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ChemistryThermochemistryBond EnthalpyMedium2 minQB

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The enthalpy change at 298 K for decomposition is given in following steps-

Step-1: H₂O(g) → H(g) + OH(g) $Δ H = 498 k J / m o l^{- 1}$

Step-2: OH(g) → H(g) + O(g) $Δ H = 428 k J / m o l^{- 1}$

Then, the value of the mean bond enthalpy of the O–H bond will be:

Answer:

Solution:

$\frac{498 + 428}{2} = \frac{926}{2} = 463 kJ / mol$

Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy

⏱ 2mℹ️ Source: QB

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