Calculate the heat of formation of diborane $[B_{2} H_{6} (g)]$ at $298 K$ if the heat of combustion of it is $- 1941 k J / m o l$ and heats of formation of $B_{2} O_{3 (s)}$ and $H_{2} O_{(g)}$ are $- 2368 k J / m o l$ and $- 241.8 k J / m o l$ respectively.

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Solution

The combustion of diborane includes following reaction :
$B_{2} H_{6 (g)} + 3 O_{2 (g)} ⟶ B_{2} O_{3 (s)} + 3 H_{2} O_{(g)}$
According to Hess Law of heat summation :
$Δ H_{c o m b u s t i o n} = [Δ H_{f o r m B_{2} O_{3}} (s) + 3 Δ H_{f o r m H_{2} O} (g)] - Δ H_{f o r m B_{2} H_{6}}$
$Δ H_{c o m b u s t i o n} =$ Heat of combustion of diborane $= - 1941 k J$
$Δ H_{f o r m B_{2} O_{3} (g)} =$ Heat of formation of borane oxide $= - 2368 k J$
$Δ H_{f o r m H_{2} O (g)} =$ Heat of formation of water $= - 241.8 k J$
$Δ H_{f o r m B_{2} H_{6} (g)} =$ Heat of formation fo diborane
Putting the values in the equation above we get :
$- 1941 = [- 2368 + 3 (- 241.8)] - Δ H_{f o r m B_{2} H_{6}}$
$Δ H_{f o r m B_{2} H_{6}} = [- 2368 - 725.4] + 1941$
$= - 3093.4 + 1941$
$= - 1152.4 k J / m o l$ .
Heat of formation of diborane is $- 1152.4 k J / m o l$

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Calculate the heat of formation of diborane [B2H6(g)] at 298K if the heat of combustion of it is −1941kJ/mol and heats of formation of B2O3(s) and H2O(g) are −2368kJ/mol and −241.8kJ/mol respectively.

Calculate the heat of formation of diborane $[B_{2} H_{6} (g)]$ at $298 K$ if the heat of combustion of it is $- 1941 k J / m o l$ and heats of formation of $B_{2} O_{3 (s)}$ and $H_{2} O_{(g)}$ are $- 2368 k J / m o l$ and $- 241.8 k J / m o l$ respectively.