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I have the following equation,

$$ K_r=\left ( \frac{P}{RT} \right )^{v}exp \left \{ \sum_{s}\left [ (\beta_{s,r}-\alpha_{s,r}) \left \langle \frac{h_s}{RT}-\frac{s_s}{R}\right \rangle \right ] \right \} $$

where $$ v=\sum_{s}\left(\beta_{s,r}-\alpha_{s,r}\right) $$ $$ \frac{h_s}{RT}=-\frac{a_1}{T^2}+a_2\frac{ln(T)}{T}+a_3+a_4\frac{T}{2}+a_5\frac{T^2}{3}+a_6\frac{T^3}{4}+a_7\frac{T^4}{5}+\frac{a_8}{T} $$ $$ \frac{s_s}{R}=-\frac{a_1}{2T^2}-\frac{a_2}{T}+a_3ln(T)+a_4T+a_5\frac{T^2}{2}+a_6\frac{T^3}{3}+a_7\frac{T^4}{4}+a_9 $$

R, P, $a_1$ to $a_9$, $\beta$ and $\alpha$ are constants.

I am interested in obtaining $$ \frac{\mathrm{d} K_r}{\mathrm{d} T} $$ But I don't know how to proceed. how do I derive a summation inside an exponential?

Thanks for your help.

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  • $\begingroup$ Use the chain rule. $\endgroup$
    – Tunococ
    May 21, 2014 at 10:23
  • $\begingroup$ Use the chain rule and linearity: $$\frac{d}{dx}e^{\sum f_i(x)}= \left(\sum\frac{df_i(x)}{dx}\right)e^{\sum f_i(x)}$$ $\endgroup$ May 21, 2014 at 10:24
  • $\begingroup$ Is this related to thermodynamics ? It looks so familiar to me ! $\endgroup$ May 21, 2014 at 10:25
  • $\begingroup$ @ClaudeLeibovici Yes. It's used to calculate the equilibrium constant using Gibb's free energy. $\endgroup$ May 21, 2014 at 10:28
  • $\begingroup$ This is what I suspected ! $\endgroup$ May 21, 2014 at 10:31

1 Answer 1

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Hint

You can make your life much easier taking first the logarithms of both sides.

The idea of using $\log(K_r)$ instead of $K_r$ comes from different places. First, from definition since $$\Delta G=\Delta H- T \Delta S=-RT \log(K_r)$$ But, to me, the most important is the numerical conditioning of the problem : very poor when using the $K_r$'s (even if they vary by only one or two orders of magnitude), very good everywhere when using the $\log(K_r)$'s. All the above apply to chemical and/or physical equilibrium.

I "wasted" more than $50$ years in this area. If you are interested, I shall be delighted to continue discussions around these topics. Cheers.

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  • $\begingroup$ you mean, do something like this: $$ln(\left ( \frac{P}{RT} \right )^{v})+\sum_{s}\left [ (\beta_{s,r}-\alpha_{s,r}) \left \langle \frac{h_s}{RT}-\frac{s_s}{R}\right \rangle \right ] $$ $\endgroup$ May 21, 2014 at 10:49
  • $\begingroup$ Yes ! You can also simplify the first term and get $$\log (K_r)=v~ \log ( \frac{P}{RT} )+\sum_{s}\left [ (\beta_{s,r}-\alpha_{s,r}) \left \langle \frac{h_s}{RT}-\frac{s_s}{R}\right \rangle \right ]$$ $\endgroup$ May 21, 2014 at 10:52

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