Q.

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

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2Hg(g)+O2(g)2HgO(s)
Heat capacity of calorimeter = 20 kJ K–1
Rise in temperature = 14.8 K
Heat evolved = 20 × 14.8 = 296 kJ
ΔH=ΔU+ΔngRTg
= –296 – 3 × 8.3 × 298 × 10–3
–303.42 kJ
ΔH=ΔHf0(HgO(s))ΔHf0(Hg(g))303.42=ΔHf0(HgO(s))2×61.32ΔHf0(HgO(s))=303.42+122.64=180.78kJΔHf0(HgO(s))=90.39kJmol1

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