Question

Silanes are silicon hydrides of general formula $S{i}_n{H}_{2n+2}$ and have several applications. From the data given below, the bond dissociation enthalpy of $Si-Si$ bond (in $kJmo{l}^{-1}$ ) is
Given:
$\Delta H$ of the reaction $2Si\left(s\right)+3H_2\left(g\right)\to S{i}_2{H}_6\left(g\right)$ is $80.3kJmo{l}^{-1}$
Bond dissociation enthalpy for $H-H=436kJmo{l}^{-1}$
Bond dissociation enthalpy for $Si-H=304kJmo{l}^{-1}$
${\left(\Delta \right)}_{f}H\left[Si\left(g\right)\right]=405kJmo{l}^{-1}$
(Note: Report your answer after rounding up to the nearest integer value.)

Answer 1

Answer: 214

$2Si\left(s\right)+3H_2\left(g\right)\to {\left(Si\right)}_2{H}_6\left(g\right)\Delta H_1=80.3{\left(kJmol\right)}^{-1}$
$H_2\left(g\right)\to 2H\left(g\right)\Delta H_2=436{\left(kJmol\right)}^{-1}$
$\Delta H_{reaction}=\sum BEofreactants-\sum BEofproducts$
$\Delta H_{reaction}=2\times \Delta H_{sublimationofSi\left(s\right)}+3\times {\left(BE\right)}_{H-H}-6\times {\left(BE\right)}_{Si-H}+{\left(BE\right)}_{si-si}$
$80.3=2\times 405+3\times 436-\left[\left(Si-Si\right)+6\left(304\right)\right]$
(Si - Si) Bond energy $=214kJmo{l}^{-1}$