(H2) = mols H2/15 L = (6.5/2)/15 = about 0.2 but this is just a close estimate and you use a more correct value.
............3 Fe(s) + 4 H2O(g) <==> Fe3O4(s) + 4 H2(g)
I..............solid.........0...................solid............0.2
C............solid......+4x.................. solid...........0.2-4x
E............solid......4x......................................0.2-4x
K = 4.6 = (H2)^4/(H2O)^4
Substitute the E line into the Keq expression and solve for x = (H2O)(g)
. The reaction of iron and water vapor results in an equilibrium reaction, 3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g) and an equilibrium constant of 4.6 at 850 degree C. What is the concentration of water present at equilibrium if the reaction is initiated with 6.5 g of H2 and excess iron oxide, Fe3O4, in a 15.0 L container?
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