Calculate pressure equilibrium constant from concentration equilibrium constant


> 
> $$\ce{N2(g) + 3H2 (g) -> 2NH3(g)}$$
>  Calculate $K\_\mathrm p$ from $K\_\mathrm c$, given that $K\_\mathrm c = \pu{4.2 M^-2}$
>  and that there is 0.2 moles of $\ce{N2}$, 0.6 moles of $\ce{H2}$ and 0.425 moles of $\ce{NH3}$. Temperature = $\pu{600K}$
> 
> 
> 


My attempt at the question:


$K\_\mathrm p= K\_\mathrm c(RT)^{\Delta n}$


$K\_\mathrm p= 4.2(8.314\times600)^{-2}$


$K\_\mathrm p= \pu{1.69\times10^{-7} Pa^-2}$


However, in our answer sheet the value of $K\_\mathrm c$ is divided by $10^6$ and the answer is $\pu{1.69x10^-13 Pa^-2}$. Could someone please explain why $K\_\mathrm c$ is divded by $10^6$?




If you intend to get an answer in Pascal, which is an SI unit, **you need to have your original values in SI units** as well.


Note that:


$$K\_\mathrm c=\pu{4.2M^-2}=4.2~\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}=4.2~\left(\frac{\text{moles}}{\pu{10^-3m^3}}\right)^{-2}$$


The conversion factor for litre (non-SI) to metre cubed (SI unit) is $10^{-3}$. Since you have $K\_\mathrm c$ in units of $\pu{M^-2}$, the final answer would be $10^{-6}$ of the answer you had arrived at.




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**Elaboration:**


Let's do dimensional analysis on the RHS factor-by-factor. 


* $K\_\mathrm c$ has the units $\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}$ as shown above.
* $RT$ has the units = $\pu{Jmol^-1T^-1}\cdot\pu{T}=\pu{J\cdot mol^-1}$
* $(RT)^{-2}$ has the units = $\pu{J^-2\cdot mol^2}$.
* Since one joule $=\pu{Pa\cdot m^3}$, hence $(RT)^{-2}$ has the units = $\pu{Pa^-2\cdot m^-6\cdot mol^2}$
* Multiply $K\_\mathrm c$ and $(RT)^{-2}$ we get 
$$\text{Units of }K\_\mathrm p=\left(\frac{\text{moles}}{\text{litre}}\right)^{-2}\cdot\pu{Pa^-2\cdot m^-6\cdot mol^2}=\pu{Pa^-2\cdot m^-6\cdot litre^2}$$


Thus your final answer is not in the units of $\pu{Pa^-2}$ but rather $\pu{Pa^-2\cdot m^-6\cdot litre^2}$. Can you now see why we need the conversion factor?


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