At 298.2 K the relationship between enthalpy of bond dissociation ($\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}$) for hydrogen ($\mathrm{E}_{\mathrm{H}}$) and its isotope, deuterium ($\mathrm{E}_{\mathrm{D}}$), is best described by:

(D) $\mathrm{E}_{\mathrm{H}} \simeq \mathrm{E}_{\mathrm{D}}-7.5$

(A) $\mathrm{E}_{\mathrm{H}}=\mathrm{E}_{\mathrm{D}}$

(B) $E_{H}=\frac{1}{2} E_{D}$

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Hydrogen

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:

(A) $E_{H} = E_{D}$
(B) $E_{H} = \frac{1}{2} E_{D}$
(C) $E_{H} = 2 E_{D}$
(D) $E_{H} ≃ E_{D} - 7.5$ ✓ Correct

Answer: (D)

Solution

The two molecules have the same electronic potential energy curve, so they differ only in the zero point energy

\frac{1}{2} h ν

they already possess at the bottom of that well. Since

ν \propto 1/ μ

and deuterium has twice the reduced mass of hydrogen,

D_{2}

has the lower zero point energy, lies deeper in the well and needs more energy to be pulled apart. The measured dissociation enthalpies bear this out:

435.9 kJ mo l^{- 1}

for

H_{2}

against

443.4 kJ mo l^{- 1}

for

D_{2}

, a gap of about

7.5 kJ mo l^{- 1}

. The relationship is therefore

E_{H} ≃ E_{D} - 7.5

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Hydrogen

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