Alculate the enthalpy of the reaction 4b(s)+3o2(g)→2b2o3(s) given the following pertinent information: b2o3(s)+3h2o(g)→3o2(g)+b2h6(g), δh∘a=+2035 kj 2b(s)+3h2(g)→b2h6(g), δh∘b=+36 kj h2(g)+12o2(g)→h2o(l), δh∘c=−285 kj h2o(l)→h2o(g), δh∘d=+4

-2312 kJ

Explanation:

By the Hess Law, when a reaction is formed by several steps, the enthalpy change (ΔH) of the global reaction can be calculated by the sum of the enthalpy changes of the steps reactions.

Besides that, when a reaction is multiplied by a constant, ΔH must be multiplied by the same constant, and when the reaction is inverted, the signal of ΔH must be inverted.

B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g) ΔH = +2035 kJ (multiply by 2 and inverted)

2B(s) + 3H₂(g) → B₂H₆(g) ΔH = +36 kJ (multiply by 2)

H₂(g) + 1/2O₂(g) → H₂O(l) ΔH = -285 kJ (multiply by 6 and inverted)

H₂O(l) → H₂O(g) ΔH = +4 kJ (multiply by 6 and inverted)

6O₂(g) + 2B₂H₆(g) → 2B₂O₃(s) + 6H₂O(g) ΔH = -4070 kJ

4B(s) + 6H₂(g) → 2B₂H₆(g) ΔH = +72 kJ

6H₂O(l) → 6H₂(g) + 3O₂(g) ΔH = +1710 kJ

6H₂O(g) → 6H₂O(l) ΔH = -24 kJ

Summing the equation, simplifying the equal terms in both sides:

4B(s) + 3O₂(g) → 2B₂O₃(s)

ΔH = -4070 + 72 + 1710 - 24

ΔH = -2312 kJ

fdvwsa gfwsargebgeebbfwfwffw gwf   vgwgw gw f gwgw fgw v v dsv dwfff