JEE Challenger
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ChemMedAdv 2022 • Paper 1

Calculate Standard Molar Enthalpy of Formation of HgO

2 mol2\text{ mol} of Hg(g)\text{Hg}(g) is combusted in a fixed volume bomb calorimeter with excess of O2\text{O}_2 at 298 K298\text{ K} and 1 atm1\text{ atm} into HgO(s)\text{HgO}(s). During the reaction, temperature increases from 298.0 K298.0\text{ K} to 312.8 K312.8\text{ K}. If heat capacity of the bomb calorimeter and enthalpy of formation of Hg(g)\text{Hg}(g) are 20.00 kJ K−120.00\text{ kJ K}^{-1} and 61.32 kJ mol−161.32\text{ kJ mol}^{-1} at 298 K298\text{ K}, respectively, the calculated standard molar enthalpy of formation of HgO(s)\text{HgO}(s) at 298 K298\text{ K} is X kJ mol−1\text{X}\text{ kJ mol}^{-1}. The value of ∣X∣|\text{X}| is ______.

[Given: Gas constant R=8.3 J K−1 mol−1\text{R} = 8.3\text{ J K}^{-1}\text{ mol}^{-1}]

Official Numerical Answer90.39

Topics & Concepts

ThermodynamicsEnthalpyFirst Law of ThermodynamicsCalorimetry

Step-by-Step Solution

The standard molar enthalpy of formation of HgO(s)\text{HgO}(s) is determined by calculating the total heat released during the combustion of 2 mol2\text{ mol} of Hg(g)\text{Hg}(g) in a constant-volume bomb calorimeter, converting the resulting change in internal energy to the reaction enthalpy, and using Hess's Law.

  1. Internal Energy Change (ΔrU\Delta_r U):
    The heat evolved during combustion at constant volume equals the heat absorbed by the calorimeter: ΔrU=−CΔT=−(20.00 kJ K−1)(312.8 K−298.0 K)=−296.0 kJ\Delta_r U = -C \Delta T = -(20.00\text{ kJ K}^{-1})(312.8\text{ K} - 298.0\text{ K}) = -296.0\text{ kJ}

  2. Reaction Enthalpy (ΔrH\Delta_r H):
    For the combustion reaction 2 Hg(g)+O2(g)→2 HgO(s)2\text{ Hg}(g) + \text{O}_2(g) \rightarrow 2\text{ HgO}(s), the change in gaseous moles is Δng=0−3=−3\Delta n_g = 0 - 3 = -3. Using ΔrH=ΔrU+ΔngRT\Delta_r H = \Delta_r U + \Delta n_g R T: ΔrH=−296.0 kJ+(−3)(8.3×10−3 kJ K−1mol−1)(298 K)≈−303.42 kJ\Delta_r H = -296.0\text{ kJ} + (-3)(8.3\times 10^{-3}\text{ kJ K}^{-1}\text{mol}^{-1})(298\text{ K}) \approx -303.42\text{ kJ}

  3. Standard Enthalpy of Formation (ΔfH∘\Delta_f H^\circ):
    Relating the reaction enthalpy to the enthalpies of formation gives: ΔrH=2ΔfH∘[HgO(s)]−2ΔfH∘[Hg(g)]\Delta_r H = 2\Delta_f H^\circ[\text{HgO}(s)] - 2\Delta_f H^\circ[\text{Hg}(g)] −303.42 kJ=2ΔfH∘[HgO(s)]−2(61.32 kJ)-303.42\text{ kJ} = 2\Delta_f H^\circ[\text{HgO}(s)] - 2(61.32\text{ kJ}) ΔfH∘[HgO(s)]≈−90.39 kJ mol−1\Delta_f H^\circ[\text{HgO}(s)] \approx -90.39\text{ kJ mol}^{-1}

Thus, X=−90.39\text{X} = -90.39, so ∣X∣=90.39|\text{X}| = 90.39.

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