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$K_p$ 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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  • $\begingroup$ Actually, the definition of partial pressure is the pressure of the gas that it would have if it was the only gas in the container. So the partial pressure ($P_j$) does have units. $\endgroup$ – LDC3 Jun 6 '15 at 16:33
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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?

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 $P_j$.

Do I assume that $\frac{P_C}{P^Θ}=1$....

That part sounds good.

leaving me with $K_p=\frac{P_{CO}}{(P_{O_2})^{1/2}}\frac{1}{P^\Theta}$

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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