy. Since G = H - TS, thus even though the
forward reaction is exothermic (ie. disfavoured by high
temperatures where enthalpy is concerned), a high temperature
multiplied by a positive change in entropy results in a more
strongly negative or favourable Gibbs free energy, thus high
temperatures actually encourage the exothermic decomposition of
ozone (O3) to dioxygen (O2).
Hence, in the context of the question, heating results
in conversion of ozone to dioxygen, which explains why an
'expansion' occurs (since more molecules of gas are formed). Again,
the iteration or choice of terminology of this question, is
dubious.
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My recommendation to students is, when stuck with
time-consuming dubious questions, skip the question and utilize
your precious time to complete the rest of the paper as quickly as
you can. All too often, students score lower than they should in
exams, because they spent too much time on a couple of problematic
questions, and fail to complete the paper in the limited time
duration allocated.
Out of curiosity, arigatoast, are you taking any Chemistry
examinations this year or the next? If so, which syllabus and level
are you sitting for?
