. 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?
(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)
To find the concentration of water present at equilibrium, we need to use the given information and apply the concept of equilibrium constants. The equilibrium constant is denoted as Kc and is calculated based on the concentrations of the reactants and products at equilibrium.
In this case, we are given:
Equilibrium reaction: 3 Fe(s) + 4 H2O(g) ⇌ Fe3O4(s) + 4 H2(g)
Equilibrium constant: Kc = 4.6
Initial amount of H2: 6.5 g
Volume of the container: 15.0 L
To solve this problem, follow these steps:
Step 1: Calculate the number of moles of H2:
Molar mass of H2 = 2 g/mol
Number of moles of H2 = mass / molar mass = 6.5 g / 2 g/mol = 3.25 mol
Step 2: Use stoichiometry to determine the moles of H2O produced at equilibrium:
According to the equation, 4 moles of H2O are produced for every 4 moles of H2 consumed.
Therefore, the number of moles of H2O produced is also 3.25 mol.
Step 3: Calculate the concentration of H2O at equilibrium:
Concentration (in mol/L) = number of moles / volume
Concentration of H2O = 3.25 mol / 15.0 L = 0.217 mol/L
Therefore, the concentration of water present at equilibrium is 0.217 mol/L.