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Thermodynamics question

2007 · Q99
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Thermodynamics question

2007 · Q99

NEETChemistryThermodynamicsMCQ+4 / −1
Given that bond energies of H −-− H and Cl −-− Cl are 430 kJ mol−-−1 and 240 kJ mol−-−1 respectively and Δ\DeltaΔHf for HCl is −-− 90 kJ mol−-−1, bond enthalpy of HCl is
  1. A
    380 kJ mol−-−1
  2. B
    425 kJ mol−-−1
  3. C
    245 kJ mol−-−1
  4. D
    290 kJ mol−-−1
View written solutionFree

Correct answer: B

12{1 \over 2}21​H2 + 12{1 \over 2}21​Cl2 →\to→ HCl

Δ\Delta ΔHf = –90 kJ mol–1

⇒\Rightarrow⇒ Δ\Delta ΔHf = [ 12{1 \over 2}21​(B.E)H2 + 12{1 \over 2}21​(B.E)Cl2] - (B.E)HCl

⇒\Rightarrow⇒ -90 = [ 12{1 \over 2}21​(430)H2 + 12{1 \over 2}21​(240)Cl2] - (B.E)HCl

⇒\Rightarrow⇒ -90 = [215 + 120] - (B.E)HCl

⇒\Rightarrow⇒ (B.E)HCl = 425 kJ mol–1

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