Problem 153 - A Complete Resource Book in Chemistry for JEE Main

Question

A redox reaction is carried out at \(127^{\circ} \mathrm{C}\). If the same reaction is carried out in presence of a catalyst at the same temperature, the rate of reaction is doubled. To what extent is the energy barrier lowered by the catalyst? [Use \(\mathrm{R}=2 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\) and \(\log 2=0.301\) ] (a) \(455 \mathrm{cal}\) (b) \(231 \mathrm{cal}\) (c) \(693 \mathrm{cal}\) (d) \(554 \mathrm{cal}\)

Step-by-Step Solution

Answer

The energy barrier is reduced by 554 cal/mol.

Step 1: Understand the Problem

We need to determine the decrease in energy barrier (activation energy) when a catalyst is used in the reaction, given that the reaction rate doubles. We're provided with the temperature, the gas constant, and the logarithm of 2.

Step 2: Write Arrhenius Equation Relation

The rate of reaction, according to the Arrhenius equation, is given by the equation \( k = A e^{-E_a/(RT)} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

Step 3: Relate the Change in Rate to Change in Activation Energy

Since the rate is doubled, we can say \(frac{k_2}{k_1} = 2\), where \(k_2\) and \(k_1\) are the rate constants with and without the catalyst, respectively. This implies:\[ frac{A e^{-E_a'/RT}}{A e^{-E_a/RT}} = 2 \]Simplifying, we get:\[ e^{-(E_a' - E_a)/RT} = 2 \]

Step 4: Take Natural Logarithm

Taking the natural logarithm on both sides of the equation:\[ -(E_a' - E_a)/RT = \ln 2 \]\[ (E_a' - E_a)/RT = - \ln 2 \]

Step 5: Solve for Energy Difference

The energy difference \( \Delta E_a = E_a - E_a' \) can be calculated as:\[ \Delta E_a = R T \ln 2 \]Substituting the given values, \( R = 2 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \), \( T = (127+273) \mathrm{~K} = 400 \mathrm{~K} \), and \( \ln 2 = 0.693 \) (since \( \log 2 = 0.301 \approx 0.693 \,\text{in natural log}\)): \[ \Delta E_a = 2 \times 400 \times 0.693 \]This yields:\[ \Delta E_a = 554 \mathrm{cal/mol} \]

Step 6: Select the Correct Answer

From the values given, option (d) \(554 \mathrm{cal/mol}\) matches our calculated energy barrier reduction value.

Key Concepts

Redox Reactions

Redox reactions are a type of chemical reaction where oxidation and reduction occur simultaneously. In these reactions, substances either gain or lose electrons. **Oxidation** refers to the loss of electrons, whereas **reduction** means the gain of electrons. Redox reactions are common in everyday processes, such as cellular respiration and photosynthesis.

A simple way to identify these reactions is by observing the change in oxidation states of the elements involved. For example, when iron rusts, it undergoes oxidation by losing electrons to oxygen. In this scenario, iron is oxidized, and oxygen is reduced. Recognizing and balancing redox equations are essential skills in chemistry.

• Oxidizing agents are substances that accept electrons, thereby getting reduced.
• Reducing agents donate electrons and are thus oxidized.
• Redox reactions can generate electricity in electrochemical cells or require electrical energy to proceed, as seen in electrolysis.

Understanding these concepts is crucial for tackling problems related to redox reactions in both academic and practical chemistry scenarios.

Arrhenius Equation

The Arrhenius Equation is a formula that describes how the rate of a chemical reaction depends on temperature and activation energy. It is expressed as:\[ k = A e^{-E_a/(RT)} \]where:

• \( k \) is the rate constant of the reaction.
• \( A \) represents the pre-exponential factor related to the frequency of collisions and the orientation of reacting molecules.
• \( E_a \) is the activation energy, the minimum energy required for the reaction to occur.
• \( R \) refers to the gas constant, which is \(2 \,\text{cal mol}^{-1} \text{K}^{-1}\) in this case.
• \( T \) is the temperature in Kelvin.

The equation shows that an increase in temperature or a decrease in activation energy results in a higher reaction rate. This means that molecules have more energy to overcome the energy barrier, participating more frequently in effective collisions. The catalyst's role is vital here, as it reduces the activation energy, allowing the reaction to proceed at a faster rate, even at lower temperatures.

Activation Energy

Activation Energy (\( E_a \)) is a crucial concept in understanding chemical reactions. It represents the amount of energy required to convert reactants into products. Imagine it as a hill reactant molecules need to climb before transitioning into products.

In a reaction, only molecules with sufficient energy will overcome this energy barrier. Catalysts play a significant role by lowering this activation energy, making it easier for the reaction to occur. They do this without being consumed in the process.

• Lower activation energy means more molecules have enough energy to reach the transition state at a given temperature.
• This leads to an increased reaction rate, which is precisely what is observed when a catalyst is used.
• Without a catalyst, many reactions would either take too long or not occur under normal conditions.

Catalysts are essential in various industrial and biological processes, facilitating reactions efficiently and economically. Understanding activation energy is key to designing and optimizing these processes.