At 298.2 K the relationship between enthalpy of bond dissociation (in kJmol -1 ) for hydrogen (EH) and its isotope, deuterium (ED), is best described by:

Question

At 298.2 K the relationship between enthalpy of bond dissociation (in kJmol -1 ) for hydrogen (EH) and its isotope, deuterium (ED), is best described by: (A) EH=(1/2) ED (B) EH=ED (C) EH ≃ ED-7.5 (D) EH=2 ED

Tardigrade Admin · Accepted Answer

Enthalpy of bond dissociation ( kJ / mole ) at 298.2 K For, hydrogen =435.88 For, Deuterium =443.35 ∴ EH ≃ ED-7.5

Q. At $298.2 K$ the relationship between enthalpy of bond dissociation (in $k J m o l^{- 1}$ ) for hydrogen $(E_{H})$ and its isotope, deuterium $(E_{D})$ , is best described by:

$E_{H} = \frac{1}{2} E_{D}$

$E_{H} = E_{D}$

$E_{H} ≃ E_{D} - 7.5$

$E_{H} = 2 E_{D}$

Enthalpy of bond dissociation $(k J / m o l e)$ at $298.2 K$
For, hydrogen $= 435.88$
For, Deuterium $= 443.35$
$∴ E_{H} ≃ E_{D} - 7.5$

Q. At 298.2K the relationship between enthalpy of bond dissociation (in kJmol−1 ) for hydrogen (EH​) and its isotope, deuterium (ED​), is best described by:

EH​=21​ED​

EH​=ED​

EH​≃ED​−7.5

EH​=2ED​