9.0 moles of pure ammonia gas were injected into a 3.0 L flask and allowed to reach equilibrium according to the equation shown below. If the equilibrium mixture was analyzed and found to contain 3.0 moles of hydrogen gas, calculate the value of the equilibrium constant.
2NH3(g) <====> 2N2(g) + 3H2(g)
To calculate the value of the equilibrium constant (Kc) for the given reaction, you need to use the stoichiometry of the reaction and the concentrations of the reactants and products at equilibrium.
First, let's define the initial and equilibrium concentrations of each species:
Initial concentrations:
NH3 = 9.0 moles / 3.0 L = 3.0 M
N2 = 0 M
H2 = 0 M
Equilibrium concentrations:
NH3 = 3.0 M - x (assuming x moles of NH3 reacted)
N2 = 2x (since the stoichiometry of the reaction is 2:2:3)
H2 = 3.0 M + 3x (since the stoichiometry of the reaction is 2:2:3)
From the given information, we know that the equilibrium mixture contains 3.0 moles of H2. Therefore, we can substitute this value into the equation for the equilibrium concentration of H2:
3.0 M + 3x = 3.0 M
3x = 0
x = 0
Since x = 0, it means that no NH3 has reacted to form N2 and H2, and thus the equilibrium concentrations of all species remain as follows:
Equilibrium concentrations:
NH3 = 3.0 M
N2 = 0 M
H2 = 3.0 M
Now, we can calculate the equilibrium constant (Kc) using the formula:
Kc = (concentration of products) / (concentration of reactants)
Kc = ([N2]^2 * [H2]^3) / [NH3]^2
Substituting the values:
Kc = (0^2 * (3.0)^3) / (3.0^2)
Kc = 0 / 9.0
Kc = 0
Therefore, the value of the equilibrium constant (Kc) for this reaction is zero.