Chemical Equilibrium for NEET: Unit 6 Kc, pH and Traps

Chemical Equilibrium for NEET: Unit 6 Kc, pH and Traps

Chemical equilibrium is the state of a reversible reaction in a closed system where the forward and reverse reactions run at the same rate, so the concentrations stop changing even though both reactions keep going. In NEET it sits in Chemistry Unit 6, "Equilibrium", together with ionic equilibrium, and the unit is still in the syllabus. This guide shows what Unit 6 asks, works each formula with numbers from your NCERT book and lists the traps that cost marks.

Is chemical equilibrium in the NEET 2027 syllabus?

Yes. Equilibrium is Unit 6 of the Chemistry section in the NMC syllabus for NEET (UG) 2026, dated 22 December 2025. It covers chemical equilibrium, physical equilibria and ionic equilibrium, and sits between Solutions (Unit 5) and Redox Reactions and Electrochemistry (Unit 7). No 2027 syllabus exists yet, so this file is the current text.

The NEET (UG) 2026 Information Bulletin answers in its FAQ that there is no change in the syllabus of NEET (UG) 2026. As of 1 November 2026, the NEET (UG) public notices list nothing about the 2027 cycle. Some websites already print 2027 dates, but none of them is NTA. Learn the current wording and check it again the day NTA releases the 2027 bulletin.

Worried a chapter was cut? Our NEET 2027 deleted syllabus guide goes through the full list. Equilibrium is not on it.

What does Unit 6 ask you to know?

Unit 6 is long, but the syllabus names nearly every chemical equilibrium idea it wants. That makes it easy to turn into a checklist. Here is the official text, split into four blocks.

Syllabus phrase (Unit 6)	What it means in practice	What to be able to do
Meaning of equilibrium, the concept of dynamic equilibrium	Both reactions continue at equal rates	Explain why concentrations stop changing
Equilibria involving physical processes, Henry's law	Melting, evaporation, dissolving, gas in liquid	Read NCERT Table 6.1; state Henry's law
Law of chemical equilibrium, Kp and Kc and their significance	The equilibrium constant expression	Write Kc and Kp for any balanced equation
Significance of ΔG and ΔG° in chemical equilibrium	ΔG° = -RT ln K	Turn K into ΔG° and back
Factors affecting equilibrium: concentration, pressure, temperature, effect of catalyst; Le Chatelier's principle	How the position shifts when you disturb it	Predict the shift and say whether K changes
Weak and strong electrolytes, concepts of acids and bases (Arrhenius, Bronsted-Lowry, Lewis)	Who ionises fully and who does not	Classify a species under each concept
Acid-base equilibria (including multistage ionisation), ionisation constants, ionisation of water, pH scale	Ka, Kb, Kw and pH	Find the pH of a strong or weak acid or base
Common ion effect, hydrolysis of salts and pH of their solutions	Shifts in ionic equilibrium	Predict whether a salt solution is acidic or basic
Solubility of sparingly soluble salts and solubility products, buffer solutions	Ksp and buffers	Convert Ksp to molar solubility; find a buffer pH

Two details matter. First, roughly half of Unit 6 is ionic equilibrium, so a chemical equilibrium plan that stops at Le Chatelier covers only half the unit. Second, Unit 20 (practical chemistry) names "acids, bases and the use of indicators" among the titrations and lists the cations and anions of salt analysis, and both lean on this unit.

What is dynamic equilibrium?

Dynamic equilibrium means the forward and reverse processes are both still running, at exactly the same rate, so nothing visible changes. It applies to physical processes such as ice melting at 273 K and to chemical reactions such as H2 and I2 forming HI. Like any chemical equilibrium, it can only be reached in a closed system.

The NCERT Equilibrium chapter, Unit 6 of Class 11 Chemistry Part 1, proves the "dynamic" part with radioactive sugar: drop it into a saturated sugar solution, and after a while radioactivity shows up in both the solution and the solid. Molecules keep swapping places even though the amount dissolved stays the same.

Its Table 6.1 lists four physical equilibria and what stays constant in each:

Process	What stays constant
Liquid and vapour, H2O(l) and H2O(g)	Vapour pressure of water at a given temperature
Solid and liquid, H2O(s) and H2O(l)	Melting point at constant pressure
Solute and solution, sugar(s) and sugar(solution)	Concentration of solute at a given temperature
Gas and liquid, CO2(g) and CO2(aq)	[gas(aq)]/[gas(g)] at a given temperature

The last row is Henry's law: the mass of a gas dissolved in a given mass of solvent is proportional to the pressure of the gas above it. That is why a soda bottle fizzes when you open it. The pressure drops, so less CO2 can stay dissolved.

How do you write Kc and Kp?

For aA + bB giving cC + dD, Kc = [C]^c[D]^d / [A]^a[B]^b, with every concentration at equilibrium. This is the law of chemical equilibrium, also called the law of mass action, proposed by Guldberg and Waage in 1864. For gases you can use partial pressures instead, which gives Kp. The two are linked by Kp = Kc(RT)^Δn.

NCERT's Problem 6.1 shows the arithmetic. For N2 + 3H2 giving 2NH3 at 500 K, with [N2] = 1.5 × 10^-2 M, [H2] = 3.0 × 10^-2 M and [NH3] = 1.2 × 10^-2 M:

Kc = (1.2 × 10^-2)² / [(1.5 × 10^-2)(3.0 × 10^-2)³] = 3.55 × 10^2

The value of K for a chemical equilibrium depends on how you write the equation, and NCERT's Table 6.4 gives the three rules:

If you...	The new constant is
Reverse the equation	1/Kc
Multiply every coefficient by n	Kc^n
Halve every coefficient	Kc^(1/2), the square root

Then the Kp side. In Kp = Kc(RT)^Δn, Δn is moles of gaseous products minus moles of gaseous reactants. NCERT uses R = 0.0831 bar L mol-1 K-1 here, with pressure in bar. The molar gas constant listed by NIST is 8.314462618 J mol-1 K-1; 0.0831 is the same constant in bar litres.

Reaction	Δn	Relation
H2(g) + I2(g) giving 2HI(g)	2 - 2 = 0	Kp = Kc, and both have no unit
N2(g) + 3H2(g) giving 2NH3(g)	2 - 4 = -2	Kp = Kc(RT)^-2
N2O4(g) giving 2NO2(g)	2 - 1 = 1	Kp = Kc(RT); Kc in mol/L, Kp in bar
CaCO3(s) giving CaO(s) + CO2(g)	1 - 0 = 1	Kp = pCO2 (solids left out)

The last row is a heterogeneous chemical equilibrium. Pure solids and pure liquids have a constant concentration, so they drop out of the expression. That is why Kp for limestone decomposition is simply the pressure of CO2.

What do K and Q tell you?

The size of K tells you how far a reaction goes: NCERT says products predominate when Kc is above 10^3, reactants predominate when it is below 10^-3, and both are present in between. The reaction quotient Qc tells you which way a mixture moves: forward if Qc is less than Kc, backward if it is greater.

NCERT's own chemical equilibrium examples span a huge range. H2 with O2 at 500 K has Kc = 2.4 × 10^47, so it runs to completion. N2 with O2 giving NO at 298 K has Kc = 4.8 × 10^-31, so almost nothing forms. H2 with I2 giving HI at 700 K has Kc = 57.0, and both sides are present.

Qc has the same form as Kc but uses the concentrations at any moment. Take that HI reaction at 700 K with [H2] = 0.10 M, [I2] = 0.20 M and [HI] = 0.40 M. Then Qc = (0.40)² / (0.10 × 0.20) = 8.0. Since 8.0 is less than 57.0, more HI forms until Qc reaches Kc.

Comparison	What happens
Qc < Kc	Net reaction goes forward, left to right
Qc > Kc	Net reaction goes backward, right to left
Qc = Kc	Already at equilibrium, no net change

How are ΔG° and K connected?

ΔG° and K are linked by ΔG° = -RT ln K. A negative ΔG° gives K greater than 1, so products are favoured. A positive ΔG° gives K less than 1. At equilibrium ΔG itself is zero, which is why the relation comes from ΔG = ΔG° + RT ln Q with Q = K.

NCERT works it both ways:

• From ΔG° to K. The phosphorylation of glucose has ΔG° = 13.8 kJ mol-1. At 298 K, ln Kc = -13,800 / (8.314 × 298) = -5.569, so Kc = 3.81 × 10^-3 (Problem 6.10).
• From K to ΔG°. Hydrolysis of sucrose has Kc = 2 × 10^13 at 300 K. ΔG° = -8.314 × 300 × ln(2 × 10^13) = -7.64 × 10^4 J mol-1 (Problem 6.11).

Watch the units. R is 8.314 J mol-1 K-1 in this equation, so ΔG° must be in joules, not kilojoules. The same equation links to cell potential through ΔG° = -nFE°, which our electrochemistry for NEET guide works through.

How does Le Chatelier's principle predict the shift?

Le Chatelier's principle states that when a system at equilibrium is disturbed, it shifts in the direction that counteracts the change. Add a reactant and it is used up; raise the pressure and the side with fewer gas moles is favoured; heat it and the endothermic direction is favoured. Only a temperature change alters the value of K.

Change	Shift	Does K change?
Add a reactant, or remove a product	Forward	No
Increase pressure by reducing volume	Towards fewer moles of gas	No
Add an inert gas at constant volume	No shift	No
Raise temperature, exothermic reaction	Backward	Yes, K decreases
Raise temperature, endothermic reaction	Forward	Yes, K increases
Add a catalyst	No shift; equilibrium reached sooner	No

The Haber process puts all of this chemical equilibrium logic in one reaction. N2 + 3H2 giving 2NH3 has ΔH = -92.38 kJ mol-1 and loses two moles of gas. Low temperature favours ammonia but makes the reaction slow, high pressure favours it, and an iron catalyst gives a usable rate. NCERT gives the working conditions as about 500 °C and 200 atm. Removing liquefied ammonia keeps Qc below Kc, so the reaction keeps running forward.

NCERT's colour experiments are favourite statement questions. NO2 (brown) forming N2O4 (colourless) has ΔH = -57.2 kJ mol-1: in ice the brown fades, in hot water at 363 K it deepens. Fe3+ and SCN- give the deep red [Fe(SCN)]2+; adding oxalic acid or HgCl2 removes a reactant, so the red fades.

One more NCERT line catches students: the oxidation of SO2 to SO3 has Kc = 1.7 × 10^26, yet it is very slow, so platinum or V2O5 is used. A large K says nothing about speed: chemical equilibrium tells you where a reaction ends, not how fast it gets there. Rates belong to our chemical kinetics for NEET guide.

What is ionic equilibrium?

Ionic equilibrium is the equilibrium between the un-ionised molecules of a weak electrolyte and its ions in solution, such as acetic acid and acetate ions in water. Strong electrolytes like NaCl ionise almost completely; weak electrolytes like acetic acid ionise only partly, so an equilibrium exists. Michael Faraday first sorted electrolytes this way.

The three acid-base concepts in the syllabus each widen the one before:

Concept	Acid	Base	Example NCERT uses
Arrhenius	Gives H+ (H3O+) in water	Gives OH- in water	HCl, NaOH
Bronsted-Lowry	Proton donor	Proton acceptor	In NH3 + H2O, water donates the proton
Lewis (1923)	Electron pair acceptor	Electron pair donor	BF3, AlCl3 and metal ions such as Mg2+ as acids

A Bronsted acid and the base it forms after losing a proton are a conjugate pair, and the stronger the acid, the weaker its conjugate base. For any conjugate pair in water, Ka × Kb = Kw.

How do Kw and the pH scale work?

The ionic product of water, Kw = [H3O+][OH-], is 1.0 × 10^-14 at 298 K, because pure water holds 1.0 × 10^-7 M of each ion. pH is -log[H3O+], so pH + pOH = 14 at 298 K. Kw is an equilibrium constant, so it changes with temperature, and pH 7 is neutral only at 298 K.

NCERT's Table 6.5 lists the pH of everyday fluids. The chart puts eight of them on one scale.

Table 6.5 also lists concentrated HCl at about -1.0 and saturated NaOH at about 15, so the 0 to 14 range is a convenience, not a limit. Remember that each pH unit is a factor of ten: gastric juice at 1.2 holds about 6,300 times more H3O+ than black coffee at 5.0 (our arithmetic: 10 to the power 3.8).

The classic trap is a very dilute acid. What is the pH of 10^-8 M HCl? Not 8, because an acid cannot make water basic. NCERT adds the H3O+ from water, solves x² + 10^-8 x - 10^-14 = 0 and gets pH 6.98.

How strong is a weak acid?

A weak acid's strength is set by its ionisation constant Ka: the larger Ka, the stronger the acid and the smaller its pKa (pKa = -log Ka). For a weak acid of concentration c, Ka = cα²/(1 - α), and when α is small, α ≈ √(Ka/c) and [H+] ≈ √(Ka × c).

NCERT's Table 6.6 lists nine weak acids at 298 K. The chart turns their Ka values into pKa, so the strongest acid sits at the left.

Work one by hand. For 0.1 M acetic acid with Ka = 1.74 × 10^-5, [H+] ≈ √(1.74 × 10^-6) = 1.32 × 10^-3 M, so pH ≈ 2.88 and only about 1.3% of the molecules are ionised (our arithmetic). Dilute the acid and α rises, because α ≈ √(Ka/c).

Weak bases work the same way with Kb. NCERT's Table 6.7 gives ammonia Kb = 1.77 × 10^-5, aniline 4.27 × 10^-10 and urea 1.3 × 10^-14. For a conjugate pair, pKa + pKb = 14 at 298 K. A dibasic acid H2X has two ionisation constants, Ka1 and Ka2, and a tribasic acid like H3PO4 has three; NCERT's Table 6.8 lists them, and each later constant is smaller. That is the "multistage ionization" in the syllabus.

Buffers, salt hydrolysis and the common ion effect

A buffer is a solution that resists a change in pH when a little acid or base is added. It is made from a weak acid and its salt, or a weak base and its salt. For the acid kind, pH = pKa + log([Salt]/[Acid]), so equal amounts give pH = pKa. NCERT's acetic acid and sodium acetate buffer sits near pH 4.76.

The common ion effect is Le Chatelier applied to ions, the same rule as in chemical equilibrium: adding an ion that is already in the equilibrium pushes the ionisation back. NCERT's example adds 0.05 M acetate to 0.05 M acetic acid and gets [H+] = 1.8 × 10^-5 M, pH 4.74, far less acidic than the acid alone.

Salt hydrolysis decides the pH of a salt solution, and the rule fits in one table:

Salt made from	Solution	Example
Strong acid and strong base	Neutral, pH 7 at 298 K	NaCl
Weak acid and strong base	Basic	CH3COONa
Strong acid and weak base	Acidic	NH4Cl
Weak acid and weak base	pH = 7 + ½(pKa - pKb)	CH3COONH4

How do you turn Ksp into solubility?

The solubility product Ksp is the equilibrium constant for a sparingly soluble salt in contact with its saturated solution, written without the solid. For a salt MxXy with molar solubility S, Ksp = (xS)^x(yS)^y = x^x y^y S^(x+y). A precipitate forms when the ionic product Qsp exceeds Ksp.

NCERT's numbers make the pattern clear:

1. A 1:1 salt. BaSO4 has Ksp = 1.1 × 10^-10 at 298 K, so S² = 1.1 × 10^-10 and S = 1.05 × 10^-5 mol L-1.
2. A 2:3 salt. For A2X3 with Ksp = 1.1 × 10^-23, Ksp = (2S)²(3S)³ = 108 S^5, so S = 1.0 × 10^-5 mol L-1 (Problem 6.26).
3. A bigger formula. Zirconium phosphate, (Zr4+)3(PO4 3-)4, gives Ksp = 6912 S^7. Practise writing the exponent before you touch the numbers.
4. Add a common ion. Ni(OH)2 has Ksp = 2.0 × 10^-15. In 0.10 M NaOH, S = 2.0 × 10^-13 M (Problem 6.28). In pure water, 4S³ = 2.0 × 10^-15 gives S ≈ 7.9 × 10^-6 M (our arithmetic), so the NaOH cuts solubility about 40 million times.

Compare salts by S, never by Ksp alone, when their formulas differ. For a 1:1 salt S = √Ksp, but for a 1:2 salt like Ni(OH)2 S = (Ksp/4)^(1/3). NCERT's Problem 6.27 does exactly this: AgCN (Ksp 6 × 10^-17) gives S = 7.8 × 10^-9 M, so Ni(OH)2 is the more soluble of the two. The same common ion idea explains why passing HCl gas through saturated brine precipitates very pure NaCl.

Two NCERT numbers worth checking yourself

Your NCERT chapter has two printed answers that our arithmetic does not reproduce. Neither changes the method, but a student copying them into notes would learn a wrong number.

• Problem 6.5. For 2NOCl giving 2NO + Cl2, Kc = 3.75 × 10^-6 at 1069 K and Δn = (2 + 1) - 2 = 1, as the book says. It then prints Kp = 0.033, but 3.75 × 10^-6 × (0.0831 × 1069) works out to about 3.3 × 10^-4.
• Problem 6.27. For Ni(OH)2 in pure water the book prints S = 0.58 × 10^-4 M. Solving 4S³ = 2.0 × 10^-15 gives about 7.9 × 10^-6 M. The conclusion, that Ni(OH)2 is more soluble than AgCN, still holds.

Checks like these are good practice: redo every NCERT example with your own calculator once, and you will catch your own slips faster in the exam.

Which lines trap students in chemical equilibrium?

Most marks lost in chemical equilibrium and ionic equilibrium come from a short list. Read it twice.

1. Putting solids and pure liquids into K. Leave them out. Kp for CaCO3 decomposing is just pCO2.
2. Counting all moles in Δn. Count only gaseous moles, products minus reactants.
3. Saying pressure or concentration changes K. They shift the position. Only temperature changes K.
4. Saying a catalyst shifts equilibrium. It lowers the activation energy of both directions by the same amount and leaves K alone.
5. Assuming a large K means a fast reaction. SO2 to SO3 has Kc = 1.7 × 10^26 and still needs a catalyst.
6. Forgetting water in very dilute acid. 10^-8 M HCl has pH 6.98, not 8.
7. Using kJ with R = 8.314. In ΔG° = -RT ln K, put ΔG° in joules.
8. Comparing Ksp values of salts with different formulas. Convert each to molar solubility S first.

A one-week revision plan for chemical equilibrium

Chemical equilibrium and ionic equilibrium build on each other, so revise them in this order. The plan fits one normal study week.

1. Copy the scope onto one page. Write the nine Unit 6 phrases from the table above. That page is everything this chapter can ask.
2. Drill K expressions. Ten balanced equations a day, each with Kc, Kp, Δn and the units, including two heterogeneous ones.
3. Do Q against K. Rework NCERT's HI example with three new sets of concentrations and say which way each mixture moves.
4. Make a Le Chatelier grid. For the Haber process, N2O4 and CaCO3, fill in the shift for each change in the table above and mark whether K changes.
5. Move to ions. Find the pH of a strong acid, a weak acid, a buffer and a salt solution, one of each per day, then do two Ksp to S conversions. Put each trap above on a card; our guide to NEET flashcards shows how.
6. Test under time. Mix chemical equilibrium questions with Units 7 and 8 and sort your wrong answers by trap number. A wrong answer costs five marks against a right one at plus four and minus one, which our NEET negative marking guide explains.

How many chemical equilibrium questions will NEET 2027 have? Nobody can tell you honestly. NTA publishes the subject-wise pattern, 45 Chemistry questions for 180 marks, not a chapter-wise split. For where Unit 6 sits in a sensible revision order, futureMBBS's NEET 2027 chapter-wise weightage guide is a good next read, and the NEET syllabus overview on futureMBBS puts all three subjects on one page.

For the whole subject plan, start with our NEET 2027 Chemistry preparation strategy, and for practice by unit see the NEET Chemistry questions guide. Bond polarity and the Lewis picture come from our chemical bonding for NEET guide. Our NEET previous year question papers guide shows how to use old papers without burning them, and futureMBBS explains why NEET leans on the NCERT book and sets out how to prepare for NEET from start to finish.

Want to see whether chemical equilibrium is where your Chemistry marks leak? NEETsheet's chapter-wise Chemistry sets and timed mocks show which trap you keep falling into, so your next revision goes there first. Start at neetsheet.in.

Sources

Every syllabus line and number above comes from these documents, opened and read on 1 November 2026.

• National Medical Commission (UGMEB), Syllabus for NEET (UG) 2026, public notice of 22 December 2025, linked from neet.nta.nic.in (Chemistry Units 6 and 20): https://cdnbbsr.s3waas.gov.in/s37bc1ec1d9c3426357e69acd5bf320061/uploads/2026/01/202601081066816297.pdf
• National Testing Agency, Information Bulletin NEET (UG) 2026 (pattern: Chemistry 45 questions and 180 of 720 marks; plus four and minus one marking; syllabus FAQ stating no change for 2026): https://cdnbbsr.s3waas.gov.in/s37bc1ec1d9c3426357e69acd5bf320061/uploads/2026/02/202602231394640855.pdf
• National Testing Agency, NEET (UG) public notices, checked on 1 November 2026 for any NEET (UG) 2027 notice: https://neet.nta.nic.in/public-notices/
• NCERT, Chemistry Class 11 Part 1, Unit 6 Equilibrium, reprint 2026-27 (Tables 6.1, 6.4, 6.5, 6.6, 6.7 and 6.8; Problems 6.1, 6.5, 6.10, 6.11, 6.26, 6.27 and 6.28; sections 6.1 to 6.13): https://ncert.nic.in/textbook/pdf/kech106.pdf
• National Institute of Standards and Technology, CODATA value of the molar gas constant, 8.314462618 J mol-1 K-1 (exact): https://physics.nist.gov/cgi-bin/cuu/Value?r