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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Rutherford on February 03, 2014, 11:52:12 AM

Title: 5th problem IChO
Post by: Rutherford 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?

Title: Re: 5th problem IChO
Post by: CrazyAssasin 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%

Title: Re: 5th problem IChO
Post by: Rutherford 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.

Title: Re: 5th problem IChO
Post by: CrazyAssasin on February 04, 2014, 12:42:37 PM

I get lnKp1=1.5

Title: Re: 5th problem IChO
Post by: Rutherford 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.

Title: Re: 5th problem IChO
Post by: CrazyAssasin 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.

Title: Re: 5th problem IChO
Post by: kristofarkas on March 07, 2014, 12:05:31 PM

How did you get the answer at 2?

I used Δ_rG°=-RTlnK_p1 at 1.
Then plotting lnK_p1 against 1/T the slope is -Δ_rH°/R. Is this correct?

Title: Re: 5th problem IChO
Post by: Rutherford on March 07, 2014, 12:45:21 PM

Yes. What did you get?

Title: Re: 5th problem IChO
Post by: kristofarkas 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?

Title: Re: 5th problem IChO
Post by: Rutherford on March 07, 2014, 01:42:28 PM

I think that first and last values should be used.

Title: Re: 5th problem IChO
Post by: Radu on March 17, 2014, 09:19:44 AM

 They corrected the problem, the values of ΔG° are positive, rather than negative.