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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 I sent 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
but the size 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.
<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
<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 (1/molar mass)
<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.
<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>
Dr. Zellmer<br>
<br>
<br>
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