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Free energy of formation

  • Thread starter fileen
  • Start date
63
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[SOLVED] free energy of formation

1. Homework Statement

Consider the reaction
2NH3(g) --> N2(g) + 3H2(g)
If the standard molar free energy of formation of NH3(g) at 298 K is –16.45 kJ·mol–1, calculate the equilibrium constant for this reaction at 298 K.
The correct answer is 1.71E-6

2. Homework Equations
DG= - RTlnK


3. The Attempt at a Solution
so, if the equation is for two moles NH3(g) then one would multiply the given DG (-16.45) by 2. Also, I converted from KJ to J giving me -32900. The equation given is for the reverse process of formation, so, do I reverse the sign of DG? I tried it both ways and got the wrong answer. In fact, I get huge numbers compared to the correct answer.
I used 8.3145 for R and 298 for T.
 

Answers and Replies

3
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Hey,

Seems your right on track. Not sure what happend to your calculation if you tried reversing the free energy.

dG= 2*-16450 j/mol
= 32900 j/mol(reversed)

32900 j/mol = -8.314*298K * ln K
K=1.71e^-6
 
63
4
haha I was pressing the anti log on my calculator by accident instead of e^x. Thank you for your help. You know youve been studying to long when...
 

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