(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant 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).

3. 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

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# Homework Help: Chem 2: Molecular Equilibria

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