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?

Question

Serenity Allen · Accepted Answer

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

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

Henry Antrim · Answer

undefined To calculate the partial pressure of PCl3 at equilibrium, you would use the equilibrium constant expression for the reaction:

Kc = [PCl3][Cl2] / [PCl5]

Since the initial pressure of PCl5 is given as 0.123 atm and assuming it decreases by x at equilibrium, the equilibrium pressure of PCl5 would be (0.123 - x) atm.

Using stoichiometry, the equilibrium pressure of PCl3 and Cl2 would both be x atm.

Now, you can plug these values into the equilibrium constant expression and solve for x. Once you find the value of x, you can use it to calculate the equilibrium partial pressure of PCl3.