Question

For the reaction, \(2Cl(g) → Cl_2 (g)\), the correct option is :

• A. \(\Delta_rH > 0 \) and\( \Delta_rS > 0\)
• B. \(\Delta_rH > 0 \) and\( \Delta_rS < 0\)
• C. \(\Delta_rH < 0 \) and\( \Delta_rS > 0\)
• D. \(\Delta_rH < 0 \) and\( \Delta_rS < 0\)

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the thermodynamic aspects of the reaction: \(2Cl(g) \rightarrow Cl_2(g)\). 

1. Understanding the Reaction: The reaction involves the formation of a chlorine molecule (\(Cl_2\)) from chlorine atoms (\(Cl\)). This is a combination reaction resulting in the formation of a diatomic molecule from monatomic species.
2. Enthalpy Change (\(\Delta_rH\)):
   • When a bond is formed, energy is released, which implies that the reaction is exothermic. Thus, the enthalpy change, \(\Delta_rH\), would be negative (\(< 0\)).
3. Entropy Change (\(\Delta_rS\)):
   • Entropy is a measure of randomness or disorder in a system.
   • The initial state involves two moles of gaseous chlorine atoms, and the final state involves one mole of chlorine gas molecules. Going from two moles of atoms to one mole of diatomic molecules reduces the disorder, resulting in a decrease in entropy.
   • This means that the entropy change, \(\Delta_rS\), would be negative (\(< 0\)).
4. Conclusion: Considering both the enthalpy and entropy changes, the correct answer is:
   • \(\Delta_rH < 0 \) (exothermic reaction, energy is released)
   • \(\Delta_rS < 0 \) (decrease in randomness, two moles of gaseous atoms forming one mole of diatomic molecule)

Therefore, the correct option is: \(\Delta_rH < 0\) and \(\Delta_rS < 0\).