Question
With the help of an example explain what is meant by pseudo first order reaction.
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Answer
Pseudo first order reaction: Reaction which appears to be a second order reaction, but actually is first order reaction is called pseudo first order reaction. This condition occurs in a chemical reaction between two substances when one reactant is present in large amount. The concentration of reactant present in excess does not get altered much during the course of the reaction. Due to this reaction behaves as first order reaction.
E.g., hydrolysis of esters.
$\text{CH}_3\text{COOHC}_2\text{H}_5+\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{CH}_3\text{COOH}+\text{C}_2\text{H}_5\text{OH}$
Rate law for this reaction is; rate $=\text{K}[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]$
But the concentration of water does not change during the course of the reaction. So; $[\text{H}_2\text{O}]$ is constant.
Therefore rate $=\text{K}_1[\text{CH}_3\text{COOC}_2\text{H}_5].$ Where $\text{K = K}\text{[H}_2\text{O}]$
The reaction behaves as a first order reaction.
E.g., hydrolysis of esters.
$\text{CH}_3\text{COOHC}_2\text{H}_5+\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{CH}_3\text{COOH}+\text{C}_2\text{H}_5\text{OH}$
Rate law for this reaction is; rate $=\text{K}[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]$
But the concentration of water does not change during the course of the reaction. So; $[\text{H}_2\text{O}]$ is constant.
Therefore rate $=\text{K}_1[\text{CH}_3\text{COOC}_2\text{H}_5].$ Where $\text{K = K}\text{[H}_2\text{O}]$
The reaction behaves as a first order reaction.
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