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?

1 Answer
Mar 6, 2017

Answer:

#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.