Question

Pentaborane b5h9(s) burns vigorously in o2 to give b2o3(s) and h2o(l). what is δh° for the combustion of 1 mol of b5h9(s)? substance δh°f (kj/mol) b2o3(s) –1273.5 b5h9(s) +73.2 h2o(l) –285.8

Answer 1

Answer: The enthalpy of burning of 1 mole of
`B_5H_9(s)` is -4543.05 kJ/mol

Step-by-step explanation:

Enthalpy change 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)]`

For the given chemical reaction:

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

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

`\Delta H^o_(rxn)=[(5* \Delta H^o_f_((B_2O_3)))+(9* \Delta H^o_f_((H_2O)))]-[(2* \Delta H^o_f_((B_5H_9)))+(12* \Delta H^o_f_((O_2)))]`

We are given:

`\Delta H^o_f_((B_2O_3))=-1273.5kJ/mol\\\Delta H^o_f_((O_2))=0kJ/mol\\\Delta H^o_f_((H_2O))=-285.8kJ/mol\\\Delta H^o_f_((B_5H_9))=73.2kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(5* (-1273.5))+(9* (-285.8))]-[(2* (73.2))+(12* 0)]\\\\\Delta H^o_(rxn)=-9086.1kJ/mol`

The enthalpy of burning of 2 moles of
`B_5H_9(s)` is coming out to be -9086.1 kJ/mol

So, enthalpy of burning of 1 mole of
`B_5H_9` will be =
`(-9086.1kJ/mol)/(2)* 1=-4543.05kJ/mol`

Hence, the enthalpy of burning of 1 mole of
`B_5H_9(s)` is -4543.05 kJ/mol

Answer 2

write a balanced chemical equation
2B5H9 + 12O2 ---> 5B205 +9 H20
From the Hess law wich state that regardless of the multiple steps of chemical reaction the total enthapy is equal to the sum of all changes
hence;
{(5 x -1273.5)+(9 x -289.8)} - { (73.2 x2) +(0 x12)}= -9086.1kj /mol