Equilibrium constant in terms of partial pressures

$K\_p$ is defined as $\Pi\_j(\frac{P\_j}{P^{\Theta}})^{\nu j}$ where $P\_j$ is the partial pressure of component j and $\nu$ is the stoichiometric coefficient. I have a couple of questions already. Firstly, $K\_p$ is meant to be dimensionless so therefore, $\frac{P\_j}{P^{\Theta}}$ must be dimensionless. However, surely $P\_j$ alone is dimensionless as it's a partial pressure defined by dividing a pressure by a pressure - leaving no units. How does that work? Or am I wrong by assuming that a partial pressure has no units?


Given that, suppose I want to write the expression for $K\_p$ in terms of partial pressures for the following reaction:


$$\ce{\frac12 O2\_{(g)} + C\_{(s)} <=>CO\_{(g)}}$$


Do I assume that $\frac{P\_{c}}{P^\Theta}=1$ leaving me with:


$$K\_p=\frac{P\_{\ce{CO}}}{(P\_{\ce{O\_2}})^{\frac12}}\frac{1}{P^\theta}$$ 





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> However, surely $P\_j$ alone is dimensionless as it's a partial pressure defined by dividing a pressure by a pressure - leaving no units. How does that work? Or am I wrong by assuming that a partial pressure has no units?
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You are wrong to assume that partial pressure has no units. Partial pressure has units of pressure, such as mmHg, atm, bar, psi, etc.


For ideal gas mixtures, $y\_j=\frac{P\_j}{P\_{tot}}$ or equivalently $y\_j P\_{tot}=P\_j$, i.e. the partial pressure $P\_j$ of a gas is equal to its mole fraction $y\_j$ times the total pressure $P\_{tot}$.


The expression you gave for $K\_p$ is dimensionless because $P^\Theta$ has the same dimension as $P\_j$.



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> Do I assume that $\frac{P\_C}{P^Θ}=1$....
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That part sounds good.



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> leaving me with $K\_p=\frac{P\_{CO}}{(P\_{O\_2})^{1/2}}\frac{1}{P^\Theta}$
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Not sure about that one. How is $K\_p$ dimensionless in what you wrote?


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The standard change in free energy, which is used to determine the equilibrium constant Kp is based on the pure reactants at 1 atm going to pure products at 1 atm. Therefore, the partial pressure must be expressed in atm in order to be consistent with Kp. This essentially normalizes the partial pressure by 1 atm.


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