Question

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

• A. first order
• B. zero order
• C. unchanged, second order
• D. third order

Answer 1

Answer: (A)

first order

Answer 2

$A + B \to$ 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.