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Thermodynamics question

2025 · 7 Apr · Shift 2 · Q14
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Thermodynamics question

2025 · 7 Apr · Shift 2 · Q14

JEE MainChemistryThermodynamicsMCQ+4 / −1
The hydration energies of K+K^+K+ and Cl−Cl^-Cl− are −x-x−x and −y-y−y kJ/mol respectively. If the lattice energy of KCl is −z-z−z kJ/mol, then the heat of solution of KCl is :
  1. A
    x+y+zx + y + zx+y+z
  2. B
    z−(x+y)z - (x + y)z−(x+y)
  3. C
    −z−(x+y)-z - (x + y)−z−(x+y)
  4. D
    x−y−zx - y - zx−y−z
View written solutionFree

Correct answer: B

  1. Write the dissolution process

When KCl\mathrm{KCl}KCl dissolves in water,

KCl(s)→K+(aq)+Cl−(aq)\mathrm{KCl(s) \rightarrow K^+(aq) + Cl^-(aq)}KCl(s)→K+(aq)+Cl−(aq)

The heat of solution is the enthalpy change for this process.

  1. Break the process into steps

Using Hess's law, we can imagine dissolution in two steps:

  • Step 1: Separate the ionic solid into gaseous ions

    KCl(s)→K+(g)+Cl−(g)\mathrm{KCl(s) \rightarrow K^+(g) + Cl^-(g)}KCl(s)→K+(g)+Cl−(g)

    If lattice energy of KCl is given as −z kJ/mol-z\,\text{kJ/mol}−zkJ/mol, this corresponds to lattice formation:

    K+(g)+Cl−(g)→KCl(s)ΔH=−z\mathrm{K^+(g) + Cl^-(g) \rightarrow KCl(s)} \qquad \Delta H = -zK+(g)+Cl−(g)→KCl(s)ΔH=−z

    Therefore, lattice dissociation requires:

    ΔH=+z\Delta H = +zΔH=+z

  • Step 2: Hydrate the gaseous ions

    K+(g)→K+(aq)ΔH=−x\mathrm{K^+(g) \rightarrow K^+(aq)} \qquad \Delta H = -xK+(g)→K+(aq)ΔH=−x Cl−(g)→Cl−(aq)ΔH=−y\mathrm{Cl^-(g) \rightarrow Cl^-(aq)} \qquad \Delta H = -yCl−(g)→Cl−(aq)ΔH=−y

    Total hydration enthalpy:

    −(x+y)-(x+y)−(x+y)

  1. Add the enthalpy changes

So, the heat of solution is:

ΔHsol=(+z)+(−x−y)\Delta H_{\text{sol}} = (+z) + ( -x - y )ΔHsol​=(+z)+(−x−y)

ΔHsol=z−(x+y)\Delta H_{\text{sol}} = z - (x+y)ΔHsol​=z−(x+y)

  1. Match with the options

This corresponds to:

B: z−(x+y)\boxed{\text{B: } z-(x+y)}B: z−(x+y)​

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