At 298.2 K the relationship between enthalpy of bond dissociation (in ) for hydrogen () and its isotope, deuterium (), is best described by:

(1) (2) (3) (4)

Question

At 298.2 K the relationship between enthalpy of bond dissociation (in ) for hydrogen () and its isotope, deuterium (), is best described by:

(1)   (2)   (3)   (4)

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Strategy:
 Analyze bond strength based on zero-point energy (ZPE). Heavier isotopes have lower ZPE, which places them deeper in the potential well and makes their bonds harder to break.

Step 1: Isotope effect
Deuterium () is heavier than protium (). The vibrational frequency is lower, so the ZPE is lower for  compared to .

Step 2: Bond dissociation energy ()
Because  starts from a lower energy level, it requires more energy to reach the dissociated state ().

Step 3: Conclusion
The established

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