Calculate the enthalpy of the reaction

2B2H6 + 6O2=2B2O3 +6H2O

given the following pertinent information:
A. B2O3(s) + 2H2) )g) = 3O2 (g) + B2H6 (g), delta H= +2035kJ
B. 2B (s) + 3H2 (g) =B2H6 (g), delta H= +36 kJ
C. H2 (g) + 1/2)2 (g) =H2O (l), delta H=-285
D. H2O (l) = H2O (g), delta H=44

To calculate the enthalpy of the given reaction, we need to use the Hess's Law principle, which states that the overall enthalpy change of a reaction is independent of the pathway between the initial and final states.

Here are the steps to calculate the enthalpy of the reaction:

Step 1: Write the given equations and their respective enthalpy changes.
i. B2O3(s) + 2H2(g) = 3O2(g) + B2H6(g), ΔH = +2035 kJ (Equation A)
ii. 2B(s) + 3H2(g) = B2H6(g), ΔH = +36 kJ (Equation B)
iii. H2(g) + 1/2O2(g) = H2O(l), ΔH = -285 kJ (Equation C)
iv. H2O(l) = H2O(g), ΔH = +44 kJ (Equation D)

Step 2: Manipulate the given equations to match the desired equation.
- Equation A: Multiply by 2 and reverse.
- Equation B: Reverse.
- Equation C: Multiply by 6 and reverse.

The modified equations are:
i. 2B2H6(g) + 6O2(g) = 4B2O3(s) + 6H2O(l), ΔH = -2 × 2035 kJ = -4070 kJ
ii. B2H6(g) = 2B(s) + 3H2(g), ΔH = -36 kJ
iii. 3H2O(l) = 3H2(g) + 3/2O2(g), ΔH = 6 × 285 kJ = +1710 kJ
iv. H2O(g) = H2O(l), ΔH = -44 kJ

Step 3: Combine the modified equations.
Adding equations i, ii, and iii, and subtracting equation iv gives us:
2B2H6(g) + 6O2(g) = 4B2O3(s) + 6H2O(l) + 2B(s) + 4H2(g) + 3/2O2(g) - 3H2O(l)

Simplifying gives:
2B2H6(g) + 6O2(g) = 4B2O3(s) + 4B(s) + 4H2(g) + 3/2O2(g)

Step 4: Calculate the overall enthalpy change.
To obtain the enthalpy of the reaction in question, we add the enthalpy changes of the modified equations:
ΔH reaction = ΔH i + ΔH ii + ΔH iii + ΔH iv
= (-4070 kJ) + (-36 kJ) + (+1710 kJ) + (-44 kJ)
= -3440 kJ

Therefore, the enthalpy of the reaction 2B2H6 + 6O2 = 2B2O3 + 6H2O is -3440 kJ.