2 mol of Hg(g) is combusted in a fixed volume bomb calorimeter with excess of \(\mathrm{O}_{2}\) 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 \(\mathrm{Hg}(\mathrm{g})\) are \(20.00 \mathrm{~kJ} \mathrm{~K}^{-1}\) and \(61.32 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at 298 K, respectively, the calculated standard molar enthalpy of formation of \(\mathrm{HgO}(\mathrm{s})\) at 298 K is \(\mathrm{X} \mathrm{kJ} \mathrm{~mol}^{-1}\). The value of \(|\mathrm{X}|\) is ____. [Given : Gas constant \(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) ]

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Chemical Thermodynamics

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 kJ mol^{- 1}

. The value of

∣ X ∣

is ____.
[Given : Gas constant

R = 8.3 J K^{- 1} mol^{- 1}

]

Answer: 90.39

Solution

The heat released in the bomb calorimeter is

Q_{rxn} = C Δ T = 20 \times 14.8 = 296

kJ for 2 moles of Hg. Therefore,

∣Δ U ∣ = 148

kJ/mol, and since the reaction is exothermic,

Δ U = - 148

kJ/mol.

For the reaction

Hg (g) + \frac{1}{2} O_{2} (g) \to HgO (s)

, we have

Δ U = - 148

kJ/mol.

Using

Δ H = Δ U + Δ n_{g} R T

, where

Δ n_{g} = - 1.5

, we get

Δ H = - 148 + (- 1.5) (8.3 \times 1 0^{- 3}) (298) = - 151.71

kJ/mol.

For the formation from Hg(l):

Δ H_{f} [HgO (s)] = - 151.71 + 61.32 = - 90.39

kJ/mol.

Therefore,

∣ X ∣ = 90.39

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Chemical Thermodynamics

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