For chemical reaction C(s)graphite + 1/2CO 2 (g) → CO(g) + x kcal , at 27ºC & 1 atm (B.E. = Bond dissociation enthalpy) List - I List - II (P) B.E.(Product) -B.E.(Reactant) − Δ H sublimation of C (1) -x kcal (Q) ( Δ H − Δ U ) (2) 0 (R) Δ H f ( CO ( g )) (3) 0.3 kcal (S) Δ H f ( O 2 ( g ) ) (4) +x kcal (5) − 0.3 kcal

Δ H = x kcal Δ H = Δ U + Δ n g RT Δ H − Δ U = 2 1 × 1000 2 × 300 = 0.3 kcal .

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ChemistryThermochemistryBond EnthalpyMedium2 minQB

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For chemical reaction

C(s)graphite + 1/2CO₂ (g) → CO(g) + x kcal , at 27ºC & 1 atm

(B.E. = Bond dissociation enthalpy)

List - I	List - II
(P) B.E.(Product) -B.E.(Reactant) $- Δ H_{sublimation of C}$	(1) -x kcal
(Q) ( $Δ H - Δ U$ )	(2) 0
(R) $Δ H_{f} (CO (g))$	(3) 0.3 kcal
(S) $Δ H_{f} (O_{2} (g))$	(4) +x kcal
	(5) $- 0.3 kcal$

Answer:

Solution:

$Δ H = x kcal$

$Δ H = Δ U + Δ n_{g} RT$

$Δ H - Δ U = \frac{1}{2} \times \frac{2 \times 300}{1000} = 0.3 kcal$ .

Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy

⏱ 2mℹ️ Source: QB

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