[justin122289]

Hi, I have a problem to do on a take home test, and I can't think of a way to do it with the information I'm given. The equilibrium CO (g) + H2O (g) equilibrium CO2 (g) + H2 (g) and the problem reads, "In an experiment, 0.35 mol of CO and 0.40 mol of H2O were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.19 mol of CO remaining. K eq at the temperature of the reaction is ______ ."
I can't ust the ideal gas law to find pressure, since I don't have the temperature, and I can't find any other way to find the needed information to find the constant given only moles and no temperature. Any help with this would be greatly appreciated.

[Borek]

Are there any changes in total number of moles of reactants? Is pressure changing during the reaction?

[madscientist]

maybe assume standard temp and pressure.

[justin122289]

Are there any changes in total number of moles of reactants? Is pressure changing druing the reaction?

The moles and pressure have to change, yes, but there's no temp to relate pressure and moles to.

[Borek]

Think again about the total number of moles. Look at the reaction equation.

[justin122289]

Think again about the total number of moles. Look at the reaction equation.

I'm thinking I can take the number of moles of CO I end up with and use stoichiometry somehow...but I just don't get how I can get the partial pressures so I can get the equilibrium constant. I need to get the pressure of each product at equilibrium, multiply them, and divide that product by the procuct of the pressures of the reactants. But I don't see how to go from only having moles (no temps) can help me get to pressure. I also thought that maybe I could get grams of something from stoichiometry, but I don't see how that would help me out either. And it doesn't help that not one of my practice problems from the book is like this one.

     K eq = (P CO2)(P H2) / (P CO)(P H2O)
I assume I have to use this equation and the partial pressures to get K eq. I don't see any other way.
PS thanks for the responses

[madscientist]

I could be wrong so wait and see what borek or someone else thinks but this is what i came up with:

equation:         CO(g)    +    H2O(g) ---------> CO2(g)    +     H2(g)
Initial (M)           0.35            0.40                   0                   0
change(M)        -0.16           -0.16                 +0.16            +0.16
equilibrium(M)     0.19           0.24                   0.16               0.16

K= [CO2][H2]/[CO][H2O] = (0.16)(0.16)/(0.19)(0.24)

Like i said though, wait and see what borek or others say about this answer, Ive had to rush it and am more than likely wrong.

cheers,

madscientist

[justin122289]

I could be wrong so wait and see what borek or someone else thinks but this is what i came up with:

equation:         CO(g)    +    H2O(g) ---------> CO2(g)    +     H2(g)
Initial (M)           0.35            0.40                   0                   0
change(M)        -0.16           -0.16                 +0.16            +0.16
equilibrium(M)     0.19           0.24                   0.16               0.16

K= [CO2][H2]/[CO][H2O] = (0.16)(0.16)/(0.19)(0.24)

Like i said though, wait and see what borek or others say about this answer, Ive had to rush it and am more than likely wrong.

cheers,

madscientist

Yeah I actually figured out how to do it. I didn't realize that I could use the moles given as part of the Initial/Change/Equil chart. But since the container is 1 liter, the moles can stand for molarity also (0.35 moles/1 liter = 0.35 M). I knew there had to be some sort of simple solution that I was overlooking. But before I realized this, I picked a random temp (STP temp) and tried it out and got the same answe, 0.56, for K eq.

                  P=nRT/v

                  P CO =  (0.35 mol)(0.08206 L-atm/mol-K)(273 K)/1.00 L   =   7.84 atm
                  P H2O = (0.40 mol)(0.08206 L-atm/mol-K)(273 K)/1.00 L   =   8.96 atm
                  P CO (equil) = (0.19 mol)(0.08206 l-atm/mol-K)(273 K)/1.00 L   =   4.26 atm

                      CO(g)        +        H2O(g)-------------->CO2(g)          +          H2(g)
Initial (atm)      7.84                     8.96                       0                              0
Change(atm)   -3.58                    -3.58                    +3.58                        +3.58
Equil (atm)       4.26                     5.38                      3.58                          3.58


                   K eq = (3.58)(3.58)/(4.26)(5.38)   =    0.56

So I ended up getting the answer either way. If you pick a different temp, it works, too. Thank all of you for your *delete me*
 

[Borek]

Note: 2 moles give 2 moles. Volume of the reactants/products is constant, no matter how the reaction proceeds. That's why you can do it this way.

[justin122289]

OK. Thanks again for your assistance.