Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

(a) O2 (g)>2/3O3(g) Kc=5.77x10^-9

(b) 3O2 (g)>2O3(g) Kc=?

The value of equilibrium constant for new reaction is

Explanation:

The given chemical equation follows:

The equilibrium constant for the above equation is

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

Hence, the value of equilibrium constant for new reaction is

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.