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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: red.orchid on March 27, 2011, 05:54:03 PM

Title: Simple Thermodynamics Question - Gibbs Free Energy
Post by: red.orchid on March 27, 2011, 05:54:03 PM

I'm sure this is simple and straightforward... but it's just not computing in my head...

Consider the specific reaction N2(g) + 3H2(g) <----> 2NH3(g) which has deltaGo = -32.90E3 J/mol at T=298K. What is Kp given R=8.314J/K*mol? Now imagine a fictitious process in which 1 mol of N2(g) reacts with three mol of H2(g) to produce 2 mol of NH3(g). If Partial Pressure N2 = 1E5Pa and Partial Pressure H2 = 0.5E5Pa initially, and the process occurs at constant pressure and temperature, then what is delta G? Will the reaction occur spontaneously?

using DeltaGo = -RTlnKp, I found Kp to be 13.279
I know the process is spontaneous, but I have trouble calculating deltaG.

deltaG = deltaGo + RTlnQ
Since we only have reactants to begin with, wouldn't Q be zero in this case? ln0... is undefined... here's where I'm stuck.

Title: Re: Simple Thermodynamics Question - Gibbs Free Energy
Post by: Rivaldo on April 13, 2011, 09:07:15 AM

I have exact same problem posted it on a different thread though.