Pseudo Order Reaction and Rate Constant

Pseudo order reaction: It is defined as the rate of reaction which seems to be of higher order but follows lower order kinetics.
For examples:
A + B → Product
The rate la for this reaction can be written as
Rate = K [A]∙[B]
If one of the reactant say reactant B is present in large amount, then its concentration almost remains constant. IN such case, the rate of reaction depends upon the reaction which is present in the smaller amount i.e., reactant A.
The rate law for the above reaction can be written as:
Rate = K'[A]
where K = K'[B]
Actually, it seems to be of 2and order kinetics, but follows 1st order kinetics. Hence, this reaction is called the pseudo order reaction.
For example:
Hydrolysis of ester: Esters like ethyl acetate upon hydrolysis in the presence of mineral acid as catalyst give acetic acid.
$$\ce{CH3COOC2H5 + $\underset{\text{Excess}}{\ce{H2O}}$ ->CH3COOH + C2H5OH}$$
Here,
Rate = K [CH3COOC2H5]. [H2O]
Rate = K [CH3COOC2H5]
Where K = k.H2O
Here, H2O is present in large excess. So, its concentration remains almost same or constant. In such case the rate of reaction depend upon that reactant which is present in small amount i.e., reactant ethyl acetate here.

Rate constant:

It is defined as the rate of reaction when the concentration of reactant is taken as unity. Consider a general reaction:
$${\text{A} \rightarrow \text{Product}}$$
$$\text{Rate} = \text{K[A]}$$
When [A] = 1,
Rate = K
When the concentration of A increases, the rate of reaction also increases and the value of K remains constant.
For zero order reaction,
$${\text{Rate} = \text{K}}$$
Rate = K
K = Rate = mol l-1s-1
For 1st order reaction, when m = 1,
$${\text{Rate} = \text{K}{{[A]}}}$$
$$\text{K} = \frac{\text{Rate}}{{[A]}} = \frac{molL^{-1}s^{-1}}{molL^{-1}} = {s^{-1}}$$
For 2nd order reaction, when m = 2,
$${\text{Rate} = \text{K}{{[A]^{2}}}}$$
$$\text{K} = \frac{\text{Rate}}{{[A]^{2}}} = \frac{molL^{-1}s^{-1}}{mol.molL^{-1}L^{-1}} = {mol^{-1}Ls^{-1}}$$
For 3rd order kinetics when m =3,
$${\text{Rate} = \text{K}{{[A]^{3}}}}$$
$$\text{K} = \frac{\text{Rate}}{{[A]^{3}}} = \frac{molL^{-1}s^{-1}}{mol.mol.mol.L^{-1}L^{-1}L^{-1}} = {mol^{-1}L^{-2}s^{-1}}$$
Zero order reaction: It is defined as the chemical reaction in which the concentration of reactant do not vary with time. For this let us consider a general reaction.

$${\text{A} \rightarrow \text{Product}}$$
$$\text{Rate} = \text{K}_{\circ}[A]^0 $$
Where,

$${\text{K}_{\circ}[A] = \text{rate constant for zero order reaction}}$$
then,
$$\frac{\text{dX}}{dT} = \text{K}_{\circ}….(i) $$
This is the differential rate law for the zero order reaction.

$${\text{A} \rightarrow \text{Product}}$$
$$\text{Rate} = \text{K}_{\circ}[A]^0 $$
Where,

$${\text{K}_{\circ}[A] = \text{rate constant for zero order reaction}}$$
then,
$$\frac{\text{dX}}{dT} = \text{K}_{\circ}….(i) $$

This is the differential rate law for the zero order reaction.

Upon integration:

$$\int{dx} = \text{K}_{\circ}\int{dt}$$

$$\text{x} = \text{K}_{\circ}{t} + I…… (ii)$$
The integration constant can be calculated as:
When x = 0, t = 0 then I = 0.

Substituting the value of I in equation (ii), we get:
$$\text{x} = \text{K}_{\circ}{t} + I$$

$$\text{K}_{\circ} = \frac{\text{X}}{t}…….(iii)$$
The equation (iii) is the integrated equation.

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