17.0 g of NH₃ completely vapourises at -33.42°C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol⁻¹. The enthalpy change for the vapourisation of 85 g of NH₃ under the same conditions is _________ kJ.

Solution

Chemistry ma’am · 4 steps

Enthalpy of vapourisation is per mole, so first count how many moles 85 g of ammonia is.

1. Molar mass. For NH₃, M = 14 + 3 × 1 = 17 g mol⁻¹. So the 17.0 g in the question is exactly 1 mol, which is why the value is quoted per mole.
2. Moles in 85 g. n = 85/17 = 5 mol.
3. Scale the enthalpy. Enthalpy is an extensive property, so it grows in proportion to the amount vapourised. The temperature and pressure are the same, so 23.4 kJ mol⁻¹ still applies.

   Δ H = n × ΔᵥₐₚH = 5 mol × 23.4 kJ mol⁻¹ = 117 kJ

4. So. Vapourising 85 g of NH₃ needs 117 kJ.

The common slip: 23.4. Reporting the molar value as it is gives 23.4. The question asks for the total for 85 g, which is 5 mol, so multiply by 5.

The idea to keep. A molar enthalpy is the value for 1 mol. For any other amount, convert mass to moles and multiply, because enthalpy is extensive.

Answer from the NTA final answer key. Solution by PrepWiser.