Question

1.15 g oh h2O (density 1.00 g/ml) is introduced into an empty 1.00 L container. The seled container is heated to 412 K and .150 moles of O2 (g) is present at equilibrium.
A. Wrte the equilibrium expression for the reaction above.
B. Calculate the equilibrium concentration (in M) of H2 (g) in the container at 412 K.
Reaction: H2O (l) <--> 2 H2 (g) + O2 (g)

Answer 1

Please look at your post and your problem. If ALL of the 1.15 g H2O decomposed to 2H2O --> 2H2 + O2
you would have (1.15g H2O x (1 mol H2O/18 g H2O) x (1 mol O2/2 mol H2O) = about 0.064/2 = about 0.032 mols O2 formed. The problem states that 0.15 mol O2 were found at equilibrium. I don't see how. You can't get more than you started with.

If you ignore that little problem and work ONLY with the mols O2 given, then (O2) at equilibrium = (0.15mol/1L) = ?
(H2) at equilibrium = (2 x 0.15/1L) = ?



part A for equilibrium expression it is
K = (H2)^2(O2)