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 a. -1486.1 kj b. -1632.5 kj c. -4396.7 kj d. -4652.85 kj e. -9086.1 kj

- None of the choice is correct. The enthalpy for the combustion of 1 mol of pentaborane is -4543.05 kJ and so none of the choice is correct.

Solution:- First of all we write the balanced equation for the combustion of one mol of pentaborane:

Let's plug in the given values in the formula:

= -3183.75 - 1286.1 - 73.2

= -4543.05 kJ

From the above calculations, the enthalpy for the combustion of 1 mol of pentaborane is -4543.05 kJ and so none of the choice is correct.

the answer for sodium is 0.186

the answer for rubidium is 0.251