Chemical Thermodynamics — Why Reactions Happen

Why does iron rust but gold doesn’t? Why does fuel burn when ignited but won’t spontaneously combust at room temperature? Why do some reactions release heat while others absorb it?

Chemical thermodynamics answers these questions. It’s the physics of energy in chemical systems — a subject that often intimidates students but rewards those who build genuine understanding. The core question: what drives chemical reactions?

Key Terms and Definitions

System: The part of the universe under study (the reacting chemicals). Surroundings: Everything else.

Open system: Can exchange both matter and energy with surroundings (e.g., open flask). Closed system: Exchanges only energy, not matter (e.g., sealed flask). Isolated system: Exchanges neither (e.g., insulated bomb calorimeter).

State function: A property that depends only on the current state of the system, not on how it got there. Temperature, pressure, volume, internal energy, enthalpy, entropy, and Gibbs free energy are all state functions. Work and heat are NOT state functions.

Internal energy (U): The total energy of all molecules in the system (kinetic + potential). We can only measure changes: $\Delta U = q + w$ (First Law).

Enthalpy (H): $H = U + PV$ . For reactions at constant pressure, $\Delta H = q_p$ (heat absorbed at constant pressure). Most lab reactions occur at constant pressure, so enthalpy is the practical thermodynamic quantity.

Entropy (S): A measure of disorder or randomness in a system. The universe tends toward maximum entropy (Second Law).

Gibbs Free Energy (G): $G = H - TS$ . The thermodynamic potential that determines spontaneity at constant $T$ and $P$ .

Spontaneous process: A process that occurs on its own without continuous external energy input. (Note: spontaneous doesn’t mean fast — diamond converting to graphite is spontaneous but extremely slow.)

First Law of Thermodynamics

The First Law is energy conservation: energy cannot be created or destroyed, only transferred.

\Delta U = q + w

where $q$ = heat absorbed by the system and $w$ = work done on the system.

Sign conventions:

$q > 0$ : heat flows into the system (endothermic)
$q < 0$ : heat flows out of the system (exothermic)
$w > 0$ : work done ON the system (compression)
$w < 0$ : work done BY the system (expansion)

Work in chemistry: At constant external pressure $P_{ext}$ :

w = -P_{ext}\Delta V

For expansion ( $\Delta V > 0$ ): $w < 0$ (system does work, loses energy)

At constant pressure: $\Delta H = \Delta U + P\Delta V = \Delta U - w = q_p$

$\Delta H = \Delta U + \Delta n_g RT$ (for ideal gases, relating $\Delta H$ and $\Delta U$ )

where $\Delta n_g$ = moles of gaseous products − moles of gaseous reactants

Exothermic: $\Delta H < 0$ (heat released) Endothermic: $\Delta H > 0$ (heat absorbed)

Hess’s Law

Hess’s Law: The enthalpy change of a reaction is the same regardless of the path taken — only the initial and final states matter (enthalpy is a state function).

This allows us to calculate $\Delta H$ for reactions we can’t directly measure.

Application: If a reaction can be written as the sum of other reactions, its $\Delta H$ is the sum of the $\Delta H$ values of those reactions.

Worked Example

Calculate $\Delta H$ for C(graphite) + $\frac{1}{2}$ O₂(g) → CO(g), given:

(1) C + O₂ → CO₂, $\Delta H_1 = -393.5$ kJ/mol

(2) CO + $\frac{1}{2}$ O₂ → CO₂, $\Delta H_2 = -283.0$ kJ/mol

Target = Equation (1) − Equation (2):

$\Delta H = \Delta H_1 - \Delta H_2 = -393.5 - (-283.0) = -110.5$ kJ/mol

In JEE and CBSE, Hess’s Law problems require manipulating given equations (reversing, multiplying by fractions) to get the target equation. When you reverse an equation, change the sign of $\Delta H$ . When you multiply by a factor $n$ , multiply $\Delta H$ by $n$ . Practice these operations until they’re automatic.

Standard Enthalpies

Standard enthalpy of formation ( $\Delta H_f°$ ): Enthalpy change when 1 mole of a compound is formed from its elements in their standard states (298 K, 1 atm). By definition, $\Delta H_f°$ for any element in its standard state = 0.

Standard enthalpy of reaction:

\Delta H°_{rxn} = \sum \Delta H_f°(\text{products}) - \sum \Delta H_f°(\text{reactants})

Standard enthalpy of combustion ( $\Delta H_c°$ ): $\Delta H$ when 1 mole of substance burns completely in excess oxygen. Always negative (exothermic) for organic compounds.

Bond enthalpy: Energy required to break one mole of a specific bond (gas phase). Used to estimate $\Delta H$ :

\Delta H \approx \sum \text{(bond enthalpies broken)} - \sum \text{(bond enthalpies formed)}

Breaking bonds requires energy (endothermic, positive). Forming bonds releases energy (exothermic, negative).

Second Law and Entropy

The Second Law of Thermodynamics: The total entropy of the universe ( $\Delta S_{universe}$ ) increases in any spontaneous process.

\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0 \text{ (spontaneous)}

Entropy increases when:

Solid → liquid → gas (disorder increases)
Dissolution of a substance in a solvent
Number of moles of gas increases in a reaction
Temperature increases (molecules have more kinetic energy, more disorder)

Entropy decreases when:

Gas → liquid → solid
Number of moles of gas decreases
Formation of a highly ordered structure

For reversible isothermal process: $\Delta S = q_{rev}/T$

For phase transitions: $\Delta S_{fusion} = \Delta H_{fusion}/T_{melting}$

For a reaction (Kirchhoff-type): $\Delta S°_{rxn} = \sum S°(\text{products}) - \sum S°(\text{reactants})$

Gibbs Free Energy — The Spontaneity Criterion

At constant temperature and pressure, the criterion for spontaneity is:

\Delta G = \Delta H - T\Delta S

$\Delta G < 0$ : Spontaneous (reaction proceeds in forward direction)
$\Delta G = 0$ : Equilibrium (no net change)
$\Delta G > 0$ : Non-spontaneous (reverse reaction is spontaneous)

JEE Main regularly tests the relationship between $\Delta G$ , $\Delta H$ , $\Delta S$ , and temperature. The four cases are classic exam material:

$\Delta H$	$\Delta S$	$\Delta G = \Delta H - T\Delta S$	Spontaneous?
−	+	Always −	Yes at all T
−	−	− at low T, + at high T	Only at low T
+	+	+ at low T, − at high T	Only at high T
+	−	Always +	Never spontaneous

Relation to equilibrium constant

At equilibrium: $\Delta G°= -RT\ln K$

Where $K$ is the equilibrium constant and $R = 8.314$ J/mol·K.

This is one of the most important equations in thermodynamics — it connects the thermodynamic spontaneity ( $\Delta G°$ ) to the equilibrium position ( $K$ ).

$\Delta G° < 0 \Rightarrow K > 1$ (products favored at equilibrium)
$\Delta G° = 0 \Rightarrow K = 1$ (reactants and products equally favored)
$\Delta G° > 0 \Rightarrow K < 1$ (reactants favored at equilibrium)

Solved Examples

Example 1 (CBSE Level)

Q: The formation of water is exothermic ( $\Delta H < 0$ ) and the product (liquid water) is more ordered than gaseous H₂ and O₂ ( $\Delta S < 0$ ). Is the reaction spontaneous at all temperatures?

Solution: With $\Delta H < 0$ and $\Delta S < 0$ , $\Delta G = \Delta H - T\Delta S$ .

At low T: $|T\Delta S|$ is small, so $\Delta G \approx \Delta H < 0$ → spontaneous.

At high T: $T\Delta S$ becomes large and positive (since $\Delta S < 0$ , $-T\Delta S$ becomes large and positive), potentially making $\Delta G > 0$ → non-spontaneous.

The reaction is spontaneous only at low temperatures. At very high temperatures, water decomposes back to H₂ and O₂ (this is indeed what happens above ~2000°C).

Example 2 (JEE Main Level)

Q: For a reaction, $\Delta H° = -40$ kJ/mol and $\Delta S° = -100$ J/mol·K. Find (a) $\Delta G°$ at 300 K, (b) the temperature above which the reaction becomes non-spontaneous.

Solution: (a) At 300 K: $\Delta G° = \Delta H° - T\Delta S° = -40000 - (300)(-100) = -40000 + 30000 = -10000$ J/mol = -10 kJ/mol

$\Delta G° < 0$ , so spontaneous at 300 K. ✓

(b) Reaction becomes non-spontaneous when $\Delta G° = 0$ :

$0 = \Delta H° - T\Delta S°$

$T = \frac{\Delta H°}{\Delta S°} = \frac{-40000}{-100} = 400$ K

Above 400 K, $\Delta G° > 0$ and the reaction is non-spontaneous.

Example 3 (JEE Advanced Level)

Q: The equilibrium constant for a reaction is $K = 10^{-5}$ at 298 K. Calculate $\Delta G°$ .

Solution: $\Delta G° = -RT\ln K = -(8.314)(298)\ln(10^{-5})$

$= -(8.314)(298)(-5 \times 2.303)$

$= (8.314)(298)(11.515) = 28,529$ J/mol ≈ 28.5 kJ/mol

Since $\Delta G° > 0$ and $K < 1$ , reactants are favored at equilibrium — confirming consistency.

Exam-Specific Tips

CBSE Class 11 Board: Thermodynamics is one of the most important chapters. Long answer questions (5 marks) commonly ask: (1) derive $\Delta H = \Delta U + \Delta n_g RT$ , (2) explain spontaneity using Gibbs free energy, (3) state and apply Hess’s Law. These derivations have fixed steps — practice writing them until fluent.

JEE Main weightage: Chemical thermodynamics contributes 2-3 questions per shift. The most tested: (a) $\Delta G$ calculation and spontaneity, (b) bond enthalpy calculations, (c) Hess’s Law problem, (d) the relationship $\Delta G° = -RT\ln K$ . Focus on numerical problems — conceptual MCQs are rare compared to calculation-based ones.

Common Mistakes to Avoid

Mistake 1: Confusing $\Delta H$ and $\Delta U$ . Enthalpy ( $\Delta H$ ) includes work done against or by atmospheric pressure ( $P\Delta V$ ). Internal energy ( $\Delta U$ ) does not. For reactions with no gaseous species (solid-solid or solution reactions): $\Delta n_g = 0$ , so $\Delta H \approx \Delta U$ . For reactions producing/consuming gases, they differ significantly.

Mistake 2: Using wrong units in $\Delta G° = -RT\ln K$ . Always use $R = 8.314$ J/(mol·K) and T in Kelvin. If $\Delta H$ is given in kJ, convert to J before using this equation. A common error: mixing kJ and J gives answers off by a factor of 1000.

Mistake 3: Assuming that exothermic reactions are always spontaneous. Spontaneity requires $\Delta G < 0$ , not just $\Delta H < 0$ . An exothermic reaction ( $\Delta H < 0$ ) with a large decrease in entropy ( $\Delta S \ll 0$ ) can be non-spontaneous at high temperatures.

Mistake 4: In bond enthalpy calculations, the formula is: bonds broken (reactants) MINUS bonds formed (products). Students sometimes subtract in the wrong order. Remember: breaking bonds costs energy (+), forming bonds releases energy (−). Net $\Delta H$ = energy in − energy out = $\sum$ (bonds broken) − $\sum$ (bonds formed).

Mistake 5: Forgetting that standard enthalpy of formation for an element in its standard state is zero. $\Delta H_f°[O_2(g)] = 0$ , $\Delta H_f°[C(\text{graphite})] = 0$ , $\Delta H_f°[Fe(s)] = 0$ . These are often not listed in tables but are implicitly zero.

Practice Questions

Q1: For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ : $\Delta H° = -92$ kJ/mol and $\Delta S° = -198$ J/(mol·K). At what temperature does the reaction become non-spontaneous?

At transition temperature, $\Delta G = 0$ :

$T = \frac{\Delta H°}{\Delta S°} = \frac{-92000}{-198} = 465$ K

Above 465 K (192°C), the reaction becomes non-spontaneous. This is why the Haber process for ammonia synthesis operates at 400-500°C (where the reaction is non-spontaneous!) but uses high pressure and catalysts to achieve acceptable rates and conversions.

Q2: Calculate $\Delta H$ for $C(s) + 2H_2(g) \to CH_4(g)$ given bond enthalpies: C=C (347 kJ/mol), H-H (436 kJ/mol), C-H (413 kJ/mol).

For C(s) + 2H₂(g) → CH₄(g), we need to break 2 H-H bonds and form 4 C-H bonds (graphite to gaseous carbon requires sublimation energy, but for this simplified approach):

Bonds broken: 2 × H-H = 2 × 436 = 872 kJ

Bonds formed: 4 × C-H = 4 × 413 = 1652 kJ

$\Delta H \approx 872 - 1652 = -780$ kJ/mol

(Note: The actual value is -74.8 kJ/mol, but this simplified calculation doesn’t account for C sublimation energy — it illustrates the method.)

FAQs

Q: What is the difference between spontaneous and thermodynamically favorable? They mean the same thing in modern usage. A spontaneous process is one where $\Delta G < 0$ — it proceeds without continuous external energy input. Note that “spontaneous” doesn’t say anything about the rate of the reaction. Diamond converting to graphite is thermodynamically spontaneous ( $\Delta G < 0$ ) but practically never happens at room temperature because of the enormously high activation energy.

Q: Can a non-spontaneous reaction be made to happen? Yes — by coupling it to a spontaneous reaction that releases enough energy to drive it. In biology, the hydrolysis of ATP ( $\Delta G \approx -30$ kJ/mol) drives thousands of non-spontaneous biochemical reactions. Electrolysis of water uses electrical energy to drive the non-spontaneous formation of H₂ and O₂.

Q: Why is the sign convention for work confusing? Two conventions exist: chemistry convention uses $w = -P\Delta V$ (work done ON the system is positive — our convention here), while physics sometimes uses $w = P\Delta V$ (work done BY the system is positive). In CBSE and JEE, the IUPAC chemistry convention ( $w = -P_{ext}\Delta V$ ) is standard. When in doubt, state your convention explicitly.