# 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.
• $\pu{0.0125mol}~\ce{AsCl5}$ is the initial amount of $\ce{AsCl5}$before the system reaches the equilibrium .The actual amount of $\ce{AsCl5}$ at equlibrium equal $\pu{0.001mol}$. – Adnan AL-Amleh Mar 16 '19 at 21:49
• $$K_\mathrm{p}= K_\mathrm{C}\cdot{(\mathrm{R}}\cdot{\mathrm{T}})^{\Delta{n}}$$ – Adnan AL-Amleh Mar 16 '19 at 21:50