Problem 192 - A Complete Resource Book in Chemistry for JEE Main

Question

In gaseous reactions important for understanding the upper atmosphere, \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}\) react bimolecularly to form two \(\mathrm{OH}\) radicals. \(\Delta \mathrm{H}\) for this reaction is \(72 \mathrm{~kJ}\) at \(500 \mathrm{~K}\) and \(\mathrm{Ea}=77 \mathrm{~kJ} \mathrm{~mol}^{-1}\), then \(\mathrm{E}\) for the bimolecular recombination of \(2 \mathrm{OH}\) radicals to form \(\mathrm{H}_{2} \mathrm{O} \& \mathrm{O}\) at \(500 \mathrm{~K}\) is

Step-by-Step Solution

Answer

The bimolecular recombination energy is 5 kJ mol^{-1} for forming H_2O and O from 2 OH radicals at 500 K.

Step 1: Understanding the Reaction and Given Data

We are dealing with the reaction where \( ext{H}_2 ext{O}\) and \( ext{O}\) react to form two \( ext{OH}\) radicals. The given data includes \(\Delta H = 72 \, \text{kJ}\, \text{mol}^{-1}\) and the activation energy for the forward reaction \(E_a = 77 \, \text{kJ}\, \text{mol}^{-1}\).

Step 2: Using the Forward and Reverse Reaction Energies

To find the energy for the reverse reaction, we can use the relationship between activation energy, enthalpy change of the reaction, and energies. We have: \(\Delta H = E_{a, \text{forward}} - E_{a, \text{reverse}}\).

Step 3: Rearranging the Formula

Our goal is to find \(E_{a, \text{reverse}}\). Rearrange the formula: \(E_{a, \text{reverse}} = E_{a, \text{forward}} - \Delta H\).

Step 4: Calculate \(E_{a, \text{reverse}}\)

Substitute the values into the rearranged formula: \(E_{a, \text{reverse}} = 77 \, \text{kJ}\, \text{mol}^{-1} - 72 \, \text{kJ}\, \text{mol}^{-1}\).

Step 5: Final Calculation

Perform the subtraction: \(E_{a, \text{reverse}} = 5 \, \text{kJ} \, \text{mol}^{-1}\). Therefore, the activation energy for the reverse reaction is \(5 \, \text{kJ} \, \text{mol}^{-1}\).

Key Concepts

Activation Energy

Activation energy is the minimum energy required for a chemical reaction to occur. It's a barrier that reactants must overcome to transform into products. Think of it like pushing a ball up a hill before it can roll down.

This concept is pivotal in chemical kinetics as it determines how fast a reaction can proceed. The higher the activation energy, the slower the reaction will be unless additional energy or catalysts are present.

• For instance, in the reaction between \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}\), the activation energy (\(\mathrm{E_a}\)) is 77 kJ/mol, which means that the molecules need this amount of energy to start reacting.
• Interestingly, activation energy isn't always constant; it can be lowered by catalysts, which enhance the reaction rate without being consumed in the process.
• In our example, the forward activation energy is crucial as it affects how easily the two molecules can form \(\mathrm{OH}\) radicals.

Bimolecular Reaction

A bimolecular reaction involves two reactant species colliding and reacting to form products. This type of reaction is characterized by its kinetics, often depicted as \(A + B \rightarrow \text{Products}\).

In our particular example, \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}\) react bimolecularly to produce two \(\mathrm{OH}\) radicals.

• Since this is a gas-phase reaction, the odds of both molecules coming into contact depend heavily on concentration and temperature.
• At elevated temperatures, as given in the problem (500 K), molecules move more rapidly, increasing the likelihood of collisions.
• Bimolecular reactions are often studied to understand atmospheric processes, like those occurring in the upper atmosphere involving radicals.

This knowledge helps chemists and scientists predict how such reactions contribute to broader environmental phenomena.

Reverse Reaction Calculation

Calculating the energy parameters for the reverse reaction is an essential part of understanding the equilibrium and kinetics of chemical reactions. The concept revolves around the relationship between the activation energy of the forward and reverse reactions, as well as the enthalpy change.

• The formula connecting these is: \(\Delta \mathrm{H} = E_{a, \text{forward}} - E_{a, 0\text{reverse}}\).
• Enthalpy change \(\Delta \mathrm{H}\), representing the heat absorbed or released, was given as 72 kJ/mol in the problem, which indicates an endothermic reaction (heat absorbed).
• To find the reverse reaction's activation energy, we rearrange to: \(E_{a, \text{reverse}} = E_{a, \text{forward}} - \Delta \mathrm{H}\).

Upon substituting the values, \(E_{a, \text{forward}}\) is 77 kJ/mol, and \(\Delta \mathrm{H}\) is 72 kJ/mol, which results in \(E_{a, \text{reverse}} = 5 \, \text{kJ} \, \text{mol}^{-1}\).

This low activation energy for the reverse reaction means the recombination of two \(\mathrm{OH}\) radicals to form \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}\) is relatively easier under the same conditions.