5I.25

Moderators: Chem_Mod, Chem_Admin

Daniela Shatzki 2E
Posts: 53
Joined: Sat Aug 24, 2019 12:16 am

5I.25

Postby Daniela Shatzki 2E » Sun Jan 12, 2020 3:08 pm

I'm kinda confused about this problem. It gives the initial concentrations for all components and the K and is asking for the equilibrium concentrations so I know you have to find x and then subtract/add it to the original concentrations but I can't seem to find x. I did an ICE table and set the concentrations equal to the K but I still can't find x. I think I'm setting it up wrong or something so maybe if someone could help me with the setup? Thank you!

Julie Park 1G
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

Re: 5I.25

Postby Julie Park 1G » Sun Jan 12, 2020 3:56 pm

For the ICE table, you should have something that looks like....
SO2(g) + NO2(g) --> NO(g) + SO3(g)
0.0200, 0.0400, 0.0200, 0.0300
-x , -x , +x , +x
0.0200 -x, 0.0400 - x, 0.0200 + x, 0.0300 + x

Were you able to get to this step? Remember that you first have to convert the moles of each substance into mol/L to find the concentrations (which you'll use in the ICE table)

Daniela Shatzki 2E
Posts: 53
Joined: Sat Aug 24, 2019 12:16 am

Re: 5I.25

Postby Daniela Shatzki 2E » Sun Jan 12, 2020 4:59 pm

yes, I got that then I did:
85.0 = (0.02+x)(0.03+x)/(0.02-x)(0.04-x)
but then I think I'm just not solving it correctly or something. probably more of a math issue than chemistry haha


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 2 guests