Please help me with this chemistry problem?

Question

Calculate the standard enthalpy of formation of gaseous diborane.( B2H6) given the following data.

4B(s) + 3O (g)----> 2B2O3 (s) H= -2509.1 kj
2H2(g) + O2(g) ----> 2H2O (l) H = -571.7 kj
B2H6 (g) + 3O2( g) ---> B2O3(s) + 3H2O (l) H= -2147.5 kj

Answer 1

You want the enthalpy change of the reaction:

2B + 3H2 --> B2H6, so

Divide first eq. by 2:
2B + 3/2 O2 --> B2O3 ΔH = -1254.55 kJ (1)

Multiply second eq. by 3/2:
3H2 + 3/2 O2 --> 3H2O ΔH = -857.55 kJ (2)

Invert third eq.:
B2O3 + 3H2O --> B2H6 + 3O2 ΔH = 2147.5 kJ (3)

Now add (1), (2) and (3) and you get:

2B + 3H2 --> B2H6, ΔH = 33.4 kJ

So the standard enthalpy of formation of B2H6 is 35.4 kJ.