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

Question

Find the bond enthalpy (in kJ/mol) of one "three centre two electron bond" in $B_{2} H_{6} {B - H - B \to 2 B (g) +$ $H (g)}$ from the given data

$Δ H_{f}^{\circ} [B H_{3} (g)] = 100 k J / m o l$

$Δ H_{f}^{\circ} [B_{2} H_{6} (g)] = 36 k J / m o l$

$Δ H_{a}^{\circ} t m [B (s)] = 565 k J / m o l$

$Δ H_{a}^{\circ} t m [H_{2} (g)] = 436 k J / m o l$

TestHub · Accepted Answer

For the formation of B₂H₆(g): 2 B ( s ) + 3 H 2 ​ ( g ) → B 2 ​ H 6 ​ ( g ) Reactants atomization energy: 2 ( 565 ) + 3 ( 436 ) = 2438 k J Formation enthalpy: 36 = 2438 − ( e n er g yo f b o n d s ) Total bond energy in B₂H₆: = 2438 − 36 = 2402 k J In diborane: 4 terminal B–H bonds + 2 three-center bonds 4 ( 400 ) + 2 E = 2402 E = 455 k J / m o l

Chemistry - Thermochemistry Question with Solution | TestHub

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