Question

A mixture of water and graphite is heated to 600 K . When the system comes to equilibrium it contains 0.11 mol of H 2 , 0.11 mol of CO , 0.31 mol of H 2 O , and some graphite. Some O 2 is added to the system and a spark is applied so that the H 2 reacts completely with the O 2 . Find the amount of CO in the flask when the system returns to equilibrium.

Answer 1

C (s) + H2O (g) <==> H2 (g) + CO (g), Kc = [H2] [CO] / [H2O] = (0.1)(0.1)/0.31 = 0.03903
Since Kc is based on concentrations, your value is correct only if the vessel in which the reaction is taking place is 1.0 L; the question does not give this information. But, continuing on as if this is correct, when the O2 is added and ignition takes place, the reaction
H2 (g) + ½ O2 (g) ==> H2O (g)
occurs, meaning the 0.11 moles of H2 becomes 0.11 moles of H2O. The ICE table becomes:
............. [H2O]............ [H2]........... [CO]
I........ 0.31+0.11........... 0............. 0.11
C............. -x................. +x............. +x
E........ 0.44-x................ x........... 0.11+x
Substituting the E line into the Kc equation:
0.03903 = (x)(0.11+x)/(0.44-x)
Which rearranges to the quadratic
x² + 0.14903x - 0.017292 = 0.
Using the quadratic formula, the only positive x is 0.076635 M. Therefore at the reestablished equilibrium,
[H2] = x = 0.076635 M
[CO] = 0.11+x = 0.1866 M = 0.1866 mol (we have taken 1 L flask ==> M = moles/liter)
[H2O] = 0.44-x = 0.3634 M.
CO molar mass = 28.01 g/mol , 0.1866 mol x 28.01 g/mol = 5.2267 g

The amount of CO in the flask when the system returns to equilibrium = 5.2267 g
Of course, if the vessel has some other volume, all this is wrong, but it can be adjusted.
I have read that when coal gets wet, it can sponaneously ignite; perhaps this mechanism is the explanation.