The Hook: Why Does a Hand Warmer Get Hot?
You’re at a cricket match on a cold winter evening. You crack open a hand warmer packet, shake it, and within seconds it feels warm in your hands. No battery, no external heat source - just a chemical reaction inside the packet.
The questions:
- Where does this heat come from?
- Why do some reactions release heat (like burning) while others absorb heat (like melting ice)?
- How can we predict and measure heat changes in reactions without actually performing them?
The answer lies in Enthalpy - the heat content of substances and the heat exchanged during reactions at constant pressure.
Real-world connections: Hand warmers, instant cold packs for injuries, cooking food, car engines, industrial chemical plants - all involve enthalpy changes!
The Core Concept
What is Enthalpy?
Enthalpy (H) is the total heat content of a system at constant pressure.
$$\boxed{H = U + PV}$$where:
- U = Internal energy
- P = Pressure
- V = Volume
Why do we need Enthalpy when we have Internal Energy?
Most chemical reactions happen at constant atmospheric pressure (open beakers, not sealed bombs). At constant pressure, heat change equals enthalpy change, making calculations easier!
$$\boxed{\Delta H = q_P}$$In simple terms: Enthalpy is the heat absorbed or released by a reaction at constant pressure.
Relation Between ΔH and ΔU
Starting from $H = U + PV$:
$$\Delta H = \Delta U + \Delta(PV)$$At constant pressure:
$$\boxed{\Delta H = \Delta U + P\Delta V}$$For reactions involving gases (using ideal gas law):
$$PV = nRT$$ $$P\Delta V = \Delta n_g RT$$where $\Delta n_g$ = (moles of gaseous products) - (moles of gaseous reactants)
$$\boxed{\Delta H = \Delta U + \Delta n_g RT}$$When to use:
- When converting between ΔH and ΔU
- When reactions involve gaseous species
- Count only gaseous moles, not liquids or solids!
Sign of Δn_g:
- More gas moles in products: Δn_g > 0 → ΔH > ΔU
- More gas moles in reactants: Δn_g < 0 → ΔH < ΔU
- No change in gas moles: Δn_g = 0 → ΔH = ΔU
Exothermic vs Endothermic Reactions
Exothermic Reactions (Heat Released)
$$\Delta H < 0 \quad \text{(negative)}$$Examples:
- Combustion: $\text{CH}_4 + 2\text{O}_2 \to \text{CO}_2 + 2\text{H}_2\text{O}$, $\Delta H = -890$ kJ/mol
- Neutralization: $\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O}$, $\Delta H = -57.1$ kJ/mol
- Freezing water: $\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(s)$, $\Delta H = -6$ kJ/mol
- Respiration: $\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \to 6\text{CO}_2 + 6\text{H}_2\text{O}$
Molecular view: Products are more stable (lower energy) than reactants. Excess energy is released as heat.
Real-life: Hand warmers, burning fuels, acid-base reactions
Endothermic Reactions (Heat Absorbed)
$$\Delta H > 0 \quad \text{(positive)}$$Examples:
- Photosynthesis: $6\text{CO}_2 + 6\text{H}_2\text{O} \to \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$, $\Delta H = +2802$ kJ/mol
- Melting ice: $\text{H}_2\text{O}(s) \to \text{H}_2\text{O}(l)$, $\Delta H = +6$ kJ/mol
- Decomposition of calcium carbonate: $\text{CaCO}_3 \to \text{CaO} + \text{CO}_2$, $\Delta H = +178$ kJ/mol
- Evaporation: $\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(g)$, $\Delta H = +44$ kJ/mol
Molecular view: Products are less stable (higher energy) than reactants. Energy must be supplied from surroundings.
Real-life: Instant cold packs, cooking (breaking bonds), evaporative cooling (sweat)
Standard Enthalpy Changes
Standard State Conditions
Standard state (°):
- Temperature: 298 K (25°C) unless specified
- Pressure: 1 bar (≈ 1 atm)
- Concentration: 1 M for solutions
- Pure substances in their most stable form
- Oxygen: O₂(g) at 1 bar, not O₃(g) or O(g)
- Carbon: C(graphite), not C(diamond)
- Water: H₂O(l), not H₂O(g) at 298 K
- Mercury: Hg(l), not Hg(s) at 298 K
Types of Standard Enthalpy Changes
| Type | Symbol | Definition | Example |
|---|---|---|---|
| Formation | $\Delta_f H°$ | Heat change when 1 mole of compound forms from elements in standard states | $\text{C} + \text{O}_2 \to \text{CO}_2$, $\Delta_f H° = -393.5$ kJ/mol |
| Combustion | $\Delta_c H°$ | Heat released when 1 mole burns completely in O₂ | $\text{CH}_4 + 2\text{O}_2 \to \text{CO}_2 + 2\text{H}_2\text{O}$, $\Delta_c H° = -890$ kJ/mol |
| Neutralization | $\Delta_{neut} H°$ | Heat released when 1 mole H⁺ neutralizes 1 mole OH⁻ | $\text{H}^+ + \text{OH}^- \to \text{H}_2\text{O}$, $\Delta_{neut} H° = -57.1$ kJ/mol |
| Atomization | $\Delta_a H°$ | Heat to convert 1 mole to gaseous atoms | $\text{CH}_4(g) \to \text{C}(g) + 4\text{H}(g)$, $\Delta_a H° = +1660$ kJ/mol |
| Bond dissociation | $\Delta_{bond} H°$ | Energy to break 1 mole of specific bonds | $\text{Cl}_2(g) \to 2\text{Cl}(g)$, $\Delta_{bond} H° = +243$ kJ/mol |
| Sublimation | $\Delta_{sub} H°$ | Solid → Gas directly | $\text{I}_2(s) \to \text{I}_2(g)$, $\Delta_{sub} H° = +62.4$ kJ/mol |
| Fusion (melting) | $\Delta_{fus} H°$ | Solid → Liquid | $\text{H}_2\text{O}(s) \to \text{H}_2\text{O}(l)$, $\Delta_{fus} H° = +6.0$ kJ/mol |
| Vaporization | $\Delta_{vap} H°$ | Liquid → Gas | $\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(g)$, $\Delta_{vap} H° = +40.7$ kJ/mol |
| Ionization | $\Delta_{ion} H$ | Removing electron from gaseous atom | $\text{Na}(g) \to \text{Na}^+(g) + e^-$, $\Delta_{ion} H = +496$ kJ/mol |
| Electron gain | $\Delta_{eg} H$ | Adding electron to gaseous atom | $\text{Cl}(g) + e^- \to \text{Cl}^-(g)$, $\Delta_{eg} H = -349$ kJ/mol |
| Hydration | $\Delta_{hyd} H$ | Ion in gas → Ion in water | $\text{Na}^+(g) \to \text{Na}^+(aq)$, $\Delta_{hyd} H = -406$ kJ/mol |
| Solution | $\Delta_{sol} H$ | Dissolving substance in solvent | $\text{NaCl}(s) \to \text{Na}^+(aq) + \text{Cl}^-(aq)$, $\Delta_{sol} H = +3.9$ kJ/mol |
Key Facts About Standard Enthalpy of Formation
Standard enthalpy of formation of elements in their standard state is ZERO.
$$\Delta_f H°[\text{element in standard state}] = 0$$Examples:
- $\Delta_f H°[\text{O}_2(g)] = 0$
- $\Delta_f H°[\text{H}_2(g)] = 0$
- $\Delta_f H°[\text{C(graphite)}] = 0$
- $\Delta_f H°[\text{Br}_2(l)] = 0$
But:
- $\Delta_f H°[\text{O}_3(g)] \neq 0$ (not standard form)
- $\Delta_f H°[\text{C(diamond)}] \neq 0$ (graphite is standard form)
Calculating ΔH for Reactions
Method 1: Using Standard Enthalpies of Formation
$$\boxed{\Delta_r H° = \sum \Delta_f H°(\text{products}) - \sum \Delta_f H°(\text{reactants})}$$Remember: Multiply each $\Delta_f H°$ by the stoichiometric coefficient!
Interactive Demo: Visualize Enthalpy Changes
See energy diagrams showing exothermic and endothermic reactions.
Example: Calculate $\Delta_r H°$ for:
$$\text{CH}_4(g) + 2\text{O}_2(g) \to \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$Given:
- $\Delta_f H°[\text{CH}_4(g)] = -74.8$ kJ/mol
- $\Delta_f H°[\text{CO}_2(g)] = -393.5$ kJ/mol
- $\Delta_f H°[\text{H}_2\text{O}(l)] = -285.8$ kJ/mol
- $\Delta_f H°[\text{O}_2(g)] = 0$ (element)
Method 2: Using Combustion Data
$$\Delta_r H° = \sum \Delta_c H°(\text{reactants}) - \sum \Delta_c H°(\text{products})$$Note: Opposite sign convention compared to formation enthalpies!
Why? Because combustion data tells us heat released when substance burns. For products, they’ve already been “burned” (oxidized), so we subtract.
Hess’s Law: The Most Powerful Tool
Statement of Hess’s Law
“The total enthalpy change for a reaction is independent of the path taken.”
Or equivalently:
“If a reaction can be carried out in steps, the sum of enthalpy changes of individual steps equals the enthalpy change of the direct reaction.”
$$\Delta H_{overall} = \Delta H_1 + \Delta H_2 + \Delta H_3 + ...$$Why? Because enthalpy is a state function - depends only on initial and final states, not the path!
Analogy: Climbing a mountain
- Route A: Straight up (direct path)
- Route B: Zigzag path (multiple steps)
- Elevation gain is the same regardless of route!
How to Apply Hess’s Law
Rules for manipulating equations:
Reversing a reaction: Change sign of ΔH
$$\text{If } A \to B, \Delta H = x$$ $$\text{Then } B \to A, \Delta H = -x$$Multiplying by a factor: Multiply ΔH by same factor
$$\text{If } A \to B, \Delta H = x$$ $$\text{Then } 2A \to 2B, \Delta H = 2x$$Adding equations: Add the ΔH values
$$A \to B, \Delta H_1$$ $$B \to C, \Delta H_2$$ $$A \to C, \Delta H = \Delta H_1 + \Delta H_2$$
Hess’s Law Strategy
Step 1: Write the target reaction clearly
Step 2: Identify given reactions and their ΔH values
Step 3: Manipulate given reactions to match target:
- Reverse reactions if needed (flip ΔH sign)
- Multiply reactions if needed (multiply ΔH)
- Cancel common species on both sides
Step 4: Add manipulated equations and their ΔH values
Step 5: Verify target equation matches and calculate ΔH
Thermochemistry: Bond Energies
Bond Dissociation Energy vs Average Bond Energy
Bond Dissociation Energy: Energy to break a specific bond in a specific molecule
- Example: Breaking first O-H bond in H₂O requires different energy than breaking second O-H bond
Average Bond Energy: Average energy of that bond across many different molecules
- Used for approximate calculations
- Example: C-H bond energy ≈ 414 kJ/mol (average across all organic molecules)
Calculating ΔH from Bond Energies
$$\boxed{\Delta H_{rxn} = \sum(\text{Bonds broken}) - \sum(\text{Bonds formed})}$$or equivalently:
$$\Delta H_{rxn} = \sum(\text{Bond energies of reactants}) - \sum(\text{Bond energies of products})$$Remember:
- Breaking bonds: Endothermic (requires energy) → Positive
- Forming bonds: Exothermic (releases energy) → Negative
Mnemonic: “Break Bad” (Breaking Bonds = Positive)
Wrong: Bond energy is energy released when bond forms. Right: Bond energy is energy required to break the bond.
Forming bonds releases energy equal in magnitude but opposite in sign!
Why Reactions Release or Absorb Heat
Exothermic reaction (ΔH < 0):
- Energy released when forming product bonds > Energy needed to break reactant bonds
- Products have stronger bonds than reactants
- Net energy is released
Endothermic reaction (ΔH > 0):
- Energy needed to break reactant bonds > Energy released when forming product bonds
- Reactants have stronger bonds than products
- Net energy must be supplied
Memory Tricks & Patterns
Mnemonic for Enthalpy
“Help = U + PV” → H = U + PV
“Heat Produces Volume” → ΔH = ΔU + PΔV (at constant pressure)
Sign Convention Memory
“EXIT is negative” → Exothermic = Negative ΔH (energy exits)
“ENTER is positive” → Endothermic = Positive ΔH (energy enters)
Neutralization Pattern
For strong acid + strong base:
$$\Delta_{neut} H° = -57.1 \text{ kJ/mol} \quad \text{(constant!)}$$Why constant? The actual reaction is just:
$$\text{H}^+(aq) + \text{OH}^-(aq) \to \text{H}_2\text{O}(l)$$For weak acid or weak base: Less heat released (some energy used in ionization)
Hess’s Law Mnemonic
“Target, Given, Manipulate, Add” (TGMA)
- Target equation
- Given equations
- Manipulate (reverse/multiply)
- Add them up
Common Mistakes to Avoid
Wrong:
$$\Delta_r H° = \Delta_f H°(\text{H}_2\text{O}) - \Delta_f H°(\text{H}_2) - \Delta_f H°(\text{O}_2)$$Right: For $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$
$$\Delta_r H° = 2\Delta_f H°(\text{H}_2\text{O}) - [2\Delta_f H°(\text{H}_2) + \Delta_f H°(\text{O}_2)]$$Always multiply by the coefficient in balanced equation!
Common mistake: Using Δn (total change in moles) instead of Δn_g
Right formula:
$$\Delta H = \Delta U + \Delta n_g RT$$where $\Delta n_g$ = change in gaseous moles only!
Example: $\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)$
- Δn_g = 1 - 0 = +1 (only CO₂ is gas)
- NOT Δn = 2 - 1 = +1
Wrong: $\Delta_f H°[\text{Br}_2(g)] = 0$ Right: $\Delta_f H°[\text{Br}_2(l)] = 0$ (liquid is standard state at 298 K)
Wrong: $\Delta_f H°[\text{C(diamond)}] = 0$ Right: $\Delta_f H°[\text{C(graphite)}] = 0$ (graphite is more stable)
Always check standard states carefully!
Wrong thinking: “C-H bond energy = -414 kJ/mol” (bond formation)
Right: Bond energy is always positive (energy to break bond)
- Breaking C-H: +414 kJ/mol (endothermic)
- Forming C-H: -414 kJ/mol (exothermic)
Use the formula correctly:
$$\Delta H = \text{Bonds broken (positive)} - \text{Bonds formed (positive values, but subtract)}$$Practice Problems
Level 1: Foundation (NCERT)
Question: Calculate the standard enthalpy of formation of CO₂ from:
$$\text{C(graphite)} + \text{O}_2(g) \to \text{CO}_2(g), \quad \Delta H = -393.5 \text{ kJ/mol}$$Solution:
This reaction itself represents the formation of CO₂ from elements in standard states.
By definition:
$$\Delta_f H°[\text{CO}_2(g)] = -393.5 \text{ kJ/mol}$$Answer: -393.5 kJ/mol
Key point: Formation enthalpy is for making 1 mole from elements.
Question: For the reaction:
$$\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$$Given $\Delta H = -92.4$ kJ at 298 K, calculate ΔU.
Solution:
$$\Delta H = \Delta U + \Delta n_g RT$$Calculate Δn_g:
- Gaseous products: 2 moles NH₃
- Gaseous reactants: 1 + 3 = 4 moles
- $\Delta n_g = 2 - 4 = -2$
Answer: ΔU = -87.44 kJ
Note: ΔH is more negative than ΔU because volume decreases (Δn_g < 0).
Question: Calculate ΔH for:
$$\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \to 2\text{Fe}(s) + 3\text{CO}_2(g)$$Given:
- $\Delta_f H°[\text{Fe}_2\text{O}_3] = -824$ kJ/mol
- $\Delta_f H°[\text{CO}] = -110.5$ kJ/mol
- $\Delta_f H°[\text{CO}_2] = -393.5$ kJ/mol
- $\Delta_f H°[\text{Fe}] = 0$ (element)
Solution:
$$\Delta_r H° = \sum \Delta_f H°(\text{products}) - \sum \Delta_f H°(\text{reactants})$$Products: $2\text{Fe} + 3\text{CO}_2$
$$= 2(0) + 3(-393.5) = -1180.5 \text{ kJ}$$Reactants: $\text{Fe}_2\text{O}_3 + 3\text{CO}$
$$= -824 + 3(-110.5) = -824 - 331.5 = -1155.5 \text{ kJ}$$ $$\Delta_r H° = -1180.5 - (-1155.5) = -25 \text{ kJ}$$Answer: -25 kJ (exothermic)
Level 2: JEE Main
Question: Calculate the enthalpy of formation of ethane (C₂H₆) from:
(1) $\text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g) \to 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$, $\Delta H_1 = -1560$ kJ (2) $\text{C(graphite)} + \text{O}_2(g) \to \text{CO}_2(g)$, $\Delta H_2 = -393.5$ kJ (3) $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \to \text{H}_2\text{O}(l)$, $\Delta H_3 = -285.8$ kJ
Solution:
Target equation:
$$2\text{C(graphite)} + 3\text{H}_2(g) \to \text{C}_2\text{H}_6(g)$$Manipulate given equations:
Reverse equation (1):
$$2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \to \text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g), \quad \Delta H = +1560 \text{ kJ}$$Multiply equation (2) by 2:
$$2\text{C(graphite)} + 2\text{O}_2(g) \to 2\text{CO}_2(g), \quad \Delta H = 2(-393.5) = -787 \text{ kJ}$$Multiply equation (3) by 3:
$$3\text{H}_2(g) + \frac{3}{2}\text{O}_2(g) \to 3\text{H}_2\text{O}(l), \quad \Delta H = 3(-285.8) = -857.4 \text{ kJ}$$Add all three:
$$2\text{C} + 2\text{O}_2 + 3\text{H}_2 + \frac{3}{2}\text{O}_2 + 2\text{CO}_2 + 3\text{H}_2\text{O} \to 2\text{CO}_2 + 3\text{H}_2\text{O} + \text{C}_2\text{H}_6 + \frac{7}{2}\text{O}_2$$Cancel common terms:
$$2\text{C(graphite)} + 3\text{H}_2(g) \to \text{C}_2\text{H}_6(g)$$ $$\Delta_f H°[\text{C}_2\text{H}_6] = +1560 - 787 - 857.4 = -84.4 \text{ kJ/mol}$$Answer: -84.4 kJ/mol
Question: Calculate the C=C bond energy in ethene (C₂H₄) given:
- $\Delta_f H°[\text{C}_2\text{H}_4(g)] = +52$ kJ/mol
- $\Delta_f H°[\text{C(g)}] = +717$ kJ/mol (atomization)
- $\Delta_f H°[\text{H(g)}] = +218$ kJ/mol (atomization)
- C-H bond energy = 414 kJ/mol
Solution:
Formation of C₂H₄ from gaseous atoms:
$$2\text{C(g)} + 4\text{H(g)} \to \text{C}_2\text{H}_4(g)$$Enthalpy change = Bond energies formed:
$$\Delta H_{\text{bonds}} = -(4 \times \text{BE}_{\text{C-H}} + 1 \times \text{BE}_{\text{C=C}})$$Formation from elements:
$$2\text{C(graphite)} + 2\text{H}_2(g) \to \text{C}_2\text{H}_4(g), \quad \Delta H = +52 \text{ kJ}$$Step 1: Atomize elements
$$2\text{C(graphite)} \to 2\text{C(g)}, \quad \Delta H = 2(717) = +1434 \text{ kJ}$$ $$2\text{H}_2(g) \to 4\text{H(g)}, \quad \Delta H = 4(218) = +872 \text{ kJ}$$Step 2: Form molecule from atoms
$$2\text{C(g)} + 4\text{H(g)} \to \text{C}_2\text{H}_4(g), \quad \Delta H = ?$$By Hess’s Law:
$$+52 = +1434 + 872 + \Delta H_{\text{bonds}}$$ $$\Delta H_{\text{bonds}} = 52 - 2306 = -2254 \text{ kJ}$$This represents forming 4 C-H bonds and 1 C=C bond:
$$-2254 = -(4 \times 414 + \text{BE}_{\text{C=C}})$$ $$-2254 = -1656 - \text{BE}_{\text{C=C}}$$ $$\text{BE}_{\text{C=C}} = 2254 - 1656 = 598 \text{ kJ/mol}$$Answer: C=C bond energy = 598 kJ/mol
Question: Calculate the heat evolved when 200 mL of 0.5 M HCl is mixed with 150 mL of 0.8 M NaOH.
Given: $\Delta_{neut} H° = -57.1$ kJ/mol
Solution:
Moles of HCl = $0.2 \times 0.5 = 0.1$ mol Moles of NaOH = $0.15 \times 0.8 = 0.12$ mol
Limiting reagent: HCl (0.1 mol)
Neutralization reaction:
$$\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O}$$or in ionic form:
$$\text{H}^+ + \text{OH}^- \to \text{H}_2\text{O}$$Heat evolved = Moles of limiting reagent × $\Delta_{neut} H°$
$$= 0.1 \times 57.1 = 5.71 \text{ kJ}$$Answer: 5.71 kJ of heat is evolved.
Note: We take magnitude since question asks for heat evolved (not ΔH).
Level 3: JEE Advanced
Question: Calculate the heat of formation of methanol (CH₃OH) from:
(1) $\text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \to \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$, $\Delta H_1 = -726$ kJ (2) $\text{C(graphite)} + \text{O}_2(g) \to \text{CO}_2(g)$, $\Delta H_2 = -393.5$ kJ (3) $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \to \text{H}_2\text{O}(l)$, $\Delta H_3 = -285.8$ kJ
Also, if the standard enthalpy of combustion of methanol is -726 kJ/mol, determine whether the formation is exothermic or endothermic.
Solution:
Target:
$$\text{C(graphite)} + 2\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \to \text{CH}_3\text{OH}(l)$$Manipulation:
Reverse equation (1):
$$\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \to \text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g), \quad \Delta H = +726 \text{ kJ}$$Use equation (2) as is:
$$\text{C(graphite)} + \text{O}_2(g) \to \text{CO}_2(g), \quad \Delta H = -393.5 \text{ kJ}$$Multiply equation (3) by 2:
$$2\text{H}_2(g) + \text{O}_2(g) \to 2\text{H}_2\text{O}(l), \quad \Delta H = 2(-285.8) = -571.6 \text{ kJ}$$Add all:
$$\text{C} + \text{O}_2 + 2\text{H}_2 + \text{O}_2 + \text{CO}_2 + 2\text{H}_2\text{O} \to \text{CO}_2 + 2\text{H}_2\text{O} + \text{CH}_3\text{OH} + \frac{3}{2}\text{O}_2$$Simplify (cancel CO₂ and 2H₂O):
$$\text{C(graphite)} + 2\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \to \text{CH}_3\text{OH}(l)$$ $$\Delta_f H° = +726 - 393.5 - 571.6 = -239.1 \text{ kJ/mol}$$Answer: $\Delta_f H°[\text{CH}_3\text{OH}] = -239.1$ kJ/mol (exothermic formation)
Question: Calculate the lattice enthalpy of NaCl using:
- $\Delta_f H°[\text{NaCl}(s)] = -411$ kJ/mol
- $\Delta_{sub} H°[\text{Na}(s)] = +108$ kJ/mol (sublimation)
- $\Delta_{diss} H°[\text{Cl}_2] = +243$ kJ/mol (bond dissociation)
- $\Delta_{ion} H[\text{Na}(g)] = +496$ kJ/mol (ionization)
- $\Delta_{eg} H[\text{Cl}(g)] = -349$ kJ/mol (electron gain)
Solution:
Target formation reaction:
$$\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g) \to \text{NaCl}(s), \quad \Delta_f H° = -411 \text{ kJ}$$Born-Haber cycle steps:
(1) Sublimation: $\text{Na}(s) \to \text{Na}(g)$, $\Delta H_1 = +108$ kJ
(2) Dissociation: $\frac{1}{2}\text{Cl}_2(g) \to \text{Cl}(g)$, $\Delta H_2 = +\frac{243}{2} = +121.5$ kJ
(3) Ionization: $\text{Na}(g) \to \text{Na}^+(g) + e^-$, $\Delta H_3 = +496$ kJ
(4) Electron gain: $\text{Cl}(g) + e^- \to \text{Cl}^-(g)$, $\Delta H_4 = -349$ kJ
(5) Lattice formation: $\text{Na}^+(g) + \text{Cl}^-(g) \to \text{NaCl}(s)$, $\Delta H_5 = U$ (lattice enthalpy)
By Hess’s Law:
$$\Delta_f H° = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 + U$$ $$-411 = 108 + 121.5 + 496 - 349 + U$$ $$-411 = 376.5 + U$$ $$U = -411 - 376.5 = -787.5 \text{ kJ/mol}$$Answer: Lattice enthalpy of NaCl = -787.5 kJ/mol
Interpretation: Large negative value indicates strong ionic bonding in the crystal.
Question: Calculate the resonance energy of benzene given:
- $\Delta_f H°[\text{C}_6\text{H}_6(g)] = +83$ kJ/mol
- $\Delta_f H°[\text{C(g)}] = +717$ kJ/mol
- $\Delta_f H°[\text{H(g)}] = +218$ kJ/mol
- C-H bond energy = 414 kJ/mol
- C-C bond energy = 347 kJ/mol
- C=C bond energy = 618 kJ/mol
Assume hypothetical benzene structure has 3 C=C and 3 C-C bonds (Kekulé structure).
Solution:
Step 1: Energy to atomize elements
$$6\text{C(graphite)} \to 6\text{C(g)}, \quad \Delta H = 6(717) = +4302 \text{ kJ}$$ $$3\text{H}_2(g) \to 6\text{H(g)}, \quad \Delta H = 6(218) = +1308 \text{ kJ}$$Step 2: Formation from atoms (actual benzene)
$$6\text{C(g)} + 6\text{H(g)} \to \text{C}_6\text{H}_6(g), \quad \Delta H = ?$$By Hess’s Law (formation from elements):
$$\Delta_f H° = \Delta H_{\text{atomization}} + \Delta H_{\text{bond formation}}$$ $$+83 = 4302 + 1308 + \Delta H_{\text{bonds}}$$ $$\Delta H_{\text{bonds}} = 83 - 5610 = -5527 \text{ kJ}$$This represents forming 6 C-H bonds and 6 C-C bonds (in resonance hybrid).
Step 3: Energy if Kekulé structure (3 C=C + 3 C-C + 6 C-H)
$$\Delta H_{\text{Kekulé}} = -(3 \times 618 + 3 \times 347 + 6 \times 414)$$ $$= -(1854 + 1041 + 2484) = -5379 \text{ kJ}$$Step 4: Resonance energy
$$\text{Resonance energy} = \Delta H_{\text{actual}} - \Delta H_{\text{Kekulé}}$$ $$= -5527 - (-5379) = -148 \text{ kJ/mol}$$Answer: Resonance energy of benzene = -148 kJ/mol
Interpretation: Actual benzene is 148 kJ/mol more stable than hypothetical Kekulé structure due to delocalization (resonance).
Quick Revision Box
Essential Formulas
| Concept | Formula | Use |
|---|---|---|
| Enthalpy | $H = U + PV$ | Definition |
| Enthalpy change | $\Delta H = \Delta U + P\Delta V$ | Constant pressure |
| Gas reactions | $\Delta H = \Delta U + \Delta n_g RT$ | Ideal gas |
| Heat at const P | $\Delta H = q_P$ | Open container |
| Formation method | $\Delta_r H° = \sum \Delta_f H°(\text{prod}) - \sum \Delta_f H°(\text{react})$ | Using formation data |
| Hess’s Law | $\Delta H_{\text{total}} = \sum \Delta H_{\text{steps}}$ | Indirect calculation |
| Bond energies | $\Delta H = \sum(\text{broken}) - \sum(\text{formed})$ | From bond data |
| Gibbs equation | $\Delta G° = -RT\ln K$ | (covered in Gibbs Energy) |
Standard Enthalpies (Common Values)
| Substance | $\Delta_f H°$ (kJ/mol) |
|---|---|
| H₂O(l) | -285.8 |
| H₂O(g) | -241.8 |
| CO₂(g) | -393.5 |
| CO(g) | -110.5 |
| CH₄(g) | -74.8 |
| NH₃(g) | -46.1 |
| HCl(g) | -92.3 |
Sign Conventions Quick Reference
| Type | ΔH Sign | Example |
|---|---|---|
| Exothermic | - (negative) | Combustion, freezing |
| Endothermic | + (positive) | Melting, photosynthesis |
| Bond breaking | + (positive) | Always requires energy |
| Bond forming | - (negative) | Always releases energy |
Teacher’s Summary
1. Enthalpy is Heat at Constant Pressure
$$H = U + PV, \quad \Delta H = q_P$$Most reactions occur at constant atmospheric pressure, making ΔH more practical than ΔU.
2. Formation Enthalpies are Building Blocks
$$\Delta_r H° = \sum \Delta_f H°(\text{products}) - \sum \Delta_f H°(\text{reactants})$$Always multiply by stoichiometric coefficients! Elements in standard state have $\Delta_f H° = 0$.
3. Hess’s Law: Path Independence
Total enthalpy change is independent of path - you can calculate ΔH for any reaction by combining other reactions algebraically.
4. Bond Energies Explain Exothermic vs Endothermic
$$\Delta H = \text{Energy to break bonds} - \text{Energy from forming bonds}$$- If more energy released in forming product bonds → Exothermic (ΔH < 0)
- If more energy needed to break reactant bonds → Endothermic (ΔH > 0)
5. Sign Conventions Matter
- Exothermic: ΔH < 0 (heat exits)
- Endothermic: ΔH > 0 (heat enters)
- Bond breaking: always positive
- Bond forming: always negative (releases energy)
“Enthalpy is the chemist’s favorite because most reactions happen at constant pressure!”
JEE Weightage: 3-4 questions per paper. High-yield topic! Master Hess’s Law calculations.
Time-saving tip: For MCQs on Hess’s Law, check if intermediate species cancel - if not, you made an error!
Cross-Links to Related Concepts
Prerequisites:
- First Law of Thermodynamics - Internal energy, work, heat
- Basic Concepts - Mole concept, stoichiometry
Builds on:
- Chemical Bonding - Bond energies, molecular stability
Leads to:
- Entropy - Why some endothermic reactions are spontaneous
- Gibbs Energy - Combining enthalpy and entropy for spontaneity
- Equilibrium - Relation between ΔH° and equilibrium constant
Applications:
- Chemical Kinetics - Activation energy and rate
- Electrochemistry - Cell potential and enthalpy
Related in Physics:
- Calorimetry - Measuring heat changes experimentally
What’s Next?
Entropy and the Second Law - Learn why ice melts spontaneously even though it’s endothermic, understand disorder and randomness, and discover why you can’t unscramble an egg!
Teaser: We’ve learned WHAT happens (ΔH), now we’ll learn WHY it happens (ΔS) and WHEN it happens spontaneously (ΔG)!