Chemistry · Thermodynamics · NEET
Count the moles of GAS on both sides. Gas has the most disorder, so it decides the sign. If gas moles increase from reactants to products, ΔS is positive (+). If gas moles decrease, ΔS is negative (−). Example: 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) makes 2 moles of gas from 0 moles of gas, so ΔS is positive. Ignore solids and liquids first; only look at them when gas moles are equal on both sides.
Because the particles get more freedom to move. In a solid, particles are locked in place (low disorder). In a liquid they slide around a bit. In a gas they fly everywhere (high disorder). So the order of entropy is: gas > liquid > solid. Any change that goes toward gas (solid→liquid→gas) increases disorder, so ΔS is positive. This is why evaporation of water and sublimation of a solid always have ΔS > 0.
ΔS is negative when disorder DROPS. Three common cases: (1) gas moles decrease, like 2H(g) → H2(g) where 2 gas particles become 1, so ΔS < 0. (2) Gas turns into liquid or solid (condensation, freezing, deposition). (3) A solid or gas is cooled toward 0 K, which orders the particles. In NEET 2019 the answer to 'which has negative entropy change' was 2H(g) → H2(g) for exactly this reason: fewer gas particles.
Then look at the total number of particles and their state. If gas moles are equal, more product particles or more complex molecules usually means slightly higher entropy. But for NEET, most questions have a clear change in gas moles, so use that first. If gas moles are truly equal and it is not obvious, the entropy change is small and close to zero — you usually will not be asked to guess the sign in that case.
No. For the SIGN, you never need a formula — just count gas moles and check the physical state. Formulas like ΔS = qrev/T or ΔS = nR ln(Vf/Vi) are for calculating the VALUE of ΔS, not its sign. NEET sign questions are pure logic and take a few seconds. Save the formula for numerical problems, like isothermal gas expansion where you compute the exact entropy change.
Adding heat (q) increases random motion, so heating a system raises its entropy (ΔS > 0) and cooling lowers it (ΔS < 0). That is why lowering a crystal from 130 K toward 0 K gives ΔS < 0 — the lattice becomes more ordered. But for a chemical reaction, the state and moles of gas matter far more than a small temperature change, so always check gas moles first.
In which case is the change in entropy negative?
In which of the following processes does entropy increase? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid is lowered from 130 K to 0 K. C. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g). D. Cl2(g) → 2Cl(g).
For a sample of perfect gas when its pressure is changed isothermally from p_i to p_f, the entropy change is given by:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Count moles of gas on each side. More gas moles in products means ΔS is positive; fewer means ΔS is negative. This solves most NEET entropy questions in seconds with no calculation.
Yes. For the same substance the order is gas > liquid > solid, because gas particles have the most freedom and disorder. So any change moving toward the gas state raises entropy.
Usually yes. When a solid dissolves, its ordered lattice breaks and ions spread through the liquid, increasing disorder, so ΔS is generally positive. (A few small, highly charged ions can lower it by ordering water, but that is rare in NEET.)
Negative. Gas → liquid reduces disorder because particles lose freedom, so ΔS < 0. Freezing (liquid → solid) and deposition (gas → solid) are also negative.
No, that formula gives the numerical value of ΔS, not the sign for reactions. For predicting the sign in NEET, just compare gas moles and physical states.
