Chemistry Thermochemistry SET 2019
PMDC Verified Question 63 of 81
The given diagram is a Born-Haber cycle for the formation of KBr:

Born-Haber Cycle for Potassium Bromide (KBr) Energy / Enthalpy (H) K(s) + ½Br₂(l) KBr(s) [Solid Crystal] K(g) + ½Br₂(l) K+(g) + e- + ½Br₂(l) K+(g) + e- + Br(g) K+(g) + Br-(g) [Gaseous Ions] ΔH_f = -392 kJ/mol +112 kJ/mol +420 kJ/mol (IE) +90 kJ/mol -342 kJ/mol (EA) Lattice Energy (U) = ?


Using the given data, the lattice energy of potassium bromide is calculated to be:
A
\( -672 \text{ kJmol}^{-1} \)
B
\( -672 \text{ KCalmol}^{-1} \)
C
\( -787 \text{ kJmol}^{-1} \)
D
\( +672 \text{ Jmol}^{-1} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( -672 \text{ kJmol}^{-1} \)
Concept:

The Born-Haber cycle equates the standard enthalpy of formation of an ionic solid to the sum of the energies of sublimation, ionization, dissociation, electron affinity, and lattice energy.

Formula:

$$ \Delta H_{\text{lattice}} = \Delta H_f - \Sigma \Delta H_{\text{indirect steps}} $$

Solution:

  • From the provided diagram's standard data:
    \( \Delta H_f = -392 \text{ kJ/mol} \).
    Indirect steps sum (\( \Delta H_x \)) = \( 112 \text{ (sub)} + 420 \text{ (IE)} + 90 \text{ (dissoc)} - 342 \text{ (EA)} = 280 \text{ kJ/mol} \).


  • Apply the cycle equation:
    \( \Delta H_{\text{lattice}} = -392 - (280) \)


  • \( \Delta H_{\text{lattice}} = -672 \text{ kJ/mol} \).


Why other options are incorrect:

Option B uses the wrong units (KCal instead of kJ). Option D uses the wrong units (Joules) and incorrect sign. Option C is the lattice energy for \( NaCl \), acting as a distractor.

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