Q. Diborane is a potential rocket fuel which undergoes combustion according to the equation $B_{2}H_{6}(g) + 3O_{2}(g) \rightarrow B_{2}O_{3}(g) + 3H_{2}O(g)$ Calculate the enthalpy change for the combustion of diborane. Given (i)$2B(s) + \frac{3}{2}O_{2}(g) \rightarrow B_{2}O_{3}(s);\mathrm{\Delta}H = - 1273\$kJ per mol (ii)$H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l);\ \mathrm{\Delta}H = - 286\$kJ per mol (iii)$H_{2}O(l) \rightarrow H_{2}O(g);\ \mathrm{\Delta}H = 44\$kJ per mol (iv)$2B(s) + 3H_{2}(g) \rightarrow B_{2}H_{6}(g);\mathrm{\Delta}H = 36$ kJ per mol
A) $+ 2035$ kJ per mol
B) $- 2035$ kJ per mol
C) $+ 2167$ kJ per mol
D) $- 2167$ kJ per mol
Explanation
(b)
For the equation
$B_{2}H_{6}(g) + 3O_{2}(g) \rightarrow B_{2}O_{3}(g) + 3H_{2}O(g)$
Eqs.(i) +3(ii)+3(iii)$-$(iv)
$\mathrm{\Delta}H = - 1273 + 3( - 286) + 3(44) - 36$
$= - 1273 - 858 + 132 - 36$
$= - 2035$ kJ/mol
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.