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Chemistry > Chemical Kinetics > Pseudo First Order Reaction
Chemical Kinetics

Pseudo First Order Reaction

Sometimes adults ‘fake’ being younger when they’re actually way older. Isn’t it? Did you know we can ‘fake’ in chemistry as well? Yes, a reaction whose order of reaction is actually more than one can ‘fake’ being a first order reaction. This type of reaction is known as a pseudo first order reaction. Let’s understand what a pseudo first order reaction is and how it ‘fakes’ being a first order reaction.

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Pseudo First Order Reaction
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Pseudo First Order Reaction

We know that order of reaction depends on the dependency of the rate of reaction on the concentration of reactants. That is if the rate is independent of the concentrations of reactants, the order of reaction is zero. Similarly, if the rate of reaction is proportional to the first power of the concentration of the reactant, then the order of reaction is one.

But sometimes the order of a reaction can be made to alter by changing the concentration of the reactants, that is, increasing or decreasing the concentration of one or the other reactant.

Learn about the Temperature dependence of the Rate of reaction.

order of reaction

(Source: slideshare.net)

A reaction which is not first-order reaction naturally but made first order by increasing or decreasing the concentration of one or the other reactant is known as Pseudo first order reaction. Pseudo means ‘fake’.

Therefore, it is clear from the name itself that it is not first-order reaction by nature. The order of reaction is made one by altering certain conditions. Let’s understand more about pseudo-first order reactions with some examples.

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Learn 5 Main types of Organic Reaction here.

Examples of Pseudo First Order Reaction

  • Consider a reaction in which one reactant is in excess, say hydrolysis of ethyl acetate. During the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, the reaction goes like this,

CH3COOC2H5   + H2O     →   CH3COOH    +    C2H5OH

Time CH3COOC2H5 H2O CH3COOH C2H5OH
t = 0 0.01 mol 10 mol o mol 0 mol
t 0 mol 9.9 mol 0.01 mol 0.01 mol

As the concentration of water does not get altered much during the process ( because it is in excess), we can take it as constant. So in the rate equation, Rate = k′ [CH3COOC2H5] [H2O]  the term [H2O] can be taken constant. Thus, the equation becomes

Rate =  k [CH3COOC2H5] where k = k′[H2O]

Therefore, the order of reaction now becomes one, that is the reaction is now first order reaction. Such reactions are called pseudo-first order reactions.

  • Another example of pseudo first-order reaction is the inversion of cane sugar.

C12H22O11     +      H2O     →       C6H12O6      +      C6H12O6

Cane sugar                                    Glucose                Fructose

Rate = k[C12H22O11]

order of reaction

(Source: en.wikipedia.org)

In pseudo-first order reactions, we are basically isolating a reactant by increasing the concentration of the other reactants. When the other reactants are in excess, change in their concentrations does not affect the reaction much, Therefore, now the reaction only depends on the concentration of the isolated reactant. The concentrations of all the other reactants are taken as constant in the rate law. Thus, the order of reaction becomes one.

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Pseudo First Order Reaction

Difference Between Pseudo First Order Reaction and First Order Reaction

The main difference is obviously that, in a first order reaction, the order of reaction is one by nature. A pseudo first-order reaction is second order reaction by nature but has been altered to make it a first order reaction.

The second difference is that in a first order reaction, the rate of reaction depends on all the reactants whereas, in a pseudo-first-order reaction, the rate of reaction depends only on the isolated reactant since a difference in concentration of the reactant in excess will not affect the reaction.

A Solved Question for You

Q: C12H22O11 + H2O  →   C6H12O6  + C6H12O6

The rate law for the above equation is:

a) r = k[C12H22O11] [H2O]        b) r = k[C12H22O11]

c) r = k[H2O]                                d) r = k[C12H22O11] [H2O]2

Solution: b) r = k[C12H22O11]. In the reaction,  C12H22O11 + H2O  →   C6H12O6  + C6H12O6 , water is present in large excess. Hence, it will not appear in the rate law expression. Thus, it is an example of pseudo first order reaction.

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