[Edher]

Saludos,

     I need your help with the following problem:

Kp=6.18x10^12 AT 298 K for the reaction SO2(g) + 1/2 O2 = SO3 (g). What is Kp for this reaction at 600K?

For this problem should I use this formula?   ln K1/K2 = ?H/R (1/T2 - 1/T1)

if the answer is yes I got Kp = 1.16x10^4    but that's wrong according to my teacher.

Thank you for your help,
Edher

[tamim83]

Hey,

I believe you have the wrong equation there.  You are going to need two thermodynamic equations

First you need to calculate the Gibb's Free Energy for the reaction at 600K You can use tje equation

deltaG= deltaH-T(delta S), you can get the delta H and delta S values from tables that are in your book.  Then you will take your delta G and plug it into this equation

deltaG=-RTlnKp, then solve for the new Kp value.  

Cheers

[Hunt]

Well if we assume that ?H doesn't change with temperature, then
ln (K1/K2) = ?H/R (1/T2 - 1/T1) seems right to me.

[tamim83]

Well if we assume that ?H doesn't change with temperature, then
ln (K1/K2) = ?H/R (1/T2 - 1/T1) seems right to me.

Well, yeah, but this delta h isn't the same one you need to solve this problem.  The delta h that is needed is the reaction enethalpy, the enthalpy that is given in the above equation is the heat of vaporization.  Also, were did that equation come from?  It looks like the Classius Clapeyron equat and the kinetics analog mixed together , I have really never seen that equation before, not even in physical chemistry, although that doesn't mean it doesn't exist.  I still think you should use the thermo equations.  

[Hunt]

The Vant Hoff equation should be used for this problem given that ?H is known & const. The answer could be wrong perhaps because we're required to find ?H_600K i.e. it does change with temp ( using Kirchoff's equation if Cp's are given ) and then compute K_p. I'm not really sure .. it depends on what is given, and there seems to be none!

The derivation of the equation is straight forward.

dG = dH - TdS ( Def of Gibb's free energy at const T ) ... (1)

dH = dU + d(PV) = dU + PdV + VdP

dH = dQ - dW_in + PdV + VdP = dQ - PdV + PdV + VdP = dQ + VdP

dG = dQ + VdP - TdS =  dQ + VdP - dQ = VdP

dG = (nRT)(dP/P)

Integrate both sides, P changes from P^o to P, G from G^o to G

G= G^o + nRT LnP ( P^o = 1 atm )

?G = ?G^o + RTLnQ_p

?G^o = -RTLnK_p ... (2)

?G^o = ?H^o - T_298.15?S^o ... (1)

Equate (1) = (2)

For a rxn at T₁ : ?H₁ - T₁?S₁ = - RT₁ LnK_p1

Same rxn at T₂ : ?H₂ - T₂?S₂ = - RT₂ LnK_p2

If we assume dS and dH are independent of temp, i.e.

?H₂ = ?H₁
?S₁ = ?S₂

Divide both equations by RT, then subtract them :

ln (K_p2 / K_p1 ) = (?H/R) (1/T2 - 1/T1)

It looks like the Classius Clapeyron equat and the kinetics analog mixed together

Yes, it's similar to the Classius Clapeyron equation.

[tamim83]

OK, forgot about that equation.  It looks like we need some more information for this problem.  You know, I think this seems to complicated for a General chemistry problem