Hello UltimaOnline,
I have some questions at hand:
Q1:
http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/section_18/5ae8d514bba2df4f1a350329d94c3d8b.jpg
With reference to the second and fourth row in the diagram, what’s
the mathematical basis behind the linearising of the first and
second order concentration against time graphs? In other words, why
do I get a linear graph when I ln the concentration in a first
order reaction and when I reciprocate the concentration in a second
order reaction?
Q2: ACJC 13/P1/Q5
I find the phrasing of the choices confusing. If I accept the
answer as D, can I take “the phosphorus atoms of both oxides” to
mean
some
of the phosphorus atoms of both oxides? (Although this will also
make C and B acceptable choices too.)
Alternatively, if I take “the phosphorus atoms of both
oxides” to mean
all
of the phosphorus atoms of both oxides, there would have been no
correct answer, given that there is a trigonal pyramidal P at
P4O9?
Q3: If I mix tetrachloromethane with methanol, what are the
intermolecular forces of attraction formed (debye forces?) and
destroyed (presumably id-id and H bond). Debye forces are weaker
than pd-pd forces but how does that account for no heat being
evolved? Likewise, when I mix trichloromethane (pd-pd) with
propanone (pd-pd), why would heat be evolved?
Thank you! :)
Hi Gohby,
Q1. The linearizing of molarity-against-time graphs to determine
2nd and 3rd orders by use of ln and reciprocal functions are beyond
the H2 syllabus, and thus not required for students to know.
Cambridge will give relevant guidance if these functions are to be
used. Nonetheless, it's a useful bonus to teach your students about
using these, and will still be acceptable by Cambridge (though
they'll usually structure their questions such that the student
will need to do other within-syllabus procedures and checks that
will already be sufficient to determine the order of reaction, so
this is just an extra bonus).
To address your question more directly is in the realm of
mathematics rather than chemistry, and thus if you would like to
pursue a deeper explanation, you could start another thread
requesting the assistance of the resident mathematics experts Eagle
and WeeWS. The key to understanding the linearity of the
molarity-against-time graphs after the ln and reciprocal functions
are applied, are deriving the Integrated (ie. applying the
Integration function of Calculus unto the rate laws) 1st and 2nd
Order Rate Laws. Once these are derived using Calculus, then the
graphical linearity becomes a natural and mathematically obvious
expression of the Integrated Rate Laws. For further reading, see
links below.
https://en.wikipedia.org/wiki/Rate_equation
http://www.chm.davidson.edu/vce/kinetics/integratedratelaws.html
http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Kinetics/Methods_of_Determining_Reaction_Order/Using_Graphs_to_Determine_Rate_Laws
One last point though, is that for the H2 syllabus, Cambridge will
instead use another graphical linearizing method to hint to the
student that the reaction is of 2nd order kinetics with respect to
a particular reactant. Graphical linearizing can be achieved by
squaring (ie. power raised to 2nd order) the molarity or partial
pressure of the reactant, and plotting against initial rate (with x
and y axes interchangeable, after all it's a linear graph). This
is to test A level students to see if they are able to correctly
interpret the linear graph to mean that the rate of reaction is
*directly proportional* (hence linear) to the square (ie. raised to
the power of 2) of the molarity or partial pressure of the
reactant, and thus deduce the order of reaction is 2nd order with
respect to that reactant. Specifically, see Singapore A levels
2010 P2 Q3 for such an example.
Q2. The question's fault for not being specific. In which case you
have to consider all the options together and see which
interpretation makes the most sense (here, it is ALL atoms rather
than SOME atoms). So D is still the best answer. The "tetrahedral
arrangement" in option D refers to the electron geometry rather
than molecular geometry. Most Singapore JC school teachers (and
private tutors) unwisely teach their students 'shape' or 'geometry'
to directly mean 'molecular geometry' or 'ionic geometry', when the
correct pedagogical approach would be to teach students to first
work out the 'electron geometry', and thereafter derive the
concordant 'molecular geometry' or 'ionic geometry'.
On a related note, most Singapore JC school teachers (and private
tutors) also unwisely teach students to identify orbital
hybridization (sp, sp2, sp3, etc) in terms of single vs double vs
triple bonds, when the correct pedagogical approach would be to
teach orbital hybridization in terms of electron geometries. In
other words, do H2 Chem students truly understand the *purpose* and
*nature* of orbital hybridization (ie. in relation to VSEPR theory,
sigma & pi bonds, and lone pair residence), or do they blindly
apply the 'single vs double vs triple bonds' rules of their school
lecture notes without understanding?
An effective way to test your students on their understanding on
this, is to ask them to identify orbital hybridizations of
non-carbon atoms in a variety of molecules, and going beyond just
sp, sp2 and sp3 orbital hybridizations. For more able students, you
can bring in resonance as well, and how it relates to and/or
affects orbital hybridizations and vice-versa.
Q4. Tetrachloromethane with methanol : since H bonding exists
between methanol, and only van der Waals (a mixture of all 3 :
Keesom or pd-pd, Debye or pd-id and London Dispersion or id-id)
forces exist between tetrachloromethane with methanol, thus newer
intermolecular attractions are weaker than original intermolecular
attractions, and thus the solvation or mixing will be
endothermic.
Trichloromethane with propanone : slightly trickier, since both are
polar aprotic molecules. In such a case, at A levels, students are
not expected to state for certain (ie. without experimental data)
whether the mixing will be exothermic or not, but A level students
are only required to state that the reaction *may* or *could* be
exothermic, because the newly formed intermolecular attractions
*may* or *could* be slightly stronger than original intermolecular
attractions (which for MCQs this understanding will usually suffice
to determine the answer, especially when combined with elimination
MCQ strategy).
Not all van der Waals (ie. whether we're talking about Keesom or
pd-pd, Debye or pd-id and London Dispersion or id-id) forces are
equal in strength, before and after mixing of 2 species. The newly
formed (Keesom or Debye or even London Dispersion) van der Waals
forces could be weaker or stronger than the original van der Waals
forces (even when considering the same specific type of van der
Waals forces, eg. Keesom forces).
Why? Consider the molecular geometries and dipoles present :
trichloromethane (with longer C-Cl bonds and a shorter C-H bond ;
the Cl atoms are strongly delta negative while the C atom is
strongly delta positive) is tetrahedral while propanone is trigonal
planar (with C being strongly delta positive and O being strongly
delta negative due to both induction and resonance).
Hence, the newly formed molecular interactions (mainly and most
significantly Keesom forces, but all 3 types of van der Waals
forces are of course present) are stronger, because the geometries
and dipoles allow it : the strongly delta negative O of propanone
forms strong Keesom attractions with the strongly delta positive C
(with H posing negligible steric hindrance) of trichloromethane,
while the strongly delta negative Cl atoms of propanone forms
strong Keesom attractions with the strongly delta positive C of
propanone.
A level students are strongly advised to draw out both molecules to
illustrate (and annotate) these intermolecular Keesom van der Waals
forces, even if the Cambridge question doesn't specifically require
so. This is because Cambridge will award marks as long as relevant
content is written by the student, regardless of whether textual or
graphical.
