Diborane is a potential rocket fuel which undergoes combustion according to the reaction $B_{2} H_{6 (g)} + 3 O_{2 (g)} \to B_{2} O_{3 (s)} + 3 H_{2} O (g)$ Form the following dada, calculation the enthalpy change for the combustion of diborane. $2 B (s) + (3 / 2) O_{2 (g)} \to B_{2} O_{3 (s)}$ $Δ H = - 1273 k J m o l^{- 1}$ $H_{2 (g)} + (1 / 2) O_{2 (g)} \to H_{2} O (1)$ $Δ H = - 286 k J m o l^{- 1}$ $H_{2} O_{(} 1) \to H_{2} O_{(} g)$ $Δ H = 44 k J m o l^{- 1}$
$2 B (s) + 3 H_{2 (g)} \to B_{2} H_{6 (g)}$
$Δ H = 36 K J m o l^{- 1}$

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Solution

$2 B (s) + \frac{3}{2} O_{2} (g) \to B_{2} O_{3} (s) Δ H = - 1273 k g m o l^{- 1}$
$H_{2} (g) + \frac{1}{2} O_{2} (s) \to H_{2} (1) Δ H = 286 k g m o l^{- 1}$
$- H_{2} O (1) \to H_{2} O (g) Δ = 44 k g m o l^{- 1}$
$- 20 (s) + 3 H_{2} (g) \to B_{2} H_{6} (g) . Δ H = 36 k g m o l^{- 1}$
$B_{2} H_{6} (g) \to 2 B (g) + 3 H (g) Δ H_{1} = - 36$
$3 \times (H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)) Δ H_{2} = - 286 \times 3$
$3 \times (H_{2} O (l) \to H_{2} O (g)) Δ H_{3} = 44 \times 3$
$2 B (s) + \frac{3}{2} O_{2} (g) \to B_{2} O_{3} (s) Δ H_{4} = - 1273$
$B_{2} H_{6} (g) + 3 O_{2} \to B_{2} O_{3} + 3 H_{2} O$
$Δ H = - 36 - 286 \times 3 + 44 \times 3 - 1273$
$Δ H = - 36 - 858 + 132 - 1273 = - 2035 k g m o l^{- 1}$

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