Question

Diborane is formed from the elements as shown in equation $(i)$
$2B(s) + 3H_2( g ) \rightarrow B_2H_8(g)$ ..... ( i )
Given, that $H_2O(l) \rightarrow H_2O(g),\Delta H^{\circ}_1=44\,kJ$
$2B + \frac{3}{2}O_2(g) \rightarrow B_2O_3(s)\Delta H^{\circ}_2 = - 1273\,kJ$
$B_2H_6(g) + 3O_2( g ) \rightarrow B_2O_3(s) + 3H_2O(g),\Delta H^{\circ}_3 = - 2035\,kJ$
$H_2(g)+ \frac{1}{2}O_2(g) \rightarrow H_2O(l). \Delta H^{\circ}_4=-286\,kJ$
The $\Delta H^{\circ}$ for the reaction (i) is

• A. $36\, kJ$
• B. $520\, kJ$
• C. $509\, kJ$
• D. $-3550\, kJ$

Answer 1

Answer: (A)

Given $H_2O(l) \rightarrow H_2O(g),\Delta H^{\circ}_1=44\,kJ$ .....(i)
$2B(s)+\frac{3}{2}O_{2}\left(g\right)\rightarrow B_{2}O_{3}\left(s\right);\Delta H^{^{\circ}}_{2}=-1273\,kJ ....\left(ii\right)$
$B_2H_6(g) + 3 O_2(g)\rightarrow B_2O_3 (s)+ 3H-2O(g); \Delta H^{\circ}_3 = - 2035\,kJ...(iii)$
$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l); \Delta H^{\circ}_4=-286kJ....(iv)$
$2B(s) + 3H_2(g)\rightarrow B_2H_6(g) (1)$
We can obtain the above Eq. (i) by
$Eq.(ii)+ Eq.(iv) \times 3+ Eq.(i) \times 3 - Eq.(iii)$ $\therefore \Delta H^{\circ}_r = -1273 + (- 286) \times 3 + 44\times3 -(-2035)$
$=36\,kJ$