1.The equilibrium concentration of `[B]_(e)` for the reversible reaction A ⇌ B can be evaluated by the expression:- (1) `K_(c)[A]_(e)^(-1),` (2) `(k_(f))/(k_(b))[A]_(e)^(-1)`
(3) `k_(i)k_(b)^(-1)[A]_(e),`
(4) `k_(b)[A]^(-1)`

What is the order of a reaction which has a rate expression, i.e. rate = k[A]^(3//2)[B]^(-1) ?

Two consecutive irreversible fierst order reactions can be represented by Aoverset(k_(1))(rarr)Boverset(k_(2))(rarr)C The rate equation for A is readily interated to obtain [A]_(t)=[A]_(0).e^(-k_(1^(t))) , and [B]_(t)=(k_(1)[A]_(0))/(k_(2)-k_(1))[e^(-k_(1)^(t))-e^(-k_(2)^(t))] When k_(1)=1s^(-1) and k_(2)=500s^(-1) , select most appropriate graph

Two consecutive irreversible fierst order reactions can be represented by Aoverset(k_(1))(rarr)Boverset(k_(2))(rarr)C The rate equation for A is readily interated to obtain [A]_(t)=[A]_(0).e^(-k_(1^(t))) , and [B]_(t)=(k_(1)[A]_(0))/(k_(2)-k_(1))[e^(-k_(1)^(t))-e^(-k_(2)^(t))] At what time will B be present in maximum concentration ?

Two consecutive irreversible fierst order reactions can be represented by Aoverset(k_(1))(rarr)Boverset(k_(2))(rarr)C The rate equation for A is readily interated to obtain [A]_(t)=[A]_(0).e^(-k_(1^(t))) , and [B]_(t)=(k_(1)[A]_(0))/(k_(2)-k_(1))[e^(-k_(1)^(t))-e^(-k_(2)^(t))] Select the corret statement for given reaction:

For A+BhArrC+D , the equilibrium constant is K_(1) and for C+DhArA+B , the equilibrium constant is K_(2) . The correct relation between K_(1) and K_(2) is

For the consecutive unimolecular-type first-order reaction A overset(k_(1))rarr R overset(k_(2))rarr S , the concentration of component R, C_( R) at any time t is given by - C_(R ) = C_(OA)K_(1)[e^(-k_(1)t)/((k_(2)-k_(1))) +e^(-k_(2)t)/((k_(1)-k_(2)))] if C_(A) = C_(AO), C_(R ) = C_(RO) = 0 at t = 0 The time at which the maximum concentration of R occurs is -

For a reaction of reversible nature, net rate is ((dx)/(dt))=k_(1)[A][B]-k_(2)[C] hence, given reaction is :