Gibbs free energy and equilibrium constant at a high T

  • #1
9
0
I posted this earlier, but I just realized it might have been in the wrong section. Sorry

Okay, so consider you have system in which ΔG<0 and ΔS>0. Using Gibbs free energy (ΔG=ΔH-TΔS), you'll know that it will always be negative. As the temperature increases, it will actually become more and more negative. This means that as the temperature increases to a higher T, ΔG will become even more negative. making the system favor products much more than reactants.

Now consider the equation for the equilibrium constant, Keq=e^-ΔG/TR. Using this definition, as T gets very high, Keq seems to be going to 1, meaning that there will be an equal amount of product and reactants at a very high temperature.

These two definitions are both correct, yet they seem to contradict one another. Why is that? I feel like I'm missing something very definition based.

Thanks in advance to anyone who helps :)
 

Answers and Replies

  • #2
Borek
Mentor
28,514
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Without getting into any other arguments - Keq doesn't go to 1.

[tex]K_{eq} = e^{\frac {-\Delta G}{TR}} = e^{\frac {-(\Delta H - T\Delta S)}{TR}} = e^{\frac {-\Delta H}{TR}}e^{\frac {\Delta S}{R}}[/tex]
 
  • #3
9
0
Without getting into any other arguments - Keq doesn't go to 1.

[tex]K_{eq} = e^{\frac {-\Delta G}{TR}} = e^{\frac {-(\Delta H - T\Delta S)}{TR}} = e^{\frac {-\Delta H}{TR}}e^{\frac {\Delta S}{R}}[/tex]

Oh wow, I can't believe I did that. Thank you.
But still, Keq seems to be going to a smaller value, while ΔG is becoming more negative, which doesn't make sense.
 
  • #6
DrDu
Science Advisor
6,052
769
It shows that the change of K with T depends on the sign of ##\Delta H##. This is in line with the argument of Borek.
 

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