How do you apply law of mass action to a gaseous reversible reaction?

Explanation

If the reactants and products are in gaseous state, then for the reaction,
`A+Bunderset(V_(b))overset(V_(f))(hArr)C+D`
The total pressure `P=P_(A)+P_(B)+P_(C)+P_(D)`
If `P_(A),P_(B),P_(C)andP_(D)` represent the partial pressures exerted by A, B, C and D respectively, then according to the law of mass action,
Velocity of forward reaction `V_(f)` is proportional to `P_(A) P_(B)` or
`V_(f)=K_(f)P_(A)P_(B)" ".....(3)`
where `K_(f)` is the velocity (proportionality) constant of forward reaction.
Velocity of backward reaction `V_(b)propP_(C)P_(D)or`
`V_(b)=K_(b)P_(C)P_(D)" "....(4)`
where `K_(b)` is the velocity (proportionality) constant of backward reaction. At equilibrium, velocity of the forward reaction is equal to the velocity of the backward reaction.
i.e., `V_(f)=V_(b)` then from (3) and (4)
`K_(f)P_(A)P_(B)=K_(b)P_(C)P_(B)orK_(f)/K_(b)=K_(p)=(P_(C)P_(D))/(P_(A)P_(B))`
Where `K_(p)` is equilibrium constant with respect to the partial pressures of the reactants.