How do you calculate equilibrium concentrations of species when the concentration of one specie is unknown?

Here's the question:
At temperatures near 800ºC, steam passed over hot coke ( a form of carbon obtained from coal) reacts to form CO and H2:

    C(s)  +  H2O(g)  ↔  CO(g)  +  H2(g)

The mixture of gases that results is an important industrial fuel called water gas. (a) At 800 ºC, the equilibrium constant for this reaction is Kc = 0.16. What are the equilibrium concentrations of H2O, CO, and H2 in the equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H2O in a 1.00-L vessel?

I tried to make an ICE chart using a variable for the initial concentration of the carbon, but I got a very strange answer and I do not think I'm doing it correctly.

1 Answer
Truong-Son N.
Mar 5, 2018

#x = "0.0697 M"#

Now evaluate each equilibrium concentration from there.

You still know enough to get the equilibrium concentrations.

#"C"(s) + "H"_2"O"(g) rightleftharpoons "CO"(g) + "H"_2(g)#

#"I"" "-" "" "0.100" "" "" "0" "" "" "0#
#"C"" "-" "-x" "" "" "+x" "" "+x#
#"E"" "-" "0.100-x" "" "x" "" "" "x#

Remember, solids show up as #1#, and the coefficients multiply by the change in concentration. Here we have #"0.100 mols H"_2"O"(g)# in #"1.00 L"#, which is #"0.100 M"#.

#K_c = 0.16 = (["CO"]["H"_2])/(["H"_2"O"])#

#= x^2/(0.100 - x)#

The value of #K_c# is not small, so this will have to be solved in full.

#0.016 - 0.16x = x^2#

#x^2 + 0.16x - 0.016 = 0#

From this, #x = "0.0697 M"# by evaluating the quadratic formula. Now evaluate the equilibrium concentrations.

#["H"_2"O"]_(eq) = "0.100 M" - x = ?#

#["CO"]_(eq) = ["H"_2]_(eq) = x = ?#

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