.924moles of A(g) is placed in 1L at 700C where 38.8% dissociated when equilibrium was established.
3A <--> 5B + 2C
What is the value of the equilibrium constant at the same temperature?
I keep getting .022
To find the value of the equilibrium constant at the given temperature, you need to use the concentration values of the reactants and products at equilibrium.
First, calculate the initial moles of A(g) placed in the 1L reactor:
Initial moles of A(g) = 0.924 moles
Next, determine the moles of A(g) that dissociated at equilibrium:
Dissociation of A(g) = 38.8% of initial moles = 0.388 * 0.924 moles = 0.358512 moles
After dissociation, you have the following molar quantities at equilibrium:
[A(g)] = (0.924 - 0.358512) moles = 0.565488 moles
[B] = 5 * 0.358512 moles = 1.79256 moles
[C] = 2 * 0.358512 moles = 0.717024 moles
Now, substitute the concentrations into the equilibrium expression for the given reaction:
Kc = ([B]^5 * [C]^2) / [A]^3
Kc = (1.79256^5 * 0.717024^2) / 0.565488^3
Calculating this expression will give you the value of the equilibrium constant (Kc) at 700°C.