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 kJmol - 1 . The

Step-by-step 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 ° = ΔU ° + Δn g RT = - 296 - 3 × 8 . 3 × 298 × 10 - 3 = - 303 . 42   kJ ΔH ° = 2 ΔH f ° HgO s - 2 ΔH f ° Hg g - 303 . 42 = 2 ΔH f ° HgO s - 2 × 61 . 32 2 ΔH f ° HgO s = - 180 . 78   kJ ΔH f ° HgO s = 90 . 39   kJ   mo