For the reaction A overset(K_(1))underset(K_(2))(hArr)B, if 'a' is the initial concentration of A and n is the number of moles of A reacting if initially no B was present then :

For the reaction A overset(K_(1))underset(K_(2))(hArr)B, if 'a' is the

1.	For the reaction A overset(K_(1))underset(K_(2))(hArr)B, if 'a' is the initial concentration of A and n is the number of moles of A reacting if initially no B was present then :
Answer» `(DX)/(DT)=[K_(1)(a-x)-K_(2)x]` initially `K_(1)(a-x_(e ))-K_(2)x_(e )=0` at equilibrium `K_(1)+K_(2)=K_(1).(a)/(xe)` `(K_(1)+K_(2))=(1)/(t) ln.(x_(e ))/(x_(e)-x)` Solution :Initially, `(dx)/(dt)=K_(1)(a-x)-K_(2)x`………`(1)` Hence choice `(A)` is correct. If `x_(e)=` equilibrium concentration of B, where net rate `=0` `K_(1)(a-x_(e ))-K_(2)x_(e)=0`………..`(2)` Hence choice (B) is correct. From EQUATION `(2)` , `K_(2)=(K_(1)(a-x_(e )))/(x_(e ))` Substituting the value of `K_(2)` in equation `(1)`, we get `(dx)/(dt)=K_(1)(a-x)-K_(1)((a-x_(e ))x)/(x_(e ))` `=K_(1)((x_(e )-x)a)/(x_(e ))` `IMPLIES(dx)/((x_(e )-x))=(a)/(x_(e ))K_(1)dt` Hence by integrating and further calculation, we get `(K_(1)+K_(2))=K_(1)(a)/(x_(e ))=(1)/(t)ln.(x_(e ))/((x_(e )-x))`