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$$^{-1}$$]

The combustion taking place inside the bomb calorimeter is
$$Hg(g)+\frac12\,O_2(g)\rightarrow HgO(s)$$

Step 1 : Heat evolved at constant volume (ΔU)
The bomb calorimeter provides the change in internal energy because its volume is fixed.

Temperature rise $$\Delta T = 312.8\,\text{K} - 298.0\,\text{K} = 14.8\,\text{K}$$

Heat capacity of the bomb $$C_{\text{cal}} = 20.00\;\text{kJ K}^{-1}$$

Heat released for 2 mol Hg(g):
$$q_v = -\,C_{\text{cal}}\;\Delta T = -\,(20.00)\,(14.8) = -296\;\text{kJ}$$

Hence for 1 mol Hg(g):
$$\Delta U^{\circ}_{\text{rxn}} = \frac{-296\;\text{kJ}}{2} = -148\;\text{kJ mol}^{-1}$$

Step 2 : Convert ΔU to ΔH for the gaseous-Hg reaction
For any reaction, $$\Delta H = \Delta U + \Delta n_{\text{gas}}\,R\,T$$, where $$\Delta n_{\text{gas}}$$ is the change in moles of gaseous species.

In the reaction, gaseous moles change as
$$\Delta n_{\text{gas}} = n_{\text{products,\,gas}} - n_{\text{reactants,\,gas}} = 0 - \left(1 + \frac12\right) = -1.5$$

Therefore
$$\Delta H^{\circ}_{1} = -148\;\text{kJ mol}^{-1} + (-1.5)(8.3\times10^{-3}\;\text{kJ mol}^{-1}\text{K}^{-1})(298\;\text{K})$$

$$\Delta H^{\circ}_{1} = -148\;\text{kJ mol}^{-1} - 3.71\;\text{kJ mol}^{-1} = -151.71\;\text{kJ mol}^{-1}$$

This is the enthalpy change for
$$Hg(g)+\frac12\,O_2(g)\rightarrow HgO(s)$$

Step 3 : Bring the reactants to their standard states
In the standard enthalpy of formation, mercury must start as liquid Hg(l).
The given data supply the enthalpy of formation of gaseous Hg:

$$Hg(l)\rightarrow Hg(g)\;;\qquad \Delta H^{\circ} = +61.32\;\text{kJ mol}^{-1}$$

Adding this step to the gaseous-Hg reaction yields the desired formation reaction:
$$Hg(l)+\frac12\,O_2(g)\rightarrow HgO(s)$$

Hence
$$\Delta H_f^{\circ}\bigl(HgO,s\bigr)= +61.32\;\text{kJ mol}^{-1} + (-151.71\;\text{kJ mol}^{-1}) = -90.39\;\text{kJ mol}^{-1}$$

Answer
The standard molar enthalpy of formation of $$HgO(s)$$ at 298 K is
$$\boxed{\Delta H_f^{\circ}(HgO,s)= -90.39\;\text{kJ mol}^{-1}}$$
Thus, $$|X| = 90.39$$.