Problem 137 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction a \(\mathrm{A} \longrightarrow \mathrm{xP}\) when \([\mathrm{A}]=2.2 \mathrm{mM}\) the rate was found to be \(2.4 \mathrm{~m} \mathrm{M} \mathrm{s}^{-1}\) On reducing concentration of \(\mathrm{A}\) to half, the rate changes to \(0.6 \mathrm{~m} \mathrm{M} \mathrm{s}^{-1}\). The order of reaction with respect to \(\mathrm{A}\) is (a) \(1.5\) (b) \(2.0\) (c) \(2.5\) (d) \(3.0\)

Step-by-Step Solution

Answer

The order of reaction with respect to \( \mathrm{A} \) is \( 2 \).

Step 1: Define the Rate Equation

For the reaction \( a \mathrm{A} \longrightarrow \mathrm{xP} \), the rate is given by the equation: \( \text{Rate} = k [\mathrm{A}]^n \), where \( k \) is the rate constant and \( n \) is the order of the reaction with respect to \( \mathrm{A} \). The task is to find the value of \( n \).

Step 2: Use First Conditions

With \([\mathrm{A}] = 2.2 \text{ mM}\) and the rate \( = 2.4 \text{ mM} \cdot \text{s}^{-1} \), the rate law becomes:\[ 2.4 = k (2.2)^n \]

Step 3: Use Second Conditions

Reducing the concentration of \( \mathrm{A} \) to half, \([\mathrm{A}] = 1.1 \text{ mM}\), changes the rate to \( 0.6 \text{ mM} \cdot \text{s}^{-1} \): \[ 0.6 = k (1.1)^n \]

Step 4: Solve for Order of Reaction

Divide the second equation by the first equation to eliminate \( k \):\[ \frac{0.6}{2.4} = \frac{(1.1)^n}{(2.2)^n} \] \[ 0.25 = \left(\frac{1.1}{2.2}\right)^n \] \[ 0.25 = \left(0.5\right)^n \] Solving for \( n \) using logarithms, we get \( n = 2 \).

Step 5: Cross-Verify the Calculation

With calculated \( n = 2 \), plug it back into the concentration ratios to verify: \( (0.5)^2 = 0.25 \), which matches the given ratio.

Key Concepts

Rate Law

The rate law is an equation that connects the rate of a chemical reaction to the concentration of its reactants. It helps us understand how changes in the concentration of reactants affect the reaction speed.
The general form of the rate law is:

• Rate = k[A]ⁿ

Here, "k" is the rate constant and "n" is the order of reaction with respect to reactant A. The value of "n" determines how sensitive the reaction rate is to changes in concentration. A higher "n" means the rate is very responsive to concentration changes.
In the exercise, you were given changes in concentration and reaction rate. These inputs are used to find "n" and understand the reaction's dependency on the reactant's concentration. For instance, by reducing the concentration of A, you can observe the significant effect on the rate when "n" is large.

Reaction Kinetics

Reaction kinetics is a branch of chemistry that studies the speed at which chemical reactions proceed. It dives into understanding the factors affecting these speeds, such as concentration, temperature, and presence of catalysts. These factors are crucial in both chemical synthesis and industrial applications.
Using kinetics, chemists can predict reaction behavior, optimize conditions for desired reaction speeds, and scale reactions safely for industrial use. For the given reaction in the exercise, kinetics allows us to mathematically find out how doubling or halving the concentration of "A" impacts the rate directly. It shows that the reaction follows second-order kinetics, where the rate change corresponds to the square of concentration change.
The real power of reaction kinetics lies in its ability to provide insights not noticed by simple observation, such as understanding the underlying reaction mechanism.

Logarithms in Chemistry

In chemistry, logarithms often help us simplify calculations involving exponential numbers, especially those seen in rate laws and reaction kinetics. When we deal with large ranges of values like concentrations and reaction rates, logarithms make these computations more manageable.
In the provided example, logarithms are critical for solving the order of reaction. When dividing two rate equations to solve for "n," logarithms help us extract the value of "n" from the equation

• 0.25 = (0.5)ⁿ

This application of logarithms allows chemists to solve otherwise complex algebraic equations easily.
Even though logarithms might feel abstract, their utility in solving practical problems in chemistry is invaluable. Understanding how to use them effectively expands your ability to tackle a wide range of chemical problems.