Chemical Kinetics and Nuclear ChemistryHard

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

The hypothetical reaction : 2A + B → C + D is catalyzed by E as indicated in the possible mechanism below -Step-I ...The colloid is :...$A \rightarrow D$ is an endothermic reaction occurring in three steps (elementary).(i) $A \rightarrow B\Delta H_{i} = + ...Consider the following consecutive first-order reaction. $A\overset{\quad K_{1}\quad}{\rightarrow}B\overset{\quad K_{2}\...The following statement (s) is (are) correct...