1. The following equilibrium is established when solutions of
and
are mixed.
 + SCN^- (aq) \leftrightharpoons Fe(SCN)^{2-} (aq))
is colourless.
is colourless.
^{2-})
is deep red.
The solution is diluted with an equal volume of water. The colour becomes paler red, the addition of the water has caused "
concentration and the number of moles of to decrease.[The bolded is the answer]
Now I understand this perfectly but it's just when I start to think more deeply (and perhaps more than I need to) that I get confused.
First is adding water will dilute the solution hence concentration of the entire solution decreases but shouldn't the system NOT partially oppose this change? Because the left hand side has 2 molecules but can't the right hand side also be considered to have 2 molecules? Since
is in (aq) it is the same as saying
(aq) and
^{4-})
(aq)? [Just like how in water
(aq) is the same as
(aq)
(aq) but it is just we are lazy so we don't need to write them separately?] So since there are the same number of molecules, the system would do nothing to oppose the change?
Secondly, how does the mole of
decrease? I know the concentration decreases because the solution is diluted by adding more water. So let
be the respective mol, concentration and volume of
when the water is not added. Now let
be the respective mol, concentration and volume of
when the water is added.
Looking
, the concentration is a number and
is the initial volume which again is a number, but once the water is added
decreases to
)
and
increases to
)
, so how do you determine whether now
is smaller than
or larger? You don't know how much
decrease or
increased so you can't find out whether the final mol is larger or smaller than the original?
2. Oxygen reacts with haemoglobin in the lungs to form a complex. This complex is moved around the body to take oxygen to the tissues. The equilibrium reaction is represented by
(haemoglobin)
 \leftrightharpoons Hb_4(O_2)_4)
. Low concentrations of carbon monoxide is poisonous. It reacts with haemoglobin and thus prevents oxygen from being carried around the body. The body is deprived of oxygen. The equation that best represents this poisoning is?
 + Hb_4(O_2)_4 \leftrightharpoons Hb_4(CO)_4 + 4O_2 (g))
[the answer]
But how does that make sense? The reaction [
] means that more
is produced hence the reaction [
(haemoglobin)
] should shift right (ie more
_4)
is produced), this is the way that the system partially opposes the change (by reducing the amount of
), however this is still PARTIALLY opposing the change, the amount of
at equilibrium should be higher than the
at the original equilibrium so how is the body deprived of oxygen?
I thought the reaction would be
 + Hb_4(O_2)_4 \rightleftharpoons Hb_4(CO)_4 + 8CO_2 (g))
? Why isn't this reaction the right answer?
3. Consider the following reaction at equilibrium.
 \rightleftharpoons 3F_2 (g) + Cl_2 (g))
,
= negative
For a particular equilibirum mixture the temperature is lowered and the amount of
changes by 0.01 mol. The changes occuring would be?
: decrease by 0.01 mol. 
: increase by 0.015 mol. 
: increase by 0.005 mol (answer)
I understand how to do this question but as I think again I dono why this way doesn't work.
Let the initial mol of
at the equilibrium before the temperature was lowered be 2 mol.
So as temperature is lowered the reaction would favour the forward reaction -> more
and
are formed.
But if I use the mol ratios...
 = \frac{3}{2} \times n(ClF_3) = \frac{3}{2} \times (2-0.010) = 2.985)
mol. However this is a decrease of 0.015 mol not an increase?
Home
Help
Search
Login
