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JEE Advanced Chemistry Thermodynamics (C) 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Chemistry Question (2022) — Solution

Question

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 value of X is___[Given: Gas constant  R = 8 . 3   J   K - 1   mol - 1 ]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

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   mol - 1

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