Question

Diborane (B2H6) is a highly reactive boron hydride, which was once considered as a possible rocket fuel for
the U.S. space program. Calculate ∆H for the synthesis of diborane from its elements, according to the
equation
2 B (s) + 3 H2 (g) → B2H6 (g)
using the following data:
∆H
2 B (s) + 3/2 O2 (g) → B2O3 (s) −1273 kJ
B2H6 (g) + 3 O2 (g) → B2O3 (s) + 3 H2O (g) −2035 kJ
H2 (g) + 1/2 O2 (g) → H2O (l) −286 kJ
H2O (l) → H2O (g) +44 kJ
Here's my answer:
a. 2 B (s) + 3/2 O2 (g) → B2O3 (s) ∆H = −1273 kJ
b. 3 H2 (g) + 3/2 O2 (g) → 3 H2O (l) ∆H = 3(−286 kJ)
C. B2O3 (s) + 3 H2O (g) → B2H6 (g) + 3 O2 (g) ∆H = +2035 kJ
d. H2O (l) → H2O (g) ∆H = 44 kJ
∆H = -52KJ, why it's not correct? correct answer is 36KJ. Thanks a lot!

Answer 1

The equation you have as d must be multiplied by 3. You've already added 1 d rxn as 44, add 2*44 more and you will have it.

Answer 2

Thank you very much!

Answer 3

How is it 36kj