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)
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)
A. The equilibrium expression for the given reaction can be written as follows:
Kc = [H2]^2 * [O2] / [H2O]
Where:
[H2] refers to the concentration of H2 gas (in M),
[O2] refers to the concentration of O2 gas (in M), and
[H2O] refers to the concentration of H2O liquid (in M).
B. To calculate the equilibrium concentration of H2 gas in the container at 412 K, we need to consider the initial amount of H2O introduced into the 1.00 L container.
Given:
Mass of H2O = 1.15 g
Density of H2O = 1.00 g/ml
Initial volume of H2O = Mass / Density = 1.15 g / 1.00 g/ml = 1.15 ml = 0.00115 L
Since the initial moles of H2O is not provided, we can use the given density to calculate it.
Molar mass of H2O = 18.015 g/mol
Moles of H2O = Mass / Molar mass = 1.15 g / 18.015 g/mol ≈ 0.0638 mol
Now, let's consider the stoichiometry of the reaction. According to the balanced equation, 1 mole of H2O produces 2 moles of H2 gas. Therefore, the initial amount of H2 gas is:
Initial moles of H2 = 2 * Moles of H2O = 2 * 0.0638 mol = 0.1276 mol
At equilibrium, the change in moles of H2 can be written as:
Change in moles of H2 = 2x
Where x represents the change in the moles of H2 at equilibrium.
Hence, the equilibrium moles of H2 = Initial moles of H2 + Change in moles of H2 = 0.1276 mol + 2x
To calculate the equilibrium concentration of H2, we need to divide the equilibrium moles by the final volume of the container (1.00 L).
Equilibrium concentration of H2 = (0.1276 mol + 2x) / 1.00 L
However, we are given the number of moles of O2 at equilibrium, which is 0.150 moles. According to the stoichiometry, 0.150 moles of O2 is equivalent to x moles of H2 gas.
0.150 moles O2 = x moles H2
From this, we can solve for x:
x = 0.150 moles of O2
Substituting this value into the equilibrium expression, we get:
Equilibrium concentration of H2 = (0.1276 mol + 2 * 0.150 mol) / 1.00 L = 0.4276 M
Therefore, the equilibrium concentration of H2 gas in the container at 412 K is approximately 0.4276 M.