Numerical/Formula
1. TITLE
How to solve Born-Haber Cycle numericals in NEET — Step-by-step
2. PROBLEM TYPE INTRO
This numerical falls under the Chemical Thermodynamics category, specifically applying Hess's Law through the Born-Haber Cycle to calculate Lattice Enthalpy. We use the formula linking the standard enthalpy of formation with the enthalpies of sublimation, ionization, bond dissociation, and electron gain. A typical mistake students make is forgetting to halve the bond dissociation enthalpy for diatomic molecules.
3. MCQ
Question:
Using the Born-Haber cycle, calculate the Lattice Enthalpy ($\Delta_{lattice}H^{\ominus}$) for $\text{NaCl}(s)$ given the following standard enthalpy values at $298 \text{ K}$:
* Standard enthalpy of formation of $\text{NaCl}(s)$ ($\Delta_{f}H^{\ominus}$) = $-411.2 \text{ kJ mol}^{-1}$
* Enthalpy of sublimation of $\text{Na}(s)$ ($\Delta_{sub}H^{\ominus}$) = $+108.4 \text{ kJ mol}^{-1}$
* Ionization enthalpy of $\text{Na}(g)$ ($\Delta_{i}H^{\ominus}$) = $+496.0 \text{ kJ mol}^{-1}$
* Bond dissociation enthalpy of $\text{Cl}{2}(g)$ ($\Delta{bond}H^{\ominus}${bond}H^{\ominus}$) = $+242.0 \text{ kJ mol}^{-1}$
* Electron gain enthalpy of $\text{Cl}(g)$ ($\Delta_{eg}H^{\ominus}$) = $-348.6 \text{ kJ mol}^{-1}$
Options:
(a) $+909.0 \text{ kJ mol}^{-1}$
(b) $+788.0 \text{ kJ mol}^{-1}$
(c) $-788.0 \text{ kJ mol}^{-1}$
(d) $+376.8 \text{ kJ mol}^{-1}$
4. STEP-BY-STEP SOLUTION
SOLUTION
Given:
* $\Delta_{f}H^{\ominus} = -411.2 \text{ kJ mol}^{-1}$
* $\Delta_{sub}H^{\ominus} = +108.4 \text{ kJ mol}^{-1}$
* $\Delta_{i}H^{\ominus} = +496.0 \text{ kJ mol}^{-1}$
* $\Delta_{bond}H^{\ominus} = +242.0 \text{ kJ mol}^{-1}$
* $\Delta_{eg}H^{\ominus} = -348.6 \text{ kJ mol}^{-1}$
To Find:
* Lattice Enthalpy ($\Delta_{lattice}H^{\ominus}$) of $\text{NaCl}(s)$
Formula:
$\Delta_{f}H^{\ominus} = \Delta_{sub}H^{\ominus} + \Delta_{i}H^{\ominus} + \frac{1}{2}\Delta_{bond}H^{\ominus} + \Delta_{eg}H^{\ominus} - \Delta_{lattice}H^{\ominus}$
Use this formula when applying Hess’s Law to an ionic solid, remembering that standard Lattice Enthalpy is defined for the dissociation of the solid into gaseous ions (making its value positive), so its formation step is subtracted.
Step 1: Calculate the enthalpy required to prepare the gaseous atoms.
We need to halve the bond dissociation enthalpy of $\text{Cl}{2}$ because only one mole of $\text{Cl}$ atoms is needed for one mole of $\text{NaCl}$.
$\frac{1}{2}\Delta{bond}H^{\ominus} = \frac{1}{2} (+242.0 \text{ kJ mol}^{-1}) = +121.0 \text{ kJ mol}^{-1}${bond}H^{\ominus} = \frac{1}{2} (+242.0 \text{ kJ mol}^{-1}) = +121.0 \text{ kJ mol}^{-1}$
Step 2: Sum the enthalpies of all intermediate steps (Sublimation, Ionization, Bond Dissociation, Electron Gain).
$\Sigma H_{\text{steps}} = \Delta_{sub}H^{\ominus} + \Delta_{i}H^{\ominus} + \frac{1}{2}\Delta_{bond}H^{\ominus} + \Delta_{eg}H^{\ominus}$
$\Sigma H_{\text{steps}} = 108.4 + 496.0 + 121.0 + (-348.6)$
$\Sigma H_{\text{steps}} = 725.4 - 348.6 = +376.8 \text{ kJ mol}^{-1}$
Step 3: Apply Hess's Law to find the Lattice Enthalpy.
Rearranging our main formula: $\Delta_{lattice}H^{\ominus} = \Sigma H_{\text{steps}} - \Delta_{f}H^{\ominus}$
$\Delta_{lattice}H^{\ominus} = 376.8 - (-411.2)$
$\Delta_{lattice}H^{\ominus} = 376.8 + 411.2 = 788.0 \text{ kJ mol}^{-1}$
Answer: $+788.0 \text{ kJ mol}^{-1}$ (Option b)
5. WHERE STUDENTS GO WRONG
List 2-3 common mistakes:
- ❌ Mistake 1: Forgetting to halve the bond dissociation enthalpy for $\text{Cl}2$ gas. Doing so leads to Option (a) $+909.0 \text{ kJ mol}^{-1}$.
- ❌ Mistake 2: Using the wrong sign for Lattice Enthalpy. Lattice Enthalpy is defined as the energy required to dissociate the lattice, so it is always a positive endothermic value. Choosing a negative value leads to Option (c) $-788.0 \text{ kJ mol}^{-1}$.
- ❌ Mistake 3: Forgetting to include the enthalpy of formation ($\Delta{f}H^{\ominus}${f}H^{\ominus}$) when applying Hess's Law, calculating only the intermediate steps. This leads to Option (d) $+376.8 \text{ kJ mol}^{-1}$.
6. QUICK FORMULA REFERENCE
Formulas Used:
* $\Delta_{f}H^{\ominus} = \Sigma H_{\text{steps}} - \Delta_{lattice}H^{\ominus}$ — Use when calculating the overall energy of an ionic compound utilizing the Born-Haber Cycle.
* $\Delta_{a}H^{\ominus} = \frac{1}{2}\Delta_{bond}H^{\ominus}$ — Use when finding the enthalpy of atomization for a diatomic gas like $\text{Cl}_2$ or $\text{H}_2$.
7. SIMILAR PROBLEM HINT
"If NEET changes the values, the method stays the same. Practice with: Finding the Lattice Enthalpy of $\text{MgCl}_2$ (Remember to account for two ionization enthalpies for Mg and the full bond dissociation enthalpy for $\text{Cl}_2$!)."
8. NCERT REFERENCE
Based on: NCERT Class 11, Chapter 5, Example Section 5.5 (d) Lattice Enthalpy
9. CTA
More solved numericals → Chemical Thermodynamics Complete Guide
Practice problems → Mock exams