The following equilibrium concentrations were observed for the Haber process at 127 oC:
[NH3] = 3.1 x 10-2 mol L-1
[N2] = 8.5 x 10-1 mol
[H2] = 3.1 x 10-3 mol
Calculate the value of K at 127 oC for this reaction
for the concentrations I meant
[NH3] = 3.1 x 10^-2 mol
[N2] = 8.5 x 10^-1 mol
[H2] = 3.1 x 10^-3 mol
Concentration is not measured in moles.
I meant mol/L^-1 or whatever the unit is. That's what the question had.
Umm can you please help me with this question?
To calculate the value of K at 127°C for the Haber process, we need to use the equilibrium concentrations of the reactants and products.
The balanced equation for the Haber process is:
N2 + 3H2 ⇌ 2NH3
The equilibrium constant expression, K, is given by:
K = [NH3]^2 / ([N2] * [H2]^3)
Now, let's substitute the given equilibrium concentrations into the expression:
[NH3] = 3.1 x 10^-2 mol L^-1
[N2] = 8.5 x 10^-1 mol
[H2] = 3.1 x 10^-3 mol
K = (3.1 x 10^-2)^2 / ((8.5 x 10^-1) * (3.1 x 10^-3)^3)
K = (9.61 x 10^-4) / (2.63 x 10^-4)
K = 3.65
Therefore, the value of K at 127°C for the Haber process is approximately 3.65.