Question

For second order kinetics:
(A) $ K=\dfrac{2.303}{t}\log \dfrac{b(a-x)}{a(b-x)} $
(B) $ K=\dfrac{2.303}{t(a-b)}\log \dfrac{b(a-x)}{a(b-x)} $
(C) $ K=\dfrac{2.303}{t(a-b)}\log \dfrac{a(a-x)}{b(b-x)} $
(D) $ K=\dfrac{2.303}{t(b-a)}\log \dfrac{a(a-x)}{b(b-x)} $

Answer 1

Hint Rate is defined as the speed at which a chemical reaction occurs. Rate is generally expressed in the terms of concentration of reactant which is consumed during the reaction in a unit of time or the concentration of product which is produced during the reaction in a unit of time.

Complete step by step solution:
A reaction is a second order reaction if the rate of the reaction is determined but the variations of the two concentration terms. For the second order kinetics, when the concentration of both reactants is different.
Consider a reaction:
\[A+B\to product\]
The initial concentration of both the reactant and product be a and b respectively, the concentration after time t will be $ (a-x)\text{ and (b-x)} $ . Hence, the expression for k will be $ K=\dfrac{2.303}{t(a-b)}\log \dfrac{b(a-x)}{a(b-x)} $

Hence, the correct answer is option (B).

Additional information:
We can also calculate the activation energy by using various methods. It can be calculated using the Arrhenius equation and also when then two temperatures and the rate constant at both temperatures are known. The temperature should be converted to kelvin while calculating activation energy using the Arrhenius equation.

Note: If the reaction is a third order reaction, the unit for third order reaction is $ {{M}^{-2}}h{{r}^{-1}} $ . The negative and positive sign in the expression of the rate or reaction only means the change in concentration. A negative charge indicates that the concentration of the reactant is decreasing, similarly a positive charge means that the concentration of product is increasing.