For the second order reaction, A + B → Products When a moles of A reacts with b moles of B , the rate equation is given by k2 t = (1/(a-b)) ln (b(a-x)/a(b-x)) When a b ,the rate expression becomes that of

Question

For the second order reaction, A + B → Products When a moles of A reacts with b moles of B , the rate equation is given by k2 t = (1/(a-b)) ln (b(a-x)/a(b-x)) When a > > b ,the rate expression becomes that of (A) first order (B) zero order (C) unchanged, second order (D) third order

Tardigrade Admin · Accepted Answer

A + B → 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.

Q. 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
$k_{2} t = \frac{1}{( a - b )} l n \frac{b ( a - x )}{a ( b - x )}$
When $a >> b$ ,the rate expression becomes that of

first order

zero order

unchanged, second order

third order

$A + B \to$ 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.

Q. For the second order reaction, A+B→ Products When a moles of A reacts with b moles of B , the rate equation is given by k2​t=(a−b)1​lna(b−x)b(a−x)​ When a>>b ,the rate expression becomes that of