Chemistry Equilibrium Partial Pressures

[Madelin Pierce]

Homework Statement

Kc = 4.15 x 10-2 at 356°C for PCl5(g) ↔ PCl3(g) + Cl2(g). A closed 2.00 L vessel initially contians 0.100 mol PCl5. Calculate the total pressure in the vessel (in atm to 2 decimal places) at 356°C when equilibrium is achieved.

Homework Equations

PV=nRT
Kp= Kc(RT)^change in n
Kp= Pressure products/pressure reactants
Small x approximation

The Attempt at a Solution

The answer key says 4.09 for Ptotal, but I keep getting a wrong answer

 

[Borek]

Show how you arrive at the wrong answer then.

 

[Madelin Pierce]

Show how you arrive at the wrong answer then.

I use the equation Kp=Kc(RT)^change in n. I plug in Kp=(.0415)(.08206)(629.15)^1= 2.1425.
Then, I solve for PV=nRT for PCl5: P(2.00L)=(.100mol)(.08206)(629.15K), equaling 2.5814atm. I use Kp= Products/Reactants which is 2.1425=x^2/(2.5814-x) where the Xs are from an ice table. I was told by my instructor to use small x approximation for this problem, so I eliminated the x on the bottom, so 2.1425*2.5814= x^2. Times them and square root and I got 2.3517 for X. Plugged Ptotal=P(PCl5)+P(PCl3)+P(Cl2) which equaled 4.935 for total pressure. I don’t know how I’m not getting the answer right, I just don’t. The answer’s supposed to be 4.09 Ptotal.

 

[Borek]

You got 2.3517 for x - does it justify the small x approximation?