a) you can't just assume that any reactants, like ethanoic acid
as a limiting reagent. But we can assume that there are no products
CH3COOCH2CH3 and H2O initially, since it's not stated in the
qn.
as this is a homogenous reaction(all reactants and products have
the same state), H20 is included in the Kc equation
Kc = [H2O][CH3COOC2H5] / [CH3COOH][C2H5OH]
for unknowns, we can use algebra, letting the initial no.of
moles of CH3COOH be x, and volume of the solution be V (for
calculating the concentration of the reactants and products). as
you know, 1.5 moles of CH3COOH react with 1.5 moles, so the
equilibirum no.of moels of CH3COOH is (x - 1.5). substitute the eqm
concentration values in the Kc eqn, and you will get the
answer!
b) i'm not sure, but it seems that there would be a backward
reaction of the previous eqn since there are more no.of moles of
original products than the original reactants. the Kc value of the
backward reaction is actually the inverse of Kc of the forward
reaction under the same temperature (values of Kc are also
temperature dependent). you can see it from here:
foward reaction's eqn: Kc = [H2O][CH3COOC2H5] /
[CH3COOH][C2H5OH]
backward reaction's Kc eqn: K'c = [CH3COOH][C2H5OH]/
[H2O][CH3COOC2H5]
Kc = 1/ K'c
letting the change in no.moles of CH3COOC2H5 be -x, and substuiting
them into the eqn, you will get 2 different values of x. (x = -5 or
x= 1/3). you should reject x= -5 because it meant that there will
be increase in the no.of moles of CH3COOC2H5.
then u shld be able to get answer, eqm no. of moles of CH3COOC2H5
is 2.67mol and not 2.67g haha.. other situations of rejecting a
value of x is that when it makes the no. of moles of a reactant or
product negative.
Yes, qdtimes2 is correct
in his helpful explaination (including pointing out my typo
error regarding the units. It should indeed be "moles", not
"grams").
Audi, based on qdtimes2's guidance, are
you able to solve the questions yet?
Whenever you're given non-zero quantities or molarities of both
LHS reactants and RHS products, you should determine the Qc value,
and compare it to Kc.
Qc, known as Equilibrium Quotient (in contrast to Kc,
Equilibrium Constant), has the same formula as Kc (ie. product of
molarities of RHS vs molarities of LHS, each raised to the power of
their stoichiometric coefficients), except that it can be applied
at any time, eg. initial. In contrast, Kc is a value derived from
the molarities of reactants and products only at equilibrium point
itself.
Notice in this question, Qc = 9.0 > Kc = 4.0. Imagine you are
Qc, standing on a number line with Kc. Do you turn your head to the
left or to the right, to look at Kc? Kc represents position of
equilibrium. Notice that in this question, since Qc > Kc, you
(ie. Qc), have to turn your head to the left to look at Kc (ie.
position of equilibrium). Therefore we say "position of
equilibrium lies to the left."
When we say "position of equilibrium lies to the left", it means
the rate of the backward reaction will exceed the rate of the
forward reaction, until equilibrium is reached, ie. Qc = Kc.
Hence in your ICE table, your Change will see "+x, +x, -x, -x"
instead of the usual "-x, -x, +x, +x".
In most scenarios, when we add, or take away, some
reactants or some products, or increase or decrease total pressure
(by changing volume of container), and we say "position
of equilibrium has shifted to the...", do you realize it is *not*
actually Kc which has changed, but Qc?
So if Qc shifts to the right (eg. when you add some product),
the Kc value or position of equilibrium has appeared to
have "shifted to the left", *relative* to Qc (even though the
Kc value has not changed at all).
The only time when Kc value actually changes, rather than the Qc
value, is when temperature changes. If the forward reaction is
endothermic (ie. heat may be regarded as a reactant), increasing
the temperature will increase the Kc value (th Qc value remains
unchanged). And therefore, it is indeed the Kc value or "position
of equilibrium, has shifted to the right", relative to Qc.
H2 Chemistry students need not feel concerned if their
JC did not teach them the above discussion, because there is much
that JCs do not teach (because they simply do not have the time or
opportunity to do so in a school setting; I know because
I'm an ex-MOE teacher), which I teach my tuition students, in
order to help them to understand, appreciate and apply Chemistry
better.
If the above paragraphs confuse you, ignore them. If you
understand them, then good for you. Use it to help you understand
Chemistry better.
Audi, with qdtimes2's helpful post
above, can you solve your questions yet? Or do yo need further help
from us?