Calculate K at 298 K for the following reactions.
NO(g) + 1/2 O2(g)......> NO2(g)
Nog 86.60
No2 g 51
o2.... O
I got 6.81 x 10^-7........thank you.
To calculate the equilibrium constant (K) at 298 K for the given reaction, we first need to determine the molar concentrations of each species at equilibrium.
Let's assume that the initial concentration of NO (g) is x mol/L, the initial concentration of O2 (g) is y mol/L, and the initial concentration of NO2 (g) is z mol/L.
From the balanced equation:
NO (g) + 1/2 O2 (g) → NO2 (g)
We can see that the stoichiometric coefficient of NO is 1, the stoichiometric coefficient of O2 is 1/2, and the stoichiometric coefficient of NO2 is 1.
At equilibrium, the concentrations become:
NO (g): x - Δx
O2 (g): y - Δ(1/2)x
NO2 (g): z + Δx
Where Δx is the change in concentration at equilibrium.
Now, let's use the given values:
Initial concentration of NO (g) = 86.60 mol/L
Initial concentration of NO2 (g) = 51 mol/L
Initial concentration of O2 (g) = not provided
Since the initial concentration of O2 (g) is not provided, we cannot directly calculate K. We need the initial concentration of all reactants to calculate the equilibrium constant.
Therefore, we cannot determine the value of K at 298 K for the given reaction with the information provided.
If you have the initial concentration of O2 (g), please provide it so we can assist you further in calculating the equilibrium constant.