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Standard XII

Chemistry

Ozone

At 298.2 K th...

Question

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

E_{H} ≃ E_{D} - 7.5

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E_{H} = 2 E_{D}

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

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E_{H} = E_{D}

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Solution

The correct option is A $E_{H} ≃ E_{D} - 7.5$
$Enthalpy of bond dissociation of hydrogen = 435.88 k J m o l^{- 1}$
$Enthalpy of bond dissociation of deuterium = 443.35 k J m o l^{- 1}$
$E_{H} = E_{D} - 7.47$
$E_{H} ≃ E_{D} - 7.5$

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

The correct option is A EH≃ED−7.5 Enthalpy of bond dissociation of hydrogen = 435.88 kJmol−1 Enthalpy of bond dissociation of deuterium = 443.35 kJmol−1 EH=ED−7.47 EH≃ED−7.5

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

The correct option is A $E_{H} ≃ E_{D} - 7.5$
$Enthalpy of bond dissociation of hydrogen = 435.88 k J m o l^{- 1}$
$Enthalpy of bond dissociation of deuterium = 443.35 k J m o l^{- 1}$
$E_{H} = E_{D} - 7.47$
$E_{H} ≃ E_{D} - 7.5$