In a pseudo first order reaction, the rate constant-

The correct answer is Option (4) → depends on concentration of reactants present in excess. ##

Consider a second-order reaction:

$A + B \rightarrow \text{Products}$

The actual rate law is:

$\text{Rate} = k[A][B]$

In a pseudo first-order reaction, one of the reactants (say B) is present in large excess. Because there is so much of it, its concentration $[B]$ remains practically constant throughout the reaction.

We then combine the true rate constant ($k$) and the concentration of the excess reactant ($[B]$) into a new "pseudo" rate constant ($k'$):

$\text{Rate} = (k[B])[A] = k'[A]$

Here, $k'$ is the pseudo first-order rate constant.

The Pseudo Rate Constant ($k'$): As shown above, $k' = k[B]$. This means the value of the rate constant that we measure ($k'$) is directly proportional to the concentration of the reactant in excess. If you double the amount of the excess reactant, the pseudo rate constant will also double.