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प्रश्न
The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.
\[\ce{H2(g) + Br(g) -> 2HBr(g)}\]
Given that Bond energy of \[\ce{H2, Br2}\] and \[\ce{HBr}\] is 435 kJ mol–1, 192 kJ mol–1 and 368 kJ mol–1 respectively.
उत्तर
\[\ce{Δ_rH^Θ = ΣB.E_{(reactant)} - ΣB.E_{(product)}}\]
= \[\ce{B.E._{H2} + B.E._{BR2} - 2 × B.E._{HBr}}\]
= 435 + 192 – (2 × 368) kJ mol–1
= – 109 kJ mol–1
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संबंधित प्रश्न
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\[\ce{CCl_4 (g) → C(g) + 4Cl(g)}\]
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ΔfHθ (CCl4) = –135.5 kJ mol–1.
ΔaHθ (C) = 715.0 kJ mol–1, where ΔaHθ is enthalpy of atomisation
ΔaHθ (Cl2) = 242 kJ mol–1
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(a) \[\ce{C (g) + 4 H (g) -> CH4 (g); ∆_rH = xkJ mol^{-1}}\]
(b) \[\ce{C (graphic) + 2H2 (g) -> CH4 (g); ∆_rH = ykJ mol^{-1}}\]
One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?
The enthalpy of atomisation for the reaction \[\ce{CH4(g) -> C(g) + 4H(g)}\] is 1665 kJ mol–1. What is the bond energy of \[\ce{C – H}\] bond?
Use the following data to calculate \[\ce{∆_{lattice}H^Θ}\] for \[\ce{NaBr}\].
\[\ce{∆_{sub}H^Θ}\] for sodium metal = 108.4 kJ mol–1
Ionization enthalpy of sodium = 496 kJ mol–1
Electron gain enthalpy of bromine = – 325 kJ mol–1
Bond dissociation enthalpy of bromine = 192 kJ mol–1
\[\ce{∆_fH^Θ}\] for \[\ce{NaBr (s)}\] = – 325 kJ mol–1
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