From the following data, determine the Heat of formation of

Question

From the following data, determine the Heat of formation of

diborane, B2H6 (g), at standard conditions (all heat in kJ/mol):
(1) B2H6 (g) + 3O2 (g) → B2O3 (s) + 3H2O (g) ∆𝐻𝑓
° = -1941
(2) 2B (s) + 3/2O2 (g) → B2O3 (s) ∆𝐻𝑓
° = -2368
(3) H2 (g) + ½O2 (g) → H2O (g) ∆𝐻𝑓
° = -241.8

Answer 1

To determine the heat of formation of diborane (B2H6), we can use the heats of formation of the other compounds involved in the reaction.

The balanced equation for the reaction is:
B2H6 (g) + 3/2 O2 (g) → B2O3 (s) + 3 H2O (g)

We can use the given heats of formation to calculate the heat of formation of diborane.

The heat of formation of B2O3 (s) is given as -1941 kJ/mol.
The heat of formation of H2O (g) is given as -241.8 kJ/mol.
The heat of formation of O2 (g) is assumed to be 0 kJ/mol since it is in its standard state.
The heat of formation of B (s) is not given directly, but we can use the heat of formation of B2O3 (s) from the second equation to calculate it.
The heat of formation of B2O3 (s) is given as -2368 kJ/mol.

Using the given data, we can write the equation:
B2H6 (g) + 3/2 O2 (g) → B2O3 (s) + 3 H2O (g)

The heat of formation of diborane can be calculated as:
ΔHf°(B2H6) = ΣΔHf°(products) - ΣΔHf°(reactants)
= (-1941 kJ/mol) - (2*(-2368 kJ/mol) + 3*(-241.8 kJ/mol))
= -1941 kJ/mol + 4736 kJ/mol - 725.4 kJ/mol
= 3069.6 kJ/mol

Therefore, the heat of formation of diborane (B2H6) at standard conditions is 3069.6 kJ/mol.