Calculation of ΔH and ΔU for the Reaction NH4NO3 (S) → N2O (G) + H2O (L)

Given Data:
Molar heat of formation of NH4NO3 (S) = -367.54 kJ/mol
Molar heat of formation of N2O (G) = 81.46 kJ/mol
Molar heat of formation of H2O (L) = -285.78 kJ/mol

Step 1: Write the Balanced Chemical Equation
NH4NO3 (S) → N2O (G) + H2O (L)

Step 2: Calculate the Change in Enthalpy (ΔH)
ΔH = Σ(nΔHf(products)) - Σ(nΔHf(reactants))

ΔH = (1 mol)(81.46 kJ/mol) + (1 mol)(-285.78 kJ/mol) - (1 mol)(-367.54 kJ/mol)
ΔH = 81.46 kJ/mol - 285.78 kJ/mol + 367.54 kJ/mol
ΔH = 163.22 kJ/mol

Therefore, the change in enthalpy (ΔH) for the reaction is 163.22 kJ/mol.

Step 3: Calculate the Change in Internal Energy (ΔU)
ΔU = ΔH - ΔnRT

Where:
Δn = (Σn(products)) - (Σn(reactants))
R = Gas constant = 8.314 J/mol·K
T = Temperature = 25°C = 298 K (Convert to Kelvin)

Δn = (1 mol + 1 mol) - (1 mol)
Δn = 1 mol

ΔU = 163.22 kJ/mol - (1 mol)(8.314 J/mol·K)(298 K)
ΔU = 163.22 kJ/mol - 2463.2 J/mol
ΔU = 163.22 kJ/mol - 2.4632 kJ/mol
ΔU = 160.7568 kJ/mol

Therefore, the change in internal energy (ΔU) for the reaction is 160.7568 kJ/mol.

Explanation:
The calculation of ΔH involves using the molar heat of formation values for the products and reactants. By subtracting the sum of the molar heat of formation of the reactants from the sum of the molar heat of formation of the products, we can determine the change in enthalpy (ΔH) for the reaction.

The calculation of ΔU involves using the equation ΔU = ΔH - ΔnRT. Δn represents the change in the number of moles of gas molecules, which can be determined by comparing the coefficients of the gaseous products and reactants in the balanced chemical equation. R is the gas constant, and T is the temperature in Kelvin.

In this case, the change in enthalpy (ΔH) for the reaction is 163.22 kJ/mol, indicating that the reaction is exothermic (releases heat). The change in internal energy (ΔU