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Since $K_c$ can already be used for calculating the equilibrium constant for a reaction involving gases, but $K_p$ cannot be used for calculating the equilibrium constant for a reaction involving aqueous solutions, what is the use for $K_p$?

It doesn't seem like there is ever a situation where $K_c$ cannot be used and $K_p$ must be used, especially since partial pressures can be readily converted to concentration through the ideal gas law. If anything, it just seems like the existence of two equilibrium constants that more often than not don't equal each other could cause more confusion.

Thoughts?

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  • $\begingroup$ It's easier to use $\mathrm{K_p}$ rather than $\mathrm{K_c}$ in cases pertaining to gaseous equilibrium. $\endgroup$
    – Avyansh Katiyar
    Apr 16, 2018 at 6:40
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    $\begingroup$ chemistry.stackexchange.com/questions/40567/… highly descripted. $\endgroup$
    – akgnitd
    Apr 16, 2018 at 7:22
  • $\begingroup$ It depends on what units you use, if pressure is measured then use $K_p$, if concentration then $K_c=K_p(RT)^{-\Delta n}$ where $\Delta n$ is the change in the number of moles of gaseous substances as the reaction goes from left to right and $p$ is the total pressure. If mole fraction $K_x=K_pp^{-\Delta n}$. $\endgroup$
    – porphyrin
    Apr 16, 2018 at 7:31
  • $\begingroup$ @porphyrin isn't that just derived from the ideal gas law, like I mentioned in the question? $\endgroup$
    – Frank
    Apr 16, 2018 at 8:16
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    $\begingroup$ Ah... so it actually does make a difference. Thanks @akgnit. $\endgroup$
    – Frank
    Apr 16, 2018 at 8:16

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