Question

+ 3 H2O (g) → 3 O2 (g) + B2H6 (g), ΔHoA = +2035 kJ 2. 2 B (s) + 3 H2 (g) → B2H6 (g), ΔHoB = +36 kJ 3. H2 (g) + 1/2 O2 (g) → H2O (l), ΔHoC = −285 kJ 4. H2O (l) → H2O (g), ΔHoD = +44 kJ Express your answer numerically, to four significant figures and in terms of kJCalculate the enthalpy of the reaction: 4 B(s) + 3 O2(g) --> 2 B2O3(s) given the following pertinent information: 1. B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g), ΔHoA = +2035 kJ 2. 2 B (s) + 3 H2 (g) → B2H6 (g), ΔHoB = +36 kJ 3. H2 (g) + 1/2 O2 (g) → H2O (l), ΔHoC = −285 kJ 4. H2O (l) → H2O (g), ΔHoD = +44 kJ Express your answer numerically, to four significant figures and in terms of kJ

Answer 1

Answer:

ΔH = -2552kJ

Explanation:

Using Hess's law, it is possible to obtain ΔH of a reaction by the sum of similar reactions, thus:

1. B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g), ΔHoA = +2035 kJ

2. 2 B (s) + 3 H2 (g) → B2H6 (g), ΔHoB = +36 kJ

3. H2 (g) + 1/2 O2 (g) → H2O (l), ΔHoC = −285 kJ

4. H2O (l) → H2O (g), ΔHoD = +44 kJ

Subtracting of 2*(2) - 2*(1):

4 B (s) + 6 H2 (g) +  6 O2 (g) → 2 B2O3 (s) + 6 H2O (g)

ΔH' = 2*36kJ - 2*2035kJ = -3998kJ

Now, this reaction - 6*(3):

4 B (s) + 3 O2(g) + 6 H2O(l) → 2 B2O3 (s) + 6 H2O (g)

ΔH'' = -3998kJ - (6*-285kJ)

ΔH'' = -2288kJ

The last reaction - 6*(4):

4 B (s) + 3 O2(g) → 2 B2O3 (s)

ΔH = -2288kJ - 6*44kJ

ΔH = -2552kJ