Enthalpy of formation

Question

Given the following thermochemical equations:

4B(s) + 3O2(g) → 2B2O3(s) ΔH = –2509.1 kJ

2H2(g) + O2(g) → 2H2O(l) ΔH = –571.70 kJ

B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(l) ΔH = –2147.5 kJ

Calculate the standard enthalpy of formation (in kJ.mol–1) of B2H6(g).

Answer 1

To find the ∆H_f for B₂H₆(g), we would need to know, or find, the ∆H_f for H₂O(l), and B₂O₃(s). We could then use B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(l) to find the ∆H_f for B₂H₆(g) knowing ΔH_rxn = –2147.5 kJ

Step 1: Find ∆H_f H₂O(l) using 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = –571.70 kJ

∆H_f H₂O(l) = -571.70 kJ/2 = -285.85 kJ/mol

Step 2: Find ∆H_f B₂O₃(s) using 4B(s) + 3O₂(g) → 2B₂O₃(s) ΔH = –2509.1 kJ

∆H_f B₂O₃(s) = -2509.1 kJ/2 = -1254.85 kJ/mol

Step 3: Use the above values to find ∆Hf B2H6(g):

B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(l) ... ΔH_rxn = -2147.5 kJ

∑products - ∑reactants = ∆H_rxn

(-1254.85 + 3x -285.85) - (B₂H₆ + 0) = -2147.5

-2112.4 - X = -2147.5 where X = ∆H_f B₂H₆(g)

-X = -35.1

∆H_f B₂H₆(g) = 35.1 kJ/mol

(as always..check my math)