questions about 2015 Chem Olympiad Local exam

[robert zellmer]

I received some questions (12, 24, 28, 36, 40 and 41) on the 2015 Chem 
Olympiad Local exam.

Here are the answers to those questions.

#12
These are all nonmetal molecular compounds.  The b.p. depends on the 
attractive forces
between the particles.  For molecular cmpds that's London Forces, 
Dipole-Dipole
and H-bonding.  LF inc with inc size (molecular wt. will do for this).  
So the LF inc
left to right for the molecules as listed.  DD AF inc with inc 
polarity.  HF is the most
polar and would have the greatest DD AF.  The others also have DD AF but 
this doesn't
change much because the electronegativity doesn't change much as you 
move down the
group from Cl to I.  Thus, the DD AF forces isn't much different for 
HCl, HBr and HI.
However, the LF do inc in that order.  So for those three molecules the 
AF inc in the
order HCl < HBr < HI.  The HF has HB.  HB is very strong. Much stronger 
than normal
DD AF.  It's LF would be less than those for HCl but the HF is more 
polar and has stronger
DD AF and HB.  Thus, the HF has stronger AF than HCl and a higher b.p.  
Actually, HF
has a higher b.p. than all the others listed.  In terms of the question 
the HCl would have
the lowest b.p.

#24
You need to remember the Clausius-Clapeyron Eqn, Arrhenius Eqn and van't 
Hoff Eqn.
They all have a similar form.  The CC eqn is from ch 11, the Arr Eqn is 
from ch 14
and I gave the van't Hoff eqn in the notes for ch 19.  This is using the 
van't Hoff eqn.

#28
This is like the reaction being 80% complete.  If 80% of the sample has 
decayed then
20% remains.  You need to use 0.20 not 0.80 in the eqn.

#36
At the first eq. pt. the H_2 SeO_3 is converted to HSeO_3 ^- .  While 
some of this will
react with the water to form a small amount of SeO_3 ^2^- (because the 
HSeO_3 ^- is
still acting as a weak acid, as shown by the pH being below 7 at the 
first eq. pt.) most
of the Se will still be in the form of HSeO_3 ^- .

#40
Do you know how to take the given info and find the moles of lets say 
Pb^^2 ^+ from
a 1 M soln of Pb(NO_3 )_2 ?  Assume they all have the same volume, we'll 
chose a
volume of 1 L which means 1 mole of each substance given and 1 mole of 
metal ion
since each cmpd has 1 metal ion per formula unit.  I'm doing this to 
show you this
all depends on the number of electrons being transferred and the molar 
mass of each
metal.  Lets see the calculation for Pb^^^2 ^+ and you'll see this.

?g Pb = (3000 s) x (1.5 C/s) x (1 mol e-/96,500 C) x (1 mol Pb^^2 ^+ /2 
mole e-) x (207.2 g Pb^^2 ^+ /1 mol Pb^^2 ^+ )

One can see that the mass of metal in this case depends on the ratio of  
molar mass (MM)
divided by the # of e- each metal ion would need to acquire to form the 
metal.  The only
difference for the other metals given would be the # e- being 
transferred and the molar
mass.

Tl^+           1 mol e-    MM = 204.4        ratio = (204.4/1) = 204.4
Pb^^^2 ^+ 2 mol e-    MM = 207.2        ratio = (207.2/2) = 103.6
Zn^^^2 ^+ 2 mol e-    MM = 65.4          ratio = (65.4/2) = 32.7
In^^^3 ^+ 3 mol e-    MM = 114.8        ratio = (114.8/3) = 38.3

Thus, TlNO_3 would produce the largest mass of metal deposited.

#41

Need to use the Nernst eqn. (using the one given on the eqn. sheet for 
this exam).

E = E^o - (RT/nF)*lnQ

Q = 1/([H^+ ]^4*P_O2 ) = 1/[H^+ ]^4 since P_O2 = 1 atm (std is 1 atm for 
a gas).

E = E^o - (RT/nF)ln(1/[H^+ ]^4)

     = E^o - (RT/nF)*ln[H^+ ]^-4

     = E^o + 4(RT/nF)*ln[H^+ ]    (since the ln[H^+ ]^-4 = -4*ln[H^+ ])

     = E^o + 4(RT/4F)*ln[H^+ ]    (since n = 4)

     = E^o + 0.025674 V*ln[H^+ ]    (since R = 8.314, T = 298, F = 96500)

The [H^+ ] = 1 M for std conditions, pH = 0.  If the pH goes up by one
unit the equates to a decrease of ten fold in the [H^+ ]. In this case that
would mean a pH = 1 and [H^+ ] = 0.1 M.  Plug this into the eqn above.

E = E^o + 0.025674 V*ln(0.1)

E = E^o - 0.0591 V

So the cell potential decreases by 0.0591 V which is 59 mV.
For every 1 unit inc in pH the voltage dec by (0.0591) * (# pH units).

(0.025674*ln(10^-1 )) = (- 0.025674*ln(10)) = (-0.025674*2.3026) = -0.05912

pH = 1, [H^+ ] = 0.1,      ln(0.1) = -1*(2.3026),     E = E^o - (1) 
(0.0591 V)
pH = 2, [H^+ ] = 0.01,    ln(0.01) = -2*(2.3026),   E = E^o - (2) (0.0591 V)
pH = 3, [H^+ ] = 0.001,  ln(0.001) = -3*(2.3026), E = E^o - (3) (0.0591 V)

Another way to look at this is to use the Nernst eqn at 25 C the way we
usually see it,

E = E^o - (0.0592/n)*log(Q)

E = E^o - (0.0592/n)log(1/[H^+ ]^4)

     = E^o - (0.0592/n)log[H^+ ]^-4

     = E^o - (-4)(0.0592/n)log[H^+ ]
^

    = E^o - (-4)(0.0592/4)log[H^+ ]

    = E^o - (-0.0592)log[H^+ ]

    = E^o - 0.0592 * pH    (since -log[H^+ ] = pH)

Now it's easy to see what happens to E when pH changes.  As above
this shows for every 1 unit pH changes the E changes by a 0.0592 V.
As pH goes up 1 unit (more basic) the E decreases by 0.0592 V.
As pH goes down 1 unit (more acidic) the E inc by 0.0592 V (relative
to the E for a higher pH).



Dr. Zellmer

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