Chemical Kinetics and Nuclear ChemistryMedium

Question

For the second-order reaction A + B → Products, the rate constant k, is given as$k = \frac{2.303}{(a - b)t}\log\frac{b(a - x)}{a(b - x)}$, where, a and b are the initial concentrations of ‘A’ and ‘B’ and x is the change in concentration after time t. If b >> a, the reaction reduces to

Options

A.first-order with respect to ‘A’.
B.zero-order with respect to ‘A’.
C.first-order with respect to ‘B’.
D.overall zero-order.

Solution

$(b - a) \simeq b\text{ and }(b - x) \simeq b$

$\therefore Kt = \frac{2.303}{a - b}.\ln\frac{b(a - x)}{a(b - x)} \simeq \frac{2.303}{( - b)}.\ln\frac{b(a - x)}{ab} \text{Or, }(b.K).t = 2.303.\ln\frac{a}{a - x}$

∴ Reaction becomes first-order with respect to A.

Create a free account to view solution

View Solution Free
Topic: Chemical Kinetics and Nuclear Chemistry·Practice all Chemical Kinetics and Nuclear Chemistry questions

More Chemical Kinetics and Nuclear Chemistry Questions

Read the following industrial methods for the preparation of H2SO4 and answer the question at the end. Professor Molina ...The rate constant of a reaction depends on :...For the first-order reaction t99% = x × t90%. The value of ‘x’ will be...t1/4 can be taken as the time taken for the concentration of a reactant to drop to of its initial value. If the rate con...The order of reaction A → Products, may be given by which of the following expression(s)?...