Problem 239 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}\left[\mathrm{H}^{+}\right]^{\mathrm{n}}\). If \(\mathrm{pH}\) of the reaction medium changes from two to one, rate becomes 100 times that of the value at \(\mathrm{pH}=2\). The order of reaction is

Step-by-Step Solution

Answer

The order of the reaction is 2.

Step 1: Understand the Given Reaction Rate Equation

The given rate equation is \( \frac{dx}{dt} = k[H^+]^n \), where \( k \) is the rate constant, \( [H^+] \) is the concentration of hydrogen ions, and \( n \) is the order of the reaction with respect to \( [H^+] \).

Step 2: Relate pH to Hydrogen Ion Concentration

The pH is related to the hydrogen ion concentration by the formula \( \text{pH} = -\log_{10}[H^+] \). Therefore, when \( \text{pH} = 2 \), \( [H^+] = 10^{-2} \) mol/L, and when \( \text{pH} = 1 \), \( [H^+] = 10^{-1} \) mol/L.

Step 3: Write Rate Expressions at Two pH Levels

At \( \text{pH} = 2 \), the rate is \( R_2 = k(10^{-2})^n \). At \( \text{pH} = 1 \), the rate is \( R_1 = k(10^{-1})^n \).

Step 4: Set Up the Ratio of Rates

According to the problem, \( R_1 = 100 \times R_2 \). Substitute the rate expressions: \( k(10^{-1})^n = 100 \times k(10^{-2})^n \). This gives \( 10^{-n} = 100 \times 10^{-2n} \).

Step 5: Solve for the Order of Reaction \( n \)

Rearrange the equation from Step 4: \( 10^{-n} = 10^2 \times 10^{-2n} \). Simplify to find \( 10^{-n} = 10^{2 - 2n} \). This implies that \( -n = 2 - 2n \). Solving for \( n \), we get \( n = 2 \).

Key Concepts

Reaction Rate Equation

The reaction rate equation is a fundamental concept in chemistry that describes how fast a reaction proceeds. It usually relates the change in concentration of a reactant or product over time to the concentration of reactants raised to a power.

In this exercise, the reaction rate equation is given by \( \frac{dx}{dt} = k[H^+]^n \). Here,

• \( \frac{dx}{dt} \) denotes the rate of change of concentration of the reactant \( x \) with time, \( t \).
• \( k \) is the rate constant, a proportionality factor that is specific to each reaction at a given temperature.
• \([H^+]^n\) indicates that the reaction rate is dependent on the hydrogen ion concentration raised to the power \( n \), which is the order of the reaction with respect to \([H^+]\).

Understanding this equation is crucial because it helps predict how changes in concentration impact reaction speeds.

pH and Hydrogen Ion Concentration

pH is a measure of the acidity or basicity of a solution, and it is directly linked to the concentration of hydrogen ions, \([H^+]\). The relationship is governed by the formula:
\[\text{pH} = -\log_{10}[H^+] \]A decrease in pH by one unit means the hydrogen ion concentration increases tenfold.

For example, in the given exercise, when the pH changes from 2 to 1:

• At \( \text{pH} = 2 \), \([H^+] = 10^{-2} = 0.01\) mol/L.
• At \( \text{pH} = 1 \), \([H^+] = 10^{-1} = 0.1\) mol/L.

This increase in \([H^+]\) significantly impacts the reaction rate due to the reaction's dependency on \([H^+]^n\), as seen in the rate equation \( \frac{dx}{dt} = k[H^+]^n \).

Rate Constant

The rate constant, denoted as \( k \), is a crucial component in the reaction rate equation. It reflects the intrinsic kinetics of a reaction. Here's what it entails:

• \( k \) quantifies how swiftly a reaction progresses at a specific temperature. It's influenced by numerous factors including temperature, presence of a catalyst, and the nature of reactants.
• While \( k \) is constant for a given reaction under constant conditions, it helps us compare the effects of different concentrations or other reaction conditions.

In the specific reaction considered in the exercise, the order of reaction with respect to \([H^+]\) is determined by how the rate expression \( R = k[H^+]^n \) changes with different \([H^+]\). Changes in pH affect \([H^+]\), altering the rate by multiplying it by ten for each unit fall in pH, leading us to mathematical evaluation of \( n \). This interplay allows us to solve for the reaction order \( n = 2 \), understanding the significant role of \( k \) in anchoring this balance.