What is the Kc for the following equilibrium at 500∘C,
N2(g)+3H2(g)⇌2NH3(g)
[H2] = 0.50M , [N2] = 0.50M , [NH3] = 2.0M ?
To find the value of Kc for the given equilibrium reaction, you need to use the concentrations of reactants and products. The general expression for the equilibrium constant (Kc) is:
Kc = ([NH3]^2) / ([N2] * [H2]^3)
Substituting the given concentrations:
Kc = (2.0M)^2 / (0.50M * (0.50M)^3)
Simplifying:
Kc = 4.0M / (0.125M^3)
Kc = 4.0M / (0.125M * 0.125M * 0.125M)
Kc = 4.0M / (0.001953125M)
Kc = 2048
Therefore, the value of Kc for the given equilibrium at 500∘C is 2048.
To find the value of Kc for the given equilibrium, you need to know the concentrations of all the reactants and products at the given temperature. Given that [H2] = 0.50 M, [N2] = 0.50 M, and [NH3] = 2.0 M, you can use these values to calculate the Kc.
The formula for Kc is calculated as follows:
Kc = ([NH3]^2)/([N2][H2]^3)
Substituting the given values into the equation, we get:
Kc = (2.0^2)/((0.50)(0.50^3))
Kc = 4/((0.50)(0.50^3))
Kc = 4/(0.50)(0.125)
Kc = 4/(0.0625)
Kc = 64
Therefore, the value of Kc for the given equilibrium at 500∘C is 64.
(Eh? Gas activity measured as molar concentration instead of partial pressure? Odd, but...)
Kc = {NH3]^2 / [H2]^3 [N2]
= (2.0M)^2 / (0.50M)^3 (0.50M)
= 64 M^(-2)