Autocatalysis is the catalysis of a reaction by the products. For example, for a reaction A -> P it may be found that the rate law is v = k[A] [P] and the reaction rate is proportional to the concentration of P. The reaction gets started because there are usually other reaction routes for the formation of some P initially, which then takes part in the autocatalytic reaction proper.
(a) Integrate the rate equation for an autocatalytic reaction of the form A -> P, with rate law v = k[A][P], and show that where a = ([A]o + [P]o)k and b= [P]o/[A]o Hint. Starting with the expression v =-d[A]/dt= k[A][P], write [A] = [A], -x, [P] = [P]o + x and then write the expression for the rate of change of either species in terms of x. To integrate the resulting expression, the following relation will be useful:
(b) Plot [Pl/[P]o against at for several values of b. Discuss the effect of autocatalysis on the shape of a plot of [P]/[P]o against t by comparing your results with those for a first-order process, in which [P]/[P]o = 1- e-kt.
(c) Show that, for the autocatalytic process discussed in parts (a) and (b), the reaction rate reaches a maximum at t max = -(l/a) In b.
(d) An autocatalytic reaction A -> P is observed to have the rate law d[PJ/dt= k[A]2[p]. Solve the rate law for initial concentrations [A], and [P]o. Calculate the time at which the rate reaches a maximum.
(e) Another reaction with the stoichiometry A -> P has the rate law d[P]/dt= k[A] [p]2; integrate the rate law for initial concentrations [A]o and [P]o. Calculate the time at which the rate reaches a maximum.