Question

$2 \,mol$ of $Hg ( g )$ is combusted in a fixed volume bomb calorimeter with excess of $O _2$ at $298 K$ and $1 \,atm$ into $HgO ( s )$. During the reaction, temperature increases from $298.0 \,K$ to $312.8 \, K$. If heat capacity of the bomb calorimeter and enthalpy of formation of $Hg ( g )$ are $20.00 \, kJ \,K ^{-1}$ and $61.32 \, kJ \,mol ^{-1}$ at $298 \,K$, respectively, the calculated standard molar enthalpy of formation of $HgO ( s )$ at $298 K$ is $X kJ mol { }^{-1}$. The value of $| X |$ is [Given : Gas constant $R =8.3 \,J K ^{-1} \, mol ^{-1}$ ]

Answer 1

$ Q _{ rxn }= C \Delta T $
$ |\Delta U | \times 2=20 \times 14.8 $
$|\Delta U |=148 \,kJ / mol $
$ \Delta U =-148\, kJ / mol $
$ Hg ( g )+\frac{1}{2} O _2( g ) \longrightarrow HgO ( s ): \Delta U =-148 \,kJ / mol $
$ \Delta H =\Delta U +\Delta n _{ g } RT$
$ =-148-\frac{3}{2} \times \frac{8.3}{1000} \times 298=-151.7101 $
$ Hg (l)+\frac{1}{2} O _2( g ) \longrightarrow HgO ( s ) $
$ \Delta H =-151.7101+61.32=-90.39\, kJ / mol $