Question

Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is 2B5H9(l) + 12O2(g) → 5B2O3(s) + 9H2O(l) Calculate the kilojoules of heat released per gram of the compound reacted with oxygen. The standard enthalpy of formations of B5H9(l), B2O3(s), and H2O(l) are 73.2, −1271.94, and −285.83 kJ/mol, respectively.

Answer 1

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as
`\Delta H^o`

The equation used to calculate enthalpy change is of a reaction is:

`\Delta H^o_(rxn)=\sum [n* \Delta H^o_f(product)]-\sum [n* \Delta H^o_f(reactant)]`

The equilibrium reaction follows:

`2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)`

The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(n_((B_2O_3))* \Delta H^o_f_((B_2O_3)))+(n_((H_2O))* \Delta H^o_f_((H_2O)))]-[(n_((B_5H_9))* \Delta H^o_f_((B_5H_9)))+(n_((O_2))* \Delta H^o_f_((O_2)))]`

We are given:

`\Delta H^o_f_((B_5H_9(l)))=73.2kJ/mol\\\Delta H^o_f_((O_2(g)))=0kJ/mol\\\Delta H^o_f_((B_2O_3(s)))=-1271.94kJ/mol\\\Delta H^o_f_((H_2O(l)))=-285.83kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(5* -1271.94)+(9* -285.83)]-[(2* 73.2)+(12* 0)]=-9078.57kJ/mol`

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

So, 1 moles of compound released heat =
`(-9078.57)/(2)=-4539.28kJ`

For per gram of compound:

Molar mass of
`B_5H_9` = 63.12 g/mole

`\Delta H^o_(rxn)=(-4539.28)/(63.12)=-71.915kJ/g`

Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g