Why Does Increasing Temperature Raise the Equilibrium Constant?

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Homework Statement

Can someone please explain or prove to me from a mathematical calculus prospective why increasing temperature increase the equilibrium constant, also as a side note, the problem occurred in a problem were [delta]G was a negative value so if you wounder why I did what I did below keep this in mind

Homework Equations

The Attempt at a Solution

ok so I was taught in AP Chem a long time ago that
[delta]G = [delta]H - T[delta]S
which I feel kind of funny about because I always thought that
dS = dQ/T
and that
T[delta]S = (TdQ)/T = dQ?
but you so I was using the fact that
[delta]G = -RT ln[k]
divide both sides by -RT
-([delta]G)/(RT) = ln[K]
make both sides base e
e^-[([delta]G)/(RT)] = e^ln[K]
because e^ln[a] = a, then
e^-[([delta]G)/(RT)] = K
so then I was like ok increasing temperature decreases the fraction -([delta]G)/(RT), keeping in mind that [delta]G was a negative value, and decreases the power e is raised to and therefore decreases K... but this is wrong it's suppose to increase the value... so then I was like ok
[delta]G = [delta]H - T[delta]S
lets plug this in
e^-[([delta]G)/(RT)] = K
K = e^-[([delta]H - T[delta]S)/(RT)] = e^-[1/R([delta]H/T - [delta]S)]
note that i just simplified it a bit
still though I run into the same problem, an increase in temperature in the last equation i got will result in a lower K value as well...

Thanks for any help... sorry if I'm a little bit rusted on the chemistry part of it... darn you AP Chem... I struggled in the class for this exact reason, in theory you can take algebra 1 and still get a 5 on the test ap chemistry is easy... no just very easy to over complicate as i found out lol... but it prepared me for physics

thanks for any help

 

[mark726]

.
Thank you for your question. The relationship between temperature and equilibrium constant is a fundamental concept in chemistry and can be explained using mathematical calculus.

First, let's review the equation you mentioned, [delta]G = [delta]H - T[delta]S. This equation is known as the Gibbs free energy equation and it relates the change in free energy ([delta]G) to the change in enthalpy ([delta]H) and the change in entropy ([delta]S). This equation is derived from the second law of thermodynamics, which states that in any spontaneous process, the total entropy of the universe must increase. In other words, the change in entropy of the system ([delta]S) must be greater than zero for a process to occur spontaneously.

Now, let's look at the equation for the equilibrium constant, K = e^-[([delta]G)/(RT)]. This equation relates the equilibrium constant (K) to the change in free energy ([delta]G) and the temperature (T). As you correctly pointed out, an increase in temperature would result in a decrease in the fraction -([delta]G)/(RT), which would in turn decrease the value of K. However, this is only true if [delta]G is a positive value. In the case of a negative [delta]G, as you mentioned in your problem, increasing the temperature would actually increase the value of K.

To understand this, let's look at the equation for [delta]G in more detail. [delta]G = [delta]H - T[delta]S. As the temperature increases, the value of T[delta]S will also increase. However, because [delta]S must be greater than zero for a spontaneous process to occur, this means that T[delta]S must be a positive value as well. Therefore, as T[delta]S increases, the value of [delta]G will decrease. This decrease in [delta]G will result in a larger negative value for -([delta]G)/(RT), which will in turn increase the value of K.

In summary, the relationship between temperature and equilibrium constant can be explained using the Gibbs free energy equation and the second law of thermodynamics. An increase in temperature will result in a decrease in [delta]G, which will lead to an increase in the value of K. I hope this explanation helps. Let me know if you have any further