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

The enthalpy of burning of 1 mole of is -4543.05 kJ/mol

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

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

For the given chemical reaction:

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

We are given:

Putting values in above equation, we get:

The enthalpy of burning of 2 moles of is coming out to be -9086.1 kJ/mol

So, enthalpy of burning of 1 mole of will be =

Hence, the enthalpy of burning of 1 mole of is -4543.05 kJ/mol

it is stable as it is in group 0 which have full outer shells of electrons