Problem 14 - A Complete Resource Book in Chemistry for JEE Main

Question

For a chemical reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), the rate of reaction doubles when the concentration of \(\mathrm{A}\) is in creased four times. The order of reaction for \(\mathrm{A}\) is (a) zero (b) one (c) two (d) half

Step-by-Step Solution

Answer

The order of the reaction for A is half.

Step 1: Understanding the Rate Law Expression

The rate of a reaction is often expressed as: \[\text{Rate} = k [A]^n,\]where \(k\) is the rate constant, \([A]\) is the concentration of reactant A, and \(n\) is the order of the reaction with respect to A. We need to find the value of \(n\).

Step 2: Applying the Given Changes

According to the problem, if the concentration of \(A\) is increased four times, the rate of the reaction doubles. This can be written mathematically as:\[\frac{k (4[A])^n}{k [A]^n} = 2.\]

Step 3: Simplifying the Equation

Using the expression for the changed rate we simplify:\[\frac{4^n [A]^n}{[A]^n} = 2.\]Cancel \([A]^n\) from top and bottom:\[4^n = 2.\]

Step 4: Solving for the Order \(n\)

We have the equation \(4^n = 2\). This is equivalent to writing it as \(2^{2n} = 2\), resulting in:\[2n = 1.\]Solving for \(n\) gives:\[n = \frac{1}{2}.\]

Key Concepts

Rate Law Expression

In chemical reactions, a rate law expression helps to describe the relationship between the concentration of reactants and the speed of the reaction. It provides valuable information about how changes in concentration can impact the rate.
For a generic reaction \[ ext{Rate} = k[A]^n[B]^m, \]where \([A]\) and \([B]\) are the concentrations of the reactants, \(k\) is the rate constant, and \(n\) and \(m\) represent the order of the reaction with respect to each reactant.
These exponents (\(n\) and \(m\)) are typically determined experimentally and may not necessarily correspond to the stoichiometric coefficients of the balanced chemical equation.

• If \(n = 0\), the reaction rate does not depend on the concentration of \(A\).
• If \(n = 1\), the rate is directly proportional to the concentration of \(A\).
• If \(n = 2\), the rate is proportional to the square of the concentration of \(A\).

Understanding rate laws allows chemists to predict how a particular system will behave when conditions change.

Chemical Kinetics

Chemical kinetics is the study of how quickly reactions occur and involves investigating the factors that influence the rate. It helps chemists understand processes from which they can control and optimize reactions for industrial applications and more.
Several factors affect reaction rates, including:

• Concentration of reactants: Changes in concentration can speed up or slow down a reaction.
• Temperature: Generally, increasing temperature increases the rate of a reaction.
• Presence of a catalyst: Catalysts speed up reactions without being consumed.
• Surface area: Greater surface area increases the rate by offering more space for collisions.

By studying kinetics, scientists can provide insights and develop methods to adjust each of these factors to achieve desired results in various fields like pharmaceuticals and materials science.

Rate Constant

The rate constant, represented as \(k\), is a crucial part of the rate law expression. It is a proportionality constant that remains constant for a given reaction at a specific temperature. However, it can change with temperature or when catalysts are involved.
For the expression \[ k[A]^n = ext{Rate}, \]\(k\) provides quantitative insight into how fast a reaction proceeds, offering important data for comparing different reactions under similar conditions.
It's important to remember:

• The units of the rate constant vary depending on the order of the reaction. For a first-order reaction, \(k\) has units of \(s^{-1}\), while for a second-order reaction, it typically has units of \(M^{-1}s^{-1}\).
• The larger the value of \(k\), the faster the reaction is at a given concentration of reactants.

Understanding \(k\) helps in evaluating reactivity and efficiency of chemical reactions under various conditions.

Concentration Effect on Reaction Rate

Concentration plays a vital role in dictating the rate of a chemical reaction. In the given exercise, we saw that increasing the concentration of reactant \(A\) affected the rate. The relationship between concentration and rate is encapsulated in the reaction order.
A practical exploration showed that quadrupling the concentration of \(A\) led to a doubling of the reaction rate. Mathematically, this was expressed as \[4^n = 2,\] leading to determining the reaction order \(n = \frac{1}{2}\).

Understanding this effect allows us to:

• Predict how the rate changes when the concentration of a reactant is altered.
• Identify suitable conditions for operating chemical processes more effectively.
• Design experiments to explore further how each reactant influences the overall rate.

Thus, manipulating concentrations enables control over reaction speed, which is essential in industrial and laboratory settings.