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 \(\mathrm{R}=2\) cal \(\mathrm{mol}^{-1} \mathrm{~K}^{-1}\) and \(\left.\log 2=0.301\right]\) (a) \(455 \mathrm{cal}\) (b) \(231 \mathrm{cal}\) (c) \(693 \mathrm{cal}\) (d) \(554 \mathrm{cal}\)

Step-by-Step Solution

Answer

The catalyst lowers the energy barrier by \(554 \ \mathrm{cal}\).

Step 1: Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin by adding 273 to the Celsius temperature.\[ T(K) = 127 + 273 = 400 \, \text{K} \]

Step 2: Apply Arrhenius Equation

The Arrhenius equation relates the rate constants \( k_1 \) and \( k_2 \) of the reaction before and after the addition of a catalyst:\[ \frac{k_2}{k_1} = e^{\left(\frac{-Ea_{\text{cat}} + Ea}{RT}\right)} \]We know that the rate doubles due to adding the catalyst, so \( \frac{k_2}{k_1} = 2 \), and the relationship between the activation energies \( Ea_{\text{cat}} \) and \( Ea \).

Step 3: Take the Natural Logarithm

Take the natural logarithm of both sides of the ratio equation.\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{-Ea_{\text{cat}} + Ea}{RT} \]Substituting \( \ln(2) \approx 0.693 \) gives:\[ 0.693 = \frac{-Ea_{\text{cat}} + Ea}{800} \]

Step 4: Rearrange to Find the Energy Change

Rearrange the equation to solve for the difference in activation energy (\(Ea - Ea_{\text{cat}}\)):\[ Ea - Ea_{\text{cat}} = 0.693 \times R \times T \]Substitute the values:\[ Ea - Ea_{\text{cat}} = 0.693 \times 2 \, \mathrm{cal/mol/K} \times 400 \, K \]

Step 5: Calculate the Energy Barrier Lowered

Calculate the reduced energy barrier using the rearranged equation:\[ Ea - Ea_{\text{cat}} = 0.693 \times 800 = 554.4 \]

Step 6: Determine the Closest Option

From the calculation result, the energy barrier lowered is approximately 554.4 cal. Choose the closest option from the given choices, which is 554 cal.

Key Concepts

Activation Energy

Activation energy is like the starting line in a race. It represents the minimum energy that reacting particles must have to start a chemical reaction. Imagine it as a hill that molecules need to climb over to turn into products.
Without sufficient energy, the reactants just can't transform, similar to how a car won't move if it doesn't have enough gas.
In the context of redox reactions, understanding activation energy can explain why some reactions occur quickly while others do not. A high activation energy means the reaction is slow because not many particles have enough energy to "cross the hill." Conversely, a low activation energy indicates a faster reaction.

• This concept helps chemists control and tweak reaction rates, especially in industrial processes.
• It also sets the stage for introducing catalysts, which lower this energy barrier.

Recognizing activation energy can help you predict how altering temperature or adding catalysts will change reaction speed, an aspect that's crucial for both academicians and industrial scientists.

Catalyst Effect

A catalyst is like a magician for chemical reactions! While a magician uses tricks, a catalyst uses chemistry to speed up reactions without being consumed. It works by lowering the activation energy, which means reactants need less energy to start transforming into products.
This magical effect is observed in the given exercise where adding a catalyst doubled the reaction rate.
Here's how catalysts provide their magic:

• They align reactants in a more favorable configuration for the reaction.
• They provide an alternative pathway with a lower energy requirement.

Because of this, more reactant molecules can "cross the hill" we've discussed earlier, speeding up the reaction. This is why catalysts are invaluable in chemical industries—making processes faster without increasing the temperature.

Arrhenius Equation

The Arrhenius equation is a powerful tool used to understand temperature's effect on reaction rates and to determine how catalysts alter these rates. Named after the Swedish chemist Svante Arrhenius, it is expressed as:\[ k = A \cdot e^{-\frac{Ea}{RT}} \]
Here, \( k \) is the rate constant, \( A \) is the frequency factor, \( Ea \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
In the exercise problem, this equation helps compare the rates before and after a catalyst is introduced. By calculating the ratio of these rates, we understand how much a catalyst lowers the activation energy.
This equation is essential because:

• It helps predict how changes in temperature influence reaction rates.
• It quantifies the catalyst's effectiveness, as seen by the decreased energy barrier.

Understanding the Arrhenius equation allows us to explore deeper insights into reaction dynamics and optimize industrial processes efficiently.