Question

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?

Answer 1

`PCl_5(g) rightleftharpoons PCl_3(g) + Cl_2(g)`

At equilibrium, `P_(PCl_3)=P_(Cl_2)=0.0330*atm`

Explanation:

`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.