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I received some questions (12, 24, 28, 36, 40 and 41) on the 2015
Chem Olympiad Local exam.<br>
<br>
Here are the answers to those questions.<br>
<br>
#12 <br>
These are all nonmetal molecular compounds. The b.p. depends on the
attractive forces <br>
between the particles. For molecular cmpds that's London Forces,
Dipole-Dipole <br>
and H-bonding. LF inc with inc size (molecular wt. will do for
this). So the LF inc <br>
left to right for the molecules as listed. DD AF inc with inc
polarity. HF is the most <br>
polar and would have the greatest DD AF. The others also have DD AF
but this doesn't <br>
change much because the electronegativity doesn't change much as you
move down the <br>
group from Cl to I. Thus, the DD AF forces isn't much different for
HCl, HBr and HI. <br>
However, the LF do inc in that order. So for those three molecules
the AF inc in the <br>
order HCl < HBr < HI. The HF has HB. HB is very strong. Much
stronger than normal <br>
DD AF. It's LF would be less than those for HCl but the HF is more
polar and has stronger <br>
DD AF and HB. Thus, the HF has stronger AF than HCl and a higher
b.p. Actually, HF <br>
has a higher b.p. than all the others listed. In terms of the
question the HCl would have <br>
the lowest b.p. <br>
<br>
#24 <br>
You need to remember the Clausius-Clapeyron Eqn, Arrhenius Eqn and
van't Hoff Eqn. <br>
They all have a similar form. The CC eqn is from ch 11, the Arr Eqn
is from ch 14 <br>
and I gave the van't Hoff eqn in the notes for ch 19. This is using
the van't Hoff eqn.<br>
<br>
#28 <br>
This is like the reaction being 80% complete. If 80% of the sample
has decayed then <br>
20% remains. You need to use 0.20 not 0.80 in the eqn. <br>
<br>
#36 <br>
At the first eq. pt. the H<sub>2</sub>SeO<sub>3</sub> is converted
to HSeO<sub>3</sub><sup>-</sup>. While some of this will <br>
react with the water to form a small amount of SeO<sub>3</sub><sup
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</sup><sup>-</sup>
(because the HSeO<sub>3</sub><sup>-</sup> is <br>
still acting as a weak acid, as shown by the pH being below 7 at the
first eq. pt.) most <br>
of the Se will still be in the form of HSeO<sub>3</sub><sup>-</sup>
.<br>
<br>
#40 <br>
Do you know how to take the given info and find the moles of lets
say Pb<sup><span class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</span></sup><sup>+</sup>
from <br>
a 1 M soln of Pb(NO<sub>3</sub>)<sub>2</sub>? Assume they all have
the same volume, we'll chose a <br>
volume of 1 L which means 1 mole of each substance given and 1 mole
of metal ion<br>
since each cmpd has 1 metal ion per formula unit. I'm doing this to
show you this <br>
all depends on the number of electrons being transferred and the
molar mass of each <br>
metal. Lets see the calculation for Pb<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span></sup><sup><span
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</span></sup><sup>+</sup>
and you'll see this. <br>
<br>
?g Pb = (3000 s) x (1.5 C/s) x (1 mol e-/96,500 C) x (1 mol Pb<sup><span
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</span></sup><sup>+</sup>/2
mole e-) x (207.2 g Pb<sup><span class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</span></sup><sup>+</sup>/1
mol Pb<sup><span class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</span></sup><sup>+</sup>)
<br>
<br>
One can see that the mass of metal in this case depends on the ratio
of molar mass (MM)<br>
divided by the # of e- each metal ion would need to acquire to form
the metal. The only <br>
difference for the other metals given would be the # e- being
transferred and the molar <br>
mass. <br>
<br>
Tl<sup>+</sup> 1 mol e- MM = 204.4 ratio =
(204.4/1) = 204.4 <br>
Pb<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span></sup><sup><span
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</span></sup><sup>+</sup>
2 mol e- MM = 207.2 ratio = (207.2/2) = 103.6 <br>
Zn<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span></sup><sup><span
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</span></sup><sup>+</sup>
2 mol e- MM = 65.4 ratio = (65.4/2) = 32.7 <br>
In<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span></sup><sup><span
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</span></sup><sup>+
</sup>3 mol e- MM = 114.8 ratio = (114.8/3) = 38.3 <br>
<br>
Thus, TlNO<sub>3</sub> would produce the largest mass of metal
deposited. <br>
<br>
#41 <br>
<br>
Need to use the Nernst eqn. (using the one given on the eqn. sheet
for this exam).<br>
<br>
E = E<sup>o</sup> - (RT/nF)*lnQ <br>
<br>
Q = 1/([H<sup>+</sup>]<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>4</sup>*P<sub>O2</sub>)
= 1/[H<sup>+</sup>]<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>4</sup>
since P<sub>O2</sub> = 1 atm (std is 1 atm for a gas). <br>
<br>
E = E<sup>o</sup> - (RT/nF)ln(1/[H<sup>+</sup>]<sup
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>4</sup>)
<br>
<br>
= E<sup>o</sup> - (RT/nF)*ln[H<sup>+</sup>]<sup
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>-4</sup>
<br>
<br>
= E<sup>o</sup> + 4(RT/nF)*ln[H<sup>+</sup>] (since the ln[H<sup>+</sup>]<sup
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>-4</sup>
= -4*ln[H<sup>+</sup>]) <br>
<br>
= E<sup>o</sup> + 4(RT/4F)*ln[H<sup>+</sup>] (since n = 4) <br>
<br>
= E<sup>o</sup> + 0.025674 V*ln[H<sup>+</sup>] (since R =
8.314, T = 298, F = 96500) <br>
<br>
The [H<sup>+</sup>] = 1 M for std conditions, pH = 0. If the pH
goes up by one <br>
unit the equates to a decrease of ten fold in the [H<sup>+</sup>].
In this case that <br>
would mean a pH = 1 and [H<sup>+</sup>] = 0.1 M. Plug this into the
eqn above. <br>
<br>
E = E^o + 0.025674 V*ln(0.1) <br>
<br>
E = E^o - 0.0591 V <br>
<br>
So the cell potential decreases by 0.0591 V which is 59 mV.<br>
For every 1 unit inc in pH the voltage dec by (0.0591) * (# pH
units).<br>
<br>
(0.025674*ln(10<sup>-1</sup>)) = (- 0.025674*ln(10)) =
(-0.025674*2.3026) = -0.05912<br>
<br>
pH = 1, [H<sup>+</sup>] = 0.1, ln(0.1) = -1*(2.3026), E = E<sup>o</sup>
- (1) (0.0591 V)<br>
pH = 2, [H<sup>+</sup>] = 0.01, ln(0.01) = -2*(2.3026), E = E<sup>o</sup>
- (2) (0.0591 V)<br>
pH = 3, [H<sup>+</sup>] = 0.001, ln(0.001) = -3*(2.3026), E = E<sup>o</sup>
- (3) (0.0591 V)<br>
<br>
Another way to look at this is to use the Nernst eqn at 25 C the way
we<br>
usually see it,<br>
<br>
E = E<sup>o</sup> - (0.0592/n)*log(Q)<br>
<br>
E = E<sup>o</sup> - (0.0592/n)log(1/[H<sup>+</sup>]<sup
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>4</sup>)<br>
<br>
= E<sup>o</sup> - (0.0592/n)log[H<sup>+</sup>]<sup
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>-4</sup>
<br>
<br>
= E<sup>o</sup> - (-4)(0.0592/n)log[H<sup>+</sup>]<sup
class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden"><br>
^</span><br>
</sup><br>
= E<sup>o</sup> - (-4)(0.0592/4)log[H<sup>+</sup>]<br>
<br>
= E<sup>o</sup> - (-0.0592)log[H<sup>+</sup>]<br>
<br>
= E<sup>o</sup> - 0.0592 * pH (since -log[H<sup>+</sup>] = pH)<br>
<br>
Now it's easy to see what happens to E when pH changes. As above<br>
this shows for every 1 unit pH changes the E changes by a 0.0592 V.<br>
As pH goes up 1 unit (more basic) the E decreases by 0.0592 V.<br>
As pH goes down 1 unit (more acidic) the E inc by 0.0592 V (relative<br>
to the E for a higher pH).<br>
<br>
<span style="display:inline-block;width:0;height:0;overflow:hidden"><br>
</span><br>
Dr. Zellmer<br>
<br>
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