Question

Part B

Calculate 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=+44 kJ

Express your answer with the appropriate units.

Hints

Answer 1

B2O3(s)+3H2O(g)→3O2(g)+B2H6(g),    ΔH∘A=+2035 kJ (1)

2B(s)+3H2(g)→B2H6(g),                            ΔH∘B=+36 kJ (2)

H2(g)+12O2(g)→H2O(l),                ΔH∘C=−285 kJ   (3)

H2O(l)→H2O(g),                                          ΔH∘D=+44 kJ (4)

Express your answer with the appropriate units.

Multiplying Eq.1 with 2 gives

2B2O3+6H2O(g) ------->6O2+ 2B2H6(g), deltaH= 2*2035= 4070 Kj (1A)

Multiplying Eq.2 with 2 and reversing it

2B2H6(g) --------->4B(s)+ 6H2(g), deltaH= -36*2=-72Kj          (2A)

Addition of Eq.1A and 2A gives

2B2O3+ 6H2O(g) ------>6O2+4B+ 6H2 (3A), deltaH= 4070-72= 3998 Kj (5)

Addition of Eq. 3 and 4 gives

H2(g)+0.5O2(g) ---> H2O(g), deltaH=-285+44=-241 Kj (6)

Multiplying Eq.6 with 6 gives

6H₂+3O₂------> 6H2O(g), deltaH= -241*6= -1446 KJ (7)

Addition of Eq.5 and Eq.7 gives

2B2O3 ----> 3O2+ 4B, deltaH=3998-1446= 2552Kj (8)

Reversing Eq.8 , 3O2+4B------->2B2O3, deltaH=-2552Kj