At 298.2 K,the relationship between the enth | vedclass

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(D) The enthalpy of bond dissociation at $298.2 \ K$ for hydrogen $(H_2)$ is approximately $435.88 \ kJ \ mol^{-1}$.The enthalpy of bond dissociation

Vedclass · Accepted Answer

$E_{H} \simeq E_{D} - 7.5$

Vedclass · Answer

(D) The enthalpy of bond dissociation at $298.2 \ K$ for hydrogen $(H_2)$ is approximately $435.88 \ kJ \ mol^{-1}$.The enthalpy of bond dissociation at $298.2 \ K$ for deuterium $(D_2)$ is approximately $443.35 \ kJ \ mol^{-1}$.Comparing these values,we find that $E_{H} \approx 435.88$ and $E_{D} \approx 443.35$.Calculating the difference: $E_{D} - E_{H} = 443.35 - 435.88 = 7.47 \ kJ \ mol^{-1} \approx 7.5 \ kJ \ mol^{-1}$.Therefore,the relationship is $E_{H} \simeq E_{D} - 7.5$.

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

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