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.

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

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