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← JEE Advanced 2022 · Paper 1
ChemistryNumerical+3Question 37 of 54

JEE Advanced 2022 · Paper 1, Chemistry question 1

2 mol of Hg(gg) is combusted in a fixed volume bomb calorimeter with excess of OX2\ce{O2} at 298 K and 1 atm into HgO(ss). 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(gg) are 20.00 kJ K−1^{-1} and 61.32 kJ mol−1^{-1} at 298 K, respectively, the calculated standard molar enthalpy of formation of HgO(ss) at 298 K is X kJ mol−1^{-1}. The value of ∣X∣|\mathrm X| is _____.

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

Answer and worked solution

Official answer: 89

Heat released (constant volume):
qV=−CΔT=−20.00×14.8=−296 kJ for 2 mol,q_V=-C\Delta T=-20.00\times14.8=-296\ \text{kJ for 2 mol},
so ΔU=−148\Delta U=-148 kJ per mol of Hg.

Convert to ΔH\Delta H. For Hg(g)+12 OX2(g)→HgO(s)\ce{Hg(g) + 1/2 O2(g) -> HgO(s)}, Δng=−1.5\Delta n_g=-1.5:
ΔH=ΔU+ΔngRT=−148−1.5×8.3×10−3×298=−151.71 kJ mol−1.\Delta H=\Delta U+\Delta n_gRT=-148-1.5\times8.3\times10^{-3}\times298=-151.71\ \text{kJ mol}^{-1}.

Formation enthalpy. This reaction starts from Hg(g), not Hg(l), so add the formation enthalpy of Hg(g):
ΔfH(HgO)=−151.71+61.32=−90.39 kJ mol−1,∣X∣=90.39.\Delta_fH(\ce{HgO})=-151.71+61.32=-90.39\ \text{kJ mol}^{-1},\qquad|X|=90.39.

Topics:CalorimetryHess's law and enthalpies of reactionFirst law and enthalpy
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