[Catalyst67]

Hello, I am new to these forums. It's the weekend and I can't get a hold of my professor, so I thought I could get help from chemists off of some forums. It is pretty late, I'm frustrated with this problem, it is getting the best of me.

Anyways, here is the problem.

K_c = 0.040 for the system below at 450°C. If a reaction is initiated with 0.20 mole of Cl₂ and 0.20 mole of PCl₃ in a 1.0 liter container, what concentration of PCl₃ will be present at equilibrium?

PCl₅  PCl₃ + Cl₂ (all of them are in the gas phase)

If you could somehow explain it to me so I can understand what to do would be awesome. I am given the answer, which is 0.07M, but I am getting something totally different.

[AWK]

Write the expression for K_c, and build ICE table.

[Dan]

I am given the answer, which is 0.07M, but I am getting something totally different.

Show your work so that we can see where you're going wrong. This is a Forum Rule.

[Catalyst67]

Ok, the first thing that I do is convert the moles of PCl₃ and Cl₂ to molarity by dividing it by the liters given, which is one. So, the molarity that I get is .20 M PCl₃ and .20 M Cl₂. From this, I set up my ICE tables.

I   0.20 PCl₃          0.20 Cl₂                      0 M PCl₅

C  -x                                     -x                                             +x

E 0.20-x                               0.20-x                                          x


So far, am I doing this right?

[Borek]

Looks OK so far. Just note you have reversed the reaction.

[Catalyst67]

Ok, from the ICE table I get this.

0.040 = [(0.2-x)²]/x

Then I try to solve for X. Which results in this.

0.040x = [(0.2-x)²]/x

Is this correct so far?

[Borek]

Ok, from the ICE table I get this.

0.040 = [(0.2-x)²]/x

This is OK.

0.040x = [(0.2-x)²]/x

This is not.

[Catalyst67]

I am sorry, this is what I meant to put.

0.040x = [(0.2-x)2]

[Borek]

This is correct, but it has nothing to do with solving for x. You have to rearrange the equation till it is obvious its a quadratic function, then solve it by completing the square (or any other equivalent method).