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

1.	How do you apply law of mass action to a gaseous reversible reaction?
Answer» Solution :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.

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