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$$\ce{N2(g) + 3H2 (g) -> 2NH3(g)}$$ 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 $10^6$?

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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.


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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    $\begingroup$ Please note that unit symbols and unit names shall not be mixed and that unit names are not mathematical entities. Therefore, it is not permissible to write $K_\mathrm c=4.2~\left(\frac{\text{moles}}{10^{-3}\ \mathrm{m^3}}\right)^{-2}$. $\endgroup$ – Loong Jul 14 '18 at 12:51
  • $\begingroup$ @Loong So, I should write $K_\mathrm c=4.2~\left(\frac{\text{mols}}{10^{-3}\ \mathrm{m^3}}\right)^{-2}$ instead? $\endgroup$ – Gaurang Tandon Jul 14 '18 at 13:40

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