Answer:
-2546 kJ
Explanation:
It is possible to obtain the enthalpy of a reaction from the sum of different intermediate reactions.
For the reaction:
4B(s) + 3O₂(g) → 2B₂O₃(s)
The intermediate reactions are:
A- B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g), ΔH°A= +2035 kJ
B- 2B(s) + 3H₂(g) → B₂H₆(g), ΔH°B= +36 kJ
C- H₂(g) + 1/2O₂(g) → H₂O(l), ΔH°C= -285 kJ
D- H₂O(l) → H₂O(g), ΔH°D= +44 kJ
2B = 4B(s) + 6H₂(g) → 2B₂H₆(g) ΔH°2B= +78 kJ
-2A = 6O₂(g) + 2B₂H₆(g) → 2B₂O₃(s) + 6H₂O(g) ΔH°-2A= -4070 kJ
-6C = 6H₂O(l) → 6H₂(g) + 3O₂(g) ΔH°-6C= +1710 kJ
-6D = 6H₂O(g) → 6H₂O(l) ΔH°-6D = -264 kJ
The sum of 2B - 2A - 6C - 6D produce:
4B(s) + 3O₂(g) → 2B₂O₃(s)
And the enthalpy is: ΔH°2B + ΔH°-2A + ΔH°-6C + ΔH°-6D = <em>-2546 kJ</em>
I hope it helps!
