[CHM 112] Burdge Problem

[Andrew J. Pounds]

On 07/19/2015 04:02 PM,  wrote:
>
> Hi Dr. Pounds,
>
> I got stuck on one of the textbook problems. On number 76 in Burdge, I 
> saw the solution and I don't understand why you added 2x and x and 
> equaled it to .318, and I also don't understand what equation you used 
> to find kp? Because kp = kc(RT)^change in moles, but I don't see the 
> temperature or R constant used in the equation?
>
> Thanks so much!
>
> Srishti
>

Yeah -- this is a good one.  So in the reaction you have a solid 
decomposing to become two different gases. Therefore the ONLY thing that 
is responsible for the pressure is the pressure of the independent 
gases.  In other words,

$P_{\mathrm{tot}} = P_{\mathrm{NH_3}} + P_{\mathrm{CO_2}}$

 From the stoichiometry of the reaction, for each molecule of the solid 
that decomposes you have to make 2 molecules of ammonia and one molecule 
of carbon dioxide.  Since these are the ONLY places where the pressure 
can come from, and I am told that the pressure inside the vessel is 
0.318 atm, then it is possible to write the following:


$P_{\mathrm{tot}} = P_{\mathrm{NH_3}} + P_{\mathrm{CO_2}}$

$P_{\mathrm{tot}} = 2x + x$

$P_{\mathrm{tot}} = 3x$

$0.318 {\ \mathrm{atm}} = 3x$

which I can then use to solve for $x$ and then plug back in to solve for 
the partial pressure of each compound.

In the solution I posted I then plugged in the values for each of the 
partial pressures into the law of mass action for the chemical 
reaction.  I think the book misappropriately uses the term $K_p$ because 
it is in fact a heterogeneous equilibrium involving solids and gases.   
As such it should be written as...

$K = \frac{(P_{\mathrm{NH_3}})^2(P_{\mathrm{CO_2}})}{1}$

where the 1 in the denominator is due to the fact that the reactant is a 
solid.


-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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