2 mol of Hg (g) is combusted in a fixed volume bomb calorimeter with excess of O₂ 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]

The correct answer is 90.39.

Step 1: Understand the reaction and given data
The combustion reaction for 2 moles of gaseous mercury, $\text{Hg}\left(\text{g}\right)$, taking place in a bomb calorimeter (fixed volume) is given by:

$2\text{Hg}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)\to 2\text{HgO}\left(\text{s}\right)$

We are given:
• Number of moles of Hg(g) combusted, $n=2\text{mol}$
• Initial temperature, $T_1=298.0\text{K}$
• Final temperature, $T_2=312.8\text{K}$
• Change in temperature, $\Delta T=312.8-298.0=14.8\text{K}$
• Heat capacity of bomb calorimeter, $C=20.00\text{kJ}{\text{K}}^{-1}$
• Enthalpy of formation of Hg(g), ${\Delta }_{\text{f}}{H}^{○}\left[\text{Hg}\right(\text{g}\left)\right]=61.32\text{kJ}{\text{mol}}^{-1}$
• Gas constant, $R=8.3\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}=0.0083\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}$

Step 2: Calculate the heat released and internal energy change (ΔU_rxn)
Since the reaction takes place at a constant volume, the heat evolved during the reaction is equal to the internal energy change:

$q_{\text{v}}=-C\times \Delta T$

Substitute the values:

$q_{\text{v}}=-20.00\text{kJ}{\text{K}}^{-1}\times 14.8\text{K}=-296\text{kJ}$

This heat is released for the combustion of 2 moles of Hg(g). Thus, for the reaction as written:

$\Delta U_{\text{rxn}}=-296\text{kJ}$

Step 3: Calculate enthalpy change (ΔH_rxn) of the reaction
The relation between enthalpy change and internal energy change is given by:

$\Delta H_{\text{rxn}}=\Delta U_{\text{rxn}}+\Delta n_{\text{g}}RT$

For the balanced reaction $2\text{Hg}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)\to 2\text{HgO}\left(\text{s}\right)$:

$\Delta n_{\text{g}}=n_{\text{g(products)}}-n_{\text{g(reactants)}}=0-(2+1)=-3$

Now, calculate $\Delta n_{\text{g}}RT$ at $T=298\text{K}$:

$\Delta n_{\text{g}}RT=-3\times 8.3\times {10}^{-3}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}\times 298\text{K}=-7.4202\text{kJ}$

Therefore, the enthalpy change for the reaction is:

$\Delta H_{\text{rxn}}=-296+(-7.4202)=-303.4202\text{kJ}$

Step 4: Calculate the enthalpy of formation of HgO(s)
The standard enthalpy change of the reaction can also be expressed in terms of standard enthalpies of formation:

$\Delta H_{\text{rxn}}=2\times {\Delta }_{\text{f}}{H}^{○}\left[\text{HgO}\right(\text{s}\left)\right]-\{2\times {\Delta }_{\text{f}}{H}^{○}[\text{Hg}\left(\text{g}\right)]+{\Delta }_{\text{f}}{H}^{○}[{\text{O}}_2\left(\text{g}\right)\left]\right\}$

Since ${\text{O}}_2\left(\text{g}\right)$ is in its elemental standard state, ${\Delta }_{\text{f}}{H}^{○}\left[{\text{O}}_2\right(\text{g}\left)\right]=0$.
Substituting the known values:

$-303.4202=2\times X-2\times 61.32$

$-303.4202=2X-122.64$

Solve for $X$:

$2X=-303.4202+122.64=-180.7802$

$X=\frac{-180.7802}{2}=-90.3901\text{kJ}{\text{mol}}^{-1}$

Taking the absolute value as requested:

$\left|X\right|=90.39$