Chemistry - Thermochemistry Question with Solution | TestHub
Find the bond enthalpy (in kJ/mol) of 'three centre two electron bond' in B2H6 from the following data: Δf H° [BH3(g)] = 100 kJ/mol; Δf H° [B2H6(g)] = 36 kJ/mol; ΔHAtomization [B(s)] = 565 kJ/mol; ΔHAtomization [H2(g)] = 436 kJ/mol.
Answer:
Solution:
First, let's calculate the bond energy for the B–H bond in BH₃.
The reaction is:
BH₃(g) tarrow B(g) + 3H(g)
The enthalpy change for this reaction is given by:
Δ H = 565 + 3(218) - 100 = 1119 kJ
Since there are three B–H bonds in BH₃, the average B–H bond energy is:
B–H = 1119/3 = 373 kJ/mol
Next, let's consider the dissociation of B₂H₆:
B₂H₆(g) tarrow 2B(g) + 6H(g)
The enthalpy change for this reaction is:
Δ H = 2(565) + 3(436) - 36 = 2402 kJ
In B₂H₆, there are four terminal B–H bonds. The energy contributed by these bonds is:
4 × 373 = 1492 kJ
The remaining energy is attributed to the two three-center, two-electron (3c-2e) bonds (also known as banana bonds). This energy is:
2402 - 1492 = 910 kJ
Since this energy is shared by two 3c-2e bonds, the energy for each 3c-2e bond is:
910/2 = 455 kJ/mol
Therefore, the answer is 0455.
