Question

given the following thermochemical equations:
4B(s) + 3O2(g) -> 2B2O3(s) deltaH=-2509.1 kJ
2H2(g) + O2(g) -> 2H2O(l) deltaH=-571.70 kJ
B2H6(g) + 3O2(g) -> B2O3(s) + 3H2O(l) deltaH=-2147.5 kJ
calculate the standard enthalpy of formation (in kJ.mol^-1) of B2H6(g)

Answer 1

Use eqn 1 as is.
Multiply eqn 2 by 3 and add to eqn 1.
Multiply eqn 3 by 2, reverse it, add to eqn 1 and 2.
Cancel those materials that appear on both sides of the equation.
If I've not goofed that will give you
4B + 6H2 ==> 2B2H6 but that is twice the eqn you want so divide the dH value by 2 to give you the value for 2B + 3H2 ==> B2H6.....dH = ?
When multiplying an eqn remember to multiply the dH value also.
When reversing an eqn remember to change the sign of the dH value also.
MOST IMPORTANT: I did this in my head and could have made a mistake so put all of this down on paper, make sure all of those extra items such as H2O and O2 cancel and that the equation really is what I found. If I've made an error, you see how it's done. Finally, post your work if you run into trouble and need more help.