Question

400 gm 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 K to 313 K. If heat capacity of the bomb calorimeter and enthalpy of formation of Hg(g) are 20.00 kJ K⁻¹ and 61.57 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 J K⁻¹ mol⁻¹, Hg = 200 gm/mol]

Answer 1

92.006

Answer 2

The combustion of 400 gm of Hg(g) in a fixed volume bomb calorimeter with excess O₂ at 298 K into HgO(s) leads to a temperature increase from 298 K to 313 K.

The number of moles of Hg(g) is given by mass/molar mass = 400 gm / 200 gm/mol = 2 mol. The reaction occurring is 2 Hg(g) + O₂(g) → 2 HgO(s).

The heat absorbed by the calorimeter is given by the product of its heat capacity and the temperature change: Q_calorimeter = C_calorimeter × ΔT Q_calorimeter = (20.00 kJ K⁻¹) × (313 K - 298 K) = 20.00 kJ K⁻¹ × 15 K = 300 kJ.

Since the reaction is carried out in a bomb calorimeter (fixed volume), the heat evolved by the reaction at constant volume is equal to the negative of the heat absorbed by the calorimeter: Q_reaction = ΔU = -Q_calorimeter = -300 kJ. This is the change in internal energy for the reaction 2 Hg(g) + O₂(g) → 2 HgO(s) at 298 K.

We need to find the standard molar enthalpy of formation of HgO(s) at 298 K, which is the enthalpy change for the reaction: Hg(l) + 1/2 O₂(g) → HgO(s) ΔH°_f(HgO, s) = X kJ mol⁻¹.

We can relate the change in internal energy (ΔU) to the change in enthalpy (ΔH) using the relation: ΔH = ΔU + Δn_g RT where Δn_g is the change in the number of moles of gas in the reaction.

For the reaction 2 Hg(g) + O₂(g) → 2 HgO(s): Initial moles of gas = moles of Hg(g) + moles of O₂(g) = 2 + 1 = 3 (assuming O₂ is in excess, but its stoichiometry in the reaction is 1 mole for 2 moles of Hg). Final moles of gas = moles of HgO(s) = 0 (HgO is solid). Δn_g = (moles of gaseous products) - (moles of gaseous reactants) = 0 - 3 = -3 mol.

Now, we calculate ΔH for the reaction 2 Hg(g) + O₂(g) → 2 HgO(s): ΔH = ΔU + Δn_g RT Given R = 8 J K⁻¹ mol⁻¹ = 8 × 10⁻³ kJ K⁻¹ mol⁻¹. ΔH = -300 kJ + (-3 mol) × (8 × 10⁻³ kJ K⁻¹ mol⁻¹) × (298 K) ΔH = -300 kJ - (3 × 8 × 298 × 10⁻³) kJ ΔH = -300 kJ - (7152 × 10⁻³) kJ ΔH = -300 kJ - 7.152 kJ ΔH = -307.152 kJ.

This is the enthalpy change for the formation of 2 moles of HgO(s) from 2 moles of Hg(g) and 1 mole of O₂(g). So, the enthalpy change for the formation of 1 mole of HgO(s) from Hg(g) and O₂(g) is: ΔH(Hg(g) + 1/2 O₂(g) → HgO(s)) = ΔH / 2 = -307.152 kJ / 2 mol = -153.576 kJ mol⁻¹.

We are given the standard enthalpy of formation of Hg(g) at 298 K, which is the enthalpy change for the process: Hg(l) → Hg(g) ΔH°_f(Hg, g) = 61.57 kJ mol⁻¹.

We want to find the standard molar enthalpy of formation of HgO(s), which is for the reaction: Hg(l) + 1/2 O₂(g) → HgO(s) ΔH°_f(HgO, s) = X.

We can use Hess's Law to relate these reactions:

1. Hg(l) + 1/2 O₂(g) → HgO(s) ΔH°_f(HgO, s) = X (Target reaction)
2. Hg(g) + 1/2 O₂(g) → HgO(s) ΔH₂ = -153.576 kJ mol⁻¹ (Calculated)
3. Hg(l) → Hg(g) ΔH₃ = 61.57 kJ mol⁻¹ (Given)

To obtain Reaction 1 from Reactions 2 and 3, we can add Reaction 3 and Reaction 2: (Hg(l) → Hg(g)) + (Hg(g) + 1/2 O₂(g) → HgO(s)) Hg(l) + Hg(g) + 1/2 O₂(g) → Hg(g) + HgO(s) Cancelling Hg(g) from both sides gives: Hg(l) + 1/2 O₂(g) → HgO(s)

According to Hess's Law, the enthalpy change for Reaction 1 is the sum of the enthalpy changes for Reaction 3 and Reaction 2: ΔH₁ = ΔH₃ + ΔH₂ X = 61.57 kJ mol⁻¹ + (-153.576 kJ mol⁻¹) X = 61.57 - 153.576 X = -92.006 kJ mol⁻¹.

The calculated standard molar enthalpy of formation of HgO(s) at 298 K is X = -92.006 kJ mol⁻¹. The question asks for the value of |X|. |X| = |-92.006| = 92.006.