oo Ultima Online posted first, saying 12 is D o.O since teacher
i suppose im wrong then >_>
>>> oo Ultima Online posted first, saying 12 is D (o.O)
since teacher i suppose im wrong then >_> <<<
No, I didn't say it's D. What I (mistakenly) said, was that
Chocoholicxx got 3 out of 3 correct. I correct
myself : he/she got 3 out of 4 correct.
But this is where the real difficulty actually
begins. There is actually a heck lot more to this
question that almost all H2/H3 candidates (and probably
many JC teachers as well, and possibly even some of the
SEAB-Cambridge folks themselves!) realize.
Ok, here... we... GO!
BOTH Options A and C are BOTH correct and wrong,
simultaneously!
(Disclaimer : Read everything below, all the way to the end. Do
NOT expect this question to be nullified by SEAB-Cambridge and
everyone to get 'free marks'.)
Why option A appears to be correct.
Equilibrium constant is, by it's mathematical
definition, the ratio of the rate of forward reaction
(compared) to the rate of the backward reaction. If both rates are
independent of temperature, equilibrium constant must therefore be
independent of temperature.
Why option C also appears to be correct.
If the forward reaction is endothermic, increasing temperature
increases equilibrium constant (ie. higher yield at
equilibrium).
If the forward reaction is exothermic, increasing temperature
decreases equilibrium constant (ie. lower yield at
equilibrium).
If the enthalpy change of the reaction is zero (ie. both forward
and backward enthalpies are equivalent),
increasing temperature has no effect on equilibrium constant.
It only increases the rate of reaction (for both forward and
backward reaction).
Why option A appears to be wrong :
Can such a reaction (in which the rate constants for either the
forward or backward reaction, or both, are truly independent
of temperature) actually exist in this Universe? When by
definition, the rate of any reaction is dependent on how fast the
particles move, which is in turn dependent on temperature!
Why option C also appears to be wrong :
What if we have an equilibrium reaction in which enthalpy
change is totally zero, but let's say there is an increase in
entropy for the forward reaction? Gibbs Free Energy predicts
that increasing the temperature would shift the position of
equilibrium to the right and hence increase the equilibrium
constant!
So is this question unsolvable? Let's take a closer look.
From the above discussion ("why option C also wrong"), if we
have a reaction in which enthalpy change is zero and entropy change
is not zero, that means you can never reach equilibrium! But
equilibrium certainly exists in real life.
So what's the deal? For all reactions that have achieved
equilibrium, Gibbs Free Energy must be equal for both forward and
backward reaction. Which means that enthalpy change vs entropy
change must be zero. Which means that if your enthalpy change is
zero, your entropy change must also be zero (for a non-zero
equilibrium constant to exist).
Otherwise (if not both of them are zeros), equilibrium can only
be reached if enthalpy change is non-zero, and entropy change
is non-zero, and at that particular temperature, they cancel
out exactly. But changing temperature would therefore change
equilibrium constant!
Note that equilibrium does NOT mean we have equal moles of
reactant vs product. What it does mean, is that the ratio of
concentration of products to reactants, is a fixed value (but there
will almost always be either more products for some
reactions, or more reactants for other reactions).
A Gibbs Free Energy change which favours the forward reaction
(under specified conditions including temperature), will of course
result in greater concentration of products over reactants. But as
long as the reaction is reversible, then there will come a point in
time, where too much products will cause Gibbs Free Energy to
favour the backward reaction, and there'll be a natural see-saw
until equilibrium is reached.
Let's consider the following reversible reaction :
CaCO3(s) <---> CaO(s) + CO2(g)
Under 'standard conditions' and at 298K, enthalpy change = 177.8
kJ/mol, and entropy change = 160.5J/K.mol. Therefore
(standard) Gibbs Free Energy at 298K for this reaction =
130kJ/mol.
In other words, equilibrium lies to the left under 'standard
conditions' and at 298K.
At point of equilibrium, where the rate of formation of
limestone = rate of decomposition of limestone, Gibbs Free Energy =
zero = Enthalpy Change - Temperature (Kelvins) x Entropy Change
Such a temperature (Kelvins), for the limestone reaction to be
at equilibrium, can then be calculated to be 1113K (or 840 deg
C).
However, (as if everything wasn't complicated
enough), as it turns out, the 'enthalpy change' and 'entropy
change' values themselves (for all
reactions), also change with temperature, pressure
and molarity! Hence the above calculations are actually
invalid and inaccurate (but to apply all of these considerations
correctly in practice, would be beyond the scope of the H2/H3
syllabus).
But it is important to appreciate that enthalpy
(change) and entropy (change) values, are themselves
subject to change at different temperatures, pressures and
molarities. This is at the crux of the reason why
equilibriums exist. To go into a proper discussion of these,
however, is beyond the H2/H3 syllabus.
For irreversible reactions, enthalpy certainly needn't be
zero (and it is never actually zero), and entropy certainly needn't
be zero (and it is never actually zero), yet
equilibrium constant is unaffected by temperature! Because this is
a non-reversible reaction, which can never reach a non-zero or
non-infinity equilibrium constant.
Notice the question states that GIVEN that this equilibrium
constant is independent of temperature. So there we have it - such
a reaction MUST be an irreversible reaction.
But look at the options again :
A. The rate constants for the forward and reverse reactions do
not vary with temperature.
C. The enthalpy change is zero.
For irreversible reactions, increasing the temperature would
increase the rate constant for the forward reaction (since
temperature increases the speed of which particles move, hence rate
of forward reaction would increase). But the rate constant for the
backward reaction remains zero (since it is irreversible). So
option A is wrong.
For non-reversible reactions, the enthalpy change isn't
necessarily zero either. (Just think of any irreversible
reaction that has non-zero enthalpy change, and you've proven your
case). So option C can't always be correct, so therefore it
must be wrong too.
------------------------------------------
Q12) Which statement is correct about a reaction for which
equilibrium constant is independent of temperature?
A. The rate constants for the forward and reverse reactions do
not vary with temperature.
B. the activation energies for both forward and reverse
reactions are zero.
C. The enthalpy change is zero.
D. There are equal numbers of moles of reactants and
products.
------------------------------------------
Finally, back to my initial disclaimer :
The above 'lecture', 'discourse', 'treatise' or 'thesis',
is really to encourage a deeper appreciation in the
student for the wonderful complexity of the Universe. It
discusses Q12 of the 2008 H2 Chem P2, from the author's
(that's me) choice of interpretation of the question
and the question's available 4 quirky options (or at
least 2 of them are).
SEAB-Cambridge reserves the right to interpret their own
question the way they see fit, and by their own reckoning, there
can only be one 'right' answer for which they will give credit only
to this 'right' option.
It's obviously either A or C. Which of these 2 options
SEAB-Cambridge chooses to be the 'correct' answer (ie. depending on
how they choose to interpret the question, from their
own point-of-view), we'll know soon enough (or at least your
JC teachers will, when sometime next year they receive an
official document from SEAB-Cambridge
examiners/markers, detailing the MCQ answers and comments on
the common errors that students of this 2008 cohort made).
