Question

Diborane is formed from the elements as shown in equation $(i)$
$2B(s)+3H_2(g)\to B_2{H}_8(g)$ ..... ( i )
Given, that $H_{2}O(l)\to H_{2}O(g),\Delta H_1^{\circ }=44kJ$
$2B+\frac{3}{2}{O}_2(g)\to B_2{O}_3(s)\Delta H_2^{\circ }=-1273kJ$
$B_2{H}_6(g)+3O_2(g)\to B_2{O}_3(s)+3H_{2}O(g),\Delta H_3^{\circ }=-2035kJ$
$H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l).\Delta H_4^{\circ }=-286kJ$
The $\Delta H^{\circ }$ for the reaction (i) is

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

Answer 1

Answer: (A)

Given $H_{2}O(l)\to H_{2}O(g),\Delta H_1^{\circ }=44kJ$ .....(i)
$2B(s)+\frac{3}{2}{O}_2\left(g\right)\to B_2{O}_3\left(s\right);\Delta H_2^{{}^{\circ }}=-1273kJ....\left(ii\right)$
$B_2{H}_6(g)+3O_2(g)\to B_2{O}_3(s)+3H-2O(g);\Delta H_3^{\circ }=-2035kJ...(iii)$
$H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l);\Delta H_4^{\circ }=-286kJ....(iv)$
$2B(s)+3H_2(g)\to B_2{H}_6(g)(1)$
We can obtain the above Eq. (i) by
$Eq.(ii)+Eq.(iv)\times 3+Eq.(i)\times 3-Eq.(iii)$ $\therefore \Delta H_r^{\circ }=-1273+(-286)\times 3+44\times 3-(-2035)$
$=36kJ$