Calculate the equilibrium constant (K) for the ammonia synthesis reaction at
25°C and show how K is related to the partial pressures of the species at
equilibrium. Consider that the overall pressure is low enough for the gases to be treated as perfect.
N2(g) + 3H2(g) -> 2NH2(g)
you can work out K, but notice the concentraton of the species in the term: concentration is n/V which is equal to P/RT
so concentration is directly a function of Pressure.
To calculate the equilibrium constant (K) for the ammonia synthesis reaction, you need to use the expression of K with respect to the partial pressures of the species involved.
The equilibrium constant (K) expression for the reaction is written as follows:
K = (P(NH3)^2) / (P(N2) * P(H2)^3)
Here, P(NH3), P(N2), and P(H2) represent the partial pressures of NH3, N2, and H2, respectively, at equilibrium.
To calculate K, we need to know the partial pressures of NH3, N2, and H2 at equilibrium.
In this case, the reaction involves the formation of 2 moles of NH3 from 1 mole of N2 and 3 moles of H2. So, the stoichiometric ratio of NH3 to N2 is 2:1, and the stoichiometric ratio of NH3 to H2 is 2:3.
Let's assume that the equilibrium partial pressure of NH3 is P(NH3), the equilibrium partial pressure of N2 is P(N2), and the equilibrium partial pressure of H2 is P(H2).
Since the overall pressure is low enough to treat the gases as perfect, we can say that the partial pressure of each gas is directly proportional to their respective molar amounts.
Therefore, we can write:
P(NH3) = 2x (x is the molar amount of NH3 at equilibrium)
P(N2) = 1x (x is the molar amount of N2 at equilibrium)
P(H2) = 3x (x is the molar amount of H2 at equilibrium)
Now substitute these partial pressure values into the equation for K:
K = (P(NH3)^2) / (P(N2) * P(H2)^3)
K = (2x)^2 / (1x * (3x)^3)
K = (4x^2) / (27x^3)
K = 4/27x (cancel out x)
So, the equilibrium constant (K) for the ammonia synthesis reaction is given by K = 4/27x, where x is the molar amount of NH3, N2, or H2 at equilibrium.