At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol⁻¹) for hydrogen (E H ) and its isotope, deuterium (E D ), is best described by :

Solution

Chemistry ma’am · 4 steps

Same bond, same well, but heavier D vibrates more slowly and sits deeper, so it needs a little more energy.

1. Same electrons. H and D differ only by a neutron, so H₂ and D₂ have the same electron cloud and the same potential energy well. The bond strength itself does not change, so options with a factor of 2 make no sense.
2. Zero-point energy. A molecule never rests at the very bottom of the well. It keeps a minimum vibrational energy 1/2hν, and the vibration frequency depends on the reduced mass μ.

   ν = 1/2π√(k/μ), μ_(H₂) = 0.5 u, μ_(D₂) = 1 u

3. Compare H and D. Doubling μ lowers ν by a factor of √2. The zero-point energy of H₂ is about 26 kJ mol⁻¹ and that of D₂ is about 18.5 kJ mol⁻¹. So D₂ sits about 7.5 kJ mol⁻¹ deeper and must climb that much further to break.
4. So. Measured values are about 436 kJ mol⁻¹ for H₂ and 443.5 kJ mol⁻¹ for D₂. That gives E_H ≃ E_D - 7.5, option (c).

The idea to keep. Isotopes share the same bond, so the difference in bond energy comes only from zero-point energy. The heavier isotope vibrates more slowly, sits lower, and has a slightly larger dissociation energy.

Answer from the NTA final answer key. Solution by PrepWiser.