Question

The rate equation for the hydrolysis of an ester in the presence of $NaOH$ is given as, $rate=k\left[ester\right]\left[NaOH\right]$. If the concentration of $NaOH$ is increased by $100$ times than that of ester, the order of the reaction will be:

• A. 1
• B. 2
• C. 0
• D. 3

Answer 1

Answer: (A)

1

The rate expression is $rate=k\left[ester\right]\left[NaOH\right]$.

Thus, the reaction is first order with respect to the concentrations of ester and $NaOH$.

When the concentration of $NaOH$ is $100$ times the concentration of ester, it remains almost constant and the reaction becomes zero order with respect to $\left[NaOH\right]$.

Hence, it is a pseudo-first-order reaction.

$Rate=k\left[ester{]}^1\right[NaOH{]}^0$.

Option A is correct.