The following reaction is at equilibrium at a particular temperature: `AB(g)hArr A(g)+B(g)`. Show that when the compound AB gets 50% dissociated, then the total pressure of the system becomes three times the numerical value of `K_(P)`.

Explanation

Let the initial number of moles of `AB(g)=1`. If AB(g) gets 50% dissociated at equilibrium, then according to the equation, at equilibrium the number of moles of AB will (1-0.5)=0.5 and the number of moles of each of A and B will be 0.5.
Therefore, total number of moles at the equilibrium mixture=0.5+0.5+0.5=1.5. if the total pressure of the mixture at equilibrium be P, then the partial pressures of different substances in the equilibrium mixture will be: `p_(AB)=((0.5)/(1.5))P(P)/(3),p_A=((0.5)/(0.3))P=(P)/(3),p_(B)=((0.5)/(1.5))P=(P)/(3)`
Equilibrium constant,, `K_(p)=(p_(A)xxp_(B))/(p_(AB))=((P)/(3)xx(P)/(3))/((P)/(3))=(P)/(3)`
`therefore P=3xxK_(p)`
So, the total pressure of the equilibrium mixture is 3 times the numerical value of `K_(p)`.