For the dissociation reaction, H 2 ( g ) → 2 H ( g ) Δ H = 162 Kcal heat of atomisation of H is

Δ H = Δ H (product) − Δ H (reactant) 162 = 2 × Δ H H − Δ H H 2 Δ H H = 2 162 ( ∵ Δ H H 2 = 0 ) Δ H H = 81 Kcal

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ChemistryThermochemistryEnthalpy of ReactionMedium2 minPYQ_2013

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For the dissociation reaction, $H_{2} (g) \to 2 H (g) Δ H = 162 Kcal$ heat of atomisation of $H$ is

Answer:

Solution:

$Δ H = Δ H_{(product)} - Δ H_{(reactant)}$ $162 = 2 \times Δ H_{H} - Δ H_{H_{2}}$ $Δ H_{H} = \frac{162}{2} (∵ Δ H_{H_{2}} = 0)$ $Δ H_{H} = 81 Kcal$

Stream:BITSATSubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction

⏱ 2mℹ️ Source: PYQ_2013

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