1. Tardigrade
  2. Question
  3. Chemistry
  4. For the second order reaction, A + B → Products When a moles of A reacts with b moles of B , the rate equation is given by k2 t = (1/(a-b)) ln (b(a-x)/a(b-x)) When a > > b ,the rate expression becomes that of

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Q. For the second order reaction,
$ A + B \to $ Products
When $ a $ moles of $ A $ reacts with $ b $ moles of $ B $ , the rate equation is given by
$ k_{2} t = \frac{1}{\left(a-b\right)} \,ln \,\frac{b\left(a-x\right)}{a\left(b-x\right)} $
When $ a > > b $ ,the rate expression becomes that of

AMUAMU 2012Chemical Kinetics

Solution:

$A + B \to$ Products
When $a > > b$, i.e., reactant '$A$' is present in large excess, rate of reaction does not depend upon its concentration. Hence, it will be of first order.