Find the bond enthalpy (in kJ / mol ) of one "three-center two-electron bond" in B₂H₆ {B–H–B → 2B(g) + H(g)} from the given data: Δ H f ∘ [ B H 3 ( g )] = 100 k J / m o l Δ H f ∘ [ B 2 H 6 (

Question

Find the bond enthalpy (in kJ / mol ) of one "three-center two-electron bond" in B₂H₆ {B–H–B → 2B(g) + H(g)} from the given data: Δ H f ∘ ​ [ B H 3 ​ ( g )] = 100 k J / m o l Δ H f ∘ ​ [ B 2 ​ H 6 ​ ( g )] = 36 k J / m o l Δ H a t m ​ [ B ( s )] = 565 k J / m o l Δ H a t m ​ [ H 2 ​ ( g )] = 436 k J / m o l

TestHub · Accepted Answer

B.E. of B–H: 3 E B − H ​ = 565 + 654–100 E C − H ​ = 373 2 E B − H − B ​ = 2 × 565 + 6 × 218–36–4 × 373 E B − H − B ​ = 455 k J / m o l = 4.55 × 1 0 2 k J / m o l e

Chemistry - Thermochemistry Question with Solution | TestHub

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