Pentaborane-9, BsHs, is a colorless, high reactive liquid that will burst into flame or even explode when exposed to oxygen
The reaction is 2 BsH9()12 02(g)5 B0 (s)+9 H20 (I) Pentaborane-9 was considered as a potential rocket fuel in the 1950s because it produces a large amount of heat per gram. However, the solid B20s formed by the combustion of BsO is an abrasive that would quickly destroy the nozzle of the rocket, and so the idea was abandoned Calculate the kilojoules of heat released per gram of the compound reacted with oxygen.
The standard enthalpy of formation of BsH9 is 73.2 KJ/mol, B203is-1263.6 KJ/mol, and H20 is-285.8 KJ/mol
To calculate the kilojoules of heat released per gram of Pentaborane-9 reacted with oxygen, we need to use the balanced equation and the enthalpy of formation values provided.
First, let's calculate the amount of heat released per mole of Pentaborane-9 reacted. Looking at the balanced equation:
2 BsH9 (l) + 12 O2 (g) -> 5 B2O3 (s) + 18 H2O (l)
We can see that for every 2 moles of BsH9 reacted, 5 moles of B2O3 are formed. So, the heat released per mole of BsH9 reacted can be calculated as follows:
ΔH = (5 mol B2O3) * (-1263.6 KJ/mol B2O3) - (2 mol BsH9) * (73.2 KJ/mol BsH9)
ΔH = -4531.8 KJ
Next, we need to calculate the molar mass of Pentaborane-9 (BsH9). From the molecular formula, we can see that it contains 1 boron atom, 9 hydrogen atoms, and a total molecular weight of 9 * (1.01 g/mol) + 1 * (10.81 g/mol) = 18.9 g/mol.
Now, to calculate the heat released per gram of BsH9, we divide the enthalpy change by the molar mass:
Heat released per gram = (-4531.8 KJ) / (18.9 g/mol) = -239.9 KJ/g
Therefore, the kilojoules of heat released per gram of Pentaborane-9 reacted with oxygen is approximately 239.9 KJ/g.