1.

For the second order reaction , A + B toProducts When a moles of A reacts with b moles of B , the rate equation is given by k_(2) t= (1)/((a-b)) "ln" (b(a-x))/(a(b-x)) When a gt gt b , the rate expression becomes that of

Answer»

First ORDER
Second order
Unchanged , second order
Third order

Solution :`k_(2) t = (1)/((a-b))` ln `(b(a-x))/(a(b-x))`
` GT gt b` means (a-x) = a and (a-b) =a
i.e., `k_(2) t = (1)/(a)` ln `(ba)/(a(b-x)) implies k_(2) t = (1)/(a)` ln `(b)/((b-x))`
(SINCE 'a' being very large , may be treated as constant after the change) . Thus , the reaction FOLLOWS first order kinetics .


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