Use total entropy for direction, and Gibbs energy when temperature and pressure are fixed.
JEE Main · JEE Advanced · NEET UG38 min study guide9 indexed sections
BY THE END, YOU CAN
Explain why spontaneity is a thermodynamic direction rather than a reaction speed.
Calculate entropy changes using a reversible reference path.
Apply the second law to the system plus surroundings.
Use ΔG=ΔH−TΔS with correct conditions and consistent units.
Distinguish reaction ΔG from standard ΔG° and connect both to Q and K.
Predict temperature effects while recognizing heat-capacity and phase limits.
BEFORE YOU START
First-law energy balance and heat conventions.
Ideal gases, logarithms and equilibrium quotients.
Moles and standard reaction quantities.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Why energy conservation does not choose a direction
Energy conservation permits many rearrangements that are not observed to happen spontaneously. An isolated ideal gas can expand into an empty chamber without changing its internal energy, yet it does not spontaneously collect back into the original half of the vessel. The first law accounts for energy; the second law supplies a direction criterion.
A spontaneous process is thermodynamically favored under specified constraints and at the current state. It may still be extremely slow because a high activation barrier prevents rapid rearrangement. A catalyst changes the accessible reaction pathway and rate but does not change the reaction’s thermodynamic equilibrium. “Spontaneous” therefore does not mean immediate, violent or complete.
The conditions matter. A phase change can be favored at one temperature and opposed at another; a reaction can change direction as the mixture composition changes. A standard-state free-energy value is not a universal verdict that applies to every mixture. Always distinguish the chosen reference state from the actual state being discussed.
We first examine entropy of the system plus surroundings. We then derive a more convenient system-only criterion, Gibbs energy, for the important case of fixed temperature and pressure. The shortcut is powerful precisely because its constraints are stated; it should not be used as an unrestricted rule for every isolated or fixed-volume process.
02 / UNDERSTAND & APPLY
2. Entropy counts accessible distributions, not visual messiness
Entropy is a thermodynamic state function connected to the number of microscopic arrangements compatible with a macroscopic state. In a simple statistical description, S=k_B lnΩ, where Ω counts accessible microstates under the specified constraints. More possible distributions of energy and particles generally correspond to greater entropy.
The word “disorder” can be a rough mnemonic but is unreliable by itself. A visually neat gas in a larger container has more positional possibilities even if nothing looks messy. Warming a sample usually increases the range of accessible energy distributions. A complex molecular solid can have greater molar entropy than a simpler solid even though both are crystalline.
For the same substance under comparable conditions, the gas phase commonly has much greater molar entropy than the liquid, and the liquid usually exceeds the ordered solid. Increasing the number of gas particles in a reaction often suggests positive reaction entropy. These are qualitative clues, not substitutes for tabulated calculations when phase, molecular complexity or competing effects matter.
Mixing distinguishable ideal gases increases entropy even if temperature and internal energy do not change. Mixing identical gases under identical conditions produces no new distinguishable composition state and should not be assigned the same ideal mixing entropy. The physical identity of the species is part of the counting problem.
S = k_B lnΩ
Statistical interpretation with Ω representing accessible microstates under the chosen constraints.
03 / UNDERSTAND & APPLY
3. Calculate entropy using a reversible path between the same states
The thermodynamic definition is dS=δq_rev/T. For a finite change, integrate reversible heat divided by the boundary temperature along a reversible reference path. The actual process need not be reversible: because entropy is a state function, any convenient reversible route between the same equilibrium endpoints gives the same ΔS.
This is why ΔS=q_actual/T is not a general formula. In an insulated free expansion, q_actual=0 but the gas entropy increases. A reversible isothermal expansion between the same ideal-gas endpoints would absorb heat q_rev=nRT ln(V₂/V₁), giving ΔS=nR ln(V₂/V₁). The fictitious reversible route is a calculation device, not a claim that the actual expansion was reversible.
For heating at constant pressure with constant molar heat capacity, ΔS=nC_P ln(T₂/T₁); at constant volume use C_V. For a general fixed-composition ideal-gas change, ΔS=nC_V ln(T₂/T₁)+nR ln(V₂/V₁), or equivalently nC_P ln(T₂/T₁)−nR ln(P₂/P₁). Temperatures must be kelvin and ratios dimensionless.
These gas expressions assume ideal behavior and the stated heat-capacity approximation. If C varies significantly with temperature, integrate C(T)/T. If the route crosses a phase transition, add the transition entropy rather than extending one heat-capacity formula through the discontinuity.
ΔS = ∫δq_rev/T
Entropy is a state function even when the actual process is irreversible.
ΔS_ideal gas = nC_V ln(T₂/T₁) + nR ln(V₂/V₁)
Fixed amount and composition, ideal gas, constant molar C_V.
ORIGINAL WORKED EXAMPLE
Free expansion has entropy change without heat input
One mole of ideal gas freely expands in an insulated vessel from volume V to 3V. Find ΔU and ΔS using R=8.314 J mol⁻¹ K⁻¹.
Vacuum expansion gives w=0 and insulation gives q=0, so ΔU=0.
For an ideal gas of fixed composition, unchanged U means unchanged temperature.
Choose a reversible isothermal reference expansion between the same endpoints.
ΔS=nR ln3=8.314×1.09861≈9.13 J K⁻¹.
Answer ΔU=0; ΔS≈+9.13 J K⁻¹ despite actual q=0.
04 / UNDERSTAND & APPLY
4. Phase changes and standard molar entropies
At a phase-transition temperature and pressure where two phases coexist reversibly, the transition entropy is ΔS_trans=ΔH_trans/T_trans. Melting and vaporization generally have positive enthalpy and entropy changes; the reverse transitions have equal-magnitude negative changes at the same coexistence conditions. Use the actual transition temperature in kelvin.
A transition occurring away from coexistence is not itself a reversible isothermal phase change. To calculate its entropy from reference data, construct a path that heats or cools to a coexistence condition, changes phase reversibly, and returns to the desired final temperature. Simply dividing a latent enthalpy measured at one temperature by an unrelated temperature can be wrong.
The third law assigns zero entropy to a perfect crystal with a unique ground state at absolute zero. It provides a reference for absolute molar entropies obtained by integrating heat capacities and including phase changes. Real orientational disorder or ground-state degeneracy can give residual entropy, so “every material has zero entropy at 0 K” is too broad.
Standard molar entropy S° at an ordinary temperature is not zero for an element in its reference form. This differs from the convention that standard formation enthalpy and formation Gibbs energy of a reference-state element are zero. Confusing those conventions can erase major terms from a reaction-entropy calculation.
ORIGINAL WORKED EXAMPLE
Entropy of an equilibrium melting step
A hypothetical pure solid melts reversibly at 400 K with molar enthalpy of fusion 8.00 kJ mol⁻¹. Find the entropy change for melting 0.500 mol.
Molar transition entropy is 8000/400=20.0 J mol⁻¹ K⁻¹.
Multiply by 0.500 mol to obtain the sample entropy change.
The sign is positive for melting and would reverse for freezing at the same coexistence temperature.
Answer ΔS=+10.0 J K⁻¹.
05 / UNDERSTAND & APPLY
5. The second law concerns the total entropy change
For the system plus surroundings treated as an isolated whole, a spontaneous irreversible process has ΔS_total>0. A reversible process has ΔS_total=0. A negative total would violate the second law for that isolated whole. The system’s entropy alone may decrease if the surroundings gain more entropy.
If the surroundings act as a large thermal reservoir at constant temperature T, their entropy change is ΔS_surr=q_surr/T=−q_sys/T. For a constant-pressure system with only PV work and appropriate thermal contact, q_sys=ΔH_sys, so ΔS_surr=−ΔH_sys/T. An exothermic system can therefore increase surroundings entropy even while becoming more internally ordered.
If reservoir temperature changes appreciably, use an integral or an appropriate heat-capacity expression rather than one fixed q/T. If heat is transferred between two large reservoirs at different constant temperatures, their entropy changes must be calculated separately. The same energy transferred at a lower temperature corresponds to a larger entropy change.
An adiabatic system has no entropy transfer through heat, but it can generate entropy internally through irreversibility. Reversible adiabatic processes are isentropic; irreversible adiabatic processes generally are not. This distinction is the entropy counterpart of the first-law warning that adiabatic does not automatically mean reversible.
ORIGINAL WORKED EXAMPLE
A system entropy decrease can be allowed
At 300 K, a constant-pressure process with only PV work has ΔH_sys=−12.0 kJ and ΔS_sys=−20.0 J K⁻¹. Treat the surroundings as a reservoir. Is the forward process thermodynamically favored?
ΔS_surr=−ΔH_sys/T=12000/300=+40.0 J K⁻¹.
ΔS_total=−20.0+40.0=+20.0 J K⁻¹.
The total is positive even though the system contribution is negative.
This is a direction criterion under the stated reservoir conditions, not a prediction of speed.
Answer Yes; ΔS_total=+20.0 J K⁻¹.
06 / UNDERSTAND & APPLY
6. Gibbs energy packages the reservoir bookkeeping
Define G=H−TS. For changes between states at the same temperature, ΔG=ΔH−TΔS. Under fixed temperature and pressure with the usual closed-system constraints, spontaneous change proceeds in a direction that lowers G, and equilibrium corresponds to a minimum with respect to allowed changes. The reaction Gibbs energy is zero at equilibrium, while the system’s total G need not be zero.
Using ΔS_surr=−ΔH_sys/T gives ΔS_total=ΔS_sys−ΔH_sys/T=−ΔG_sys/T for the appropriate constant-T,P setting without additional useful-work extraction in that simple bookkeeping. This explains why negative ΔG corresponds to positive total entropy. The Gibbs criterion is a compact way to include the surroundings, not a denial that their entropy matters.
Keep energy units consistent. If ΔH is given in kJ mol⁻¹ and ΔS in J mol⁻¹ K⁻¹, convert one before evaluating TΔS. Temperature must be kelvin. A missing factor of one thousand can reverse a sign and produce a physically absurd crossover temperature.
At constant temperature and pressure, the decrease in Gibbs energy bounds the useful non-expansion work obtainable, with equality in a reversible limit under the relevant work convention. Actual devices deliver less useful work because irreversibility produces entropy. Free energy is therefore not energy that is literally costless or wholly convertible into any desired form.
ΔG = ΔH − TΔS at fixed T
Use compatible energy units and temperature in kelvin.
At fixed T and P: forward reaction Δ_rG<0 is favored; Δ_rG=0 at equilibrium
The criterion concerns the allowed reaction direction at the current composition.
ORIGINAL WORKED EXAMPLE
Check the units before the sign
For a reaction at 350 K, ΔH=+18.0 kJ mol⁻¹ and ΔS=+60.0 J mol⁻¹ K⁻¹. Estimate ΔG using these values at that temperature.
Convert ΔS to 0.0600 kJ mol⁻¹ K⁻¹.
TΔS=350×0.0600=21.0 kJ mol⁻¹.
ΔG=18.0−21.0=−3.00 kJ mol⁻¹.
The entropy term outweighs the positive enthalpy under the specified conditions.
Answer ΔG=−3.00 kJ mol⁻¹; the forward direction is thermodynamically favored in this stated calculation.
07 / UNDERSTAND & APPLY
7. Temperature changes the balance of enthalpy and entropy
If ΔH<0 and ΔS>0, both terms favor negative ΔG at positive temperatures within the range where those signs and the chemical states remain valid. If ΔH>0 and ΔS<0, both oppose it. When ΔH and ΔS share the same sign, temperature controls which contribution dominates.
For ΔH>0 and ΔS>0, the forward process becomes favored above the approximate crossover T*=ΔH/ΔS. For ΔH<0 and ΔS<0, it is favored below that positive crossover. A negative value of ΔH/ΔS from opposite signs is not a physical positive-temperature crossover; the two terms already favor the same direction or oppose it throughout the assumed range.
The crossover estimate assumes ΔH and ΔS are approximately temperature independent over the interval. Heat capacities generally change them, and phase transitions or altered chemical speciation can change the situation sharply. Therefore a statement such as “spontaneous at all temperatures” should be understood as a sign-model conclusion within the specified state and range, not a prediction to arbitrarily high temperatures.
For a phase boundary at a specified pressure, the two phases have equal molar Gibbs energy at the transition. For a chemical reaction, composition also matters. A standard-state crossover does not guarantee that a particular nonstandard mixture has zero reaction driving force at that same temperature.
ORIGINAL WORKED EXAMPLE
Find and interpret a crossover
Approximate ΔH°=−40.0 kJ mol⁻¹ and ΔS°=−100 J mol⁻¹ K⁻¹ as constant. Find the standard-state crossover and the favored side of the temperature range.
Convert ΔS° to −0.100 kJ mol⁻¹ K⁻¹.
T*=ΔH°/ΔS°=(−40.0)/(−0.100)=400 K.
ΔG°=−40.0+0.100T in kJ mol⁻¹.
Below 400 K it is negative; above 400 K it is positive, within the constant-property approximation.
Answer T*=400 K; standard-state forward preference below 400 K.
08 / UNDERSTAND & APPLY
8. Standard data and the actual mixture answer different questions
Reaction entropy from tabulated molar values is Δ_rS°=ΣνS°(products)−ΣνS°(reactants), including all phases and coefficients. Apply the same products-minus-reactants structure to formation Gibbs energies for Δ_rG°. Do not omit a reference-state element’s S° merely because its standard formation energy is defined as zero.
The actual reaction driving force is Δ_rG=Δ_rG°+RT lnQ. At equilibrium Q=K and Δ_rG=0, so Δ_rG°=−RT lnK. A negative standard reaction Gibbs energy means K>1, but a mixture with Q>K still has a positive forward Δ_rG and tends to react backward. Standard preference and current direction are not interchangeable.
The reaction quotient uses dimensionless activities raised to stoichiometric powers. In ideal approximations, pressures and concentrations appear relative to their standard values. The logarithm must not be applied to a dimensional quantity without its reference convention. Pure solid or liquid standard phases normally have unit activity in the introductory model.
Multiplying a reaction by a positive factor multiplies Δ_rG° and Δ_rS° by that factor and raises K to that power. Reversing changes their signs and inverts K. These relationships provide a consistency check across thermochemistry, equilibrium and electrochemistry.
For coupled processes whose reaction equations genuinely add, Gibbs energy changes add. A favorable process can drive an unfavorable one only through an actual physical or chemical coupling under the relevant conditions. Merely writing two unrelated reactions beside each other does not transfer their driving force.
ORIGINAL WORKED EXAMPLE
A favorable standard reaction can reverse in the current mixture
At 300 K a reaction has Δ_rG°=−5.00 kJ mol⁻¹ and current Q=100. Find Δ_rG using R=8.314 J mol⁻¹ K⁻¹ and decide direction.
The positive value means the forward direction is not favored at this composition; the reverse direction lowers G.
The standard value remains negative; it has not changed because Q changed.
Answer Δ_rG≈+6.49 kJ mol⁻¹; net reverse direction is favored.
09 / UNDERSTAND & APPLY
9. Original entropy and free-energy practice
These original questions use the stated ideal or constant-property models. Report sample entropy changes in J K⁻¹ and molar reaction entropy changes in J mol⁻¹ K⁻¹, keeping their meanings distinct. Do not infer kinetics from a calculated free-energy sign.
For every result ask: does it refer to the system, surroundings or their total; is it standard-state or actual-state; what temperature and pressure constraints apply; and was heat divided by temperature along a reversible reference path or an appropriate reservoir transfer?
ORIGINAL PRACTICE · 1
Can a spontaneous process have ΔS_system<0?
Reveal answer & reasoning
Yes, if the surroundings entropy increase is larger.
The second-law criterion applies to the isolated total. Freezing under suitable conditions can lower the material’s entropy while released heat raises surroundings entropy enough.
ORIGINAL PRACTICE · 2
Two moles of ideal gas expand isothermally from V to 2V. Find ΔS.
Reveal answer & reasoning
2R ln2≈11.53 J K⁻¹.
Use ΔS=nR ln(V₂/V₁). This state-function result applies between the specified ideal-gas endpoints regardless of whether the actual expansion is reversible.
ORIGINAL PRACTICE · 3
A reversible phase transition absorbs 6.00 kJ at 300 K. Find the sample entropy change.
Reveal answer & reasoning
+20.0 J K⁻¹.
ΔS=q_rev/T=6000/300. The formula applies at the stated reversible transition condition.
ORIGINAL PRACTICE · 4
Is the standard molar entropy of O₂(g) at 298 K zero because oxygen is an element?
Reveal answer & reasoning
No.
The zero convention applies to formation enthalpy/Gibbs energy of the reference-state element, not to its ordinary-temperature absolute molar entropy.
ORIGINAL PRACTICE · 5
At 300 K, ΔH=−10.0 kJ mol⁻¹ and ΔS=−20.0 J mol⁻¹ K⁻¹. Find ΔG.
Reveal answer & reasoning
−4.00 kJ mol⁻¹.
TΔS=300(−0.0200)=−6.00 kJ mol⁻¹, so ΔG=−10.0−(−6.00)=−4.00.
ORIGINAL PRACTICE · 6
If ΔH>0 and ΔS<0 remain unchanged in sign over a positive-temperature range, can ΔG become negative in that range?
Reveal answer & reasoning
No in that stated sign model.
ΔH is positive and −TΔS is also positive, so their sum is positive. The conclusion presumes the same states and signs remain applicable.
ORIGINAL PRACTICE · 7
If K=1 at a given temperature, what is Δ_rG°? Does every mixture then have Δ_rG=0?
Reveal answer & reasoning
Δ_rG°=0; only mixtures with Q=1 have Δ_rG=0.
Δ_rG°=−RT ln1=0, but actual Δ_rG=RT lnQ can be positive or negative when composition differs from equilibrium.
ORIGINAL PRACTICE · 8
A reaction A→2B has S°(A)=120 and S°(B)=80 J mol⁻¹ K⁻¹. Find Δ_rS° for the equation as written.
Reveal answer & reasoning
+40 J mol⁻¹ K⁻¹.
Products minus reactants gives 2(80)−120=40. The coefficient two must multiply the product entropy.
KEEP THE BIG IDEAS
Quick revision
Spontaneity describes a favored direction under conditions, not speed.
Entropy is a state function; calculate it with reversible heat even for irreversible endpoint changes.
The second law concerns total entropy, so system entropy may decrease.
Gibbs energy provides the fixed-temperature, fixed-pressure criterion.
Convert J and kJ consistently before evaluating TΔS.
Δ_rG° determines K; the actual quotient Q determines the current driving force.
Temperature crossover and ideal-gas formulas carry explicit assumptions.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.