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?

Md Ariful Islam

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Md Ariful Islam

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$K_p=(P_(PCl_3)P_(Cl_2))/P_(PCl_5)=0.0121$

Initially, $P_(PCl_5)=0.123*atm$, if a quantity $x$ dissociates, then,

$K_P=0.0121=x^2/(0.123-x)$

(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 $0.123-x~=0.123$.

Thus $x_1=sqrt(0.0121xx0.123)=0.0386*atm$. This result is indeed small compared to $0.123$, but we will recycle it to give a second approximation:

$x_2=0.0319*atm$

$x_3=0.0332*atm$ (the result is converging)

$x_4=0.0330*atm$ and finally,

$x_5=0.0330*atm$

And thus $P_(PCl_3)=P_(Cl_2)=0.0330*atm;$ $P_(PCl_5)=(0.123-0.0330)*atm=0.0900*atm$.

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.

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