Calculate bond enthalpy of H − Cl bond from following reaction. H 2 ( g ) + Cl 2 ( g ) → 2 HCl ( g ) , Δ r H ∘ = − 185 kJ (Given bond enthalpies of H − H , Cl − Cl bonds are 435.0 kJ mol − 1 and 244 kJ mol − 1 respectively)

Question

Calculate bond enthalpy of H − Cl bond from following reaction. H 2 ​ ( g ) + Cl 2 ​ ( g ) → 2 HCl ( g ) , Δ r ​ H ∘ = − 185 kJ (Given bond enthalpies of H − H , Cl − Cl bonds are 435.0 kJ mol − 1 and 244 kJ mol − 1 respectively), 340 kJ, 432 kJ, 370 kJ, 864 kJ

MARKS App · Accepted Answer

​ Δ H r ∘ ​ = ( B ⋅ E ) in ​ − ( B ⋅ E ) out ​ Δ H r ∘ ​ = B ⋅ E H − H ​ + B ⋅ E Cl − Cl ​ − 2 × B ⋅ E H − Cl ​ − 185 = 435 + 244 − 2 x 2 x = 435 + 244 + 185 x = 432 KJ ​

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