The enthalpy change for the conversion of 1/2 Cl₂( g) to Cl⁻(aq) is (-) ___________ kJ mol⁻¹ (Nearest integer)
Given : Δ_(dis) H_(Cl_(2( g))^θ)⊖=240 kJ mol⁻¹, Δ_(eg) H_(Cl_((g)))⊖=-350 kJ mol⁻¹,
Δ _(hyd)H_(Cl_((g)) -)^Θ = - 380 kJ mol⁻¹

Solution

Chemistry ma’am · 3 steps

Break the change into three steps you have data for, then add their enthalpies.

1. Build the path. Going from 1/2Cl₂(g) to Cl⁻(aq) happens in three steps. First break the bond to get a gaseous atom, then add an electron, then hydrate the ion. By Hess's law, the total enthalpy is the sum of the three.

   Step	Process	Δ H / kJ mol⁻¹
   Atomisation	1/2Cl₂(g) → Cl(g)	1/2 × 240 = +120
   Electron gain	Cl(g) + e⁻ → Cl⁻(g)	-350
   Hydration	Cl⁻(g) → Cl⁻(aq)	-380

2. Why half. The dissociation enthalpy of 240 is for a whole mole of Cl₂, which gives two moles of Cl atoms. You start with only half a mole of Cl₂, so you need only half of it.

   Δ H = 120 + (-350) + (-380) = -610 kJ mol⁻¹

3. So. The enthalpy change is -610 kJ mol⁻¹. The blank already has the minus sign, so the answer is 610.

The common slip: 490. Using the full 240 for the bond-breaking step instead of half of it gives 490, not 610.

The idea to keep. For gas-to-aqueous ion questions, go atomise, then ionise, then hydrate, and add the three. Always scale the bond enthalpy to the amount of molecule you actually start with.

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