The question involves finding the enthalpy of combustion (Δ_cH) for ethanol from bomb calorimeter data. A bomb calorimeter measures the change in internal energy (ΔU) for a reaction at constant volume. The relationship between ΔH (enthalpy change) and ΔU is given by:

$\Delta H=\Delta U+\Delta ngRT$

where Δn_g is the change in the number of moles of gaseous products and reactants, R is the gas constant, and T is the temperature in Kelvin.

Step 1: Identify the given values

From the bomb calorimeter: ΔU = -1364.47 kJ mol⁻¹ (negative because heat is released)
Temperature, T = 25°C = 25 + 273 = 298 K
Gas constant, R = 8.314 J mol⁻¹ K⁻¹ = 0.008314 kJ mol⁻¹ K⁻¹ (since values are in kJ)

Step 2: Calculate Δn_g

Reaction: C₂H₅OH(ℓ) + 3O₂(g) → 2CO₂(g) + 3H₂O(ℓ)
Gaseous reactants: 3 moles O₂
Gaseous products: 2 moles CO₂ (H₂O is liquid, not gas)
Δn_g = Moles of gaseous products - Moles of gaseous reactants = 2 - 3 = -1

Step 3: Apply the formula

ΔH = ΔU + Δn_gRT
Substitute the values:
ΔH = -1364.47 kJ mol⁻¹ + (-1) × (0.008314 kJ mol⁻¹ K⁻¹) × (298 K)
First, calculate Δn_gRT: (-1) × 0.008314 × 298 = -2.477572 kJ mol⁻¹ (approximately -2.478 kJ mol⁻¹)
Then, ΔH = -1364.47 + (-2.478) = -1366.948 kJ mol⁻¹ ≈ -1366.95 kJ mol⁻¹

Final Answer: –1366.95 kJ mol⁻¹

Related Topics and Formulae

Thermodynamics: Enthalpy (H) is a state function related to internal energy (U) by H = U + PV. For reactions, ΔH = ΔU + Δ(PV). At constant pressure, this simplifies to ΔH = ΔU + PΔV. For ideal gases, PΔV = Δn_gRT, leading to ΔH = ΔU + Δn_gRT.

Calorimetry: Bomb calorimeters operate at constant volume, measuring ΔU. Coffee-cup calorimeters operate at constant pressure, measuring ΔH directly.

Combustion: Combustion reactions are exothermic, so ΔH and ΔU are negative. The magnitude depends on the phases of reactants and products due to Δn_g.