Why this chapter is bookkeeping
A typical NEET question gives ΔH and ΔS and asks whether a reaction is spontaneous at a stated temperature. The physical chemistry is one substitution. The work is converting ΔS from J/K to kJ/K, and handling the double negative when ΔS itself is negative - because subtracting a negative TΔS adds to ΔH rather than subtracting from it.
Almost every error in this chapter is one of those two. That makes it unusually tractable: a student who writes the convention down and converts units before substituting will get most of these questions right without any deeper insight.
How NEET actually asks Thermodynamics
NTA publishes no chapter-wise weightage for NEET, so figures circulating online are coaching estimates from past papers; confirm the current paper structure in the official NTA information bulletin.
The recurring types are: identify a state function, apply the first law to a described process, use Hess's law or enthalpies of formation, decide spontaneity from ΔH and ΔS, and answer conceptual questions on the second and third laws. Numericals are single-step, which is the main difference from the engineering-entrance version.
Key concepts, compressed
- ✓State functions depend only on the current state; path functions depend on the route taken. Heat and work are the two path functions that matter.
- ✓The first law is conservation of energy for a thermodynamic system.
- ✓Enthalpy is defined so that ΔH equals the heat exchanged at constant pressure - the usual laboratory condition.
- ✓Entropy measures the dispersal of energy, and the second law requires the universe's total entropy to increase in any spontaneous process.
- ✓Gibbs free energy combines enthalpy and entropy into one spontaneity criterion at constant temperature and pressure.
- ✓Extensive properties depend on the amount of substance; intensive properties do not.
Relations you need before the questions
| Quantity | Formula | Note |
|---|---|---|
| First law | ΔU = q + w | work on the system is positive |
| Work of expansion | w = −P(ext)ΔV | |
| Enthalpy | H = U + PV | ΔH = heat at constant pressure |
| ΔH and ΔU | ΔH = ΔU + Δn(g)RT | gas moles only |
| Hess's law | ΔH is independent of route | enthalpy is a state function |
| From formation enthalpies | ΔH = Σ products − Σ reactants | elements are zero |
| From bond enthalpies | ΔH = broken − formed | order matters |
| Entropy change | ΔS = q(rev)/T | T in kelvin |
| Second law | ΔS(universe) > 0 | for a spontaneous process |
| Gibbs free energy | ΔG = ΔH − TΔS | |
| Spontaneity | ΔG < 0 | |
| Free energy and K | ΔG° = −RT ln K |
The five mistakes that cost the most marks
- ✓Switching sign conventions mid-problem. Choose ΔU = q + w or ΔU = q − w and stay with it throughout.
- ✓Combining ΔH in kilojoules with ΔS in joules. Convert first - this is the chapter's most common failure.
- ✓Reversing the bond-enthalpy subtraction, which gives the right magnitude with the wrong sign.
- ✓Treating spontaneous as meaning fast. ΔG describes feasibility only.
- ✓Applying the second law to the system alone. It constrains the universe, which is why a system's entropy can fall spontaneously.
Practice set 1: systems and the first law
1. What are open, closed and isolated systems?
An open system exchanges both matter and energy with its surroundings, a closed system exchanges energy only, and an isolated system exchanges neither. A cup of tea is open, a sealed flask closed, and an ideal thermos approximates isolated. Which quantities can change depends entirely on this classification.
2. What is the difference between a state function and a path function?
A state function depends only on the initial and final states - internal energy, enthalpy, entropy and free energy are examples. A path function depends on how the change occurred, and heat and work are the two. This distinction is what makes Hess's law valid, since enthalpy is a state function.
3. State the first law of thermodynamics.
ΔU = q + w - the change in internal energy equals the heat added to the system plus the work done on it. It is conservation of energy expressed for a thermodynamic system: energy can be transferred as heat or work but never created or destroyed.
4. In the convention ΔU = q + w, what signs apply?
Heat absorbed by the system is positive and heat released negative; work done on the system is positive and work done by the system negative. Some texts write ΔU = q − w with w meaning work done by the system, which is identical physics with reversed bookkeeping. Mixing the two is the standard error.
5. How is expansion work calculated?
w = −P(ext)ΔV, work done on the system against a constant external pressure. The negative sign means expansion costs the system energy. For a free expansion into a vacuum the external pressure is zero, so no work is done at all - a case NEET asks about specifically.
6. What is held constant in isothermal, adiabatic, isochoric and isobaric processes?
Temperature, heat exchange, volume and pressure respectively. Each simplifies the first law: an adiabatic process has q = 0 so ΔU = w, and an isochoric process has no expansion work so ΔU = q. Recognising which condition applies is usually the first step of a first-law question.
7. What does the heat absorbed at constant volume equal?
The change in internal energy ΔU, because no expansion work is possible. At constant pressure the heat absorbed equals ΔH instead. This is why bomb calorimetry, which holds volume fixed, measures ΔU while ordinary open-vessel calorimetry measures ΔH.
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Practice set 2: enthalpy
8. How is enthalpy defined and why is it the useful quantity?
H = U + PV. Its usefulness is that ΔH equals the heat exchanged at constant pressure, which is the condition most reactions occur under in an open vessel. So ΔH is effectively what a thermometer measures in ordinary laboratory work.
9. How do ΔH and ΔU differ for a reaction involving gases?
By Δn(g)RT, where Δn(g) is the change in the number of gas moles. They are equal when the gas-mole count is unchanged. Only gases are counted, because solids and liquids contribute negligibly to volume change.
10. What do exothermic and endothermic mean?
Exothermic means ΔH is negative and the system releases heat, warming the surroundings. Endothermic means ΔH is positive and the system absorbs heat. The sign is always from the system's perspective, and the surroundings experience the opposite - which is worth stating explicitly to avoid confusion.
11. State Hess's law and explain why it holds.
The enthalpy change of a reaction is the same whether it occurs in one step or several. It holds because enthalpy is a state function, so ΔH depends only on the initial and final states. Practically, it allows an unmeasurable enthalpy change to be calculated by combining reactions that can be measured.
12. What is the standard enthalpy of formation, and what is its value for an element?
The enthalpy change when one mole of a compound forms from its elements in their standard states. For an element in its standard state it is zero by definition, since forming it from itself involves no change. This convention makes reaction enthalpies calculable as products minus reactants.
13. What is the enthalpy of combustion?
The enthalpy change when one mole of a substance burns completely in excess oxygen. It is always negative, since combustion is exothermic. It is measured by calorimetry and is the route by which many formation enthalpies are determined indirectly, using Hess's law.
14. How is a reaction enthalpy found from bond enthalpies?
ΔH = sum of bond enthalpies of bonds broken minus sum of those formed. Breaking bonds absorbs energy and forming them releases it, which fixes the order. Reversing it produces the correct magnitude with the wrong sign, converting an exothermic reaction into an endothermic one.
Practice set 3: entropy
15. What does entropy measure?
The dispersal of energy among the available arrangements of a system, commonly described as disorder. Higher entropy means energy is spread over more accessible states. Unlike enthalpy it has an absolute value, which is a consequence of the third law.
16. How does entropy change during melting and boiling?
It increases in both, since particles gain freedom of movement. The increase on boiling is much larger, because the change from liquid to gas opens far more accessible arrangements than the change from solid to liquid. Freezing and condensation have negative ΔS for the system.
17. State the second law of thermodynamics.
The total entropy of the universe increases in any spontaneous process. It applies to system plus surroundings, not to the system alone - which is precisely why a system's entropy can decrease spontaneously if the surroundings gain more, as when water freezes.
18. State the third law and what it makes possible.
The entropy of a perfect crystalline substance is zero at absolute zero, because only one arrangement is possible. It provides the reference point that makes absolute entropies measurable, which is why tables of standard molar entropy exist while tables of absolute enthalpy do not.
19. How is entropy change calculated for a reversible process?
ΔS = q(rev)/T, the reversibly exchanged heat divided by the absolute temperature. Because entropy is a state function, the value applies even when the real process is irreversible - you compute it along a reversible path between the same states. Temperature must be in kelvin.
20. How do the entropies of the three states compare?
Gas is far greater than liquid, which exceeds solid. Gas particles have the most freedom of position and motion and therefore the most accessible arrangements. This ordering lets you predict the sign of ΔS for many reactions simply by counting gas moles on each side.
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Practice set 4: Gibbs free energy
21. Write the Gibbs free energy relation.
ΔG = ΔH − TΔS, with T in kelvin. It combines the enthalpy and entropy contributions into a single criterion valid at constant temperature and pressure. The two terms can oppose one another, which is why spontaneity often depends on temperature.
22. What is the criterion for spontaneity?
ΔG < 0. A negative free energy change means the process can occur without external input at constant temperature and pressure. ΔG = 0 means equilibrium, and ΔG > 0 means the reverse process is the spontaneous one.
23. Describe the four sign combinations of ΔH and ΔS.
Negative ΔH with positive ΔS is spontaneous at all temperatures. Positive ΔH with negative ΔS is never spontaneous. Both negative is spontaneous only at low temperature. Both positive is spontaneous only at high temperature. Learning this as a four-case table answers most spontaneity questions immediately.
24. What does ΔG = 0 indicate?
Equilibrium - no net drive in either direction. It is also the condition for a phase transition, such as melting, where two phases coexist. Setting ΔH − TΔS = 0 gives the transition temperature T = ΔH/ΔS, which is a standard calculation.
25. How does standard free energy relate to the equilibrium constant?
ΔG° = −RT ln K. A negative ΔG° corresponds to K greater than 1, meaning products are favoured. Note that ΔG° is fixed for a reaction while ΔG varies with composition and reaches zero at equilibrium - the two are distinct quantities.
26. Does a negative ΔG mean the reaction is fast?
No. ΔG determines thermodynamic feasibility only and says nothing about rate. The conversion of diamond to graphite has a negative ΔG and proceeds imperceptibly slowly because the activation energy is enormous. Rate belongs to kinetics, which is an entirely separate consideration.
Practice set 5: calculations and properties
27. A reaction has ΔH = −100 kJ and ΔS = −200 J/K. Is it spontaneous at 300 K?
Yes - ΔG = −40 kJ. Convert ΔS to kJ/K first, giving −0.200. Then ΔG = −100 − (300)(−0.200) = −100 + 60 = −40 kJ. Note the double negative: subtracting a negative TΔS adds 60 kJ. The reaction is spontaneous, though only because the temperature is low enough.
28. Above what temperature does that reaction stop being spontaneous?
500 K. Setting ΔG = 0 gives T = ΔH/ΔS = (−100)/(−0.200) = 500 K. Both quantities are negative, so the ratio is positive. Below 500 K the favourable enthalpy dominates; above it the unfavourable entropy term takes over. This is the both-negative case from question 23.
29. What does calorimetry measure, and how do the two types differ?
It measures the heat exchanged in a reaction. A bomb calorimeter holds volume constant and therefore measures ΔU; a constant-pressure calorimeter measures ΔH directly. Converting between the two requires ΔH = ΔU + Δn(g)RT, which is why question 9 matters practically.
30. What is the difference between extensive and intensive properties?
Extensive properties depend on the amount of substance - mass, volume, internal energy, enthalpy and entropy. Intensive properties do not - temperature, pressure, density and molar quantities. Dividing an extensive property by the amount produces an intensive one, which is why molar enthalpy is intensive while enthalpy is not.
How to study this chapter efficiently
- ✓Write your sign convention at the top of every solution and never switch mid-problem.
- ✓Convert ΔS to kilojoules before combining with ΔH. Make it automatic rather than a check.
- ✓Write 'broken minus formed' explicitly before substituting bond enthalpies.
- ✓Memorise the four sign combinations as a small table - they answer most spontaneity questions instantly.
- ✓Keep ΔG and ΔG° distinct: the standard value fixes K, the actual value is zero at equilibrium.
- ✓Stay at NCERT level. NEET's numericals here are single-step, so extended practice on complex calculations is misdirected effort.
Turn this into active practice
Because the errors are mechanical rather than conceptual, they are invisible when reading. A worked solution shows the double negative already resolved and the units already converted - only producing one yourself reveals whether you handle both reliably.
The NEET Thermodynamics quiz on QUFF generates fresh questions across the first law, enthalpy, entropy and free energy, marks them instantly and explains each answer. Do timed sets and classify each error as a sign error, a unit error or a conceptual one. The first two will dominate, and both are entirely fixable.