Class 11 Chemistry · Original EZT practice
Equilibrium (Class 11 Chapter 6): summary and practice
Equilibrium is a balance that never sleeps: forward and reverse reactions keep running at equal rates while the concentrations hold steady. Kc and Kp measure where that balance sits, Q tells you which way a mixture must move, and Le Chatelier predicts how it responds when pushed. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Dynamic equilibrium
Physical processes (solid-liquid, liquid-vapour, dissolution) and chemical reactions both reach a state where forward and reverse rates match, so macroscopic properties stay constant while the microscopic exchange continues.
2. Law of chemical equilibrium and Kc
For aA + bB <=> cC + dD, Kc = [C]^c [D]^d / [A]^a [B]^b at equilibrium. K is constant at a given temperature; reversing the reaction inverts K, and pure solids and liquids do not appear in the expression.
3. Kp and Kc
Gaseous equilibria can be written in partial pressures, and Kp = Kc (RT)^(delta n) where delta n is the change in moles of gas. Homogeneous equilibria have all species in one phase; heterogeneous ones span phases.
4. Applications: extent, direction, concentrations
A large K means products dominate at equilibrium. Comparing the reaction quotient Q with K predicts the direction of change, and the ICE table turns K into actual equilibrium concentrations. The unit also relates K, Q and Gibbs energy.
5. Le Chatelier's principle
A system at equilibrium counteracts a disturbance: adding a reactant shifts it forward, pressure favours the side with fewer gas moles, inert gas at constant volume changes nothing, temperature shifts K itself, and a catalyst speeds both directions without moving the balance.
6. Ionic equilibrium in solution
The unit then applies the same ideas to acids and bases: Arrhenius, Bronsted-Lowry and Lewis definitions, ionization constants, the pH scale, buffer solutions and the solubility product with the common ion effect.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Compute Kc from equilibrium concentrations
Choose the method: Products over reactants, each raised to its coefficient.
EZT practice question
For H2 + I2 <=> 2HI, the equilibrium concentrations are [H2] = 0.1, [I2] = 0.1 and [HI] = 0.8 mol/L. Find Kc.
Show the worked solution
- Kc = [HI]^2 / ([H2] [I2]).
- = 0.8^2 / (0.1 x 0.1) = 0.64 / 0.01.
Answer: Kc = 64.
Check: A K well above 1 matches the product-heavy mixture: far more HI than H2 or I2.
Watch for this: Square HI because its coefficient is 2 - forgetting exponents is the common slip.
Now try it yourself
For N2O4 <=> 2NO2, [NO2] = 0.04 and [N2O4] = 0.02 - find Kc.
Check your answer
0.08.
Kc = 0.04^2 / 0.02 = 0.0016 / 0.02 = 0.08.
Handle the reversed reaction
Choose the method: K' = 1/K for the reverse reaction.
EZT practice question
Kc for H2 + I2 <=> 2HI is 64. What is Kc for 2HI <=> H2 + I2?
Show the worked solution
- Reversing swaps numerator and denominator of the K expression.
- K' = 1/64.
Answer: K' = 0.0156.
Check: A reaction strong in one direction is weak in the reverse - the two K values are reciprocals.
Watch for this: Do not copy 64 for the reverse reaction; invert it.
Now try it yourself
Kc for N2O4 <=> 2NO2 is 0.08 - Kc for the reverse?
Check your answer
12.5.
1 / 0.08 = 12.5.
Solve an ICE table
Choose the method: Initial, Change, Equilibrium - then apply K.
EZT practice question
H2 + I2 <=> 2HI has Kc = 50. Starting with 1 mol each of H2 and I2 in 1 L (no HI), find the equilibrium [HI].
Show the worked solution
- Let x mol/L of H2 react: equilibrium values are 1 - x, 1 - x and 2x.
- Kc = (2x)^2 / (1 - x)^2 = 50, so 2x / (1 - x) = sqrt(50) = 7.071.
- 2x = 7.071 - 7.071x gives 9.071x = 7.071, so x = 0.7795.
Answer: [HI] = 2x = 1.559 mol/L.
Check: Back-substitute: (1.559)^2 / (0.2205)^2 = 2.43 / 0.0486 = 50 - consistent.
Watch for this: Take the square root of both sides before solving - skipping it makes the algebra much harder.
Now try it yourself
In the same setup, what are [H2] and [I2] at equilibrium?
Check your answer
0.2205 mol/L each.
1 - x = 1 - 0.7795 = 0.2205.
Convert between Kp and Kc
Choose the method: Kp = Kc (RT)^(delta n).
EZT practice question
A reaction has delta n = -2 at 500 K. By what factor do Kp and Kc differ (R = 0.0821 L atm/(mol K))?
Show the worked solution
- Kp = Kc (RT)^(delta n) = Kc (RT)^(-2).
- RT = 0.0821 x 500 = 41.05, so (RT)^(-2) = 1 / 41.05^2.
Answer: Kp = Kc x 5.93 x 10^-4.
Check: Fewer gas moles on the product side shrinks Kp relative to Kc here.
Watch for this: Use the sign of delta n carefully - products minus reactants, gas moles only.
Now try it yourself
For a reaction with delta n = 0, how do Kp and Kc compare?
Check your answer
They are equal.
(RT)^0 = 1.
Predict direction with the reaction quotient
Choose the method: Compare Qc with Kc.
EZT practice question
For 2SO2 + O2 <=> 2SO3, Kc = 40. A mixture has [SO2] = 0.2, [O2] = 0.1 and [SO3] = 0.4 mol/L. Which way does the reaction move?
Show the worked solution
- Qc = [SO3]^2 / ([SO2]^2 [O2]) = 0.4^2 / (0.2^2 x 0.1) = 0.16 / 0.004 = 40.
- Qc equals Kc.
Answer: It does not move - the mixture is already at equilibrium.
Check: Qc < Kc would mean forward (more product needed); Qc > Kc would mean backward.
Watch for this: Compute Q with the same exponents as K - the only difference is that Q uses current, not equilibrium, values.
Now try it yourself
Same reaction, Kc = 40, but the mixture gives Qc = 10 - which way?
Check your answer
Forward, making more SO3.
Qc < Kc, so products must increase.
Apply Le Chatelier's principle
Choose the method: The system counteracts the change imposed.
EZT practice question
For N2 + 3H2 <=> 2NH3 (exothermic), predict the effect of (a) adding N2, (b) raising the pressure, (c) raising the temperature.
Show the worked solution
- Adding N2 shifts the equilibrium forward, consuming the added reactant.
- Higher pressure favours fewer gas moles: 4 to 2, so forward.
- Raising the temperature favours the endothermic direction, which is backward here - and it also lowers K.
Answer: Forward, forward, backward.
Check: Temperature is special: it is the only factor among these that changes the value of K, as the chapter states.
Watch for this: A catalyst changes neither the position nor K - it only gets the system there faster.
Now try it yourself
What does adding an inert gas at constant volume do to this equilibrium?
Check your answer
Nothing.
Partial pressures of the reactants and products are unchanged.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Kc for aA + bB <=> cC + dD?
Show answer and reason
[C]^c [D]^d / [A]^a [B]^b.
Products over reactants, coefficients as exponents.
2. K for the reversed reaction?
Show answer and reason
1 / K.
Numerator and denominator swap.
3. Kp in terms of Kc?
Show answer and reason
Kp = Kc (RT)^(delta n).
delta n counts gas moles only.
4. Qc < Kc means?
Show answer and reason
Reaction moves forward.
Products must rise to reach K.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- For N2O4 <=> 2NO2 with [NO2] = 0.04 and [N2O4] = 0.02, find Kc.
- Kc for H2 + I2 <=> 2HI is 64: find Kc for the reverse reaction.
- For 2SO2 + O2 <=> 2SO3, Kc = 40 and a mixture gives Qc = 10: which way does it move?
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Questions about this practice page
- What does a large K mean?
- Products dominate at equilibrium - the reaction goes far forward before the balance is reached. A small K means reactants dominate.
- How is the reaction quotient Q used?
- Q has the same form as K but uses current concentrations. Q < K means the reaction moves forward, Q > K backward, and Q = K means it is already at equilibrium.
- What changes the value of K?
- Only temperature, as the chapter states. Concentration, pressure, inert gas and catalysts can shift the position of equilibrium but leave K unchanged.
- Why do pure solids and liquids not appear in K?
- Their concentrations stay effectively constant during the reaction, so they are absorbed into the value of K itself.
- Are these the NCERT exercise questions for Equilibrium (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.