Need help with Chemistry, thanks in advance!
For the system: 2H2O(g) ↔ 2H2(g) + O2(g) ; at equilibrium 27.0g of H2O, 0.25g of H2 and 16.0g of O2 was found in a 5.0L flask. Calculate the equilibrium constant, Keq.
To calculate the equilibrium constant, Keq, you need to use the concentration of each compound at equilibrium.
First, we need to convert the given masses of each compound to their respective molar amounts. We can do this by dividing the mass of each compound by their molar masses:
Molar mass of H2O = 18 g/mol
Molar mass of H2 = 2 g/mol
Molar mass of O2 = 32 g/mol
The molar amounts (in moles) are:
n(H2O) = 27.0 g / 18 g/mol = 1.5 mol
n(H2) = 0.25 g / 2 g/mol = 0.125 mol
n(O2) = 16.0 g / 32 g/mol = 0.5 mol
Next, we need to calculate the molar concentrations (in mol/L) by dividing the moles of each compound by the volume of the flask:
c(H2O) = n(H2O) / V = 1.5 mol / 5.0 L = 0.3 mol/L
c(H2) = n(H2) / V = 0.125 mol / 5.0 L = 0.025 mol/L
c(O2) = n(O2) / V = 0.5 mol / 5.0 L = 0.1 mol/L
Now, we have the molar concentrations of each compound at equilibrium. For the reaction equation 2H2O(g) ↔ 2H2(g) + O2(g), the equilibrium constant, Keq, is defined as:
Keq = (c(H2)^2 * c(O2)) / (c(H2O)^2)
Substituting the values we calculated:
Keq = (0.025 mol/L)^2 * 0.1 mol/L / (0.3 mol/L)^2
Simplifying:
Keq = 0.000625 mol^2/L^3 / 0.09 mol^2/L^3
Calculating:
Keq = 0.00694
So, the equilibrium constant, Keq, for the given system is approximately 0.00694.
M H2O - (27.0/18/5) = ?
M H2 = (0.25/2/5) = ?
M O2 = (16.0/32/5) = ?
But note that I don't think those gases will dissolve in that little water.
....................2H2O(g) ↔ 2H2(g) + O2(g)
E........................?..............?..............? from above.
Write the Kc expression and substitute the concns into Kc and calculate.