Problem 8 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Two reactions \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have identical pre- exponential factors. Activation energy of \(\mathrm{R}_{1}\) exceeds that of \(\mathrm{R}_{2}\) by \(10 \mathrm{~kJ} \mathrm{~mol}^{-1}\). If \(k_{1}\) and \(k_{2}\) are rate constants for reactions \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) respectively at \(300 \mathrm{~K}\), then \(\ln \left(k_{2} / k_{1}\right)\) is equal to : \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) 8 (b) 12 (c) 6 (d) 4

Step-by-Step Solution

Answer

(d) 4

Step 1: Use the Arrhenius Equation

The Arrhenius equation states that the rate constant \(k\) is given by: \[ k = Ae^{-E_a/(RT)} \]where \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. For both reactions \(R_1\) and \(R_2\), the equation can be written as:\[ k_1 = Ae^{-E_{a1}/(RT)}\]\[ k_2 = Ae^{-E_{a2}/(RT)}\]

Step 2: Determine the Relation Between Activation Energies

We are given that the activation energy \(E_{a1}\) of \(R_1\) exceeds that of \(R_2\) by \(10\, \text{kJ mol}^{-1}\). Thus, we can write:\[ E_{a1} = E_{a2} + 10000 \, \text{J mol}^{-1} \]

Step 3: Set up the Equation for \(k_2/k_1\)

Using the modified activation energy relation in the Arrhenius equation, the expression for the ratio of rate constants \(\frac{k_2}{k_1}\) becomes:\[ \frac{k_2}{k_1} = \frac{Ae^{-E_{a2}/(RT)}}{Ae^{-E_{a1}/(RT)}} \]

Step 4: Simplify the Equation

Since the pre-exponential factors \(A\) are identical for both reactions, they cancel out, leading to:\[ \frac{k_2}{k_1} = e^{(E_{a1} - E_{a2})/(RT)} \]

Step 5: Substitute Values into the Equation

We substitute \(E_{a1} - E_{a2} = 10000 \, \text{J mol}^{-1}\), \(R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}\), and \(T = 300 \, \text{K}\) into the equation:\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{10000}{8.314 \times 300} \]

Step 6: Calculate the Result

Perform the calculations using the substitution:\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{10000}{2494.2} \approx 4.01 \]This value rounds to \(4\).

Step 7: Choose the Correct Answer

The value obtained is approximately \(4.01\), which is closest to option (d) 4.

Key Concepts

Activation Energy

The concept of activation energy is pivotal in understanding chemical reactions. In essence, activation energy is the minimum energy required for a reaction to occur. It signifies the barrier that reactants must overcome to convert into products.

During a chemical reaction, bonds in the reactants need to weaken and eventually break, while new bonds form in the products. This process requires energy, which is provided by the activation energy. If the molecules do not have enough energy, they cannot surpass the energy barrier, and the reaction will not proceed.

Factors affecting activation energy include:

• The nature of the reactants: Different reactants have varying bond strengths, thus requiring different amounts of activation energy.
• The presence of catalysts: Catalysts can lower the activation energy, facilitating a quicker reaction without being consumed in the process.

The Arrhenius equation helps us quantify the relationship between activation energy and the rate constant in chemical reactions.

Rate Constant

The rate constant, represented as \(k\), figures prominently in the Arrhenius equation and is crucial for determining the speed of a chemical reaction. This constant integrates the effects of all reaction-specific factors that dictate how quickly a reaction occurs under given conditions.

In mathematical terms and under a fixed set of conditions, the rate constant can be expressed as follows using the Arrhenius equation:
\[ k = Ae^{-E_a/(RT)} \]
This equation establishes that the rate constant depends on:

• Activation energy \(E_a\): Higher activation energy leads to a lower rate constant, slowing down the reaction.
• Temperature \(T\): As temperature increases, reactant molecules gain energy, increasing \(k\) and accelerating the reaction.
• Gas constant \(R\): A universal constant crucial for maintaining dimensional consistency.
• Pre-exponential factor \(A\): A term that we will explore more in the next section.

Understanding the rate constant assists in accurately predicting the kinetics of a reaction, which is essential in fields like chemical engineering and biochemistry.

Pre-Exponential Factor

The pre-exponential factor, often denoted by \(A\), is a component of the Arrhenius equation that represents the frequency of collisions resulting in a successful reaction. It's sometimes called the frequency factor because it accounts for the number of times that reactants approach each other per unit of time.

Here's how it fits in the Arrhenius equation:
\[ k = Ae^{-E_a/(RT)} \]
The pre-exponential factor is influenced by:

• Orientation of molecules: Collisions must be oriented correctly for a reaction to occur.
• Physical state and nature of the reactants: Different states and characteristics affect how often reactants collide.

Although \(A\) remains constant under fixed conditions, changes in these factors can alter reaction rates overall.

Understanding \(A\) is important because, unlike the variables \(E_a\), \(R\), and \(T\), which are influenced externally, \(A\) provides insight into the intrinsic properties and behavior of the reactants themselves. It's a key to decoding the more nuanced details of reaction dynamics.