Answer:
3,956 KJ
Explanation:
Using the formula below, we can obtain the enthalpy of the reaction:
ΔH = ∑H products ± ∑H reactants. --------------------------------(Formula 1)
∑H products = Sum total of all enthalpies at the product side (left side) of the reaction.
∑H reactants = Sum total of all enthalpies at the reactant side (right side) of the reaction.
∑H products can be obtained by checking the standard enthalpy of reaction of [tex]B_{2}O_{3}[/tex]. This can be obtained from the chemical reaction given in (A) above Δ[tex]H^{0}[/tex] (A) = +2035 KJ
∑H reactants can be obtained by checking the standard enthalpy of reaction of Boron and Oxygen.
The enthalpy of formation of [tex]O_{2}[/tex] = 0. This is a law in thermodynamics. The standard enthalpy of formation of an element in its naturally ocurring state is always = 0 KJ
The enthalpy of formation of Boron can be obtained from the reaction given in equation (b) as Δ[tex]H^{0}[/tex] (B) = +36 KJ
Applying the variables to formula (1) above, we have that
ΔH = 2×(2035) - 4×(36) + 3×(0)
ΔH = 4070 - 114
ΔH = 3,956 KJ
