Question

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.

Answer 1

Correct Answer: 117

To find the enthalpy change for the vaporization of 85 g of NH₃ under the given conditions, we first need to establish the relationship between the enthalpy change and the amount of substance. Given enthalpy change for NH₃: 23.4 kJ mol^–1.

Step 1: Determine the molar mass of NH₃:
The molar mass of NH₃ = 14.01 g/mol (N) + 3 × 1.01 g/mol (H) = 17.04 g/mol.

Step 2: Calculate the number of moles in 85 g of NH₃:
Number of moles = mass/molar mass = 85 g / 17.04 g/mol ≈ 4.99 mol.

Step 3: Calculate the total enthalpy change for 85 g of NH₃:
Total enthalpy change = moles × enthalpy change per mole = 4.99 mol × 23.4 kJ/mol ≈ 116.87 kJ.

This calculated value, 116.87 kJ, falls within the expected range of 117,117 kJ, ensuring the correctness of our solution. Therefore, the enthalpy change for the vaporization of 85 g of NH₃ is approximately 117 kJ.