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This question already has an answer here:

In my chemistry textbook, it is written that :

Gibbs energy for a reaction in which all reactants and products are in standard state, $\Delta_\mathrm{r}G^\circ$ is related to the equilibrium constant of the reaction as follows: $$\Delta_\mathrm{r}G^\circ=-RT\ln K$$

A solved example was given there for the above equation.

Calculate $\Delta_\mathrm{r}G^\circ$ for conversion of of oxygen to ozone, $$\ce{3/2O2(g)->O3(g)} $$ at $298\ \mathrm{K}$ if $K_p$ for this conversion is $2.47\times 10^{-29}$.

The answer to above problem is simple. Just substitute the values in the gibbs equation and get the value of $\Delta_\mathrm{r}G^\circ$ but the thing that I can not understand is that in the equation they substituted the value of $K_p$ directly in the equation without converting into $K_c$ for equilibrium constant.

So my question is: In the gibbs equation what does equilibrium constant stand for? ($K_c$ or $K_p$).

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marked as duplicate by bon, M.A.R., Martin - マーチン Jul 11 '15 at 19:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Found it, chemistry.stackexchange.com/q/5811/16683 It's a very good answer but it might be a bit too detailed, not sure. $\endgroup$ – orthocresol Jul 11 '15 at 12:24
  • $\begingroup$ Well the answer is really good even though I have to take time to understand it. I am sure i will sort out my problem. Thank you very much for the link. $\endgroup$ – Kartoos Jul 11 '15 at 12:33
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    $\begingroup$ No problem, tell me if you want any explanation. Thermodynamics is one of the (unfortunately many) topics where you learn that whatever you were taught before was a lie... $\endgroup$ – orthocresol Jul 11 '15 at 13:17