Amixture of water and graphite is heated to 600 k in a 1 l container. when the system comes to equilibrium it contains 0.15 mol of h2, 0.15 mol of co, 0.57 mol of h2o, and some graphite. some o2 is added to the system and a spark is applied so that the h2 reacts completely with the o2.

(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???

when the reaction equation is:

C(s) + H2O(g)  ↔ H2(g) + CO(g) 

∴ Kc = [H2] [CO] / [H2O]

and we have Kc = 0.0393 (given missing in the question)

when the O2 is added so, the reaction will be:

2H2(g) + O2(g) → 2H2O(g)

that means that 0.15 mol H2 gives 0.15 mol of H2O

∴ by using ICE table:

            [H2O]          [H2]        [CO]

initial 0.57 + 0.15      0               0.15

change  -X                +X              +X

Equ   (0.72-X)             X            (0.15+X)

by substitution:

0.0393 = X (0.15+X) / (0.72-X)  by solving for X

∴ X = 0.098 

∴[H2] = X = 0.098 M

∴[CO] = 0.15 + X
 
           = 0.15 + 0.098 = 0.248 M

∴[H2O] = 0.72 - X

             = 0.72 - 0.098

             = 0.622 M

The answer is a) carbon-12