Problem 214 - A Complete Resource Book in Chemistry for JEE Main

Question

The energies of activation for forward and reverse reactions for \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}\) are \(180 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(200 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by \(100 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The enthalpy change of the reaction \(\left(\mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}\right)\) in the presence of catalyst will be (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) \(\quad\) (a) 120 (b) 280 (c) 20 (d) 300

Step-by-Step Solution

Answer

The enthalpy change with the catalyst is \(-20 \ \mathrm{kJ/mol}\), so the answer is (c) 20.

Step 1: Determine Original Enthalpy Change

The enthalpy change \( \Delta H \) for a reaction can be calculated using the difference between the activation energies for the forward \( E_{af} \) and reverse reactions \( E_{ar} \): \[ \Delta H = E_{af} - E_{ar} \] Given that \( E_{af} = 180 \ \text{kJ/mol} \) and \( E_{ar} = 200 \ \text{kJ/mol} \), the original enthalpy change is \[ \Delta H = 180 - 200 = -20 \ \text{kJ/mol} \] This indicates the reaction is exothermic by 20 kJ/mol.

Step 2: Calculate Activation Energies with Catalyst

The presence of a catalyst reduces the activation energy for both the forward and reverse reactions by \( 100 \ \text{kJ/mol} \). Therefore: * New forward activation energy \( E_{af}' = 180 - 100 = 80 \ \text{kJ/mol} \) * New reverse activation energy \( E_{ar}' = 200 - 100 = 100 \ \text{kJ/mol} \)

Step 3: Determine New Enthalpy Change with Catalyst

The catalyst changes the activation energies but does not change the overall enthalpy change of the reaction. Therefore, \[ \Delta H' = E_{af}' - E_{ar}' = 80 - 100 = -20 \ \text{kJ/mol} \] The enthalpy change \( \Delta H \) with the presence of the catalyst remains the same as the original one, \(-20 \ \text{kJ/mol} \).

Key Concepts

Enthalpy Change

Enthalpy Change, often represented as \( \Delta H \), plays a crucial role in understanding chemical reactions. It refers to the total heat absorbed or released during a reaction at constant pressure. When calculating the enthalpy change for a reaction, you subtract the activation energy of the reverse reaction from the activation energy of the forward reaction:\[\Delta H = E_{af} - E_{ar}\]The sign of \( \Delta H \) helps us understand whether the reaction is exothermic or endothermic. If \( \Delta H \) is negative, as in \( \Delta H = -20 \, \text{kJ/mol} \), the reaction releases energy and is exothermic. Conversely, a positive \( \Delta H \) indicates an endothermic process where energy is absorbed.

• Understanding how to calculate and interpret \( \Delta H \) is essential for predicting reaction behavior.
• The enthalpy change does not alter with the presence of a catalyst.

In our example, the reaction remains exothermic with the catalyst, confirming that it releases 20 kJ/mol of energy.

Catalysts in Reactions

Catalysts are fascinating substances that accelerate reactions by lowering the activation energy required for the reaction to proceed. Importantly, they achieve this without being consumed themselves, effectively participating without undergoing permanent change. In the given exercise, the catalyst reduces both forward and reverse activation energies by \( 100 \, \text{kJ/mol} \), which is significant.

• The new activation energy for the forward reaction, \( E_{af}' \), becomes \( 80 \, \text{kJ/mol} \) after the catalytic effect.
• The reverse reaction activation energy, \( E_{ar}' \), shifts to \( 100 \, \text{kJ/mol} \).

This reduction makes the reaction faster, as less energy is needed to start the process. However, it is crucial to note that while catalysts change the rate of reactions, they do not affect the enthalpy change \( \Delta H \) of the reaction. The overall energy difference between reactants and products remains constant.

Exothermic Reactions

Exothermic reactions are those that release energy to the surroundings, typically in the form of heat. These reactions have a negative enthalpy change \( \Delta H \). In our context, the reaction described is exothermic with \( \Delta H = -20 \, \text{kJ/mol} \), indicating it releases 20 kJ of energy per mole as the reaction proceeds from reactants to products.

• This energy release often results in an increase in temperature, observable in the reaction environment.
• Exothermic reactions are favorable as they are driven by the energy release.

Such reactions can be seen in everyday processes like combustion, where energy release is apparent. It's vital to recognize that the presence of a catalyst in exothermic reactions like ours inspires no change in the amount of energy released; it merely facilitates reaching the energetic threshold more efficiently. Understanding this can help predict how conditions influence reaction rates and outcomes.