Thermodynamics: Class 11 Chemistry · Chapter 5

Chemistry Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

130 min
1 simulation
28 practice questions
30 MCQs

What you will learn in Thermodynamics

Distinguish system, surroundings and boundary, and classify systems as open, closed or isolated
Explain state functions and apply the first law, ΔU = q + w, with IUPAC sign conventions
Calculate pressure–volume work for free, constant-pressure and reversible isothermal expansions of an ideal gas
Define enthalpy, relate ΔH to ΔU through Δn_gRT, and use heat capacities, including Cp − Cv = R
Describe bomb and constant-pressure calorimetry and calculate ΔU and ΔH from calorimetric data
Calculate reaction enthalpies from formation enthalpies, Hess's law, bond enthalpies and Born–Haber cycles
Explain spontaneity with entropy and the second and third laws, and predict it from ΔG = ΔH − TΔS
Relate the standard Gibbs energy change to the equilibrium constant and calculate one from the other

Original companion text aligned to NCERT Chemistry Class 11, Chapter 5, pp. 136–167. Not affiliated with or endorsed by NCERT.

A pressure cooker's weight lifts and whistles, a matchstick flares, an ice cube vanishes into a glass of nimbu-paani. Each is a change that trades energy with its environment, as heat, as work, or both. Thermodynamics (ऊष्मागतिकी) is the bookkeeping of those trades.

What makes thermodynamics powerful is what it ignores. It never follows individual molecules, only quantities a laboratory can measure on a whole sample of some $10^{23}$ of them, such as pressure, volume and temperature. It never asks how fast a change goes or by what mechanism; it only compares where the change starts with where it ends. The price is that it describes samples at equilibrium, whose bulk properties have stopped changing, and changes between such states.

5.1 Thermodynamic Terms

Before any calculation we need a small, precise vocabulary. Most of these words are everyday words given sharper meanings.

5.1.1 The System and the Surroundings

Drop a piece of zinc into dilute hydrochloric acid in a test tube. If that reaction is what you want to study, the zinc, the acid and the products form the system (निकाय): the part of the universe we watch and measure. Everything else is the surroundings (परिवेश). Together they make up the universe.

\begin{array}{c} universe = system \\ + surroundings \end{array}

In practice only the nearby surroundings matter: the glass of the tube, the air and the bench, which can actually exchange energy or matter with the reaction. A distant galaxy is technically part of the surroundings, but no measurement of ours will ever notice it.

Where the system ends is your choice, and the dividing surface is called the boundary. It can follow a real wall, such as the glass of the tube, or be an imaginary surface, say one drawn around just the reacting liquid. Whatever passes through it is counted as heat, work or matter exchanged with the surroundings, so a careless choice muddles the bookkeeping.

Definition

System

The part of the universe chosen for study; the rest of the universe that can interact with it forms the surroundings.

ExampleThe contents of a reaction flask.

5.1.2 Types of the System

We sort systems by what is allowed to cross their boundary (Fig. 5.1).

An open system exchanges both energy and matter with its surroundings. A pan of water boiling on a stove takes heat from the flame and loses water as steam.
A closed system exchanges energy but not matter. A sealed can of food warmed in a water bath takes in heat, but nothing enters or leaves the can.
An isolated system exchanges neither. A well-stoppered vacuum flask comes close: for a few hours its tea neither leaks nor cools much.

4/4Isolated: neither energy nor matter crosses.

Fig. 5.1What may cross the boundary decides the type of system.NCERT Fig. 5.2 · p. 137

A pan of water on a flame, a sealed can and a vacuum flask stand side by side. Step one: a dashed boundary around the pan separates the system from the surroundings. Step two: a dashed arrow shows steam (matter) leaving the pan and a solid arrow shows energy crossing; it is labelled open. Step three: the can gets an energy arrow and a cross over its lid for matter; it is labelled closed. Step four: the flask gets crosses for both and is labelled isolated.

Real walls always leak a little, so 'isolated' is an approximation that holds over some stretch of time. Most reactions we analyse in this chapter are treated as closed systems.

5.1.3 The State of the System

To calculate anything, we must say exactly what the system is like before and after a change. In mechanics you described a body by the position and velocity of every part, which is hopeless for $10^{23}$ molecules. Thermodynamics is content with a few bulk properties that a thermometer, a pressure gauge or a measuring cylinder can read: pressure $p$ , volume $V$ , temperature $T$ , amount $n$ and composition.

These are state variables. Their values describe the present condition of the system and say nothing about its history. They are not all independent. For a fixed amount of an ideal gas, once you choose $p$ and $T$ , the volume is settled by $p V = n R T$ . Choose values for enough independent variables (two, for a fixed amount of a pure gas) and all the rest follow.

A quantity whose value depends only on the present state, and not on how that state was reached, is a state function (अवस्था फलन). Its change depends only on the initial and final states. A hill makes the idea concrete (Fig. 5.2): the height you gain between two points is fixed, however far you walk to get there.

4/4Both routes gain exactly +600 m. Height behaves like $U$ ; distance walked behaves like $q$ or $w$ .

Fig. 5.2Height gained is a state function; distance walked depends on the path.

A hillside with point A at 2000 m near the bottom and point B at the summit, 2600 m. In the first step only the two points and their heights are shown. In the second step a straight, steep route 1 of 1.2 km joins them. In the third step a dashed zig-zag route 2 of 3.5 km is added. In the fourth step a double arrow on the right shows the height gain, +600 m, the same for both routes, while the distances walked are different.

Definition

State function

A property whose value depends only on the present state of a system, so that its change depends only on the initial and final states.

Example $p$ , $V$ , $T$ , $U$ and $H$ are state functions; $q$ and $w$ are not.

Temperature is a familiar example. Heat a pot of milk from 20 °C to 60 °C directly, or let it cool to 15 °C first and then heat it: either way $Δ T = + 40$ °C. A mobile wallet works the same way. If its balance goes from ₹200 to ₹500, it has gained ₹300, whatever mix of top-ups and payments happened in between. The balance is like a state function; the list of transactions is like the path.

5.1.4 The Internal Energy as a State Function

A system stores energy in many forms: the kinetic energy of molecules flying about and spinning, the energy of vibrating bonds, the chemical energy of electrons and nuclei. We lump all of it together as the internal energy $U$ (आंतरिक ऊर्जा). It can change in only three ways: heat flows in or out, work is done on the system or by it, or matter carries energy across the boundary. Matter crosses only in open systems, so for closed systems heat and work are enough.

Work

A wall that blocks heat completely is called adiabatic (रुद्धोष्म); the double wall of a good vacuum flask comes close. Any change in a system enclosed by such walls is an adiabatic process. With no heat able to cross, work is the only way left to change the system's energy.

Now change the state of 200 g of water at 25.0 °C in two different ways (Fig. 5.3). First, churn it with a wooden churner, like the mathani used for buttermilk, until 2.50 kJ of mechanical work has gone in. Second, starting again from the same state, run a 50 W heating coil in it for 50 s, which is 2.50 kJ of electrical work. Both times the water ends at 28.0 °C. The same adiabatic work, done in any way at all, produces the same change of state.

4/4Conducting walls and no work: 2.50 kJ of heat gives the same 28.0 °C, so $Δ U = q .$

Fig. 5.3Equal energy in, by any route, gives the same final state.NCERT Fig. 5.3 · p. 138

A container of 200 g of water at 25.0 °C with a thermometer. In the first step nothing has been added yet. In the second step, with insulated walls, a churner does 2.50 kJ of work and q is zero; the thermometer shows 25.0 rising to 28.0 °C. In the third step, starting again, an electric heater does 2.50 kJ of work and the rise is the same. In the fourth step the walls are conducting, no work is done, and 2.50 kJ of heat flows in from a hot bath, shown by arrows; the rise is again 25.0 to 28.0 °C.

This experimental fact lets us define $U$ . If equal adiabatic work always produces the same change, the work must be measuring the change in a property that belongs to the state itself:

Δ U = U_{2} - U_{1} = w_{ad}

So $U$ is a state function. We follow the IUPAC sign convention: work done on the system counts as positive, because it raises the system's energy, and work that the system does on its surroundings counts as negative.

Heat

Return to the starting state and reach 28.0 °C a third way. Swap the insulated walls for thin steel ones, which let heat through, and stand the container in a hot water bath. Energy flows in because of the temperature difference. Energy transferred like this is heat, $q$ . With no work done and no change in volume, all of it raises the internal energy, so $Δ U = q .$

By the same convention, $q$ is positive when heat flows into the system and negative when the system loses heat to its surroundings.

The general case

Usually heat and work act together, and the change in internal energy is their sum:

First law of thermodynamics

Δ U = q + w

(5.1)

Here is the subtle point. Take a system between the same two states by two different routes, and the split between heat and work can come out quite differently, just as the distance walked did on the hill; yet the total always agrees. And if nothing crosses the boundary at all, as in an isolated system, $q = w = 0$ and $U$ cannot change.

LawFirst law of thermodynamics

The total energy of an isolated system never changes. Energy can change its form or move between a system and its surroundings, but it is never created or destroyed.

5/5500 J of heat in and 120 J of work done by the system: $Δ U = + 380 J .$

Fig. 5.4Every sign is counted from the system's point of view.

A box labelled system holds energy U. Heat arrows sit on the left, heat in above and heat out below; dashed work arrows sit on the right, work on above and work by below. Steps one to four each highlight one arrow and give its sign: heat in makes q positive, heat out makes q negative, work on makes w positive and work by makes w negative. In the fifth step heat in (500 J) and work by (120 J) are highlighted together, and the result Delta U = 500 − 120 = +380 J is shown.

Example 5.1NCERT Problem 5.1 · p. 140

Reading the first law

Write $Δ U$ for each change and say what kind of boundary the system must have. (a) Gas in a sealed, rigid steel bottle is warmed by 300 J of heat. (b) Gas in an insulated cylinder pushes its piston out, doing 150 J of work. (c) Gas in a cylinder is compressed by 400 J of work while it loses 250 J of heat to cooling water.

Try it yourself first. Then reveal the solution one step at a time.

5.2 Applications

Fuel burning in an engine cylinder makes hot gas that shoves a piston; the same fuel burning under a pan simply heats the food. To put numbers on both with the first law, we need working formulas for $w$ and for $q$ .

5.2.1 Work

We consider only pressure–volume work, done when a gas expands or is compressed against an external pressure. Picture one mole of an ideal gas in a cylinder closed by a weightless, frictionless piston.

DerivationWork done against a constant external pressure

Let the piston have area $A$ , with a constant external pressure $p_{ex}$ acting on it, so the force on it is $F = p_{ex} A .$ If the piston moves in by a distance $l$ , the volume changes by $Δ V = V_{f} - V_{i} = - l A,$ negative because the gas is squeezed.

\begin{aligned} w & = F \times l \\ = p_{ex} A l \\ = - p_{ex} Δ V \end{aligned}

w = - p_{ex} (V_{f} - V_{i})

(5.2)

The minus sign makes the convention come out right. On compression $V_{f} < V_{i}$ , so $w > 0$ : work is done on the gas. On expansion $V_{f} > V_{i}$ and $w < 0$ : the gas does work on the surroundings.

Now let the external pressure change during the process. If it changes in a few finite jumps, the total work is the sum of $- p_{ex} Δ V$ over the stages. Make the jumps smaller and smaller. In the limit the external pressure differs from the gas pressure only infinitesimally at every moment, and each stage changes the volume by an infinitesimal $d V$ :

w = - \int_{V_{i}}^{V_{f}} p_{ex} d V

(5.3)

Such an idealised process, carried out in endless tiny steps with the system never more than a whisker away from equilibrium, is a reversible process (उत्क्रमणीय प्रक्रम): nudge $p_{ex}$ the other way by a hair and it runs backwards, retracing its path. Every real process, which takes a finite time, is irreversible. On a $p V$ graph the work is an area, which makes the comparison easy to see (Fig. 5.5).

Definition

Reversible process

An idealised process that passes through an unbroken chain of equilibrium states, so that a tiny change in conditions would send it back along the same path.

ExampleA gas expanding as the weight on its piston is removed grain by grain.

4/4A reversible expansion delivers the most work; a reversible compression needs the least.

Fig. 5.5The more stages, the closer the area gets to the reversible maximum.NCERT Fig. 5.5 · p. 141

Pressure in bar against volume in litres for 1.00 mol of ideal gas at 300 K, with a dashed isotherm from 5.0 bar at 5 L to 1.25 bar at 20 L. Step one: one stage against 1.25 bar; the shaded rectangle, the work, is 1.87 kJ. Step two: two stages, against 2.49 bar to 10 L and then 1.25 bar; the staircase is 2.49 kJ. Step three: reversible expansion; the whole area under the isotherm, 3.46 kJ, is shaded. Step four overlays the stage outlines on it: reversible gives the most work.

DerivationReversible isothermal expansion of an ideal gas

In a reversible process $p_{ex} = p \pm d p,$ where $p$ is the gas pressure. The product $d p d V$ is negligibly small, so $p_{ex}$ can be replaced by $p$ :

w_{rev} = - \int_{V_{i}}^{V_{f}} p d V

(5.4)

Unlike $p_{ex}$ , the gas pressure is known: for $n$ mol of ideal gas $p = n R T / V$ . At constant temperature (an isothermal process) $n R T$ is constant and comes out of the integral, so the reversible work is

\begin{aligned} w & = - n R T \int_{V_{i}}^{V_{f}} \frac{d V}{V} \\ = - n R T \ln \frac{V_{f}}{V_{i}} \end{aligned}

\begin{aligned} w_{rev} & = - 2.303 n R T \\ \times \log \frac{V_{f}}{V_{i}} \end{aligned}

(5.5)

Putting Eq. 5.2 into the first law gives $Δ U = q - p_{ex} Δ V$ for a change against a constant pressure. Several special cases follow.

Free expansion. A gas expanding into a vacuum pushes against nothing: $p_{ex} = 0,$ so $w = 0$ . Joule found that an ideal gas expanding freely at constant temperature absorbs no heat either, so $Δ U = 0.$
Constant volume. With $Δ V = 0$ there is no pressure–volume work and $Δ U = q_{V};$ the subscript records that the heat was supplied at constant volume.
Isothermal change of an ideal gas. The internal energy of an ideal gas depends only on its temperature, so $Δ U = 0$ and $q = - w$ . Against a constant pressure, $q = p_{ex} (V_{f} - V_{i});$ reversibly, $q = n R T \ln (V_{f} / V_{i}) .$
Adiabatic change. $q = 0$ , so $Δ U = w_{ad} .$

Fig. 5.6Set the temperature, the start and end volumes and the number of stages, then press Play to run the piston and watch the work add up.

Example 5.2NCERT Problem 5.2 · p. 142

Expansion into a vacuum

1.00 mol of an ideal gas at 300 K occupies 5.0 L. It expands isothermally into an evacuated space until its volume is 20.0 L. Find $w$ , $q$ and $Δ U$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 5.3NCERT Problem 5.3 · p. 142

Expansion against a steady pressure

Repeat Example 5.2, but let the gas expand against a constant external pressure of 1.0 bar.

Try it yourself first. Then reveal the solution one step at a time.

Example 5.4NCERT Problem 5.4 · p. 142

The same expansion, done reversibly

Now let the gas of Example 5.2 expand reversibly and isothermally from 5.0 L to 20.0 L at 300 K.

Try it yourself first. Then reveal the solution one step at a time.

5.2.2 Enthalpy, H

A useful new state function

A sealed, rigid vessel is not how most chemistry happens. In a school laboratory, reactions bubble away in open beakers under whatever pressure the atmosphere provides, and any gas they make has to shove the air aside to make room, which costs energy (Fig. 5.7). So the heat measured on an open bench is not $Δ U$ , and we need a new state function that this heat does measure.

DerivationHeat at constant pressure

At constant pressure the only work is $w = - p Δ V,$ so the first law reads $Δ U = q_{p} - p Δ V,$ with $q_{p}$ the heat absorbed. Label the initial state 1 and the final state 2:

\begin{aligned} q_{p} & = (U_{2} - U_{1}) \\ + p (V_{2} - V_{1}) \end{aligned}

\begin{aligned} q_{p} & = (U_{2} + p V_{2}) \\ - (U_{1} + p V_{1}) \end{aligned}

(5.6)

The combination $U + p V$ appears at both ends, so we give it a name, the enthalpy $H$ (एन्थैल्पी):

H = U + p V

(5.7)

\begin{aligned} q_{p} & = H_{2} - H_{1} \\ = Δ H \end{aligned}

This is a remarkable result. Heat is normally path-dependent, yet the heat taken in at constant pressure equals the change in a state function, since $U$ , $p$ and $V$ each depend only on the state. For a finite change at constant pressure,

Δ H = Δ U + p Δ V

(5.8)

So whenever you measure heat at constant pressure, you are measuring $Δ H$ . A reaction that releases heat is exothermic and has $Δ H < 0;$ one that absorbs heat is endothermic, with $Δ H > 0.$ At constant volume $Δ V = 0,$ and Eq. 5.8 gives $Δ H = Δ U = q_{V} .$

Definition

Enthalpy

$H = U + p V$ , a state function whose change equals the heat absorbed by a system at constant pressure.

ExampleMelting 1 mol of ice at 0 °C: $Δ H = + 6.01 k J .$

3/3For the same $Δ U$ , the constant-pressure heat is larger by $p Δ V$ . That heat is $Δ H$ .

Fig. 5.7At constant pressure, part of the heat pays for pushing the air back.

A bolted rigid vessel of gas, labelled fixed volume, and a cylinder whose free piston is pushed down by the air, labelled fixed pressure. Step one: the rigid vessel is heated while the cylinder is faded. Step two: the cylinder is heated and its piston rises. Step three: both are heated for comparison, with q V = Delta U under the vessel and q p = Delta U + p Delta V under the cylinder.

How big is the $p Δ V$ term? A mole of a solid or liquid occupies only tens of millilitres, so for reactions among them $p Δ V$ is a few joules and $Δ H \approx Δ U .$ A mole of gas near room temperature occupies about 25 L, and then $p Δ V$ runs to kilojoules. The ideal gas law gives it neatly.

Derivation

Δ H

and

Δ U

for reactions of gases

At constant $T$ and $p$ , let the gaseous reactants contain $n_{A}$ mol in a volume $V_{A}$ , and the gaseous products $n_{B}$ mol in $V_{B}$ :

\begin{aligned} p V_{A} & = n_{A} R T \\ p V_{B} & = n_{B} R T \end{aligned}

Subtracting the first from the second,

p Δ V = Δ n_{g} R T

(5.9)

where $Δ n_{g}$ is the change in the amount of gas: moles of gas on the product side less moles of gas on the reactant side. Substituting in Eq. 5.8 gives

Δ H = Δ U + Δ n_{g} R T

(5.10)

Only gases count in $Δ n_{g}$ . Watch the units: $R = 8.314 J K^{- 1} m o l^{- 1},$ so convert $Δ n_{g} R T$ to kJ before adding it to a value in kJ.

Example 5.5NCERT Problem 5.5 · p. 143

From $Δ H$ to $Δ U$

For $N_{2} (g) + 3 H_{2} (g) \overset{}{\to} 2 N H_{3} (g)$ at 298 K, $Δ H = - 91.8 k J .$ Treating the gases as ideal, find $Δ U$ .

Try it yourself first. Then reveal the solution one step at a time.

Extensive and intensive properties

Pour a 1.0 L jug of hot chai into two cups (Fig. 5.8). Each cup holds half the volume, half the mass and half the enthalpy, yet the chai is just as hot and just as dense as before. A property that scales with the amount of matter, such as mass, volume, $U$ , $H$ or heat capacity, is an extensive property (विस्तीर्ण गुणधर्म). One that does not, such as temperature, pressure or density, is an intensive property (गहन गुणधर्म).

2/2Each cup: 0.5 L, still at 80 °C.

Fig. 5.8Splitting a sample halves its extensive properties and leaves its intensive ones unchanged.NCERT Fig. 5.6 · p. 144

In the first step a jug holds 1.0 L of chai at 80 °C. In the second step the jug is empty and two cups each hold 0.5 L at 80 °C; text below says extensive properties such as volume, mass and enthalpy are halved, while intensive properties such as temperature and density stay the same.

Dividing one extensive property by another gives an intensive one. Density is mass divided by volume, and a molar property $X_{m} = X / n$ is the value of an extensive property $X$ per mole: molar volume $V_{m}$ , molar heat capacity $C_{m}$ , molar enthalpy $H_{m}$ . None of these depends on how much substance you have.

Heat capacity

When heat flows into a system that is not changing phase or reacting, its temperature rises in proportion: $q = C Δ T .$ The constant of proportionality, $C$ , is the heat capacity: the heat needed per kelvin of temperature rise. The bigger it is, the more sluggishly the temperature responds. Water's is unusually big, one reason why seaside Mumbai has much milder swings of temperature than inland Delhi.

$C$ is extensive, so we quote it per amount. The molar heat capacity $C_{m} = C / n$ is the heat needed to warm 1 mol by 1 K. The specific heat capacity $c$ is the heat needed to warm 1 g by 1 K; for liquid water $c = 4.18 J g^{- 1} K^{- 1} .$ For a sample of mass $m$ ,

q = c m Δ T = C Δ T

(5.11)

The relation between Cp and CV for an ideal gas

Heat capacity depends on the conditions of heating. At constant volume $q_{V} = C_{V} Δ T = Δ U;$ at constant pressure $q_{p} = C_{p} Δ T = Δ H .$

Derivation

C_{p} - C_{V}

for one mole of ideal gas

For 1 mol of ideal gas $p V = R T$ , so $Δ (p V) = R Δ T$ and

Δ H = Δ U + R Δ T

(5.12)

Substitute $Δ H = C_{p} Δ T$ and $Δ U = C_{V} Δ T,$ then divide by $Δ T$ :

C_{p} - C_{V} = R

(5.13)

At constant pressure the gas must also do expansion work, so it needs more heat for the same rise in temperature: $C_{p}$ exceeds $C_{V}$ by $R = 8.314 J K^{- 1} m o l^{- 1} .$ For argon, $C_{V} = 12.5$ and $C_{p} = 20.8 J K^{- 1} m o l^{- 1} .$

Quick check

1 of 4

For which reaction is

Δ H

equal to

Δ U

H_{2} (g) + C l_{2} (g) \overset{}{\to} 2 H C l (g)

2 C O (g) + O_{2} (g) \overset{}{\to} 2 C O_{2} (g)

C a C O_{3} (s) \overset{}{\to} C a O (s) + C O_{2} (g)

N_{2} O_{4} (g) \overset{}{\to} 2 N O_{2} (g)

Show answer

Answer (A) $H_{2} (g) + C l_{2} (g) \overset{}{\to} 2 H C l (g)$

Δ H = Δ U

when

Δ n_{g} = 0 :

two moles of gas on each side for

H_{2} + C l_{2} \overset{}{\to} 2 H C l .

The others have

Δ n_{g} = - 1,

+ 1

and

+ 1

5.3 Measurement of ΔU and ΔH: Calorimetry

To measure heat, let it warm (or cool) something whose heat capacity you know, and read a thermometer. That is calorimetry. The 'something' is a calorimeter, a vessel together with the liquid around it, calibrated so that each kelvin of temperature change stands for a known number of joules: $q = C Δ T .$ Whether the reaction is sealed in or open to the air decides what that heat tells us (Fig. 5.9).

ΔU measurements

The point of a bomb calorimeter is that its reaction vessel cannot expand. A pellet of the sample sits in a small cup inside a thick-walled steel 'bomb' charged with oxygen at about 30 atm. The bomb is lowered into a stirred, insulated bucket of water and the sample is fired electrically through a fine wire. With the volume fixed, no $p V$ work is possible even if gases are produced, so all the energy released ends up as heat in the steel and the water: the heat measured is $q_{V}$ , and it equals $- Δ U$ of the reaction. A thermometer gives $Δ T$ , and the calorimeter's heat capacity, found beforehand by burning a known mass of benzoic acid, turns it into joules.

ΔH measurements

To measure $q_{p}$ , the reaction runs open to the atmosphere. For reactions in solution, two nested foam cups with a loose lid work surprisingly well in a school laboratory. Heat measured this way, at constant pressure, is the enthalpy of reaction, $Δ_{r} H .$ Its sign follows the system: if the solution warms up, the reaction gave heat away and $Δ_{r} H < 0;$ if the solution cools, the reaction took heat in and $Δ_{r} H > 0.$

3/3A reaction in the cup runs at atmospheric pressure, so the heat measures $Δ H$ .

Fig. 5.9Constant volume gives

Δ U

; constant pressure gives

Δ H

.NCERT Fig. 5.7 · p. 145

Left: a sealed steel bomb with a sample cup and ignition wires in a stirred, insulated water bath with a thermometer. Right: two nested foam cups with a lid, stirrer and thermometer in a solution. Step one shows both before any reaction. Step two: a flame appears in the bomb, wavy arrows carry heat into the water, its thermometer rises, and a label says this measures Delta U. Step three: the cup's thermometer rises, a note says its loose lid is open to air, and a label says this measures Delta H.

Example 5.6NCERT Problem 5.6 · p. 146

Burning naphthalene in a bomb

A 0.500 g sample of naphthalene, $C_{10} H_{8}$ (molar mass 128.2 g mol⁻¹), burns completely in a bomb calorimeter whose heat capacity is 10.0 kJ K⁻¹. The temperature rises by 2.01 K near 298 K. Find $Δ_{c} U$ and $Δ_{c} H$ per mole of naphthalene.

Try it yourself first. Then reveal the solution one step at a time.

Try this

Measure a reaction enthalpy in foam cups

You’ll need

two nested foam cups with a lid
a thermometer reading to 0.1 °C
50 mL of 1.0 M hydrochloric acid
50 mL of 1.0 M sodium hydroxide solution
a measuring cylinder

Predict first

Mixing equal volumes of the two solutions will

cool the mixture by a few degreeswarm the mixture by a few degreesleave the temperature unchanged

Measure the temperature of each solution; they should agree within 0.2 °C.
Pour the acid into the cups, add the alkali quickly, fit the lid and stir.
Record the highest temperature reached.
Find $Δ T$ and $q = (100 g) (4.18 J g^{- 1} K^{- 1}) Δ T,$ treating the mixture as water.

Table 5.1

Reading	Value
Starting temperature (°C)
Highest temperature (°C)
Rise ΔT (K)
Heat q (J)

Your entries are saved automatically.

Quick check

1 of 3

A bomb calorimeter measures the heat released when a fuel burns. This heat gives

Δ H

, because the reaction is fast

Δ U

, because the volume is fixed

Δ H

, because oxygen is in excess

Δ U

, because the temperature is fixed

Show answer

Answer (B) $Δ U$ , because the volume is fixed

The sealed bomb cannot expand, so no work is done and

q_{V} = Δ U .

The temperature is not fixed: measuring its rise is the whole point.

5.4 Enthalpy Change, ΔrH of a Reaction – Reaction Enthalpy

In a reaction, reactants become products. The enthalpy change that goes with it is the reaction enthalpy, $Δ_{r} H :$ the total enthalpy of the products minus that of the reactants, each weighted by its coefficient in the balanced equation.

\begin{aligned} Δ_{r} H & = \sum_{i} a_{i} H_{prod} \\ - \sum_{i} b_{i} H_{react} \end{aligned}

(5.14)

where

$H_{prod}, H_{react}$: molar enthalpies of the products and of the reactants

Here $a_{i}$ and $b_{i}$ are the stoichiometric coefficients of products and reactants. For the burning of hydrogen, with $H_{m}$ standing for a molar enthalpy,

\begin{aligned} 2 H_{2} (g) + O_{2} (g) \\ \overset{}{\to} 2 H_{2} O (l) \end{aligned}

\begin{aligned} Δ_{r} H & = 2 H_{m} (H_{2} O) \\ - 2 H_{m} (H_{2}) \\ - H_{m} (O_{2}) \end{aligned}

Such numbers settle practical questions, such as how large a cooling system an ammonia plant needs to stop its reactor overheating, or how much fuel a cement kiln must burn every hour.

Standard enthalpy of reactions

Burn methane so that its water comes out as vapour instead of liquid, and $Δ H$ changes by 88 kJ per mole. Values are only comparable if everyone agrees on the conditions. The agreed reference is each substance pure, at a pressure of exactly 1 bar, in a stated physical form: its standard state. An enthalpy change with every reactant and product in its standard state is the standard enthalpy of reaction, $Δ_{r} H^{\circ} .$ Temperature is not part of the definition, so it must be quoted; most tables use 298 K.

The superscript ° (often printed as a plimsoll sign, ⦵) means 'standard'. 'Standard' copper at 400 K, for instance, is simply pure solid copper at 400 K and 1 bar.

Enthalpy changes during phase transformations

Melting, boiling and sublimation also absorb or release energy, even though no bonds inside the molecules break. At constant pressure each takes place at a fixed temperature: heat goes in, but the thermometer stalls until the change is complete (Fig. 5.10). The energy is used to pull molecules apart against their mutual attraction.

5/5Only now does the steam get hotter.

Fig. 5.10During melting and boiling the heat changes the phase, not the temperature.

Temperature in degrees Celsius against heat added in kilojoules for 1 mol of water. Step one: a short, steep line shows ice warming from −20 to 0 °C. Step two: a flat line at 0 °C, labelled melting 6.0 kJ. Step three: a sloping line shows liquid water warming from 0 to 100 °C, using about 7.5 kJ. Step four: a long flat line at 100 °C, labelled boiling 40.7 kJ, far longer than the melting plateau. Step five: a short, steep line shows steam warming to 120 °C.

The standard enthalpy of fusion, $Δ_{fus} H^{\circ},$ is the enthalpy change when 1 mol of a solid melts at constant temperature and 1 bar; for ice at 273 K it is +6.01 kJ mol⁻¹. Freezing is the same change run backwards, so a freezer must pull those 6.01 kJ out of every mole of water it turns to ice. The standard enthalpy of vaporisation, $Δ_{vap} H^{\circ},$ is the enthalpy change when 1 mol of a liquid vaporises at constant temperature and 1 bar; for water at its boiling point it is +40.7 kJ mol⁻¹.

Sublimation turns a solid straight into vapour. The standard enthalpy of sublimation, $Δ_{sub} H^{\circ},$ is the enthalpy change when 1 mol of a solid sublimes at constant temperature and 1 bar. Iodine crystals need about 62 kJ mol⁻¹ at 298 K, and the camphor (kapur) tablets used at pujas slowly vanish by subliming.

All three enthalpies are positive, since melting, boiling and subliming always absorb heat. Their sizes follow the strength of the attractions between the particles (Table 5.2). Methane molecules attract each other only weakly and hydrogen sulphide molecules a little more. Ethanol and water form hydrogen bonds, so boiling a mole of either takes several times more heat than boiling a mole of methane. Mercury atoms are held by metallic bonding, so mercury has the highest vaporisation enthalpy in this list.

Table 5.2Melting and boiling at 1 bar (enthalpies in kJ mol⁻¹)NCERT Table 5.1 · p. 147

Substance	$T_{m}$ / K	$Δ_{fus} H^{\circ}$	$T_{b}$ / K	$Δ_{vap} H^{\circ}$
$C H_{4}$	90.7	0.94	111.7	8.19
$H_{2} S$	187.7	2.38	212.9	18.7
$C_{2} H_{5} O H$	159.0	4.93	351.4	38.6
$H_{2} O$	273.2	6.01	373.2	40.7
$H g$	234.3	2.29	629.9	59.1

Example 5.7NCERT Problem 5.7 · p. 148

Why hand sanitiser feels cold

A squirt of hand sanitiser leaves 2.3 g of ethanol (46.1 g mol⁻¹) on your palm, and it evaporates at 298 K. How much heat does it take from your hand, and what is the internal energy change of the ethanol? At 298 K, $Δ_{vap} H (C_{2} H_{5} O H) = 42.3 k J m o l^{- 1};$ treat the vapour as an ideal gas.

Try it yourself first. Then reveal the solution one step at a time.

Example 5.8NCERT Problem 5.8 · p. 148

Making ice cubes

An ice tray holds 90 g of water at 27 °C. How much heat must a freezer remove to turn it into ice at 0 °C? Find $Δ H$ and $Δ U$ for the change. Take $c = 4.18 J g^{- 1} K^{- 1}$ for water and $Δ_{fus} H = 6.01 k J m o l^{- 1} .$

Try it yourself first. Then reveal the solution one step at a time.

Standard enthalpy of formation

Listing $Δ_{r} H^{\circ}$ for every possible reaction would be impossible. Instead we list one special reaction per compound: its formation from its elements. The standard molar enthalpy of formation, $Δ_{f} H^{\circ}$ (मानक संभवन एन्थैल्पी), is the enthalpy change when 1 mol of a compound in its standard state forms from its elements in their reference states. An element's reference state is its most stable form at 25 °C and 1 bar: hydrogen, nitrogen and oxygen as diatomic gases, bromine and mercury as liquids, sodium as a solid, carbon as graphite and sulphur as rhombic crystals.

Definition

Standard enthalpy of formation

The enthalpy change when 1 mol of a compound in its standard state forms from its elements in their reference states.

Example $N a (s) + \frac{1}{2} C l_{2} (g) \overset{}{\to} N a C l (s),$ $Δ_{f} H^{\circ} = - 411.2 k J m o l^{- 1} .$

In a formation equation exactly one mole of product forms from elements alone. Ammonia, for which $Δ_{f} H^{\circ} = - 45.9 k J m o l^{- 1},$ is a typical case:

\begin{aligned} \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \\ \overset{}{\to} N H_{3} (g) \end{aligned}

By definition, $Δ_{f} H^{\circ}$ of an element in its reference state is zero: forming graphite from graphite involves no change. Read the fine print before calling a value a formation enthalpy: every reactant must be an element in its reference state, and exactly one mole of product must form. Oxidising $S O_{2}$ to $S O_{3}$ fails the first test; burning two moles of sodium in chlorine (−822.4 kJ) fails the second, giving $2 Δ_{f} H^{\circ}$ of NaCl.

Table 5.3Standard enthalpies of formation at 298 K, in kJ mol⁻¹NCERT Table 5.2 · p. 149

Substance	$Δ_{f} H^{\circ}$	Substance	$Δ_{f} H^{\circ}$
$A l_{2} O_{3} (s)$	−1675.7	$H_{2} O (g)$	−241.8
$C$ (diamond)	+1.9	$H_{2} O_{2} (l)$	−187.8
$C H_{4} (g)$	−74.8	$K C l (s)$	−436.7
$C_{2} H_{4} (g)$	+52.3	$M g O (s)$	−601.7
$C_{2} H_{6} (g)$	−84.7	$N H_{3} (g)$	−45.9
$C O (g)$	−110.5	$N O (g)$	+90.3
$C O_{2} (g)$	−393.5	$N O_{2} (g)$	+33.2
$F e_{2} O_{3} (s)$	−824.2	$O_{3} (g)$	+142.7
$H_{2} O (l)$	−285.8	$S O_{2} (g)$	−296.8
$N a C l (s)$	−411.2	$S O_{3} (g)$	−395.7

With such a table, any reaction enthalpy follows from

\begin{aligned} Δ_{r} H^{\circ} & = \sum a_{i} Δ_{f} H_{P}^{\circ} \\ - \sum b_{i} Δ_{f} H_{R}^{\circ} \end{aligned}

(5.15)

where

$Δ_{f} H_{P}^{\circ}, Δ_{f} H_{R}^{\circ}$: standard enthalpies of formation of the products and of the reactants

This works because enthalpy is a state function. Imagine taking the reactants apart into their elements and then building the products from those same elements: the elements cancel out, and only formation enthalpies remain.

Example

Thermite welding of rails

Railway track gangs join steel rails using the thermite reaction. Using Table 5.3, find $Δ_{r} H^{\circ}$ for $2 A l (s) + F e_{2} O_{3} (s)$ $\overset{}{\to} A l_{2} O_{3} (s) + 2 F e (s) .$

Try it yourself first. Then reveal the solution one step at a time.

Thermochemical equations

A balanced equation that carries its $Δ_{r} H,$ with the physical state (and, where it matters, the allotrope) of every species, is a thermochemical equation. Compressed natural gas burns as

\begin{aligned} C H_{4} (g) + 2 O_{2} (g) \overset{}{\to} \\ C O_{2} (g) + 2 H_{2} O (l) \end{aligned}

with $Δ_{r} H^{\circ} = - 890.3 k J m o l^{- 1} .$ Three conventions govern how to read it:

Read each coefficient as an amount in moles. That is why fractions such as ½ are perfectly acceptable: half a mole makes sense, half a molecule does not.
$Δ_{r} H^{\circ}$ belongs to the equation as written, and its unit kJ mol⁻¹ means 'per mole of reaction'. Halve every coefficient and $Δ_{r} H^{\circ}$ halves too, to −445.2 kJ mol⁻¹, because enthalpy is extensive.
Reverse the equation and $Δ_{r} H^{\circ}$ changes sign: making methane from carbon dioxide and water would need +890.3 kJ per mole.

States matter too. If the water formed were vapour instead of liquid, 2 × 44.0 = 88.0 kJ less heat would be released per mole of methane, since that much would be spent vaporising the water.

Hess's law of constant heat summation

This is the hill of Fig. 5.2 again, now for reactions. The enthalpies of the reactants and of the products are fixed points, so any sequence of steps between them, real or imagined, must add up to the same $Δ_{r} H .$ That is Hess's law, and its practical power is that it gives the enthalpies of reactions that cannot be carried out cleanly.

LawHess's law

When an overall reaction can be broken into a series of steps at one temperature, adding up the enthalpy changes of the steps gives the enthalpy change of the overall reaction.

\begin{aligned} Δ_{r} H & = Δ_{r} H_{1} + Δ_{r} H_{2} \\ + Δ_{r} H_{3} + \dots \end{aligned}

(5.16)

Burning sulphur shows this on an enthalpy diagram (Fig. 5.11). Sulphur can be oxidised straight to sulphur trioxide, or first to the dioxide, which is then oxidised further, as in the contact process for making sulphuric acid. Both routes start and end at the same levels.

3/3The steps add up to the direct value, as Hess's law requires.

Fig. 5.11Two routes, one total: the enthalpy change depends only on the end points.

An enthalpy axis points up. The top level is sulphur plus one and a half O2; the bottom level is SO3 gas. Step one: a single arrow runs straight down between them, labelled −395.7 kJ per mole. Step two: a middle level, SO2 plus half O2, appears; a dashed arrow of −296.8 runs from the top level to it, and a second dashed arrow of −98.9 runs from it to SO3. Step three: text adds the two steps, −296.8 − 98.9, and shows that they equal −395.7.

Example

Diamond to graphite

At 298 K, $Δ_{c} H^{\circ}$ is −393.5 kJ mol⁻¹ for graphite and −395.4 kJ mol⁻¹ for diamond, both burning to carbon dioxide. Find $Δ_{r} H^{\circ}$ for $C (d i a m o n d) \overset{}{\to} C (g r a p h i t e),$ a change far too slow to measure directly.

Try it yourself first. Then reveal the solution one step at a time.

5.5 Enthalpies for Different Types of Reactions

Some kinds of reaction are so common that their enthalpies have their own names and symbols. Each is a special case of $Δ_{r} H .$

Standard enthalpy of combustion

Every fuel, from the CNG in an auto-rickshaw to the liquid hydrogen in a rocket's upper stage, is rated by the heat a mole of it gives out when it burns completely in oxygen. Measured with everything in its standard state, that is the standard enthalpy of combustion, $Δ_{c} H^{\circ};$ since burning always releases heat, it is always negative. Methane, the main gas in CNG, has $Δ_{c} H^{\circ} = - 890.3 k J m o l^{- 1}$ when the water formed is liquid.

Food is fuel too. Your cells oxidise sugars to carbon dioxide and water through a long chain of enzyme-controlled steps, but by Hess's law the overall enthalpy change equals that of burning them in a calorimeter. Sucrose has $Δ_{c} H^{\circ} \approx - 5640 k J m o l^{- 1},$ so a 4 g teaspoon of sugar (0.0117 mol) supplies about 66 kJ. That is the '16 Calories' on a food label, since 1 Calorie = 1 kcal = 4.184 kJ.

Example 5.9NCERT Problem 5.9 · p. 152

Formation enthalpy of propane from its combustion

Propane, a component of LPG, has $Δ_{c} H^{\circ} = - 2220.0 k J m o l^{- 1}$ at 298 K. Using $Δ_{f} H^{\circ}$ of $C O_{2} (g)$ (−393.5) and $H_{2} O (l)$ (−285.8 kJ mol⁻¹), find $Δ_{f} H^{\circ}$ of propane.

Try it yourself first. Then reveal the solution one step at a time.

Enthalpy of atomization

The enthalpy of atomization, $Δ_{a} H^{\circ},$ is the enthalpy change when 1 mol of a substance breaks up completely into separate atoms in the gas phase. Splitting oxygen, $O_{2} (g) \overset{}{\to} 2 O (g),$ needs 498 kJ mol⁻¹; tearing ammonia fully apart into one N and three H atoms needs 1173 kJ mol⁻¹. For a metal such as potassium, whose solid is already a lattice of atoms, atomization and sublimation are one and the same change: $K (s) \overset{}{\to} K (g)$ needs 89.2 kJ mol⁻¹.

Bond enthalpy

Look at any reaction at the level of molecules and you see bonds being taken apart and new ones assembled. Taking a bond apart always needs energy and assembling one always gives energy out, so a tally of both gives, roughly, the reaction enthalpy. Chemists list bond energies in two slightly different ways.

A diatomic molecule has only one bond. Its bond dissociation enthalpy is the enthalpy change for breaking one mole of that bond, with the molecule and the fragments all in the gas phase: 431 kJ mol⁻¹ for $H C l (g) \overset{}{\to} H (g) + C l (g),$ and 945 kJ mol⁻¹ for the triple bond of $N_{2}$ . With only one bond to break, splitting a diatomic molecule into its atoms is atomization too, so the two numbers coincide.

In a polyatomic molecule, identical bonds need different energies to break one after another, because the fragment left behind keeps changing. Water shows this. Removing the first hydrogen, $H_{2} O (g) \overset{}{\to} H (g) + O H (g),$ takes about 497 kJ mol⁻¹, but removing the second, from OH, takes only about 430 kJ mol⁻¹. For calculations we use the mean bond enthalpy (आबंध एन्थैल्पी), the average over both: 463.5 kJ mol⁻¹ for O–H in water. Mean values for a given bond change only a little from one molecule to another, so one listed value per bond type works reasonably well.

For a reaction in which every species is a gas, bond enthalpies give an estimate:

\begin{aligned} Δ_{r} H^{\circ} & \approx \sum B E_{react} \\ - \sum B E_{prod} \end{aligned}

(5.17)

where

$B E_{react}, B E_{prod}$: bond enthalpies of the bonds in the reactants and in the products

Two cautions. Mean values are averages, so expect errors of several per cent (the ethene example shows a typical one). And the rule applies only when every substance is a gas, since it ignores the energy of condensing or freezing. Use it when formation enthalpies are missing. Some books list bond formation enthalpies, the same numbers with negative signs; with those, the products' sum comes first.

Example

Estimating an enthalpy from bond enthalpies

Estimate $Δ_{r} H^{\circ}$ for the hydrogenation of ethene, $C_{2} H_{4} (g) + H_{2} (g) \overset{}{\to} C_{2} H_{6} (g),$ using mean bond enthalpies (kJ mol⁻¹): C=C 612, H–H 436, C–C 348, C–H 413. Compare with the value from Table 5.3.

Try it yourself first. Then reveal the solution one step at a time.

Lattice enthalpy

The lattice enthalpy (जालक एन्थैल्पी) of an ionic solid is the enthalpy change when 1 mol of it breaks up into gaseous ions, for example $K C l (s) \overset{}{\to} K^{+} (g) + C l^{-} (g) .$ It measures how strongly the lattice holds together. No experiment does this directly, so we build a closed loop of steps whose enthalpies can be measured, a Born–Haber cycle, and use Hess's law: around any closed cycle the enthalpy changes add up to zero (Fig. 5.12).

Definition

Lattice enthalpy

The enthalpy change when 1 mol of an ionic solid separates into its gaseous ions.

Example $K C l (s) \overset{}{\to} K^{+} (g) + C l^{-} (g),$ about +718 kJ mol⁻¹.

4/4The lattice enthalpy is whatever closes the loop: +717.8 kJ mol⁻¹.

Fig. 5.12Going round a closed cycle, the enthalpy changes must add up to zero.NCERT Fig. 5.9 · p. 155

Enthalpy levels in kJ per mole. Step one: solid potassium plus half a mole of chlorine sit at zero, and a dashed arrow of −436.7 drops to solid KCl. Step two: short arrows add +89.2 for gaseous potassium and +121.7 for chlorine atoms. Step three: an arrow of +418.8 ionises potassium, reaching K+ plus an electron plus Cl at the top; a dashed arrow of −348.6 drops to gaseous K+ and Cl−. Step four: a thick arrow from solid KCl up to the gaseous ions, +717.8, is the lattice enthalpy.

Example

Lattice enthalpy of potassium chloride

Use these values in kJ mol⁻¹: $Δ_{f} H^{\circ} (K C l) = - 436.7;$ sublimation of K, +89.2; half the bond enthalpy of $C l_{2}$ , +121.7; ionization enthalpy of K, +418.8; electron gain enthalpy of Cl, −348.6. Find the lattice enthalpy of KCl and the matching internal energy change at 298 K.

Try it yourself first. Then reveal the solution one step at a time.

Enthalpy of solution

Dissolve a mole of a solute in a stated amount of solvent, and the heat absorbed at constant pressure is its enthalpy of solution, $Δ_{sol} H .$ The value drifts with the amount of solvent until the solution is so dilute that solute particles never meet one another; that limiting value is the enthalpy of solution at infinite dilution.

Dissolving an ionic solid can be split into two imaginary steps. First the lattice is pulled apart into gaseous ions, which costs the lattice enthalpy. Then water molecules crowd round each ion, which releases the hydration enthalpy, $Δ_{hyd} H :$

\begin{aligned} Δ_{sol} H & = Δ_{lattice} H \\ + Δ_{hyd} H \end{aligned}

For KCl the lattice costs +717.8 kJ mol⁻¹ and the measured enthalpy of solution is +17.2 kJ mol⁻¹, so hydrating one $K^{+}$ and one $C l^{-}$ must release about 700.6 kJ. The result is a small difference between two large numbers: KCl dissolves with slight cooling. KCl is typical, since dissolving most salts takes in a little heat, and such salts usually dissolve better in hot water than in cold. At the other extreme, a lattice held very tightly may never be repaid by hydration, and the salt hardly dissolves at all.

Try this

Feel the heat of solution

You’ll need

three small beakers or steel tumblers
a thermometer
water
5 g each of sodium chloride, ammonium chloride and anhydrous calcium chloride

Predict first

Which salt should make its water noticeably colder?

calcium chlorideammonium chloridesodium chloride

Put 50 mL of water in each container and note its temperature.
Add 5 g of one salt to each, stir until it dissolves, and note the lowest or highest temperature reached.

Table 5.4

Salt	Start (°C)	End (°C)
NaCl
NH₄Cl
CaCl₂

Your entries are saved automatically.

Enthalpy of dilution

An enthalpy of solution depends on how much solvent is used. The table gives the enthalpy change when 1 mol of pure liquid sulphuric acid is mixed with different amounts of water at 298 K, calculated from the NIST–JANAF thermochemical tables.

Table 5.5Mixing 1 mol of

H_{2} S O_{4} (l)

with water at 298 K

Water added (mol)	$Δ H$ / kJ mol⁻¹
1	−27.8
2	−41.5
4	−53.9
6.5	−61.4

The values grow in size as more water is added, but by smaller and smaller amounts, towards a limit at infinite dilution. The difference between two entries is an enthalpy of dilution: the heat exchanged when extra solvent is added to an existing solution. Taking the solution from 2 to 4 mol of water per mole of acid releases a further 12.4 kJ mol⁻¹:

\begin{aligned} Δ_{dil} H & = - 53.9 + 41.5 \\ = - 12.4 \end{aligned}

Notice that the number depends on where you start (2 mol of water per mole of acid) and on how much water you add (2 mol more), so both must be quoted. For sulphuric acid it is large and negative, which is why every laboratory follows the rule always add acid to water, slowly and with stirring. Pour water into concentrated acid and the first drops can boil on contact and spray acid out.

5.6 Spontaneity

The first law tells us that energy is conserved, but not which way things go. Yet nature is strikingly one-way. Hot tea cools in a cool room; it never warms itself by drawing heat from the room. A drop of perfume spreads through a whole room; the scent never gathers back into the bottle. Iron rusts in moist air; rust never turns back into shiny iron on its own. What decides the direction?

A process that can take place by itself, without continuing help from outside, is spontaneous (स्वतः प्रवर्तित). The word says nothing about speed. Petrol vapour mixed with air can sit for years without burning, yet the reaction is spontaneous: a single spark starts it, and then it runs by itself. Diamond turning into graphite is spontaneous at room temperature but so slow that no one has ever seen it happen. And left to itself, a spontaneous change never runs backwards. To un-mix, un-burn or re-warm something, you must pay with outside help, such as a pump, a refrigerator or a battery.

Is decrease in enthalpy a criterion for spontaneity?

A ball rolls to the bottom of a bowl; a stretched rubber band snaps back. Mechanical systems settle where their energy is lowest, so it is tempting to guess that reactions run in whichever direction lowers enthalpy. Many spontaneous reactions are indeed exothermic: methane burning (−890.3 kJ mol⁻¹), magnesium burning with its dazzling white flame, ammonia forming from nitrogen and hydrogen (−45.9 kJ per mole of ammonia). On an enthalpy diagram their products sit below their reactants.

But the idea fails for some very ordinary changes. Ice melts on a warm day, absorbing 6.01 kJ mol⁻¹. Water evaporates from a wet cloth, absorbing heat, which is why the cloth feels cool. Ammonium nitrate dissolves in water so endothermically that instant cold packs for sports injuries rely on it. All of these are spontaneous, yet the enthalpy goes up. A fall in enthalpy often helps a change along, but it cannot be the whole story.

Entropy and spontaneity

Look at a change with no enthalpy change at all. Two different ideal gases sit side by side in an isolated box, separated by a partition (Fig. 5.13). Remove the partition and they mix, spontaneously and completely, even though $Δ H = 0.$ They never unmix. Why not?

Count the chances. Each molecule wanders at random and is equally likely to be in either half of the box. The chance that one particular A molecule is in the left half is ½. The chance that all 8 A molecules in the figure are there together is $(1 / 2)^{8}$ , or 1 in 256. For a mole of gas the chance is ½ multiplied by itself $6 \times 10^{23}$ times, so small that the unmixed state will never be seen. The mixed state wins not because of energy, but because there are overwhelmingly more ways of being mixed.

3/3All 8 circles back on the left: a 1 in 256 chance. For $10^{23}$ molecules: never.

Fig. 5.13Mixing is spontaneous because mixed arrangements vastly outnumber unmixed ones.NCERT Fig. 5.11 · p. 158

An isolated box holds 8 molecules of gas A drawn as circles and 8 molecules of gas B drawn as squares. Step one: a partition divides the box, with all of A on the left and all of B on the right. Step two: the partition is removed and the molecules move until circles and squares are spread through the whole box. Step three: text below notes that the chance of all 8 A molecules returning to the left half is 1 in 256, and that for a mole of gas it is never seen.

The property that captures this is entropy, $S$ (एन्ट्रॉपी). Loosely, entropy measures the randomness or disorder of a system; more precisely, it measures how many ways its particles and energy can be arranged without changing what we observe. In an isolated system, spontaneous change always heads towards higher entropy. A crystal, with every particle in place, has the lowest entropy of a substance's states, its liquid more, and its gas the most. A reaction whose products are less ordered than its reactants, such as one that makes a gas from solids, has a positive entropy change.

Definition

Entropy

A state function, $S$ , that measures how widely a system's matter and energy are spread out, loosely its disorder; the total entropy rises in every spontaneous change.

SimulationOrder and disorder in the three states

Water particles in a closed box with a temperature slider from −20 to 120 degrees Celsius. At low temperature the particles sit in a regular lattice and only vibrate, as ice. Warming melts them into a jostling liquid at the bottom of the box, and above 100 degrees they fly through the whole box as a gas. A thermometer and a state label show the current temperature and state.

Fig. 5.14Drag the temperature from −20 °C to 120 °C and watch how freely the particles move and spread.

Try this

Start at −20 °C and warm slowly. At which two temperatures does the arrangement change most suddenly?
Rank ice, liquid water and steam by entropy, judging by how many different arrangements the particles can take.
Within one state, does warming make the motion more random? Compare 10 °C with 90 °C.

Entropy can also be measured through heat. Heat added to a system makes its particles move more vigorously, so it raises the entropy. But the same heat matters more to a cold system than to a hot one. A sneeze in a silent library is far more disruptive than the same sneeze in a crowded railway station. Likewise, a joule of heat stirs up a cold, quiet system far more than a hot, already agitated one, so the entropy it creates must shrink as the temperature rises. For a reversible process at temperature $T$ ,

Δ S = \frac{q_{rev}}{T}

(5.18)

Like $U$ and $H$ , entropy is a state function, so $Δ S$ between two states is the same however the change happens. The catch is that we must calculate it along a reversible path, using $q_{rev},$ even when the real process is irreversible. For a spontaneous process, the total entropy change of system plus surroundings is positive:

\begin{aligned} Δ S_{total} & = Δ S_{sys} \\ + Δ S_{surr} \\ > 0 \end{aligned}

(5.19)

A spontaneous change is therefore a climb up an entropy hill. Equilibrium is the summit: once there, any further small change leaves $Δ S_{total} = 0.$

Entropy can tell apart processes that energy cannot. Let 1 mol of ideal gas double its volume at a constant 298 K. Whether the expansion is free or reversible, $Δ U = 0,$ and the gas reaches the same final state, so $Δ S_{sys}$ is the same too. Along the reversible path $q_{rev} = R T \ln 2,$ so $Δ S_{sys} = R \ln 2 = + 5.76 J K^{- 1} .$ In the reversible expansion the surroundings supply that heat at the same temperature: $Δ S_{surr} = - 5.76 J K^{- 1}$ and $Δ S_{total} = 0.$ In the free expansion nothing reaches the surroundings, so $Δ S_{total} = + 5.76 J K^{- 1},$ the mark of an irreversible, spontaneous change.

Example 5.10NCERT Problem 5.10 · p. 159

Predicting the sign of $Δ S$

Does the entropy of the system increase or decrease? (a) Dry ice, solid $C O_{2}$ , turns into gas. (b) Water vapour condenses on a cold glass. (c) $C a C O_{3} (s) \overset{}{\to} C a O (s) + C O_{2} (g)$ (d) $N_{2} (g) + 3 H_{2} (g) \overset{}{\to} 2 N H_{3} (g)$

Try it yourself first. Then reveal the solution one step at a time.

Example 5.11NCERT Problem 5.11 · p. 160

Burning magnesium: the entropy of the universe

For $2 M g (s) + O_{2} (g) \overset{}{\to} 2 M g O (s)$ at 298 K, $Δ_{r} H^{\circ} = - 1203.4 k J m o l^{- 1}$ and $Δ_{r} S^{\circ} = - 216.7 J K^{- 1} m o l^{- 1} .$ The system's entropy falls, because a gas is used up. Why is the reaction spontaneous?

Try it yourself first. Then reveal the solution one step at a time.

Gibbs energy and spontaneity

Checking $Δ S_{total}$ means doing the bookkeeping for the surroundings as well as for the system, which is awkward. In the 1870s the American scientist J. Willard Gibbs found a way to fold the surroundings into a property of the system alone, a new state function now called the Gibbs energy $G$ (गिब्ज़ ऊर्जा):

G = H - T S

(5.20)

$G$ is an extensive state function. For a change at constant temperature,

Δ G = Δ H - T Δ S

(5.21)

This is the Gibbs equation. Each term belongs to the system alone, and each is an energy: $T$ in K times $Δ S$ in J K⁻¹ gives joules.

Definition

Gibbs energy

$G = H - T S$ ; at constant temperature and pressure a process is spontaneous when $Δ G < 0.$

DerivationWhy

Δ G < 0

means spontaneous

Let the system be at the same temperature $T$ as its surroundings, at constant pressure. Whatever heat the system gives out, the surroundings absorb, so $Δ H_{surr} = - Δ H_{sys}$ and

\begin{aligned} Δ S_{surr} & = \frac{Δ H_{surr}}{T} \\ = - \frac{Δ H_{sys}}{T} \end{aligned}

Substitute this into Eq. 5.19 and multiply through by $T$ :

\begin{aligned} T Δ S_{total} \\ = T Δ S_{sys} - Δ H_{sys} \\ = - Δ G_{sys} \end{aligned}

A spontaneous process has $Δ S_{total} > 0,$ so for it

\begin{aligned} Δ G & = Δ H - T Δ S \\ < 0 \end{aligned}

(5.22)

At constant temperature and pressure, then: if $Δ G < 0$ the process is spontaneous; if $Δ G > 0$ it is non-spontaneous, and the reverse process is the spontaneous one. There is an energy reading of the equation too. Of the enthalpy a reaction releases, the share $T Δ S$ is tied up in the change of the system's entropy and cannot be harnessed. What is left, $- Δ G$ , is the most useful work other than expansion work, such as electrical work in a cell, that the reaction could ever deliver. Hence the older name, free energy.

The two terms can pull together or against each other (Table 5.6, Fig. 5.15). When $Δ H$ and $Δ S$ have the same sign, temperature decides, and the changeover comes where $Δ G = 0,$ at $T = Δ H / Δ S$ (taking both as roughly constant). Take a reaction that absorbs heat but creates disorder, with $Δ H > 0$ and $Δ S > 0.$ A big entropy gain brings its changeover temperature down, perhaps below room temperature; a small one pushes it up to furnace temperatures, which is why kilns and blast furnaces run so hot.

Table 5.6Is the reaction spontaneous? How temperature decidesNCERT Table 5.4 · p. 162

$Δ H$	$Δ S$	At low $T$	At high $T$
−	+	yes	yes
−	−	yes	no
+	+	no	yes
+	−	no	no

5/5All four cases together.

Fig. 5.15

Δ G = Δ H - T Δ S

is a straight line in

T

with intercept

Δ H

and slope

- Δ S

Delta G against temperature; below the axis is shaded and labelled spontaneous. Each line starts at Delta H and has slope minus Delta S. Step one: line 1 (Delta H negative, Delta S positive) starts below zero and falls: always spontaneous. Step two: line 2 (both negative) starts below zero and rises, crossing at T = Delta H over Delta S. Step three: line 3 (both positive) starts above zero and falls through the axis. Step four: line 4 (Delta H positive, Delta S negative) starts above zero and rises: never spontaneous. Step five shows all four, numbered.

Fig. 5.16Pick a reaction or set your own

Δ H

and

Δ S

, then drag the temperature to see

Δ G

, the changeover temperature and

K

Entropy and second law of thermodynamics

Draw one boundary round the whole universe, system and surroundings together, and you have the ultimate isolated system, its energy fixed by the first law. What changes as time runs forward is its entropy, which only ever grows. That is the second law of thermodynamics. Clausius's version, that heat does not pass by itself from a colder body to a hotter one, and the engineer's version, that no engine running in a cycle can turn heat completely into work, are the same law seen from other angles.

It also explains a pattern you may have noticed: most reactions that 'just go' give out heat. The heat they dump into the surroundings is a large entropy gift, $Δ S_{surr} = - Δ H / T,$ which usually outweighs any order created in the system, as it did for burning magnesium in Example 5.11.

Absolute entropy and third law of thermodynamics

Enthalpy has no absolute zero, which is why we only ever list its changes. Entropy is different. Cool a perfect crystal of a pure substance towards 0 K and its molecules lose, one after another, their freedom to fly about (translation), spin (rotation) and stretch their bonds (vibration), until only a single, perfectly ordered arrangement is left. The third law of thermodynamics says that its entropy then falls to zero. Solutions and glassy, supercooled liquids miss out, because some randomness is frozen into them and survives even at 0 K.

With a zero to start from, an absolute entropy can be built up from heat measurements alone. Warm the crystal from near 0 K in small reversible steps, add $q_{rev} / T$ for every step and for each melting or boiling on the way, and stop at 298 K. The total is the standard molar entropy $S^{\circ}$ of the substance. Reaction entropies then come from the same products-minus-reactants sum used for enthalpies.

Table 5.7Some standard molar entropies at 298 K

Substance	$S^{\circ}$ / J K⁻¹ mol⁻¹
$C$ (diamond)	2.4
$H_{2} O (l)$	69.9
$H_{2} (g)$	130.7
$H_{2} O (g)$	188.8
$N_{2} (g)$	191.6
$N H_{3} (g)$	192.8
$O_{2} (g)$	205.1

Gases have far larger entropies than solids and liquids, and liquid water lies far below steam, just as the simulation suggested.

Example

When does ammonia synthesis stop being spontaneous?

For $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \overset{}{\to} N H_{3} (g),$ $Δ H^{\circ} = - 45.9 k J m o l^{- 1} .$ Using Table 5.7, find $Δ S^{\circ},$ then $Δ G^{\circ}$ at 298 K, and the temperature above which the reaction is no longer spontaneous under standard conditions.

Try it yourself first. Then reveal the solution one step at a time.

5.7 Gibbs Energy Change and Equilibrium

Mix nitrogen and hydrogen over a catalyst and ammonia forms; start from pure ammonia and some of it falls apart. Either way the mixture ends at the same composition and stays there. Gibbs energy explains where that resting point lies.

Picture the Gibbs energy of the whole reacting mixture as the reaction proceeds (Fig. 5.17). Pure reactants have some value and pure products another, but a mixture of the two lies lower than either, because mixing adds entropy. So $G$ passes through a minimum in between. From either side, moving towards the minimum lowers $G$ and is spontaneous; moving away would raise it. The mixture settles at the minimum, a dynamic equilibrium in which the forward and backward reactions carry on at equal rates, and

At equilibrium

Δ_{r} G = 0

3/3At the minimum $Δ_{r} G = 0$ and the composition stops changing.

Fig. 5.17From either side, a reaction mixture slides downhill in

G

to the equilibrium minimum.

The vertical axis is G and the horizontal axis runs from all reactants on the left to all products on the right. Step one: two dots mark G of pure reactants, higher, and of pure products, lower; a double arrow on the right shows the gap, the standard Gibbs energy change. Step two: a curve joins them, dipping below both because mixing lowers G. Step three: the lowest point is marked equilibrium, with a flat dashed tangent there, and arrows on the curve show the mixture sliding downhill to the minimum from both sides.

The standard Gibbs energy change, $Δ_{r} G^{\circ},$ with every reactant and product in its standard state, is linked to the equilibrium constant $K$ of the reaction:

\begin{array}{c} Δ_{r} G^{\circ} = - R T \ln K \\ = - 2.303 R T \log K \end{array}

(5.23)

Combined with the Gibbs equation under standard conditions, this bridges heat measurements and equilibrium:

\begin{aligned} Δ_{r} G^{\circ} & = Δ_{r} H^{\circ} \\ - T Δ_{r} S^{\circ} \\ = - R T \ln K \end{aligned}

(5.24)

A negative $Δ_{r} G^{\circ}$ means $K > 1$ , so products are favoured; a positive one means $K < 1$ . Because $K$ depends exponentially on $Δ_{r} G^{\circ},$ a few tens of kilojoules separate 'hardly any product' from 'essentially complete' (Table 5.8). Enthalpy usually dominates, so strongly exothermic reactions tend to run almost to completion and strongly endothermic ones barely start. When $Δ_{r} H^{\circ}$ is modest, the $T Δ_{r} S^{\circ}$ term can tip the balance either way, as it does for $N_{2} O_{4}$ in Example 5.14.

Table 5.8Standard Gibbs energy change and K at 298 K

$Δ_{r} G^{\circ}$ / kJ mol⁻¹	$K$
−50	$5.8 \times 10^{8}$
−10	57
0	1
+10	0.018
+50	$1.7 \times 10^{- 9}$

In practice Eq. 5.24 is a two-way bridge. A chemical engineer can take $Δ H^{\circ}$ and $Δ S^{\circ}$ from tables, compute $K$ at the temperature of a planned plant and see whether the yield would pay. A laboratory chemist can measure $K$ and turn it into $Δ G^{\circ} .$

Example 5.12NCERT Problem 5.12 · p. 163

Why air does not turn into nitric oxide

For $N_{2} (g) + O_{2} (g) \overset{}{⇌} 2 N O (g)$ at 298 K, $K = 4.5 \times 10^{- 31} .$ Find $Δ_{r} G^{\circ} .$

Try it yourself first. Then reveal the solution one step at a time.

Example 5.13NCERT Problem 5.13 · p. 163

From $Δ_{r} G^{\circ}$ to $K$

For $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \overset{}{⇌} N H_{3} (g)$ at 298 K, $Δ_{r} G^{\circ} = - 16.4 k J m o l^{- 1},$ as found in the ammonia example. Find $K$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 5.14NCERT Problem 5.14 · p. 163

Partly dissociated dinitrogen tetroxide

At 25 °C and a total pressure of 1.00 bar, $N_{2} O_{4} (g) \overset{}{⇌} 2 N O_{2} (g)$ reaches equilibrium with 19 % of the $N_{2} O_{4}$ dissociated. Find $K$ and $Δ_{r} G^{\circ}$ at this temperature.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Δ_{r} G^{\circ} = 0

for a reaction at some temperature, its equilibrium constant at that temperature is

01−1infinitely large

Show answer

Answer (B) 1

\ln K = - Δ_{r} G^{\circ} / R T = 0,

K = 1

K

can never be negative.

Recap

Thermodynamics: key points and summary

Thermodynamics treats matter in bulk and compares initial and final states; it says nothing about rates (Section 5.1).
A system is open, closed or isolated depending on whether matter and energy can cross its boundary (Section 5.1.2).
State functions such as $p$ , $V$ , $T$ , $U$ , $H$ , $S$ and $G$ depend only on the state; $q$ and $w$ depend on the path (Section 5.1.3).
First law: $Δ U = q + w,$ with $q$ and $w$ positive when energy enters the system ((5.1)).
Pressure–volume work is $w = - p_{ex} Δ V;$ reversible isothermal work on an ideal gas is $- n R T \ln (V_{f} / V_{i}),$ the most an expansion can deliver (Section 5.2.1).
Heat at constant volume measures $Δ U$ ; heat at constant pressure measures $Δ H$ , where $H = U + p V$ and $Δ H = Δ U + Δ n_{g} R T$ (Section 5.2.2, Section 5.3).
$q = c m Δ T = C Δ T,$ and for an ideal gas $C_{p} - C_{V} = R$ (heat capacity).
Reaction enthalpies follow from formation enthalpies, products minus reactants weighted by coefficients, or from any route of steps by Hess's law (formation, Hess's law).
Phase changes, combustion, atomization, bond breaking, lattice break-up, solution and dilution each have a named enthalpy; bond enthalpies estimate gas-phase reaction enthalpies and Born–Haber cycles give lattice enthalpies (Section 5.5).
A spontaneous process can run by itself, however slowly; the total entropy rises in every spontaneous change, and $Δ S = q_{rev} / T$ (entropy).
At constant $T$ and $p$ , a change is spontaneous when $Δ G = Δ H - T Δ S < 0;$ temperature decides when $Δ H$ and $Δ S$ share a sign (Gibbs energy).
Third law: a perfect crystal has zero entropy at 0 K, which makes absolute entropies possible (third law).
At equilibrium $Δ_{r} G = 0,$ and $Δ_{r} G^{\circ} = - R T \ln K$ links heat data to equilibrium constants (Section 5.7).

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Quick check: MCQs on Thermodynamics

The general case4 questions
5.2.1 Work3 questions
The relation between Cp and CV for an ideal gas4 questions
ΔH measurements3 questions
Hess's law of constant heat summation4 questions
Enthalpy of dilution4 questions
Absolute entropy and third law of thermodynamics5 questions
5.7 Gibbs Energy Change and Equilibrium3 questions

Glossary · 15

Key terms (English–Hindi)

adiabaticरुद्धोष्मruddhoshm: Describes a wall or process through which no heat passes between system and surroundings.First seen in the chapter →

bond enthalpyआबंध एन्थैल्पीaabandh enthalpy: The enthalpy needed to break one mole of a given bond in gaseous molecules; usually a mean over similar bonds.First seen in the chapter →

enthalpyएन्थैल्पीenthalpy: $H = U + p V$ ; its change equals the heat absorbed at constant pressure.First seen in the chapter →
entropyएन्ट्रॉपीentropy: A state function, $S$ , measuring how widely matter and energy are spread out; it rises in every spontaneous change.First seen in the chapter →
extensive propertyविस्तीर्ण गुणधर्मvistirn gundharm: A property proportional to the amount of matter, such as mass, volume or enthalpy.First seen in the chapter →

Gibbs energyगिब्ज़ ऊर्जाGibbs oorja: $G = H - T S$ ; a process at constant $T$ and $p$ is spontaneous when $Δ G < 0$ .First seen in the chapter →

Hess's lawहेस का नियमHess ka niyam: A reaction enthalpy is the same whether the reaction happens in one step or several, so step enthalpies add.First seen in the chapter →

intensive propertyगहन गुणधर्मgahan gundharm: A property independent of the amount of matter, such as temperature, pressure or density.First seen in the chapter →
internal energyआंतरिक ऊर्जाaantarik oorja: The total energy stored in a system; only its change, $Δ U$ , can be measured.First seen in the chapter →

lattice enthalpyजालक एन्थैल्पीjaalak enthalpy: The enthalpy change when 1 mol of an ionic solid separates into its gaseous ions.First seen in the chapter →

reversible processउत्क्रमणीय प्रक्रमutkramaneeya prakram: A process run through near-equilibrium states, so that an infinitesimal change could reverse it at any stage.First seen in the chapter →

spontaneous processस्वतः प्रवर्तित प्रक्रमsvatah pravartit prakram: A process that can proceed by itself without continuing outside help, however slowly.First seen in the chapter →
standard enthalpy of formationमानक संभवन एन्थैल्पीmaanak sambhavan enthalpy: The enthalpy change when 1 mol of a compound forms from its elements, all in their standard or reference states.First seen in the chapter →
state functionअवस्था फलनavastha phalan: A property that depends only on the present state, so its change depends only on the initial and final states.First seen in the chapter →
systemनिकायnikaay: The part of the universe chosen for study; everything else that can interact with it is the surroundings.First seen in the chapter →

5.1 Thermodynamic Terms

5.1.1 The System and the Surroundings

5.1.2 Types of the System

5.1.3 The State of the System

5.1.4 The Internal Energy as a State Function

Work

Heat

The general case

5.2 Applications

5.2.1 Work

5.2.2 Enthalpy, H

A useful new state function

Extensive and intensive properties

Heat capacity

The relation between Cp and CV for an ideal gas

5.3 Measurement of ΔU and ΔH: Calorimetry

ΔU measurements

ΔH measurements

5.4 Enthalpy Change, ΔrH of a Reaction – Reaction Enthalpy

Standard enthalpy of reactions

Enthalpy changes during phase transformations

Standard enthalpy of formation

Thermochemical equations

Hess's law of constant heat summation

5.5 Enthalpies for Different Types of Reactions

Standard enthalpy of combustion

Enthalpy of atomization

Bond enthalpy

Lattice enthalpy

Enthalpy of solution

Enthalpy of dilution

5.6 Spontaneity

Is decrease in enthalpy a criterion for spontaneity?

Entropy and spontaneity

Gibbs energy and spontaneity

Entropy and second law of thermodynamics

Absolute entropy and third law of thermodynamics

5.7 Gibbs Energy Change and Equilibrium

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