3.2 Quantitative approach for changes of p

Using Eq. (3.9) and taking into account that in the left hand side of Eq. (3.8) \(\Delta_r G^0\) is independent of \(p\) (since \(G^0\) is the standard potential at 1 bar), i.e. \(K_p\) is independent of pressure, we find

 \begin{equation*} \left(\frac{\partial \ln(K_p)}{\partial p}\right)_T= 0 = \left(\frac{\partial \ln(K_x)}{\partial p}\right)_T + \frac{\partial}{\partial p} \left(\sum_i \ln \left(\frac{p_{tot}}{p^0} \right)^{\nu_i} \right) \Rightarrow \left(\frac{\partial \ln(K_x)}{\partial p}\right)_T = -\frac{\sum_i \nu_i}{p_{tot}} \quad . \end{equation*}(3.10)

So if \(\sum_i \nu_i = 0\) holds \(K_x\) is independent of \(p\), but if e.g. \(\sum_i \nu_i \gt 0\) and \(\Delta p \gt 0\) than \(K_x\) becomes smaller, i.e. we have less reaction products.


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© J. Carstensen (TD Kin II)