ChemistryEasy84×since 2003Q1249At 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 :AEH=12ED{E_H} = {1 \over 2}{E_D}EH=21EDBEH=ED{E_H} = {E_D}EH=EDCEH≃ED−7.5{E_H} \simeq {E_D} - 7.5EH≃ED−7.5DEH=2ED{E_H} = 2{E_D}EH=2EDCheck answerSkip