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!

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.

Thank you very much!

How is it 36kj

To calculate ∆H for the synthesis of diborane (B2H6) from its elements, you need to use Hess's Law, which states that the enthalpy change of a chemical reaction is the same regardless of the route taken as long as the initial and final conditions are the same. Here's the correct approach:

1. You correctly identified the relevant equations:
a. 2 B (s) + 3/2 O2 (g) → B2O3 (s) ∆H = -1273 kJ
b. H2 (g) + 1/2 O2 (g) → H2O (l) ∆H = -286 kJ
c. B2H6 (g) + 3 O2 (g) → B2O3 (s) + 3 H2O (g) ∆H = -2035 kJ
d. H2O (l) → H2O (g) ∆H = 44 kJ

2. You need to manipulate these equations to obtain the desired equation:
a. Reverse equation a: B2O3 (s) → 2 B (s) + 3/2 O2 (g) ∆H = +1273 kJ
b. Scale equation b by 2 to match the coefficient of H2 in equation c: 2 H2 (g) + O2 (g) → 2 H2O (l) ∆H = 2(-286 kJ) = -572 kJ
c. Reverse and scale equation c by 2/3 to match the coefficient of B2H6 in the desired equation: 2/3 (B2O3 (s) + 3 H2O (g) → B2H6 (g) + 3 O2 (g)) ∆H = (2/3)(-2035 kJ) = -1357 kJ
d. Reverse equation d: H2O (g) → H2O (l) ∆H = -44 kJ

3. Now, add equations a, b, and c to obtain the desired equation:
B2O3 (s) + H2O (g) + O2 (g) → 2 B (s) + 3 H2 (g) + B2H6 (g) + 3 O2 (g) ∆H = 1273 kJ - 572 kJ - 1357 kJ = -656 kJ

So, the ∆H for the synthesis of diborane from its elements is -656 kJ, not -52 kJ. It seems there was an error in your calculation, possibly a sign mistake. The correct answer is -656 kJ, not 36 kJ.