- #1

- 60

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But that is not generally true!

Since:

$$\left (\frac{\partial \Delta_{R} G}{\partial p} \right )_{T,\vec{n}}=\Delta_{R} V$$

and

$$\Delta_{R} G^{0}=-RT\cdot \ln K$$

it should be:

$$\left (\frac{\partial \ln K}{\partial p} \right )_{T,\vec{n}}=-\frac{\Delta_{R} V^{0}}{RT}$$

If the standard state is ideal gas with the same temperature ##T## and pressure ##p##, it is:

$$V^{0}=\frac{RT}{p}(n_{1}+n_{2}+...+n_{N})$$

and the reaction volume is then:

$$\Delta _{R}V^{0}=\sum \upsilon _{i}\cdot \left (\frac{\partial V^{0}}{\partial n_{i}} \right )_{T,p,n_{j\neq i}}=\frac{RT}{p}\cdot \sum \upsilon _{i}$$

$$\left (\frac{\partial \ln K}{\partial p} \right )_{T,\vec{n}}=- \frac{\sum \upsilon _{i}}{p}$$

How can anybody claim that ##(\partial \ln K / \partial p)_{T,\vec{n}}=0## for any reaction in ideal gas?

Edit: Usually ##p=100000 \, Pa## thus the derivation is small enough but not principally zero. Maybe that's what the script is saying.