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Question #44177
The Question
Consider the gaseous tetrahedral molecule which undergoes dissociation into two diatomic molecules at . The single bond energy in is and the double bond energy in is . The standard entropy change for the dissociation reaction is . Which of the following correctly represents the standard enthalpy (), standard Gibbs free energy () of the reaction, and the thermodynamic stability of at ?
Options
- A.
+120 \text{ kJ mol}^{-1}, -15 \text{ kJ mol}^{-1}, \text{Unstable}
- B.
+120 \text{ kJ mol}^{-1}, +106.5 \text{ kJ mol}^{-1}, \text{Stable}
- C.
-200 \text{ kJ mol}^{-1}, -335 \text{ kJ mol}^{-1}, \text{Unstable}
- D.
+540 \text{ kJ mol}^{-1}, +405 \text{ kJ mol}^{-1}, \text{Stable}
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Numerical_Precision
Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.
Concepts Behind This Question
Bond Energy vs Bond Length|Gibbs Energy and Spontaneity|Entropy of Phase Change
Why This Question Is Tricky
Common Mistakes
This problem tests the critical distinction between kinetic stability (high total bond energy) and thermodynamic stability ().
Students often fall into the trap of assuming that a molecule with a massive total bond dissociation energy must be thermodynamically stable.
The 'Hidden Variable' is the entropy dominance () which drives the endothermic dissociation process spontaneously at room temperature.
Additionally, it tests structural geometry knowledge: a common trap is assuming a tetrahedral cluster has 4 bonds instead of 6 (Option C), or forgetting the stoichiometric coefficient of the products (Option D), or making a decimal error in the entropy unit conversion (Option B).
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Chemical Bonding and Molecular Structure - Class 11 Part 1 |
| Concept | Bond Energy vs Bond Length |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |