Can a spontaneous process have negative Delta S of system?

Yes. Freezing of water below zero has negative Delta S of system but the heat released increases entropy of surroundings by more, so Delta S universe is positive - and the process is spontaneous.

Why do we use reversible heat to calculate Delta S even for irreversible processes?

Because Delta S is a state function - it depends only on initial and final states. We use q_rev as a mathematical device to compute the state-function change; the actual irreversible heat is different.

Spontaneity

Q: Why is the first law not enough to predict spontaneity?

The first law only says energy is conserved; it doesn't distinguish forward from reverse processes. Heat flowing from cold to hot conserves energy just as heat flowing from hot to cold. Predicting direction requires the second law and entropy.

Q: What is entropy in simple physical terms?

Entropy measures disorder or randomness - the number of microscopic arrangements consistent with the observed macrostate. Gases have higher entropy than liquids, which have higher entropy than solids.

Q: State the second law of thermodynamics in your own words.

In any spontaneous process the entropy of the universe (system plus surroundings) increases. In a reversible process it stays constant. It can never decrease. Equivalently, heat cannot spontaneously flow from cold to hot, and no cyclic engine converts all heat into work.

Q: What is the entropy change for a reversible adiabatic process?

Zero. In reversible adiabatic each dq_rev is zero so Delta S of system is zero, and since it is adiabatic Delta S of surroundings is also zero. Such processes are called isoentropic.

Q: State the third law of thermodynamics. Why is it useful?

The entropy of a perfect crystal of a pure substance at 0 K is zero. This gives an absolute reference: by integrating heat capacity from 0 K upward, absolute entropy values can be tabulated - unlike U or H where only differences are known.

Q: Why does entropy increase when a gas expands into vacuum?

Expansion into a larger volume increases the number of possible molecular positions, so the number of microstates W rises. By Boltzmann's S = k ln W, entropy rises. For an ideal gas at constant T, Delta S = nR ln(V2/V1) which is positive for expansion.

ChemistryChemical ThermodynamicsFor NEET aspirants

Some processes happen on their own - ice melts above $0^{\circ}$ C, iron rusts, gases mix - while others never spontaneously reverse. The first law of thermodynamics says nothing about direction. The second law and the concept of entropy ( $S$ ) provide the criterion. A process is spontaneous if the entropy of the universe increases: $Δ S_{u ni v er se} = Δ S_{sy s t e m} + Δ S_{s u r r o u n d in g s} > 0$ . This concept covers why the first law is insufficient, the physical meaning of entropy (disorder), the second and third laws, entropy calculations for various processes, and how spontaneity is judged.

Key Formulas - Quick Reference

Entropy change (reversible process): $d S = \frac{d q _{r e v}}{T}$ , or $Δ S = \int \frac{d q _{r e v}}{T}$
Second law: $Δ S_{u ni v er se} = Δ S_{sy s t e m} + Δ S_{s u r r o u n d in g s} \geq 0$ (equality for reversible, inequality for irreversible)
Entropy change of surroundings: $Δ S_{s u r r} = - \frac{q _{sy s t e m}}{T} = - \frac{Δ H _{sy s t e m}}{T}$ (constant $T$ , $P$ )
Entropy of phase transition: $Δ S = \frac{Δ H _{t r an s i t i o n}}{T _{t r an s i t i o n}}$ (fusion, vaporization, sublimation)
Ideal gas, general: $Δ S = n C_{v} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}}$ = $n C_{p} ln \frac{T _{2}}{T _{1}} - n R ln \frac{P _{2}}{P _{1}}$
Isothermal ideal gas: $Δ S = n R ln \frac{V _{2}}{V _{1}} = n R ln \frac{P _{1}}{P _{2}}$
Third law: entropy of a perfect crystal of a pure substance at $0 K$ is zero

1. Why we need a second law

The first law says energy is conserved but tells us nothing about direction. It cannot explain why:

Heat flows spontaneously from a hot body to a cold one, but never the reverse.
Ice melts above $0^{\circ}$ C but water freezes below $0^{\circ}$ C.
Gases mix spontaneously but never spontaneously separate.
Ink diffuses through water but never gathers back into a drop.
An iron nail rusts but rust doesn't spontaneously become shiny iron.

The second law of thermodynamics provides this missing direction.

2. Spontaneous and non-spontaneous processes

A spontaneous process is one that occurs on its own once initiated - the system moves in a definite direction without external help.

Examples: (i) mixing of gases, (ii) melting of ice above

0^{\circ}

C, (iii) rusting of iron, (iv) burning of fuel, (v) spreading of a drop of ink in water.

Non-spontaneous processes are those that do not occur naturally and require continuous external energy input to proceed - e.g. electrolysis of water, refrigeration.

Is energy the only criterion?

All exothermic processes ( $Δ H < 0$ ) are spontaneous, right? Wrong. Consider these exceptions:

Melting of ice above $0^{\circ}$ C: $Δ H > 0$ but spontaneous.
Vaporization of water above $10 0^{\circ}$ C: $Δ H > 0$ but spontaneous.
Spreading of ink in water: $Δ H \approx 0$ , still spontaneous.
Endothermic dissolution of salts (like NH $_{4}$ NO $_{3}$ ): $Δ H > 0$ but spontaneous.

These show energy alone doesn't decide spontaneity. Nature also drives towards disorder (randomness).

3. Entropy - a measure of disorder

Entropy ( $S$ ) is a state function that measures the disorder or randomness of a system. Higher entropy $=$ more disorder.

Mathematical definition

The change in entropy of a system when it exchanges heat $d q_{r e v}$ reversibly with the surroundings at temperature $T$ is:

d S = \frac{d q _{r e v}}{T}, Δ S = \int_{A}^{B} \frac{d q _{r e v}}{T}

Units: J K $^{- 1}$ (SI) or cal K $^{- 1}$ . Entropy is a state function and extensive property.

Why divide by $T$ ?

At high $T$ the system is already highly disordered, so adding the same heat causes less change in randomness. Dividing by $T$ captures this: entropy change depends more on heat added at low $T$ than at high $T$ .

Entropy trends: predicting the sign of $Δ S$

Increase in $S$ : solid $\to$ liquid $\to$ gas; expansion of a gas; heating; more moles of gas produced in a reaction; dissolving crystalline solid; mixing.
Decrease in $S$ : gas $\to$ liquid $\to$ solid; compression; cooling; fewer moles of gas after reaction; precipitation; freezing.

4. Second law of thermodynamics

Several equivalent statements exist:

Kelvin-Planck statement

It is impossible to construct a cyclic engine that converts heat from a single reservoir completely into work. Some heat must always be rejected to a colder reservoir.

Clausius statement

Heat cannot flow spontaneously from a colder body to a hotter body without some external work being done on the system.

Entropy statement (most useful in chemistry)

The entropy of an isolated system (or the entropy of the universe) always increases in any spontaneous process.

Δ S_{u ni v er se} = Δ S_{sy s t e m} + Δ S_{s u r r o u n d in g s} \geq 0

Equality holds for reversible processes; strict inequality for irreversible (spontaneous) processes.

Entropy change of the surroundings

Since the surroundings are effectively an infinite reservoir, any heat exchange occurs reversibly for them. If the system releases heat $q_{sy s}$ at temperature $T$ :

Δ S_{s u r r} = - \frac{q _{sy s t e m}}{T} = - \frac{Δ H _{sy s t e m}}{T} (constant T, P)

5. Entropy of phase transitions

Phase transitions (melting, vaporization, sublimation) at constant $T$ and $P$ are isothermal reversible processes. Entropy change is simply:

Δ S_{f u s} = \frac{Δ H _{f u s}}{T _{m}}, Δ S_{v a p} = \frac{Δ H _{v a p}}{T _{b}}, Δ S_{s u b} = \frac{Δ H _{s u b}}{T _{s u b}}

Solved Example 1

Calculate the entropy change for vaporization of

1 mol

of liquid water to steam at

10 0^{\circ}

C, given

Δ H_{v a p} = 40.8 kJ mol^{- 1}

Solution:

Boiling point: $T_{b} = 373 K$ ; $Δ H_{v a p} = 40800 J mol^{- 1}$ .

$Δ S_{v a p} = \frac{Δ H _{v a p}}{T _{b}} = \frac{40800}{373} \approx 109.4 J K^{- 1} mol^{- 1}$ .

6. Entropy calculations for an ideal gas

For an ideal gas going from state $(T_{1}, V_{1})$ to $(T_{2}, V_{2})$ , using $d U = n C_{v} d T$ and $d q_{r e v} = d U + P d V$ :

Δ S_{sy s t e m} = n C_{v} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}}

Equivalently, using $P V = n R T$ :

Δ S_{sy s t e m} = n C_{p} ln \frac{T _{2}}{T _{1}} - n R ln \frac{P _{2}}{P _{1}}

Special cases

Process	$Δ S_{sy s t e m}$
Isothermal reversible or irreversible	$n R ln (V_{2} / V_{1}) = n R ln (P_{1} / P_{2})$
Isochoric ( $V$ constant)	$n C_{v} ln (T_{2} / T_{1})$
Isobaric ( $P$ constant)	$n C_{p} ln (T_{2} / T_{1})$
Adiabatic reversible	$0$ (isoentropic)
Adiabatic irreversible	$> 0$

Key insight:

Δ S_{sy s t e m}

is a state function - the value is the same for reversible and irreversible paths between the same two states. But

Δ S_{s u r r o u n d in g s}

differs, so

Δ S_{u ni v er se}

differs. This is what distinguishes reversible from irreversible.

7. Isothermal expansion: reversible vs irreversible

Reversible isothermal expansion of ideal gas

Since $Δ U = 0$ (isothermal, ideal), $q_{r e v} = - w_{r e v} = n R T ln (V_{2} / V_{1})$ .

Δ S_{sy s} = n R ln (V_{2} / V_{1}) > 0

Δ S_{s u r r} = - q_{sy s} / T = - n R ln (V_{2} / V_{1}) < 0

Δ S_{u ni v er se} = Δ S_{sy s} + Δ S_{s u r r} = 0

Reversible processes are always at the boundary of spontaneity.

Irreversible isothermal expansion against constant $P_{e x t}$

Now $q_{sy s} = - w_{sy s} = P_{e x t} (V_{2} - V_{1})$ .

Δ S_{sy s} = n R ln (V_{2} / V_{1})

(same as reversible - state function)

Δ S_{s u r r} = - P_{e x t} (V_{2} - V_{1}) / T

Δ S_{u ni v er se} = n R ln (V_{2} / V_{1}) - P_{e x t} (V_{2} - V_{1}) / T > 0

Positive because $∣ w_{r e v} ∣ > ∣ w_{i r r e v} ∣$ - irreversible expansion generates less heat, so the drop in $Δ S_{s u r r}$ is less than the gain in $Δ S_{sy s}$ , giving positive net entropy change.

Solved Example 2

One mole of an ideal gas expands isothermally and reversibly at

2 5^{\circ}

C from

10 L

20 L

. Find (i)

Δ S_{sy s t e m}

, (ii)

Δ S_{s u r r o u n d in g s}

, (iii)

Δ S_{u ni v er se}

. Also solve for irreversible expansion into vacuum.

Solution:

(a) Reversible isothermal expansion:

$Δ S_{sy s} = n R ln (V_{2} / V_{1}) = 1 \times 8.314 \times ln 2 = 5.76 J K^{- 1}$ .

$q_{r e v} = - w_{r e v} = n R T ln (V_{2} / V_{1}) = 298 \times 5.76 = 1718 J$ .

$Δ S_{s u r r} = - q / T = - 1718/298 = - 5.76 J K^{- 1}$ .

$Δ S_{u ni v er se} = 5.76 + (- 5.76) = 0$ (as expected for a reversible process).

(b) Irreversible expansion into vacuum (free expansion):

$Δ S_{sy s} = 5.76 J K^{- 1}$ (same - state function).

Free expansion: $P_{e x t} = 0 \Rightarrow w = 0 \Rightarrow q = 0 \Rightarrow Δ S_{s u r r} = 0$ .

$Δ S_{u ni v er se} = 5.76 J K^{- 1} > 0$ (irreversible - entropy of universe increases).

8. Entropy change for chemical reactions

For a chemical reaction at constant $T$ and $P$ :

Δ S_{r x n}^{\circ} = \sum ν_{i} S_{m}^{\circ} (products) - \sum ν_{i} S_{m}^{\circ} (reactants)

where $S_{m}^{\circ}$ is the standard molar entropy of each species, tabulated from third-law measurements.

Predicting the sign

If moles of gas increase, $Δ S_{r x n} > 0$ (more disorder).
If moles of gas decrease, $Δ S_{r x n} < 0$ .
Reactions producing gases from solids/liquids: $Δ S_{r x n} ≫ 0$ .
Reactions of liquids/solids only: $∣Δ S_{r x n} ∣$ is small.

9. Third law of thermodynamics

Statement. The entropy of a perfect crystal of a pure substance at absolute zero ( $T = 0 K$ ) is zero.

lim_{T \to 0} S = 0

(for a perfect crystalline solid)

Consequence. Unlike $U$ and $H$ (whose absolute values cannot be determined), the absolute entropy of any substance can be calculated by integrating heat capacity from $0 K$ to the desired temperature:

S^{\circ} (T) = \int_{0}^{T} \frac{n C d T}{T}

This gives the standard molar entropy $S_{m}^{\circ}$ used in reaction-entropy calculations.

10. Physical significance of entropy

Entropy measures molecular disorder. Higher disorder $\to$ higher entropy. Gases have higher entropy than liquids, which have higher entropy than solids.
Entropy is related to the number of microstates: Boltzmann's formula $S = k ln W$ , where $W$ is the number of microscopic arrangements consistent with the macrostate.
Spontaneous processes tend to increase disorder. Gas mixing, dissolution, diffusion all increase entropy without doing work.
Entropy is a state function. $Δ S$ depends only on initial and final states.
Entropy is extensive. Doubling the sample doubles $S$ .

Solved Example 3

A system absorbs

300 cal

of heat at

300 K

in an irreversible manner. When the same change of state is carried out reversibly,

900 cal

of heat is absorbed. Calculate the change in entropy of the system.

Solution:

$Δ S$ is a state function, so use the reversible heat:

$Δ S_{sy s} = \frac{q _{r e v}}{T} = \frac{900}{300} = 3.0 cal K^{- 1}$ .

The irreversible heat value ( $300 cal$ ) is not used for computing $Δ S_{sy s}$ ; that would give the wrong (path-dependent) answer.

Solved Example 4

Predict the sign of

Δ S

for each: (i)

H_{2} O (ℓ) \to H_{2} O (g)

, (ii)

CaCO_{3} (s) \to CaO (s) + CO_{2} (g)

, (iii)

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (ℓ)

Solution:

(i) Liquid $\to$ gas: large increase in disorder, $Δ S > 0$ .

(ii) Solid decomposes to solid + gas: gas produced, $Δ S > 0$ .

(iii) 3 mol of gas $\to$ 0 mol of gas (product is liquid): large decrease, $Δ S < 0$ .

11. Common Mistakes to Avoid

Watch out

Using $q_{i r r e v}$ to calculate $Δ S_{sy s t e m}$ . $Δ S = q_{r e v} / T$ - always use the reversible heat, even if the actual process is irreversible. This works because $Δ S$ is a state function.
Confusing $Δ S_{sy s t e m}$ with $Δ S_{u ni v er se}$ . The second law's criterion is on $Δ S_{u ni v er se}$ , not $Δ S_{sy s t e m}$ . A process can be spontaneous even if $Δ S_{sy s t e m} < 0$ (e.g. freezing water below $0^{\circ}$ C), provided $Δ S_{s u r r o u n d in g s}$ increases more.
Assuming exothermic $\Rightarrow$ spontaneous. Not always true. Endothermic processes with sufficient entropy increase can also be spontaneous.
Using $Δ H$ instead of $q$ in $Δ S_{s u r r}$ . These are equal only at constant $T$ and $P$ . In general, $Δ S_{s u r r} = - q_{sy s} / T$ .
Forgetting that entropy is defined only for equilibrium states. An irreversible process passes through non-equilibrium intermediate states where $S$ isn't defined for the system - but $Δ S$ between well-defined initial and final states still makes sense.
Applying $Δ S = q / T$ with the wrong sign of $q$ . Follow the IUPAC convention: $q_{sy s}$ is positive when the system absorbs heat, negative when it releases heat. $Δ S_{s u r r}$ has the opposite sign.
Confusing "entropy is disorder" with disorder in a colloquial sense. Entropy is precisely $k ln W$ (number of microstates), not just "how messy something looks."

Frequently Asked Questions

Q1. Why is the first law not enough to predict spontaneity?

The first law only says energy is conserved - it doesn't distinguish forward from reverse processes. Heat flowing from cold to hot conserves energy just as easily as hot to cold. To predict direction, we need the second law and entropy.

Q2. What is entropy in simple physical terms?

Entropy measures the disorder or randomness of a system - equivalently, the number of microscopic arrangements consistent with the observed macroscopic state. More disorder means more possible arrangements and higher entropy. Gases have higher entropy than liquids, which have higher entropy than solids.

Q3. State the second law of thermodynamics in your own words.

In any spontaneous process, the entropy of the universe (system plus surroundings) increases. In a reversible process it stays constant. It can never decrease. Equivalently, heat cannot flow from cold to hot spontaneously, and no cyclic engine can convert all heat into work.

Q4. Can a spontaneous process have $Δ S_{sy s t e m} < 0$ ?

Yes. Freezing of water below $0^{\circ}$ C has $Δ S_{sy s} < 0$ (liquid becomes more ordered solid), but it is spontaneous because the heat released increases $Δ S_{s u r r o u n d in g s}$ by more than the system's entropy decreases. The net $Δ S_{u ni v er se} > 0$ .

Q5. Why do we use reversible heat to calculate $Δ S$ even for irreversible processes?

Because $Δ S$ is a state function - it depends only on the initial and final states, not on the path. We use $q_{r e v}$ (the heat that would flow along a reversible path) as a mathematical device to compute the state-function change. The actual irreversible heat is different and not equal to $T Δ S$ .

Q6. What is the entropy change for a reversible adiabatic process?

Zero. In a reversible adiabatic process, $d q_{r e v} = 0$ at every step, so $Δ S_{sy s} = \int d q_{r e v} / T = 0$ . Since it is also adiabatic, $Δ S_{s u r r} = 0$ . Reversible adiabatic processes are also called isoentropic.

Q7. State the third law of thermodynamics. Why is it useful?

The entropy of a perfect crystalline pure substance is zero at absolute zero (0 K). It is useful because it provides an absolute reference for entropy: by integrating heat capacity from $0 K$ upward, we get absolute entropy values $S^{\circ}$ that can be tabulated and used in reaction entropy calculations - unlike $U$ or $H$ where only differences are known.

Q8. Why does entropy increase when a gas expands into vacuum?

When a gas expands into a larger volume, the number of possible positions for each molecule increases - so the number of microstates ( $W$ ) increases sharply. By Boltzmann's formula $S = k ln W$ , entropy rises. For an ideal gas at constant $T$ , $Δ S = n R ln (V_{2} / V_{1}) > 0$ for expansion.

Previous year questions on Spontaneity

1 question from past papers, each with a step-by-step solution.

JEE Main 2025 Jan 23 Shift 2, Chemistry Q1

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Practice questions →Take a mock test

1. Why we need a second law

2. Spontaneous and non-spontaneous processes

Is energy the only criterion?

3. Entropy - a measure of disorder

Mathematical definition

Why divide by T?

Entropy trends: predicting the sign of ΔS

4. Second law of thermodynamics

Kelvin-Planck statement

Clausius statement

Entropy statement (most useful in chemistry)

Entropy change of the surroundings

5. Entropy of phase transitions

6. Entropy calculations for an ideal gas

Special cases

7. Isothermal expansion: reversible vs irreversible

Reversible isothermal expansion of ideal gas

Irreversible isothermal expansion against constant Pext​

8. Entropy change for chemical reactions

Predicting the sign

9. Third law of thermodynamics

10. Physical significance of entropy

11. Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why is the first law not enough to predict spontaneity?

Q2. What is entropy in simple physical terms?

Q3. State the second law of thermodynamics in your own words.

Q4. Can a spontaneous process have ΔSsystem​<0?

Q5. Why do we use reversible heat to calculate ΔS even for irreversible processes?

Q6. What is the entropy change for a reversible adiabatic process?

Q7. State the third law of thermodynamics. Why is it useful?

Q8. Why does entropy increase when a gas expands into vacuum?

Previous year questions on Spontaneity

Ready to master Chemical Thermodynamics?

Why divide by $T$ ?

Entropy trends: predicting the sign of $Δ S$

Irreversible isothermal expansion against constant $P_{e x t}$

Q4. Can a spontaneous process have $Δ S_{sy s t e m} < 0$ ?

Q5. Why do we use reversible heat to calculate $Δ S$ even for irreversible processes?