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]

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answer is 90.39.

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Detailed Solution

$2 Hg (g) + O_{2} (g) ⟶ 2 HgO (s)$
Heat capacity of calorimeter = 20 kJ K^–1
Rise in temperature = 14.8 K
Heat evolved = 20 × 14.8 = 296 kJ
${ΔH}^{\circ} = {ΔU}^{\circ} + Δ {ngRT}_{g}$
= –296 – 3 × 8.3 × 298 × 10^–3
–303.42 kJ
$\begin{array}{l} {ΔH}^{\circ} = {ΔH}_{f}^{0} (HgO (s)) - {ΔH}_{f}^{0} (Hg (g)) \\ - 303.42 = {ΔH}_{f}^{0} (HgO (s)) - 2 \times 61.32 \\ {ΔH}_{f}^{0} (HgO (s)) = - 303.42 + 122.64 \\ = - 180.78 kJ \\ |{ΔH}_{f}^{0} (HgO (s))| = 90.39 kJ {mol}^{- 1} \end{array}$