# The equilibrium constant for the reaction: PCl_5 rightleftharpoons PCl_3 + Cl_2 is 0.0121. A vessel is charged with PCl_5, giving an initial pressure of 0.123 atmospheric. How do you calculate the partial pressure of PCl_3 at equilibrium?

Mar 6, 2017

$P C {l}_{5} \left(g\right) r i g h t \le f t h a r p \infty n s P C {l}_{3} \left(g\right) + C {l}_{2} \left(g\right)$

At equilibrium, ${P}_{P C {l}_{3}} = {P}_{C {l}_{2}} = 0.0330 \cdot a t m$

#### Explanation:

${K}_{p} = \frac{{P}_{P C {l}_{3}} {P}_{C {l}_{2}}}{P} _ \left(P C {l}_{5}\right) = 0.0121$

Initially, ${P}_{P C {l}_{5}} = 0.123 \cdot a t m$, if a quantity $x$ dissociates, then,

${K}_{P} = 0.0121 = {x}^{2} / \left(0.123 - x\right)$

(I have assumed that the quoted equilibrium constant is ${K}_{P}$ not ${K}_{c}$. And if I am wrong, I am wrong).

Now the given ${K}_{P}$ expression is a quadratic in $x$. We will assume that $0.123 - x \cong 0.123$.

Thus ${x}_{1} = \sqrt{0.0121 \times 0.123} = 0.0386 \cdot a t m$. This result is indeed small compared to $0.123$, but we will recycle it to give a second approximation:

${x}_{2} = 0.0319 \cdot a t m$

${x}_{3} = 0.0332 \cdot a t m$ (the result is converging)

${x}_{4} = 0.0330 \cdot a t m$ and finally,

${x}_{5} = 0.0330 \cdot a t m$

And thus P_(PCl_3)=P_(Cl_2)=0.0330*atm; ${P}_{P C {l}_{5}} = \left(0.123 - 0.0330\right) \cdot a t m = 0.0900 \cdot a t m$.

Given the circumstances of the reaction, I think I am quite justified in assuming that I was given ${K}_{p}$, and not ${K}_{c}$ in the boundary conditions of the problem.