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

Answer 1

2B5H9(s) + 12O2(g) -> 5B2O3(s) + 9H2O(l)
{5(-1273.5)+9(-285.8)}-{2*73.2+12*0}= -9086.1

e is the answer.

Answer 2

Answer:- 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:

$B_5H_9(s)+6O_2(g)\rightarrow 2.5B_2O_3(s)+4.5H_2O(l)$

$\Delta H_r_x_n^0=[\sum \Delta H_f]_p_r_o_d_u_c_t-[\sum \Delta H_f]_r_e_a_c_t_a_n_t$

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

$\Delta H_r_x_n^0=[2.5(-1273.5)+4.5(-285.8)]-[73.2+6(0)]$

$\Delta H_r_x_n^0$ = -3183.75 - 1286.1 - 73.2

$\Delta H_r_x_n^0$ = -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.