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Grade upto college level Physical Chemistry

Diborane is a potential rocket fuel which undergoes combustion according to the reaction.
B2H6 (g) +3 O2 (g) → B2O3 (s) + 3.H2O (g)
From the following data, calculate the enthalpy change for the combustion of diborane.
2B(s) + 3/2 O2 (g) → B2O3(s) ∆H = - 1273 kJ mol-1
H2(g) + 1/2 O2(g) → H2O(ℓ) ∆H = 44kJ mol-1
2B(s) + 3H2 (g) → B2H6(g) ∆H = 36 kJ mol-1

Profile image of Shane Macguire
12 Years agoGrade upto college level
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1 Answer

Profile image of Deepak Patra
12 Years ago
Hello Student,
Please find the answer to your question
The chemical reaction for combustion of diborane is
B2H6(g) 3O2(g) → B2O3 (s) + 3H2O (g), ∆H = ?
For this the enthalpy change can be calculated in the following way.
∆H =[\Delta H_{B_{2}O_{3}(S)} + 3\Delta H_{H_{2}O(g)}] - \Delta H_{B_{2}O_{6}(g)} ;
(∵\Delta H_{f}^{o}of O2 = 0)
\Delta H_{H_{2}O(g)}can be obtained by adding 3H_{H_{2}O(l)}and
\Delta H_{H_{2}O(g)}, i.e. – 286 + 44 = - 242 kJ mol-1
∆H = [- 1273 + 3 * - 242] – 36 kJ mol-1 = - 1273 – 726 – 36 = - 2035 kJ mol-1

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jitender
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