JEE Main 2020 · 5 September · evening · Single correct · +4 / −1

Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol –1 , and 4 kJ mol –1 , respectively.
The hydration enthalpy of NaCl is :

  1. (a) –780 kJ mol –1
  2. (b) –784 kJ mol –1
  3. (c) 780 kJ mol –1
  4. (d) 784 kJ mol –1
Answer and solution

Answer (b) –784 kJ mol –1

Solution

Chemistry ma’am · 3 steps

Dissolving is two steps: break the lattice, then hydrate the ions. The two enthalpies must add to 4.

  1. Two-step path. First the crystal breaks into gaseous ions, which costs the lattice enthalpy, +788 kJ mol⁻¹. Then water surrounds the gaseous ions, which releases the hydration enthalpy.

    Δ Hₛₒₗ = Δ H_(lattice) + Δ H_(hyd)

  2. Put in numbers. 4 = 788 + Δ H_(hyd), so Δ H_(hyd) = 4 - 788 = -784 kJ mol⁻¹.
  3. So. The hydration enthalpy of NaCl is -784 kJ mol⁻¹, option (b). The sign is negative because ions attracting water molecules always releases energy.

The idea to keep. Solution enthalpy = lattice enthalpy (positive, breaking the crystal) + hydration enthalpy (negative, ions grabbing water). A small positive solution enthalpy means the two nearly cancel.

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

Study Chemical Thermodynamics

Asked about 160 times in JEE Main since 2020. This question is from 2020.