Respuesta :
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.
{5(-1273.5)+9(-285.8)}-{2*73.2+12*0}= -9086.1
e is the answer.
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:
[tex]B_5H_9(s)+6O_2(g)\rightarrow 2.5B_2O_3(s)+4.5H_2O(l)[/tex]
[tex]\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[/tex]
Let's plug in the given values in the formula:
[tex]\Delta H_r_x_n^0=[2.5(-1273.5)+4.5(-285.8)]-[73.2+6(0)][/tex]
[tex]\Delta H_r_x_n^0[/tex] = -3183.75 - 1286.1 - 73.2
[tex]\Delta H_r_x_n^0[/tex] = -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.
