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Pseudo Order Reactions

Reactions that appear to be of higher order but follow first-order kinetics.

By chemca Team • Updated Jan 2026

A Pseudo First Order Reaction is a chemical reaction that behaves like a first-order reaction, even though the stoichiometry suggests a higher order (usually second order). This happens when one of the reactants is present in large excess.

1. The Concept & Condition

Why "Pseudo"?

Consider a general reaction:

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

The expected rate law is:

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

Order of Reaction = 1 + 1 = 2

The Special Case: If Reactant B is present in large excess (e.g., Solvent), its concentration remains practically constant throughout the reaction ($[B] \approx \text{constant}$).

2. Mathematical Derivation

Since $[B]$ is constant, we can combine it with the true rate constant $k$.

$$ \text{Rate} = k [A] [B]_{excess} $$ $$ \text{Let } k' = k [B]_{excess} $$ $$ \text{Rate} = k' [A] $$

Here, $k'$ is the Pseudo First Order Rate Constant.
The reaction now behaves as a First Order reaction with respect to A.

Important Relationship: $$ k_{true} = \frac{k_{pseudo}}{[B]_{excess}} $$ The unit of $k'$ is $time^{-1}$, while the unit of true $k$ is $L \, mol^{-1} \, time^{-1}$.

3. Classic Examples

A. Acid Catalyzed Hydrolysis of Ester

Hydrolysis of Ethyl Acetate in excess water.

$$ CH_3COOC_2H_5 + H_2O \text{ (excess)} \xrightarrow{H^+} CH_3COOH + C_2H_5OH $$

$$ \text{Rate} = k' [CH_3COOC_2H_5] $$ Where $k' = k[H_2O]$.

B. Inversion of Cane Sugar

Hydrolysis of Sucrose (Cane sugar) into Glucose and Fructose in excess water.

$$ C_{12}H_{22}O_{11} + H_2O \text{ (excess)} \xrightarrow{H^+} C_6H_{12}O_6 \text{ (Glucose)} + C_6H_{12}O_6 \text{ (Fructose)} $$

$$ \text{Rate} = k' [\text{Sucrose}] $$

4. Calculation & Monitoring

Monitoring the Reaction

Since these reactions often involve changes in optical rotation (sugar) or acidity (ester hydrolysis), the rate constant is calculated using physical properties.

For Ester Hydrolysis (Titration with NaOH):

$$ k' = \frac{2.303}{t} \log \left( \frac{V_\infty - V_0}{V_\infty - V_t} \right) $$

$V_t$ = Volume of NaOH used at time t.

For Inversion of Sugar (Polarimeter):

$$ k' = \frac{2.303}{t} \log \left( \frac{r_0 - r_\infty}{r_t - r_\infty} \right) $$

$r_t$ = Angle of rotation at time t.

Knowledge Check

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