2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Is Kp just reaction quotient possibly? $\endgroup$ – Graham Feb 26 '15 at 3:07
4
$\begingroup$

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}

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.