Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

The activation energies of two reactions are \(18 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and \(4.0 \mathrm{~kJ} \mathrm{~mol}^{-4}\) respectively. Assuming the pre-exponential factor to be the same for both reactions, the ratio of their rate constants at \(27^{\circ} \mathrm{C}\) is (a) \(3.656 \times 10^{-3}\) (b) \(3.624 \times 10^{-6}\) (c) \(36.52 \times 10^{-8}\) (d) \(4.656 \times 10^{-4}\)

Step-by-Step Solution

Answer

The correct answer is option (a) \(3.656 \times 10^{-3}\).

Step 1: Write the Arrhenius Equation

The Arrhenius equation relates the rate constant \(k\) of a reaction to its activation energy \(E_a\), temperature \(T\), and the pre-exponential factor \(A\). It is given by:\[ k = A e^{-E_a / (RT)} \]where \(R\) is the gas constant \(8.314 \text{ J} \text{ mol}^{-1} \text{ K}^{-1}\).

Step 2: Define the Reaction Parameters

For Reaction 1, the activation energy \(E_{a_1} = 18 \text{ kJ mol}^{-1} = 18000 \text{ J mol}^{-1}\).For Reaction 2, the activation energy \(E_{a_2} = 4.0 \text{ kJ mol}^{-1} = 4000 \text{ J mol}^{-1}\).The temperature \(T = 27^{\circ} \text{C} = 300 \text{ K}\).

Step 3: Write the Ratio of Rate Constants Formula

The ratio of the rate constants for the two reactions can be formulated as:\[ \frac{k_1}{k_2} = \frac{A e^{-E_{a_1}/(RT)}}{A e^{-E_{a_2}/(RT)}} = e^{-(E_{a_1} - E_{a_2})/(RT)} \]The pre-exponential factors \(A\) cancel out because they are assumed to be the same.

Step 4: Calculate the Exponential Term

Substitute the values into the ratio expression:\[ \frac{k_1}{k_2} = e^{-\frac{(18000 - 4000)}{8.314 \times 300}} \]Calculate the exponent:\[ \frac{(18000 - 4000)}{8.314 \times 300} = \frac{14000}{2494.2} \approx 5.614 \]

Step 5: Compute the Ratio of Rate Constants

Calculate the exponential term:\[ e^{-5.614} \approx 3.656 \times 10^{-3} \]

Step 6: Identify the Correct Answer Option

Based on the computed ratio \(\frac{k_1}{k_2} = 3.656 \times 10^{-3}\), the correct answer is option (a).

Key Concepts

Activation Energy

Activation energy is a crucial concept in understanding how chemical reactions proceed. Imagine it as the energy barrier that must be overcome for a reaction to occur. In simpler terms, it's like needing a certain amount of push to get a ball rolling down a hill. Without enough energy, the ball stays put.
In the realm of chemistry, activation energy is represented by the symbol \(E_a\). Each reaction has its own specific \(E_a\) determined by the nature of the reactants and the type of chemical bonds involved. In the given exercise, we see two different activation energies: \(18 \mathrm{~kJ} \mathrm{~mol}^{-1}\) for one reaction and \(4.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) for another.
The difference in these values shows that one reaction has a higher energy threshold than the other. This means more energy is needed to initiate the first reaction. Understanding activation energy helps us predict how a reaction's rate will change under various conditions. It explains why some reactions occur faster than others, simply because they have a lower activation barrier to cross.

Rate Constants

The rate constant \(k\) plays a pivotal role in the Arrhenius equation, defining how quickly a reaction proceeds at a given temperature. It is influenced by the reaction's activation energy, temperature, and the pre-exponential factor \(A\).
The Arrhenius equation is expressed as \( k = A e^{-E_a / (RT)} \). From this formula, it's evident that the rate constant depends exponentially on the negative ratio of activation energy to the product of the ideal gas constant \(R\) and temperature \(T\).
This dependence implies:

• A higher \(E_a\) leads to a smaller \(k\) when \(T\) is constant, meaning the reaction is slower because fewer molecules have the energy needed to overcome the activation energy barrier.
• If the pre-exponential factor \(A\) remains constant—as assumed in the given problem—comparing rate constants is simplified to observing the changes in the exponential term related to \(E_a\).

In the exercise, calculating the ratio of rate constants between the two reactions shows how varying \(E_a\) affects \(k\), even when other factors are consistent.

Temperature Effect on Reaction Rates

Temperature exerts a profound influence on reaction rates. As temperature increases, molecules move faster and collide more energetically. This leads to an increased likelihood of reactions occurring because more molecules can pierce through the energy barrier, defined by activation energy.
In the context of the Arrhenius equation, temperature \(T\) appears in the denominator of the expression \(e^{-E_a/(RT)}\). This positioning means that raising the temperature decreases the value of \(E_a/(RT)\), making the negative exponent smaller and consequently increasing the rate constant \(k\).
Some key points include:

• Higher temperatures generally accelerate chemical reactions, as more molecules reach or exceed the needed activation energy.
• The relationship is such that even small temperature changes can cause significant variations in reaction rates, particularly for reactions with high activation energies.
• In the exercise provided, a temperature of \(27^{\circ} C\) was used to determine how it affects the calculated ratio of rate constants for the given reactions.

This aspect of temperature influence is critical not only in theoretical chemistry but also in industrial processes where controlling reaction speeds is essential.