Chem 2: Molecular Equilibria

  • #1

Homework Statement


The standard Gibb's free energy for the reaction N2+3H2<--->2NH3 is -33.0 KJ at 298 K. Everything is a gas. Calculate Kp for this reaction.


Homework Equations



I have in my notes that Keq=e^(-Delta G/RT) where R is the constant 0.008314 KJ/K. I also have a formula that converts between Kp and Keq: Kp= Kc(.0821Xtemperature)^(Moles of gasseous products-moles of gasseous reactants).


The Attempt at a Solution


My attempt at a solution was to use the first formula (which I think gives Kc) and then convert it into Kp using the second formula.
My answer ended up differing from the one given in my homework though. Here is my work:

Keq= e^(33.0/(0.008314x298)= 6.09x10^5

Kp= (6.09x10^5)(0.0821x298)^(-2) = not the correct answer

The answer is: Kp=5.97x10^5
 
Last edited:

Answers and Replies

  • #2
I think here, you have to use R=8.314 joules/mol . Keq can be calculated in the same way as you showed.

That is Kc. Kp=Kc(RT)^(delta Ng)
 

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