Problem 53 - A Complete Resource Book in Chemistry for JEE Main

Question

Graph between \(\log _{10} \mathrm{~K}\) and \((1 / t)\) is linear with slope \(S\). Hence \(\mathrm{E}\) is (a) \(\mathrm{R} \times \mathrm{S}\) (b) \(\mathrm{S} / \mathrm{R}\) (c) \(\mathrm{R} / \mathrm{S}\) (d) \(2.303 \mathrm{RS}\)

Step-by-Step Solution

Answer

The correct answer is (d) \(2.303 \mathrm{RS}\).

Step 1: Understand the Relationship

We are given the relationship between \( \log_{10} K \) and \( \frac{1}{t} \), where the equation of the line is linear. The general form of the linear relationship is \( y = mx + c \), where \( m \) is the slope. In this context, \( m \) is replaced by \( S \), so the equation can be expressed as \( \log_{10} K = S \left( \frac{1}{t} \right) + c \).

Step 2: Connect to the Arrhenius Equation

The Arrhenius equation in logarithmic form is \( \log_{10} k = \log_{10} A - \frac{E}{2.303RT} \). Comparing this with the linear equation extracted from Step 1, \( S \) is equivalent to \( -\frac{E}{2.303R} \).

Step 3: Solve for Activation Energy (E)

We equate \( S = -\frac{E}{2.303R} \). To find \( E \), rearrange to get \( E = -S \times 2.303R \). Recognizing that \( S \) is negative due to the slope and considering sign conventions, \( E = 2.303RS \) matches with the convention of positive activation energy.

Key Concepts

Activation Energy (E)

Activation energy, denoted as \( E \), plays a crucial role in chemical kinetics and reaction rates. It represents the minimum energy required for a reaction to occur. Without sufficient energy to overcome this barrier, reactions either proceed very slowly or not at all.
In the context of the Arrhenius equation, activation energy is important because it directly affects the rate constant \( k \). Lower activation energies indicate that a reaction can occur more easily at a given temperature. Conversely, higher values suggest that more energy is required for the reaction to proceed.

• Activation energy is measured in joules per mole (J/mol) or kilojoules per mole (kJ/mol).
• In practice, we often determine \( E \) experimentally by measuring reaction rates at different temperatures.
• The magnitude of \( E \) gives us insight into the nature of the chemical bonds that need to be broken for the reaction to occur.

The value of \( E \) derived in the given problem is linked to the slope \( S \) of the graph between \( \log_{10} K \) and \( \frac{1}{t} \), reinforcing the concept that it can be determined graphically as well.

Logarithmic Form

The Arrhenius equation can be expressed in a logarithmic form, which is extremely useful for analyzing experimental data. This form of the equation is given by:
\[ \log_{10} k = \log_{10} A - \frac{E}{2.303RT} \]
This format allows us to linearize a nonlinear equation by plotting \( \log_{10} k \) against \( \frac{1}{T} \). The slope of the resulting line can help us determine the activation energy \( E \).

• The term \( \log_{10} A \) represents the intercept of the plot. It is related to the frequency factor \( A \), which describes the number of times molecules collide with the right orientation per second to react.
• By using the logarithmic form, we can transform complex exponential relationships into simple linear equations that are easier to interpret and analyze.
• Interpreting the inverse temperature, \( \frac{1}{T} \), on the x-axis is key, as it allows for the extraction of information about the temperature dependence of the reaction rate.

The conversion to a logarithmic form simplifies the determination of parameters like \( E \) from experimental data, making it an essential tool in chemical kinetics.

Linear Relationship

When discussing the linear relationship within the context of chemical kinetics, it's often looking at how certain variables relate to one another in a predictable, straight-line manner. In this exercise, a linear relationship was described between \( \log_{10} K \) and \( \frac{1}{t} \), where \( K \) might be a rate constant or a similar parameter.
The general form of a linear equation is:

• \( y = mx + c \), where:
• \( y \) is the dependent variable, \( mx \) represents the slope \( m \) times the independent variable \( x \), and \( c \) is the y-intercept.

In the problem at hand, \( \log_{10} K = S \left( \frac{1}{t} \right) + c \) describes the linear relationship, with \( S \) as the slope. This relates directly to the rearranged Arrhenius equation. Understanding how these variables interact linearly allows chemists to:

• Predict how changes in one variable, like temperature, can affect another variable like the reaction rate constant.
• Determine activation energy \( E \) by analyzing the slope \( S \) of the line.
• Apply graphical analysis techniques to extrapolate other key rate parameters from experimental data.

This approach underscores the power of mathematical models in understanding and predicting chemical behavior through kinetic studies.