Given that the average bond enthalpies, estimate the enthalpy of atomisation for the following diborane (H2B-BH2)

1. Break Down the Problem

To estimate the enthalpy of atomization for diborane (H₂B-BH₂), we need to consider the individual bonds that are present in the molecule and the average bond enthalpies for each type of bond. The molecule consists of B-H and B-B bonds.

2. Relevant Concepts

The enthalpy of atomization is defined as the energy required to break all the bonds in a molecule to convert it into its gaseous atoms. We will use average bond enthalpy values for B-H and B-B bonds to calculate this value.

3. Analysis and Detail

1. Identify the bonds in diborane:

   • Diborane (H₂B-BH₂) has:
     • 4 B-H bonds (2 in each BH₂ group)
     • 1 B-B bond (the bond between the two boron atoms)

2. Average bond enthalpy values:

   • B-H bond enthalpy: Approximately 400 kJ/mol (can vary in sources).
   • B-B bond enthalpy: Approximately 200 kJ/mol.

3. Calculate the total enthalpy of atomization: $\text{Enthalpy of atomization} = \text{(Number of B-H bonds × Average B-H bond enthalpy)} + \text{(Number of B-B bonds × Average B-B bond enthalpy)}$ $= (4 \times 400 \, \text{kJ/mol}) + (1 \times 200 \, \text{kJ/mol})$ $= 1600 \, \text{kJ/mol} + 200 \, \text{kJ/mol} = 1800 \, \text{kJ/mol}$

4. Verify and Summarize

The calculations have been verified, and each bond has been accounted for with the average bond enthalpies used.

Final Answer

The estimated enthalpy of atomization for diborane (H₂B-BH₂) is approximately 1800 kJ/mol.