Question

$A+B\to C+D$
Rate law for the given reaction is given as $k\left[A\right]\left[B\right]$.
If B is taken in excess then the rate of reaction will become:
Here,
$k$ is the rate constant.
$k^{\prime }=k\left[B\right]$

• A. $Rate=k\left[A\right]\left[B\right]$
• B. $Rate=k^{\prime }\left[B\right]$
• C. $Rate=k^{\prime }\left[A\right]$
• D. $Rate=k.k^{\prime }\left[A\right]$

Answer 1

The correct option is C $Rate=k^{\prime }\left[A\right]$

Pseudo order reaction:
A reaction whose order is different from the actual due to excess concentration of one of the reactants is called pseudo order reaction.
Depending upon the conditions, a second order reaction can behave as a first order reaction.

For a $2^{nd}$ order reaction,
$A+B\to C+D$
$Rate,R=k\left[A\right]\left[B\right]$

If the concentration of reactant, B, is taken in large excess. Change in concentration of B is negligible during the reaction. So, it can be considered as constant.
Now, reaction rate depends on concentration of the other reactant (A) only. Thus, the reaction becomes a first order reaction.
Hence,
$Rate,R=k^{\prime }\left[A\right]$
where,
$k^{\prime }=k\left[B\right]$
Here, the reactions is actually second order but behave as first order reactions. This reaction is called pseudo first order reaction.

Examples of pseudo first order reaction:
1. Hydrolysis of ethyl acetate
$C{H}_{3}COO{C}_2{H}_5+H_{2}O\stackrel{H^{+}}{⟶}C{H}_{3}COOH+C_2{H}_{5}OH$

Here $H_{2}O$ will be in excess so the rate of the reaction is

$Rate,R=k^{\prime }\left[C{H}_{3}COO{C}_2{H}_5\right]$
where,
$k^{\prime }=k\left[H_{2}O\right]$

2. Hydrolysis of cane sugar
$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \underset{Glucose}{C_6{H}_{12}{O}_6}+\underset{Fructose}{C_6{H}_{12}{O}_6}$
Here $H_{2}O$ will be in excess so the rate of the reaction is

$Rate,R=k^{\prime }\left[C_{12}{H}_{22}{O}_{11}\right]$
where,
$k^{\prime }=k\left[H_{2}O\right]$