Problem 119 - A Complete Resource Book in Chemistry for JEE Main

Question

For a \(1^{\text {st }}\) order reaction \(\mathrm{A} \longrightarrow \mathrm{P}\), the temperature (T) dependent rate constant (K) was found to follow the equation \(\log \mathrm{k}=-(2000) \frac{1}{\mathrm{~T}}+6\) The pre- exponential factor A and activation energy Ea are respectively? (a) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(9.2 \mathrm{~kJ} / \mathrm{M}\) (b) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(38.3 \mathrm{~kJ} / \mathrm{M}\) (c) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(16.6 \mathrm{~kJ} / \mathrm{M}\) (d) \(6 \mathrm{~S}^{-1}\) and \(16.6 \mathrm{~kJ} / \mathrm{M}\)

Step-by-Step Solution

Answer

The answer is (b): \(1 \times 10^6 \text{ s}^{-1}\) and \(38.3 \text{ kJ/mol}\).

Step 1: Identify the Given Equation

We are given the formula for the rate constant in terms of temperature: \(\log k = -(2000) \frac{1}{T} + 6\). This resembles the Arrhenius equation in logarithmic form.

Step 2: Recall the Arrhenius Equation's Log Form

The logarithmic form of the Arrhenius equation is: \(\log k = \log A - \frac{E_a}{2.303RT}\), where \(A\) is the pre-exponential factor and \(E_a\) is the activation energy.

Step 3: Compare With Given Equation

Compare the standard form \(\log k = \log A - \frac{E_a}{2.303RT}\) with \(\log k = -(2000) \frac{1}{T} + 6\). This allows us to equate constants for calculation of \(A\) and \(E_a\).

Step 4: Solve for Pre-exponential Factor \(A\)

The constant term in the equation \(6\) corresponds to \(\log A\). Thus, \(\log A = 6\). Calculating \(A\) gives \(A = 10^6\).

Step 5: Solve for Activation Energy \(E_a\)

Equate \(-\frac{E_a}{2.303R} = -2000\), leading to \(\frac{E_a}{2.303R} = 2000\). We know \(R = 8.314\, \text{J/mol K}\), thus solve \(E_a = 2000 \times 2.303 \times 8.314\).

Step 6: Calculate Activation Energy \(E_a\)

Compute \(E_a = 2000 \times 2.303 \times 8.314 = 38281.36 \text{ J/mol}\). Convert to \(\text{kJ/mol}\) to get \(38.28 \text{ kJ/mol}\), approximately \(38.3 \text{ kJ/mol}\).

Key Concepts

Activation Energy

Activation energy is a crucial concept in chemical kinetics. It represents the minimum amount of energy needed for a reaction to occur. Think of it as the energy barrier that reactants must overcome to transform into products. This energy ensures that molecules are in the correct configuration to allow bonds to be broken and new bonds to be formed. An important aspect to remember is that higher activation energy means that a reaction is less likely to happen because it's harder for reactants to gather the needed energy.

When using the Arrhenius equation, the activation energy is denoted by \(E_a\). In log form, this equation looks something like this:

• \( \log k = \log A - \frac{E_a}{2.303RT} \)

Here, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
This expression clearly outlines the inverse relationship between \(E_a\) and \(T\) – higher temperatures can lower the effective energy barrier for a reaction.
This is what makes the Arrhenius equation so valuable, allowing us to predict how temperature affects reaction rates and understand how much energy is needed to kickstart different reactions.

Pre-exponential Factor

The pre-exponential factor, often designated as \(A\), is another key term in the Arrhenius equation. It might sound complicated, but it's a crucial part of understanding the speed of chemical reactions. You can think of it as a factor that accounts for the frequency of collisions and the proper orientation of molecules when they collide.

So, how do we get \(A\) from our equation? Recall that using the log form of the Arrhenius equation:

• \( \log A \) turns out to be the constant term in the given problem.
• So if we have an equation like \( \log k = -(2000) \frac{1}{T} + 6 \), the part where it says \(6\) is actually \( \log A \).

By simple calculation, this tells us that the pre-exponential factor \(A = 10^6\).

Understanding \(A\) helps us grasp how often reaction-specific requirements—like proper molecular alignment—are met, thus aiding in estimating reaction rates under different conditions.

First Order Reaction

First-order reactions are fundamental in the study of reaction kinetics. In these reactions, the rate is directly proportional to the concentration of one reactant. This simplicity makes them easier to understand and predict in terms of how the reaction proceeds over time.

• Mathematically, the rate of a first-order reaction can be expressed as: \( \text{rate} = k[A] \)
• Here \(k\) is the rate constant, and \([A]\) is the concentration of the reactant.

Given that \( ext{A} \longrightarrow ext{P} \), we're examining a common type of transformation where one substance is converted into another.

In the context of the Arrhenius equation, the rate constant \(k\) is what changes with temperature. The relationship can help determine how fast a first-order reaction will occur at different temperatures. Comprehensively understanding this concept provides a solid foundation for grasping more complex kinetic scenarios later on.