The Born-Haber cycle for KCl is evaluated with the following data: $\Delta_{\mathrm{f}} \mathrm{H}^{\Theta}$ for $\mathrm{KCl}=-436.7 \mathrm{kJmol}^{-1}$ ; $\Delta_{\text {sub }} \mathrm{H}^{\Theta}$ for $\mathrm{K}=89.2 \mathrm{kJmol}^{-1}$; $\Delta_{\text {ionization }} \mathrm{H}^{\Theta}$ for $\mathrm{K}=419.0 \mathrm{kJmol}^{-1}$ ; $\Delta_{\text {electrongain }} \mathrm{H}^{\Theta}$ for $\mathrm{Cl}_{(\mathrm{g})}=-348.6 \mathrm{kJmol}^{-1}$; $\Delta_{\text {bond }} \mathrm{H}^{\Theta}$ for $\mathrm{Cl}_{2}=243.0 \mathrm{kJmol}^{-1}$ The magnitude of lattice enthalpy of KCl in $\mathrm{kJ} \mathrm{mol}^{-1}$ is \_\_\_\_ (Nearest integer)

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Chemical Thermodynamics

The Born-Haber cycle for KCl is evaluated with the following data:

Δ_{f} H^{Θ}

for

KCl = - 436.7 kJmol^{- 1}

;

Δ_{sub} H^{Θ}

for

K = 89.2 kJmol^{- 1}

;

Δ_{ionization} H^{Θ}

for

K = 419.0 kJmol^{- 1}

;

Δ_{electrongain} H^{Θ}

for

Cl_{(g)} = - 348.6 kJmol^{- 1}

;

Δ_{bond} H^{Θ}

for

Cl_{2} = 243.0 kJmol^{- 1}

The magnitude of lattice enthalpy of KCl in

kJ mol^{- 1}

is ____ (Nearest integer)

Answer: 718

The Born-Haber cycle for potassium chloride gives

Δ_{f} H^{⊖} = Δ_{s u b} H^{⊖} + Δ_{i o n} H^{⊖} + \frac{1}{2} Δ_{b o n d} H^{⊖} + Δ_{e g} H^{⊖} + Δ_{l a tt i ce} H^{⊖}

only half the chlorine bond enthalpy entering because one mole of KCl needs half a mole of

C l_{2}

. Substituting the data in

kJ mo l^{- 1}

- 436.7 = 89.2 + 419.0 + \frac{243.0}{2} + (- 348.6) + Δ_{l a tt i ce} H^{⊖}

Δ_{l a tt i ce} H^{⊖} = - 436.7 - 281.1 = - 717.8

The magnitude of the lattice enthalpy of KCl is

718 kJ mo l^{- 1}

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Integer answer