April 25, 2024, 12:08:54 PM
Forum Rules: Read This Before Posting


Topic: 5th problem IChO  (Read 4347 times)

0 Members and 1 Guest are viewing this topic.

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
5th problem IChO
« on: February 03, 2014, 11:52:12 AM »
http://icho2014.hus.edu.vn/document/01-24-2014%20IChO46-Preparatory.pdf
Okay, I got the following results:
2. -89.1kJ/mol although in I found in some resources that the reaction is endothermic, but how, if Kp1 is getting lower with T increase?
3. 33.12
4. p of sulfur trioxide 349Pa, p of sulfur dioxide 11849Pa, Kp3=4.14*10-4
5. 1.17%
Are my results correct? Did someone get something else? 4. part was the hardest. I did it:
Marked partial pressures at equilibrium of sulfur trioxide with x and of sulfur dioxide with y, so Kp3=x*y.
Using the expression for Kp2 I got that: y=33.94x.
Then I used the material balance:
x+2*p(O2)=y-2*p(O2), is this valid?
« Last Edit: February 03, 2014, 12:04:41 PM by Raderford »

Offline CrazyAssasin

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +4/-0
Re: 5th problem IChO
« Reply #1 on: February 03, 2014, 06:02:38 PM »
In the first approximately the same result as yours -88.7kJ/mol, they probably made mistake, it should be endothermic reaction
Later on I get K=4.50 and here, I think, is your mistake, because your answer seems to be too high
Then I get Kp3=4.44*10-3, and you doing wrong with the material balance equation, because SO3 and SO2 values of pressure must be squared.
And the percentage is 0.87%

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: 5th problem IChO
« Reply #2 on: February 04, 2014, 07:25:19 AM »
I already took the square root when I wrote y=33.94x. If you mean this equation x+2*p(O2)=y-2*p(O2), it shouldn't be squared, as it is analog to the mass balance equation in solutions.
What did you get for lnKp1 at temperature 651.33 oC? I got 1.75.

Offline CrazyAssasin

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +4/-0
Re: 5th problem IChO
« Reply #3 on: February 04, 2014, 12:42:37 PM »
I get lnKp1=1.5

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: 5th problem IChO
« Reply #4 on: February 04, 2014, 01:17:24 PM »
Okay, I didn't analyze the plot in big detail, so let's say that our values of Kp1 are good. Kp2 and Kp1 are related through this equation: Kp2=Kp12, so if I put your value for Kp1 I get Kp2=20.1. This is almost five times the value you got.

Offline CrazyAssasin

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +4/-0
Re: 5th problem IChO
« Reply #5 on: February 04, 2014, 02:55:30 PM »
Sorry, my bad, I didn't notice that the coefficients are different in the reaction. Now I get the same Kp3 like yours.

Offline kristofarkas

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: 5th problem IChO
« Reply #6 on: March 07, 2014, 12:05:31 PM »
How did you get the answer at 2?

I used ΔrG°=-RTlnKp1 at 1.
Then plotting lnKp1 against 1/T the slope is -ΔrH°/R. Is this correct?


Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: 5th problem IChO
« Reply #7 on: March 07, 2014, 12:45:21 PM »
Yes. What did you get?

Offline kristofarkas

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: 5th problem IChO
« Reply #8 on: March 07, 2014, 12:56:09 PM »
Rookie mistake. Didn't convert to Kelvin...

When calculating the slope, should I use the line of best fit, or just use the first and the last values of lnK and 1/T?
« Last Edit: March 07, 2014, 01:33:39 PM by kristofarkas »

Offline Rutherford

  • Sr. Member
  • *****
  • Posts: 1868
  • Mole Snacks: +60/-29
  • Gender: Male
Re: 5th problem IChO
« Reply #9 on: March 07, 2014, 01:42:28 PM »
I think that first and last values should be used.

Offline Radu

  • Regular Member
  • ***
  • Posts: 87
  • Mole Snacks: +2/-0
Re: 5th problem IChO
« Reply #10 on: March 17, 2014, 09:19:44 AM »
 They corrected the problem, the values of ΔG° are positive, rather than negative.

Sponsored Links