Rate Law - Chemistry's Laws Of Motion

What is the Rate Law?

The rate law is a mathematical relationship that governs the process flow of a reaction. As Newton’s laws of motion are for kinematics, this is for chemical reactions. We use differential and integrated rate laws to determine the rate at which a reaction proceeds. But you may wonder that in chemistry: ‘Where does motion exist which needs to be governed by a rate law?’ The answer is that Chemistry is all about how a substance with definite properties is transformed into another substance with different properties. This transformation, which is also known as process flow of a reaction, is what we are referring to as a motion. Now let’s try to understand how a reaction proceeds and how can the rate law be determined.

Rate Law Explanation:

For a reaction to proceed, the reactants need to come in contact with each other with some threshold energy so that the old bonds can be broken and new bonds can be formed. Thus the reaction depends on upon the collisions between the reactants. The amount of product formed depends on upon the collisions with proper orientation and activation energy, and the collisions per unit time determine the rate of reaction.

Consider two beakers A & B filled with Hydrogen gas and oxygen gas as shown:

Rate Law

As evident from the figure, the collisions in Beaker A will be more than the collisions in Beaker B. Also there is more probability of reactants to hit each other with proper energy and orientation in Beaker A, so the rate of reaction will be more in Beaker A than the Beaker B.

\(\Rightarrow ~Rate~ of ~reaction~ ∝ ~Concentration ~of~ Reactants\).

Rate Law Expression

Consider a reaction: \(R \rightarrow P\)

The rate law for a reaction can be expressed as follows:

\(Rate ∝ [R]\)

\(Rate\) = \(k[R]^n\)Eqn 1

Where, \(k\) is proportionality constant known as Rate constant,
\([R]\) is concentration of reactant
\(n\) is order of reaction which is experimentally determined.

Instantaneous rate of reaction can be written as:

\(Rate\) = \(-\frac{d[R]}{dt}\) = \(\frac{d[P]}{dt}\)Eqn 2

The concentration of reactant decreases with time so the negative sign is attached to it to make the overall rate positive. The above-mentioned expression is differential rate law. It’s a beautiful expression that shows that the amount of product formed is equal to the amount of reactant used and change in their concentration per unit time is the rate of reaction.

Integrated rate law

We need an integrated rate law to find the concentration of reactant or product as a function of time.

Combining equation 1 and 2, we will get

\(- \frac{d[R]}{dt}\) = \(k[R]^n\)

The above rate law equation can be integrated to get the concentration of reactant as a function of time. ‘n’ for a reaction is experimentally determined and is known as the order of reaction.

Example:

First order Rate law

Question: Decomposition of H2O2 in aqueous solution follows first-order reaction with rate constant, \(k\) = \(3.66 × 10^{−3} s^{−1}\). Calculate its half-life using integrated rate law approach.

\(H_2O_2~ \rightarrow~ H_2O~ +~ \frac{1}{2} ~O_2\)

Solution:
Half-life is the amount of time required for something to fall to half of its initial value. So half-life for given reaction will be the time taken for \(H_2O_2\) to reach to half of its initial value.

Let initial concentration of \(H_2O_2\) be [\(A_o\)].

At an instant,
\(\Rightarrow~~Rate\) = \(k[A]\)

\(\Rightarrow~~Rate\) = \(- \frac{d[A]}{dt}\)

\(- \frac{d[A]}{dt}\) = \(k[A]\)

Integrating both the sides, we get

\(∫_0^t~\frac{d[A]}{dt}\) =\( -∫_{A_o}^A~k[A]\)

\(\Rightarrow ~ln\frac{[A]}{[A_o]}\) = \(-kt\) ……………………….. (This is also known as integrated first order equation)

\(\Rightarrow\)For given question, \(A\) = \(\frac{A_o}{2}\)

\(\Rightarrow~ln(\frac{1}{2})\) = \(-kt\)

\(\Rightarrow~kt\) = \(0.693\)

\(t\) = \( \frac{0.693}{k}\) ……………………………….. (This is also termed as half-life)

\(\Rightarrow\) Putting value of \(k\), we get

\(\Rightarrow~t\) = \(189.344~ seconds\)

The topic of rate laws, half-life and integrated rate law are very prominent in different competitive examinations. For detailed and more interactive study of the topic, visit our site or download Byju’s learning app from play store.


Practise This Question

In Samwell's classes, the Maesters made the following statements. He had to figure out how many were right. Can you help him?

Maester 1 said: Rate constant is characteristic of a reaction
Maester 2 said: Rate constant is dependent on temperature
Maester 3 said: Rate constant is not dependent on the catalyst