JEE Advanced 2022 — Paper 1 — CHEM-I solution

JEE Advanced2022Paper 1CHEM-I

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 $^{- 1}$ ]

Your answer

Solution

Reaction: $2 Hg(g) + O_{2} (g) \to 2 HgO(s)$

Heat released in the bomb (constant volume):

$q_{V} = C_{c a l} \cdot Δ T = 20.00 \times (312.8 - 298.0) = 296$ kJ (released by reaction).

So $Δ U^{\circ} = - 296$ kJ for 2 mol of Hg(g) reacting.

Convert to enthalpy: $Δ n_{g} = 0 - 3 = - 3$ (reactants have 2 mol Hg(g) + 1 mol O $_{2}$ ; products are solid).

$Δ H^{\circ} = Δ U^{\circ} + Δ n_{g} R T = - 296 + (- 3) (8.3) (298) \times 1 0^{- 3} = - 296 - 7.42 = - 303.42$ kJ.

Relate to formation enthalpies (for 2 mol of reaction):

$Δ H^{\circ} = 2Δ H_{f}^{\circ} (HgO(s)) - 2Δ H_{f}^{\circ} (Hg(g))$

$- 303.42 = 2Δ H_{f}^{\circ} (HgO(s)) - 2 (61.32)$

$Δ H_{f}^{\circ} (HgO(s)) = \frac{- 303.42 + 122.64}{2} = - 90.39$ kJ mol $^{- 1}$ .

$∣ X ∣ = 90.39$ kJ mol $^{- 1}$ .

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