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Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: CandySun on March 31, 2014, 11:40:14 PM

Hello everyone, I am a Canadian student preparing for the upcoming CCO. When I was doing Problem Set#3 on the official website, I couldn't get the correct answer from these two following problems. Could someone help me out?
#1 By how much does the solubility of silver chloride increase when you increase the temperature of the solution from 10 to 90 degree celsius.
a. 170x b. 21x c. 76x d. 214x e. 14x...
The data give are AgCl (standard heat of formation (kJ/mol): 127 standard entropy (J/mol): 109)
Ag+ (105, 77)
Cl (167, 131)
#2 A plume of air near a smoke stack contains a partial pressure of sulfur dioxide gas of 1.0*10^7 atm, using the equilibrium reaction shown below calculate the pH of cloud droplets that form in this air plume?
SO2(g) + 2H2O(l) = HSO3(aq) + H3O+(aq) K=1.7*10^2
a. 2.79 b. 5.56 c. 3.26 d. 2.55 e. 4.38
Thank you very much if you could show me the full solution!

Forum rules (http://www.chemicalforums.com/index.php?topic=65859.0) require that you show an attempt at solving the problem.
At least try to write equations that can be applied here.

Sorry, I'm new to the forum.
Here're my attempts.
#42. From the give data, standard change of enthalpy = 105+(167)(127)=65kJ/mol
standard change of entropy = 77131(109)=55 J/mol
Standard gibbs free energy = 65000  (298)(55) = 48610J
G also = RT lnK
K at 283K = G/(RT) = e^20.66
K at 363K = G/(RT) = e^16.11
Answer: the square root of (K at 363K / K at 283K) = square root of e^4.55 = 9.75
No choice corresponds to this answer.
#45 I don't know the temperature of sulfur dioxide gas, so I couldn't convert pressure to concentration. The products are in aqueous forms, therefore should be expressed in concentrations, so I couldn't use a normal equilibrium approach.

#42. ΔG varies with temperature. You need to calculate ΔG at each temperature using ΔG = ΔH  TΔS.
alternatively you can do it without having to calculate ΔG by using
ln(K_{2}/K_{1}) = ΔH/R*(1/T_{1}  1/T_{2})
#45. I get one of the given answers by making the very naughty assumption that their K = [HSO_{3}^{}][H_{3}O^{+}]/P_{SO2}. (Notice they didn't give any units for K!) Then you don't need to know the temperature. I think this is bad practice  unless it's a sneaky trick to catch me out!

#45. I get one of the given answers by making the very naughty assumption that their K = [HSO_{3}^{}][H_{3}O^{+}]/P_{SO2}. (Notice they didn't give any units for K!) Then you don't need to know the temperature. I think this is bad practice  unless it's a sneaky trick to catch me out!
I agree it should be defined K = [HSO_{3}^{}][H_{3}O^{+}]·p° / (P_{SO2} · c°^{2}). In other words I like your equilibrium constant. But my impression from the question is that this gas is then dissolving in a cloud of water to form droplets which are solutions with some H+. How is it reasonable to assume that P_{SO2} is not changed by this reaction happening? On the other hand we cannot know how many moles of SO_{2} there are without the temperature...

#42. ΔG varies with temperature. You need to calculate ΔG at each temperature using ΔG = ΔH  TΔS.
alternatively you can do it without having to calculate ΔG by using
ln(K_{2}/K_{1}) = ΔH/R*(1/T_{1}  1/T_{2})
#45. I get one of the given answers by making the very naughty assumption that their K = [HSO_{3}^{}][H_{3}O^{+}]/P_{SO2}. (Notice they didn't give any units for K!) Then you don't need to know the temperature. I think this is bad practice  unless it's a sneaky trick to catch me out!
I tried your assumption, but I still got a pH around 7. I used quadratic formula to solve the H+ concentration. Could you show me the process?

#1 problem is so easy.
ΔG changes with temperature.
so ΔG(10°C)≠ΔG(90°C)
for 10°C ΔG=ΔH283ΔS
for 90°C ΔG=ΔH363ΔS
if you do like that you'll take true answer

But my impression from the question is that this gas is then dissolving in a cloud of water to form droplets which are solutions with some H+. How is it reasonable to assume that PSO2 is not changed by this reaction happening? On the other hand we cannot know how many moles of SO2 there are without the temperature...
I assume there is a dynamic equilibrium, or a steady state  SO_{2} is constantly being emitted by the chimney, acid water droplets are constantly being carried away by air currents etc, and a quasiequilibrium is established.
I tried your assumption, but I still got a pH around 7. I used quadratic formula to solve the H+ concentration. Could you show me the process?
[HSO_{3}^{}][H_{3}O^{+}] = K.P_{SO2} = 1.7e9
[HSO_{3}^{}] = [H_{3}O^{+}] = sqrt(1.7e9) = 4.12 e5
pH = log(4.12e5) = 4.38

I assume there is a dynamic equilibrium, or a steady state  SO_{2} is constantly being emitted by the chimney, acid water droplets are constantly being carried away by air currents etc, and a quasiequilibrium is established.
Yes, I'm sure you're right. :)
By the way I find it interesting that they chose pH = 4.38. I think that this was a great choice  for sulphurous acid has pK_{a1} = 1.89 and pK_{a2} = 7.21, and by choosing a pH so far from either of these, the problem writers made sure that their implicit assumption not to worry about protonation or deprotonation of HSO_{3}^{} is indeed valid.

But my impression from the question is that this gas is then dissolving in a cloud of water to form droplets which are solutions with some H+. How is it reasonable to assume that PSO2 is not changed by this reaction happening? On the other hand we cannot know how many moles of SO2 there are without the temperature...
I assume there is a dynamic equilibrium, or a steady state  SO_{2} is constantly being emitted by the chimney, acid water droplets are constantly being carried away by air currents etc, and a quasiequilibrium is established.
I tried your assumption, but I still got a pH around 7. I used quadratic formula to solve the H+ concentration. Could you show me the process?
[HSO_{3}^{}][H_{3}O^{+}] = K.P_{SO2} = 1.7e9
[HSO_{3}^{}] = [H_{3}O^{+}] = sqrt(1.7e9) = 4.12 e5
pH = log(4.12e5) = 4.38
Thankyou very much! I didn't get the answer last time because I haven't learnt the quasiequilibrium. Now I understand!