Calculating Equilibrium Partial Pressures Given Kp and Mass


> 
> For the reaction
>  $$\ce{AsCl5(g) <=> AsCl3(g) + Cl2(g)}$$
>  at $550$ K, the equilibrium constant ($K\_p$) is $9.81$. Suppose that $3.150 \ g$ $\ce{AsCl5}$ is placed in an evacuated $600$ ml bulb, which is then heated to $550$K.
> 
> 
> What is the partial pressure of $\ce{AsCl5}$ at equilibrium?
> 
> 
> 


So at the moment, I understand that 
$$\ce{K\_p=$$\frac{(P\_{AsCl3})(P\_{Cl\_2})}{(P\_{AsCl\_5})}$$}$$
And that using a rearranged Ideal Gas Law I can get
$$\ce{P=$$\frac{n(0.0125mol AsCl5)\*R\*T(550K)}{V (0.6L)}$$}$$
Giving me $$\ce{0.94 atm}$$
but I'm not sure where to go from here. Can somebody help by pointing me in the right direction?




Say you start with $A$ moles of $\ce{AsCl5}$, and then you reach a temperature $T$, for which the equilibrium constant is $K\_p$.


In order to achieve equilibrium, some amount of forward reaction ($\ce{AsCl5 -> AcCl3 + Cl2}$) happens to consume $x$ moles of $\ce{AsCl5}$. Based on the stoichiometry of the equation, you know that exactly $x$ new moles of the products are formed.


$$K\_p=\frac{P\_{\ce{AsCl3}}P\_{\ce{Cl2}}}{P\_{\ce{AsCl5}}}$$
Since for $V$, $R$, $T$ are constant, we can replace the above equation by:
$$K\_p=\frac{RT}{V}\frac{n\_{\ce{AsCl3}}n\_{\ce{Cl2}}}{n\_{\ce{AsCl5}}}=\frac{RT}{V}\frac{x^2}{A-x}$$
Solve this for $x$, and substitute in the ideal gas equation to get the answer.


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