At 906 K, 3.60 mol of ammonia gas is placed into 2.00 L vessel and allowed to decompose to the elements nitrogen and hydrogen. If Kc = 6.56 E-3 for this reactions at this temperature, calculate the equilibrium concentration of each component at equilibrium.
[NH3] = ________
[H2] = ________
[N2] = ________
To calculate the equilibrium concentrations of each component, we can use the concept of equilibrium constant (Kc) and the stoichiometry of the reaction.
The given reaction is the decomposition of ammonia (NH3) into nitrogen (N2) and hydrogen (H2):
2NH3(g) ⟶ N2(g) + 3H2(g)
We are provided with the equilibrium constant Kc = 6.56 E-3, which is expressed as:
Kc = ([N2] * [H2]^3) / [NH3]^2
Let's assign the equilibrium concentration of NH3 as x. Since NH3 decomposes to form N2 and H2 in a 2:3 ratio, the concentrations of N2 and H2 can be expressed as 2x and 3x, respectively.
Now we can plug these values into the Kc expression:
Kc = ([N2] * [H2]^3) / [NH3]^2
6.56 E-3 = (2x) * (3x)^3 / (x)^2
Simplifying the expression, we get:
6.56 E-3 = 54x^4
To solve for x, let's rearrange the equation:
x^4 = (6.56 E-3) / 54
x^4 = 1.213 E-4
Taking the fourth root of both sides, we find:
x ≈ 0.047
Now that we have the value of x, we can calculate the equilibrium concentrations of NH3, N2, and H2:
[NH3] = x = 0.047 mol/L
[H2] = 3x ≈ 0.141 mol/L
[N2] = 2x ≈ 0.094 mol/L
Therefore, the equilibrium concentrations for each component are:
[NH3] = 0.047 mol/L
[H2] = 0.141 mol/L
[N2] = 0.094 mol/L