In this equilibrium equation, $$K = K_p\cdot (RT)^{-\Delta n}$$ what does $K$ represent in comparison to $K_p$? It seems to me that they both are equilibrium constants, so how are they different?

  • $\begingroup$ Is Kp just reaction quotient possibly? $\endgroup$
    – Graham
    Commented Feb 26, 2015 at 3:07

2 Answers 2


The equilibrium constant is defined by the expression $$K_x=\prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}}.\tag1$$

For $x$ you can substitute a number of quantities, most commonly when working with solutions is (amount) concentration $c$ and when working with gasses often the partial pressure $p$ is used. In the latter case the equilibrium constants can be related via the ideal gas law $$pV = n\mathcal{R}T.\tag2$$

Substituting this into $(1)$ gives you $$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag3$$

This can be transformed into $$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}} \prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{3a}$$

We know that the concentration is defined as $$c_{\ce{B}}=\frac{n_{\ce{B}}}{V}\tag4$$ and from $(1)$ we can set up the concentration based equilibrium constant $$K_c=\prod_{\ce{B}} c_{\ce{B}}^{\nu_{\ce{B}}} =\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}}.\tag5$$

Substituting $5$ in $3$ we find $$K_p=K_c\prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{6}$$

From the associativity of multiplication it follows therefore $$K_p=K_c\cdot\left(\mathcal{R}T\right)^{\sum_\ce{B}\nu_{\ce{B}}}.\tag{7}$$

This is often abbreviated with $K=K_c$ and $\sum_\ce{B}\nu_{\ce{B}}=\Delta n$ and therefore $$K = K_p\cdot (\mathcal{R}T)^{-\Delta n}.\tag{7a}$$

Associative law of multiplications for exponents: \begin{align} x^{n+m} &= x^n\cdot x^m\\ x^{\sum d} &= \prod_d x^d \end{align}


K is for concentrations in Molarity. Kp is for what you're using pressures for reactions with all gases.


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