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.
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