Q. Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ mol⁻¹. If k₁ and k₂ are the rate constants of first and second reaction respectively at 300 K, then ln (k₂/k₁) will be _____. (nearest integer)
[R = 8.3 J K⁻¹ mol⁻¹]
Step-by-Step Explanation
Step 1: Write the Arrhenius Equation for both reactions.
The Arrhenius equation is given by: \( k = A \cdot e^{-E_a/RT} \)
For Reaction 1: \( k_1 = A \cdot e^{-E_{a1}/RT} \)
For Reaction 2: \( k_2 = A \cdot e^{-E_{a2}/RT} \)
(Note: Pre-exponential factor 'A' is identical for both as per the question).
Step 2: Take the ratio of the rate constants.
\( \frac{k_2}{k_1} = \frac{A \cdot e^{-E_{a2}/RT}}{A \cdot e^{-E_{a1}/RT}} \)
\( \frac{k_2}{k_1} = e^{(E_{a1} - E_{a2})/RT} \)
Step 3: Take natural logarithm (ln) on both sides.
\( \ln\left(\frac{k_2}{k_1}\right) = \frac{E_{a1} - E_{a2}}{RT} \)
Step 4: Substitute the given values.
Difference in activation energy: \( E_{a1} - E_{a2} = 20 \text{ kJ mol}^{-1} = 20,000 \text{ J mol}^{-1} \)
Gas constant: \( R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1} \)
Temperature: \( T = 300 \text{ K} \)
\( \ln\left(\frac{k_2}{k_1}\right) = \frac{20000}{8.3 \times 300} \)
Step 5: Perform the final calculation.
\( \ln\left(\frac{k_2}{k_1}\right) = \frac{20000}{2490} \)
\( \ln\left(\frac{k_2}{k_1}\right) \approx 8.032 \)
The nearest integer is 8.
Related Theory
Chemical Kinetics is a crucial chapter for JEE Main and Advanced, focusing on the speed of chemical reactions and the factors affecting them. The temperature dependence of the rate of a chemical reaction is most accurately described by the Arrhenius Equation.
1. The Arrhenius Equation Components
- Rate Constant (k): Proportionality constant in the rate law.
- Pre-exponential factor (A): Also called the frequency factor, it represents the frequency of collisions with correct orientation.
- Activation Energy (\(E_a\)): The minimum energy required by reactants to form the activated complex.
- RT: Represents the average kinetic energy of the molecules.
2. Effect of Activation Energy on Rate
From the equation \( k = A \cdot e^{-E_a/RT} \), we can observe that as the activation energy (\(E_a\)) increases, the term \( e^{-E_a/RT} \) decreases, leading to a smaller rate constant. In this problem, Reaction 1 has a higher activation energy than Reaction 2, which is why \( k_1 \) is smaller than \( k_2 \), making the ratio \( k_2/k_1 \) greater than 1.
3. Logarithmic Forms
To solve numerical problems, we often use the log forms:
- Natural Log: \( \ln k = \ln A - \frac{E_a}{RT} \)
- Base 10 Log: \( \log k = \log A - \frac{E_a}{2.303RT} \)
These equations allow for linear plotting (Arrhenius Plot) where the slope equals \( -E_a/R \) or \( -E_a/2.303R \).
4. Collision Theory Context
Activation energy acts as an energy barrier. Only those collisions that possess energy equal to or greater than \(E_a\) result in a product. Increasing the temperature increases the fraction of molecules that can cross this barrier, which explains why reactions generally speed up at higher temperatures.
5. Important JEE Tips
- Units: Always check if \(E_a\) is in Joules or kiloJoules. \(R\) is usually given in J, so you must convert kJ to J.
- R Value: Pay attention to the value of R provided. Sometimes it's 8.314, sometimes 8.3, and sometimes \(25/3\) to make calculations easier.
- Ratio Problems: When comparing two reactions at the same temperature, the 'A' and 'T' terms often cancel or simplify, focusing the problem entirely on the difference in activation energies.
Frequently Asked Questions (FAQs)
1. What is the unit of the pre-exponential factor A?
The unit of A is the same as the unit of the rate constant (k), which depends on the order of the reaction.
2. Does Activation Energy change with temperature?
For most JEE problems, \(E_a\) is considered independent of temperature over a small range.
3. What is the significance of a high activation energy?
A high \(E_a\) means the reaction rate is very sensitive to temperature changes and is generally slower at room temperature.
4. What happens if the activation energy is zero?
If \(E_a = 0\), the rate constant k equals the frequency factor A, and the rate becomes independent of temperature.
5. How does a catalyst affect activation energy?
A catalyst provides an alternative reaction pathway with a lower activation energy, thereby increasing the rate of reaction.