April 27, 2024, 01:18:29 PM
Forum Rules: Read This Before Posting


Topic: ICE/Equilibrium Question # 2  (Read 6954 times)

0 Members and 1 Guest are viewing this topic.

Offline Frederick95

  • Regular Member
  • ***
  • Posts: 82
  • Mole Snacks: +0/-12
ICE/Equilibrium Question # 2
« on: March 30, 2008, 01:35:42 AM »
When carbon dioxide is heated in a closed container, it decomposes into carbon monoxide and oxygen according to the following equilibrium equation:
2CO2(g) ---> 2CO(g) + O2(g)

When 2.0 mol of CO2(g) is placed in a 5.0-L closed container andheated to a particular temperature, the equilibrium concentration of CO2(g) is measured to be 0.039 mol/L. Use an ICE table to determine the equuilibrium concentrations of CO(g) and )2(g)


My ICE table appearrs as the following:


     2CO2(g) -------->2CO(g)  +  O2(g)
I    2.0                       0             0
C   -2x                       +2           +x
E   0.39                     +2x          +x



Im really stuck on the part with the calulcations please help.


Offline boostar

  • Regular Member
  • ***
  • Posts: 25
  • Mole Snacks: +3/-0
Re: ICE/Equilibrium Question # 2
« Reply #1 on: March 30, 2008, 02:51:53 AM »
When 2.0 mol of CO2(g) is placed in a 5.0-L closed container and heated to a particular temperature, the equilibrium concentration of CO2(g) is measured to be 0.039 mol/L. Use an ICE table to determine the equilibrium concentrations of CO(g) and )2(g)

     2CO2(g) -------->2CO(g)  +  O2(g)
I    2.0                       0             0
C   -2x                       +2           +x
E   0.39                     +2x          +x

Do you see your mistake?

Once you have corrected your mistake, you will know

1) 2 moles you began with
2) Moles of CO2 that are left.

Than check out your stoichiometry equation to figure out the rest. (hint: figure out how much CO2 has been lost, and than think about what this has to do with your x values: 2x and x in your products)

Edit: P.S I think the people here like it if you keep questions in one topic - especially questions of the same type.
« Last Edit: March 30, 2008, 03:05:11 AM by boostar »

Offline Henri

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +1/-0
Re: ICE/Equilibrium Question # 2
« Reply #2 on: March 30, 2008, 08:53:31 AM »
I would do this something like this:

     2CO2(g) -------->2CO(g)  +  O2(g)
I    2.0                       0             0
C   -2x                       +2x           +x
E   2-2x                     +2x          +x

As we know that 2-2x = 0.39 => x=0.805
And then just insert that value to x and 2x.

Offline Frederick95

  • Regular Member
  • ***
  • Posts: 82
  • Mole Snacks: +0/-12
Re: ICE/Equilibrium Question # 2
« Reply #3 on: March 30, 2008, 10:47:25 AM »
The answers are supposedly  [CO(g)] = 0.01 mol/L; [O2(g)] = 0.005 mol/L

How do I get them?

Offline Nathaniel

  • Regular Member
  • ***
  • Posts: 31
  • Mole Snacks: +0/-9
Re: ICE/Equilibrium Question # 2
« Reply #4 on: March 30, 2008, 07:15:23 PM »
inserting that value to x and 2x does not give [CO(g)] = 0.01 mol/L; [O2(g)] = 0.005 mol/L
Is there an alternative way to solve the question?

Sponsored Links