17.0 g of NH completely vapourises at –33.42 C and 1 bar pressure and the enthalpy change in the pro — Thermodynamics and Thermochemistry Chemistry Question
Question
17.0 g of NH completely vapourises at –33.42 C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol . The enthalpy change for the vapourisation of 85 g of NH under the same conditions is _____ kJ. 3 o –1 3
💡 Solution & Explanation
**Step 1: Find the molar mass of NH₃** Molar mass = 14 + (3 × 1) = 17 g/mol **Step 2: Calculate moles in 17.0 g of NH₃** Moles = mass ÷ molar mass = 17.0 g ÷ 17 g/mol = 1 mol **Step 3: Verify the given enthalpy of vaporization** Given: ΔH_vap = 23.4 kJ/mol for 17.0 g (which is 1 mol) This confirms 23.4 kJ/mol is the molar enthalpy of vaporization. **Step 4: Calculate moles in 85 g of NH₃** Moles = 85 g ÷ 17 g/mol = 5 mol **Step 5: Apply the concept that enthalpy is an extensive property** Key concept: Total enthalpy change = (moles) × (molar enthalpy of vaporization) ΔH_total = 5 mol × 23.4 kJ/mol = 117 kJ Therefore, the answer is 117.