Question

What do you understand by the ‘order of a reaction’? Identify the reaction order from each of the following units of reaction rate constant.
A) ${{\text{L}}^{ - 1}}{\text{ mol }}{{\text{s}}^{ - 1}}$
B) ${\text{L mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$

Answer 1

Hint: The sum of the powers of concentration of the reactants in the rate equation of the chemical equation is known as the order of the reaction. It is always derived experimentally not theoretically.

Complete step by step answer:
Order of reaction: The exponent to which the concentration term of a specific reactant is raised in the rate law is known as the order of the reaction with respect to that reactant. This is the partial order.
The sum of the powers of concentration of the reactants in the rate equation of the chemical equation is known as the overall order of the reaction.
Consider a general reaction,
${\text{aA + bB }} \to {\text{ cC + dD}}$
The rate law for the reaction is,
${\text{Rate}} = k{\left[ {\text{A}} \right]^{\text{a}}}{\left[ {\text{B}} \right]^{\text{b}}}$
Where $k$ is the rate constant.
The order is a with respect to A and b with respect to B. But the overall order of the reaction is,
${\text{Order}} = {\text{a}} + {\text{b}}$
Identify the reaction order from the units of reaction rate constant as follows:
The speed at which the reaction occurs is known as the rate of the reaction. The rate of any chemical reaction is expressed as the decrease in concentration of reactants per unit time or increase in the concentration of products per unit time.
${{\text{L}}^{ - 1}}{\text{ mol }}{{\text{s}}^{ - 1}}$
The units of the rate constant are,
$k ={{{\text{mol }}{{\text{L}}^{ - 1}}}}{{{{\text{s}}^{ - 1}}}}$
The units suggest that the rate constant is equal to the rate of the reaction (concentration per time) and is independent of the concentration of the reacting substance.
The reactions in which the rate of the reaction is independent of the concentration of the reacting substance is known as the zero order reaction.
Thus, the reaction order when the units of reaction rate constant are ${{\text{L}}^{ - 1}}{\text{ mol }}{{\text{s}}^{ - 1}}$ is zero order.
${\text{L mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$
The units of the rate constant are,
$k = {{\text{L}}}{{{\text{mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}}}$
The units suggest that the units of concentration are reversed from ${\text{mol }}{{\text{L}}^{ - 1}}$ to ${\text{L mo}}{{\text{l}}^{ - 1}}$. This is possible when the reaction has two reacting molecules. In such cases the unit concentration is 1/2 and thus gets reversed.
The reactions in which the rate of the reaction is proportional of the concentration of two reacting molecules is known as the second order reaction.
Thus, the reaction order when the units of reaction rate constant are ${\text{L mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$ is second order.

Note: The units for first order reaction are ${{\text{s}}^{ - 1}}$. The units of third order reaction are ${\text{mo}}{{\text{l}}^{ - 2}}{\text{ }}{{\text{L}}^2}{\text{ }}{{\text{s}}^{ - 1}}$. By knowing the rate constant values we can directly predict the order of the reaction if it is not given in the question.