Equilibrium concentrations of CO and Cl2 by dissociation of phosgene

I am working through a problem set from MIT's open courseware course [Principles of Chemical Science](https://ocw.mit.edu/courses/5-111sc-principles-of-chemical-science-fall-2014/517e9ef9c38ea95d3c03876dc53a44c9_MIT5_111F14_Lec18Prob.pdf).


The question is:



> 
> Phosgene $\ce{(COCl2)}$ is a chemical warfare agent that decomposes by the reaction:
> $$\ce{COCl2(g) <=> CO(g) + Cl2(g)}$$
> $K= 8.3 \times 10^{-4}$ (at $\pu{360 ^\circ C}$)
> 
> 
> Calculate the $\ce{[CO], [Cl2],}$ and $\ce{[COCl2]}$ when $\pu{10.0 mol}$ of phosgene decomposes at $\pu{360 ^\circ C}$ and reach equilibrium in a 5.00-L flask.
> 
> 
> 


I am a bit unsure whether equilibrium constant given is the ratio of partial pressures, or concentration of gases. ($K$ or $K\_\mathrm{c}$)


My first approach is assuming that $K$ is given (not $K\_\mathrm{c}$):
----------------------------------------------------------------------


Given: $T =\pu{360 ^\circ C} = \pu{633.15 K}$


1. Convert $K$ to $K\_\mathrm{c}$
  
$K = (RT)^{\Delta n\_\mathrm{r}} \cdot K\_\mathrm{c}$
  
$K\_\mathrm{c} = \dfrac{K}{(RT)^{\Delta n\_\mathrm{r}}}$
  
$K\_\mathrm{c} = \dfrac{8.3 \times10^{-4}} {(8.3145 \times10^{-2} \cdot 633.15)^1}$
  
$K\_\mathrm{c} = 1.577 \times10^{-5}$
2. Calculate starting concentration of $\ce{COCl2}$
  
$[\ce{COCl2}] = \dfrac{\pu{10.0 mol}}{\pu{5.00 L}} = \pu{2.00 mol L-1} $
3. Calculate equilibrium concentration of each species:   

$\begin{align} 
& [\ce{COCl2}] = 2.00 - x \\
& [\ce{CO}] = x \\
& [\ce{Cl2}] = x \\
& K\_\mathrm{c} = \dfrac{[\ce{CO}][\ce{Cl2}]}{[\ce{COCl2}]} = \dfrac{x^2}{2.00 - x}
\end{align}$
  
  

Using ($K\_\mathrm{c} = 1.577 \times10^{-5}$ above): (ignoring negative solution for $x$)
  

$x = 5.608 \times10^{-3}$
  
  

$\begin{align} 
& [\ce{COCl2}] = 2.00 - 5.608 \times10^{-3} = \pu{1.99 M} \\
& [\ce{CO}] = \pu{5.61 \times10^{-3} M }\\
& [\ce{Cl2}] = \pu{5.61 \times10^{-3} M} \\
\end{align}$


The solutions given [here](https://ocw.mit.edu/courses/5-111sc-principles-of-chemical-science-fall-2014/4d0d7db1079f4fee2bdd58d95033f36d_MIT5_111F14_Lec18Soln.pdf) are:
------------------------------------------------------------------------------------------------------------------------------------------------------------------------



> 
> $$\begin{align} 
> & [\ce{COCl2}] = \pu{1.96 M }\\
> & [\ce{CO}] = \pu{0.041 M} \\
> & [\ce{Cl2}] = \pu{0.041 M} \\
> \end{align}$$
> 
> 
> 


I'm not quite understanding why my answer is different.


My second approach is to determine the partial pressures of each species at equilibrium:
----------------------------------------------------------------------------------------


1. Calculate initial partial pressure using   

$\begin{align}
& P\_\mathrm{\ce{COCl2}} = \dfrac{n\_\mathrm{\ce{COCl2}}RT}{V} \\
& P\_\mathrm{\ce{COCl2}} = \dfrac{10.0}{5.00}RT \\
& P\_\mathrm{\ce{COCl2}} = \pu{105.2 bar}
\end{align}$
2. Calculate equilibrium pressures of each species:   

$\begin{align} 
& P\_\mathrm{\ce{COCl2}} = 105.2 - x \\
& P\_\mathrm{\ce{CO}} = x \\
& P\_\mathrm{\ce{Cl2}} = x \\
& K = \dfrac{P\_\mathrm{\ce{CO}}P\_\mathrm{\ce{Cl2}}}{P\_\mathrm{\ce{COCl2}}} = \dfrac{x^2}{105.2 - x}
\end{align}$
  
  

Using ($K = 8.3 \times10^{-4}$ above): (ignoring negative solution for $x$)
  

$x = 0.294 $
  
  

$\begin{align} 
& [\ce{COCl2}] = \dfrac{P\_\mathrm{\ce{COCl2}}}{RT} = \dfrac{105.2 - 0.294}{RT} = \pu{1.99 M} \\
& [\ce{CO}] = \dfrac{0.294}{RT} = \pu{5.59 \times10^{-3} M} \\
& [\ce{Cl2}] = \dfrac{0.294}{RT} = \pu{5.59 \times10^{-3} M} \\
\end{align}$


My last approach is to assume that the problem set has given me $K\_\mathrm{c}$
-------------------------------------------------------------------------------


I get much closer to the given answer.


1. Calculate equilibrium concentration of each species:   

$\begin{align} 
& [\ce{COCl2}] = 2.00 - x \\
& [\ce{CO}] = x \\
& [\ce{Cl2}] = x \\
& K = \dfrac{[\ce{CO}][\ce{Cl2}]}{[\ce{COCl2}]} = \dfrac{x^2}{2.00 - x}
\end{align}$
  
  

**Using ($K = 8.3 \times10^{-4}$)**: (ignoring negative solution for $x$)
  

$x = 0.0403$
  
  

$\begin{align} 
& [\ce{COCl2}] = 2.00 - 0.0403 = \pu{1.96 M} \\
& [\ce{CO}] = \pu{0.040 M }\\
& [\ce{Cl2}] = \pu{0.040 M }\\
\end{align}$


I'm still off by one significant figure.


Thanks for your help. Sorry in advance for any inconsistent formatting.


