Diborane (B 2 H 6 ) is a highly reactive boron hydride, which was once considere

Question

Diborane (B2H6) is a highly reactive boron hydride, which was once considered as a possible rocket fuel for the US space program. Calculate the DH for the combustion of diborane.

                        B2H6(g) + 3 O2(g)    ------> B2O3(s) + 3 H2O(g)                     DH = ?

                        2 B(s) + 3 H2(g)       ------> B2H6(g)                                         DH =      36kJ

                        2 B(s) + 3/2O2(g)      --------> B2O3(s)                                          DH = -1273kJ

                        H2(g) + 1/2 O2(g)      --------> H2O(l)                                           DH =   -286kJ

                                         H2O(l)        -----> H2O(g)                                          DH =      44kJ

            a) –3651kJ             b) -2035kJ            c) -1177kJ          d) 1963kJ           e) 3651kJ

I thought i knew Hess's law but i kept getting -2123 which isnt an answer. Please help.

Explanation / Answer

Answer:

2 B(s) + 3 H2(g)       ------> B2H6(g) DH1 =      36kJ ---------- (I)

2 B(s) +  3/2O2(g)      --------> B2O3(s) DH2 = -1273kJ -----------(II)

H2(g) +  1/2 O2(g)      --------> H2O(l) DH3 =   -286kJ -----------(III)

H2O(l)        -----> H2O(g) DH4 =      44kJ -----------(IV)

a) By Eq.(II) - Eq.(I) we get Eq.(5) and DH5 as,

2 B(s) +  3/2 O2(g) - 2 B(s) - 3 H2(g) ---------> B2O3(s) - B2H6(g)

i.e. 3/2 O2(g) - 3 H2(g) ---------> B2O3(s) - B2H6(g)

i.e. B2H6(g) + 3/2 O2(g) --------> B2O3(s) + 3 H2(g)

Accordingly, DH5 = DH2 - DH1 = (-1273) - (+36) = -1309 kJ

Hence,

B2H6(g) + 3/2 O2(g) --------> B2O3(s) + 3 H2(g) DH5 = -1309 kJ ----------- (V)

b) By Eq(III) x 3 - Eq.(IV) x 3 we get Eq.(VI) and DH6 as,

3 H2(g) + 3/2 O2(g) + 3 H2O(l) --------> 3 H2O(l) + 3 H2O(g)

i.e. 3 H2(g) + 3/2 O2(g) --------> + 3 H2O(g) ---------------- (VI)

Accordingly, DH6 = DH3+DH4 = 3x (-286) + 3 x (+44) = - 726 kJ

Finally,

3 H2(g) + 3/2 O2(g) --------> 3 H2O(g) DH6 = - 242kJ ---------------- (VI)

On Eq.(V) + Eq.(VI)

B2H6(g) + 3/2 O2(g) + 3 H2(g) + 3/2 O2(g) --------> B2O3(s) + 3 H2(g)  + 3 H2O(g)

I.e. B2H6(g) + 3 O2(g) --------> B2O3(s) + 3 H2O(g)

Is the conversion for which DH asked to find.

Accordinly,

DH = DH5 + DH6 = (-1309) + (-726) = -2035 kJ

DH = - 2035 kJ

Answer option : (b) - 2035 kJ.

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