# Problem: Part B. Calculate the enthalpy of the reaction4B(s) + 3O2(g) → 2B2O3(s)Given the following pertinent information:1. B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g), ΔH°A = + 2035 kJ2. 2B(s) + 3H2(g) → B2H6(g), ΔH°B = + 36 kJ3. H2(g) + 1/2 O2(g) → H2O(l), ΔH°C = − 285 kJ4. H2O(l) → H2O(g), ΔH°D = + 44 kJExpress your answer with the appropriate units.Hess's law states that "the heat released or absorbed in a chemical process is the same whether the process takes place in one or in several steps." It is important to recall the following rules:1. When two reactions are added, their enthalpy values are added.2. When a reaction is reversed, the sign of its enthalpy value changes.3. When the coefficients of a reaction are multiplied by a factor, the enthalpy value is multiplied by that same factor.

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Part B. Calculate the enthalpy of the reaction
4B(s) + 3O2(g) → 2B2O3(s)

Given the following pertinent information:

1. B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g), ΔH°A = + 2035 kJ
2. 2B(s) + 3H2(g) → B2H6(g), ΔH°B = + 36 kJ
3. H2(g) + 1/2 O2(g) → H2O(l), ΔH°C = − 285 kJ
4. H2O(l) → H2O(g), ΔH°D = + 44 kJ