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Thermodynamics question

2021 · 26 Aug · Shift 1 · Q20
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Thermodynamics question

2021 · 26 Aug · Shift 1 · Q20

JEE MainChemistryThermodynamicsNumerical+4 / −1
The Born-Haber cycle for KClKClKCl is evaluated with the following data : ΔfHΘ{\Delta _f}{H^\Theta }Δf​HΘ for KClKClKCl =−-− 436.7 kJ mol −-− 1 ; ΔsubHΘ{\Delta _{sub}}{H^\Theta }Δsub​HΘ for K = 89.2 kJ mol −-− 1 ; ΔionizationHΘ{\Delta _{ionization}}{H^\Theta }Δionization​HΘ for K = 419.0 kJ mol −-− 1 ; Δelectron gainHΘ{\Delta _{electron\,gain}}{H^\Theta }Δelectrongain​HΘ for Cl(g) =−-− 348.6 kJ mol −-− 1 ; ΔbondHΘ{\Delta _{bond}}{H^\Theta }Δbond​HΘ for Cl2Cl_2Cl2​ = 243.0 kJ mol −-− 1 The magnitude of lattice enthalpy of KClKClKCl in kJ mol −-− 1 is ‾\underline{\hspace{2cm}}​ (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 718

  1. Write the Born–Haber cycle for KCl(s)\mathrm{KCl(s)}KCl(s)

The formation reaction is

K(s)+12Cl2(g)→KCl(s)\mathrm{K(s) + \tfrac12 Cl_2(g) \to KCl(s)}K(s)+21​Cl2​(g)→KCl(s)

with

ΔfH∘=−436.7 kJ mol−1\Delta_f H^\circ = -436.7\ \text{kJ mol}^{-1}Δf​H∘=−436.7 kJ mol−1
  1. Break the formation into steps

For the Born–Haber cycle:

  • Sublimation of potassium:
K(s)→K(g)ΔH=+89.2\mathrm{K(s) \to K(g)} \qquad \Delta H = +89.2K(s)→K(g)ΔH=+89.2
  • Ionization of potassium:
K(g)→K+(g)+e−ΔH=+419.0\mathrm{K(g) \to K^+(g) + e^-} \qquad \Delta H = +419.0K(g)→K+(g)+e−ΔH=+419.0
  • Dissociation of chlorine:
12Cl2(g)→Cl(g)ΔH=243.02=121.5\mathrm{\tfrac12 Cl_2(g) \to Cl(g)} \qquad \Delta H = \frac{243.0}{2} = 121.521​Cl2​(g)→Cl(g)ΔH=2243.0​=121.5
  • Electron gain by chlorine:
Cl(g)+e−→Cl−(g)ΔH=−348.6\mathrm{Cl(g) + e^- \to Cl^-(g)} \qquad \Delta H = -348.6Cl(g)+e−→Cl−(g)ΔH=−348.6
  • Lattice formation:
K+(g)+Cl−(g)→KCl(s)ΔH=−U\mathrm{K^+(g) + Cl^-(g) \to KCl(s)} \qquad \Delta H = -UK+(g)+Cl−(g)→KCl(s)ΔH=−U

where UUU is the magnitude of lattice enthalpy.

  1. Apply Hess's law

So,

ΔfH∘=ΔsubH∘+ΔionizationH∘+12ΔbondH∘+Δelectron gainH∘−U\Delta_f H^\circ = \Delta_{sub}H^\circ + \Delta_{ionization}H^\circ + \frac12\Delta_{bond}H^\circ + \Delta_{electron\,gain}H^\circ - UΔf​H∘=Δsub​H∘+Δionization​H∘+21​Δbond​H∘+Δelectrongain​H∘−U

Substitute values:

−436.7=89.2+419.0+121.5−348.6−U-436.7 = 89.2 + 419.0 + 121.5 - 348.6 - U−436.7=89.2+419.0+121.5−348.6−U
  1. Simplify

First add the positive terms:

89.2+419.0+121.5=629.789.2 + 419.0 + 121.5 = 629.789.2+419.0+121.5=629.7

Now include electron gain:

629.7−348.6=281.1629.7 - 348.6 = 281.1629.7−348.6=281.1

Thus,

−436.7=281.1−U-436.7 = 281.1 - U−436.7=281.1−U

So,

U=281.1+436.7=717.8U = 281.1 + 436.7 = 717.8U=281.1+436.7=717.8
  1. Nearest integer
U≈718 kJ mol−1U \approx 718\ \text{kJ mol}^{-1}U≈718 kJ mol−1

Hence, the magnitude of lattice enthalpy is

718\boxed{718}718​
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