Chemistry (please help me!)

H2(g) + CO2(g)<-> H20(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2,000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured:
[H2]=0.20 M
[CO2]=0.30 M
[H2O]=[CO]=0.55M
a) What is the mole fraction of CO(g) in the equilibrium mixture?
b)Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction.
c) Determine Kp,in terms of Kc for this system.
d)When the system is cooled from 2,000 K to a lower temperature, 30.0% of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature.
e)In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0 L reaction vessel at 2,000K. Calculate the equilibrium concentration, in moles/liter, of CO(g) at this temperature.

I know this is lengthy and if i at least get a good start on this...maybe I will know how to do the rest...but please help me!! Thanks!

Oh my goodness... what grade are you in? That has got to be the most confusing problem I have ever seen in my life!! (Of course I'm only barely in middle school!)

H2(g) + CO2(g)<-> H20(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2,000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured:
[H2]=0.20 M
[CO2]=0.30 M
[H2O]=[CO]=0.55M
a) What is the mole fraction of CO(g) in the equilibrium mixture?

This looks straight forward. XCO=(mols CO/mols CO + mols H2 + mols H2O + mols CO2)

b)Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction.

This looks straight forward. Kc=(CO)(H2O)/(CO2)(H2).
Simply plug in the concentrations and calculate Kc.


c) Determine Kp,in terms of Kc for this system.

Look in your text/notes and find the conversion from Kc to Kp.

d)When the system is cooled from 2,000 K to a lower temperature, 30.0% of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature.

Subtract 30% CO to obtain the new CO concentration, add 30% to CO2 to obtain the new CO2 concentration. Same thing for H2O and H2 concentrations. Plug the new numbers into Kc expression above and calculate the new Kc.

e)In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0 L reaction vessel at 2,000K. Calculate the equilibrium concentration, in moles/liter, of CO(g) at this temperature.

Use Kc found in your earlier parts at 2000 K. Post your work for the other parts and we can help you get through this part. I suspect you will know how to do this part by the time you get here.



Check my thinking.
Check my work.

Simple, simple, simple...Just refer to your text book and it should all be there. Isn't calculating Kp so easy?!

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  1. (a) CO = f(0.55 mol, 1.6 mol) = 0.34
    (b) Kc = ([H2O][CO])/([H2][CO2]) = (0.55×0.55)/(0.20×0.30) = 5.04
    (c) since Δn = 0, Kc = Kp
    (d) [CO] = 0.55 - 30.0% = 0.55 - 0.165 = 0.385 M, [H2O]= 0.55 - 0.165 = 0.385 M, [H2]= 0.20 + 0.165 = 0.365 M
    [CO2] = 0.30 + 0.165 = 0.465 M K = (0.385)2/(0.365×0.465) = 0.87
    (e) let X = Δ[H2] to reach equilibrium
    [H2] = 0.50 mol/3.0L - X = 0.167 – X, [CO2] = 0.50 mol/3.0L - X = 0.167 – X, [CO] = +X ; [H2O] = +X
    K = X2/(0.167 - X)2 = 5.04 ; X = [CO] = 0.12 M

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