Chemical Equilibrium Formula Sheet PDF for JEE (N Awasthi Companion)

Stop searching. Here are the three PDFs you need for JEE Chemical Equilibrium: a complete formula sheet, a one-page cheat sheet for quick revision, and a list of the most common mistakes that cost marks. They are our own, they are unfiltered, and they are completely free.

What these PDFs are (and what they are not)

Let’s be clear. These aren’t scanned pages from some textbook or a random collection of formulas from the web. These are original resources created by the 1% Syndicate at AllAboutIIT. We wrote them in our own words, based on what actually works for solving problems and clearing the exam. They are designed to be a perfect companion while you are grinding problems from a book like N Awasthi (N Avasthi), helping you connect the theory to the application, fast.

This is not the book. These sheets don’t contain the text, problems, or solutions from any book. They are a strategic overlay—a system to help you study smarter, not just harder.

Not affiliated with the author or publisher of any book mentioned.

Download the Chemical Equilibrium PDFs

We’ve split the resources into three distinct files. One for deep study, one for rapid revision, and one for fixing your errors. Use them all.

Chemical Equilibrium formula sheet (PDF)

This is your master document. It contains every key formula and relationship you’ll need for this chapter, laid out cleanly. Don’t just download it; read it before you start solving problems to prime your brain on the tools you have available.

Kc, Kp, and Kx expressions with units and conditions
The core relation: Kp = Kc (RT)^Δng
The partial pressure relation: Kp = Kx (P total)^Δng
Reaction Quotient (Q) and its comparison with K
Degree of dissociation (α) and its link to vapour density
The thermodynamic link: ΔG° = -RT ln K
The integrated van’t Hoff equation for temperature effects

Download the Chemical Equilibrium formula sheet (PDF)

Chemical Equilibrium cheat sheet (PDF)

This is your one-page weapon for last-minute revision and high-pressure moments. It cuts through the noise to give you the absolute essentials. The core idea is the “if-you-see-this-think-this” section, which helps you map question patterns to solution strategies.

A short note from us: Understand Le Chatelier’s Principle deeply.
Key ideas: dynamic equilibrium, the meaning of K’s magnitude, Q vs K.
Effects of temperature, pressure, and concentration.
Inert gas addition: constant volume vs. constant pressure.
The role (and non-role) of a catalyst.
Ignoring pure solids and liquids.
Exam shortcuts: the small-alpha approximation and how to check it.
An “if-you-see-this-think-this” list for pattern recognition.

Download the Chemical Equilibrium cheat sheet (PDF)

Chemical Equilibrium mistakes and traps (PDF)

This might be the most valuable PDF of the three. We’ve compiled the 11 most common errors that students make in this chapter—the ones that separate a correct answer from a negative mark. For every wrong problem you solve, consult this list. Find your mistake, understand why you made it, and don’t make it again.

Forgetting coefficients become powers in the K expression.
Incorrectly calculating Δng (e.g., including non-gaseous species).
Using moles instead of concentration in Kc.
Confusing inert gas effects at constant volume vs. constant pressure.
Thinking a catalyst changes the equilibrium constant K.
Including concentrations of pure solids or liquids.
Using the small-alpha approximation when α is not small.
Forgetting to adjust K when a reaction is reversed or scaled.

Download the Chemical Equilibrium mistakes and traps (PDF)

The chapter ideas in plain words

Forget the dense textbook language. Here are the core ideas of Chemical Equilibrium as we see them.

The Equilibrium Constant (Kc and Kp)

Equilibrium is not when a reaction stops; it’s when the forward and backward reaction rates become equal. The equilibrium constant, K, is a number that tells you the ratio of products to reactants at this steady state. Kc uses molar concentrations [ ], while Kp uses partial pressures (P). For a general reaction aA + bB <=> cC + dD:

Kc = [C]^c [D]^d / ([A]^a [B]^b)

Kp = (PC)^c (PD)^d / ((PA)^a (PB)^b)

A large K means the reaction favors products (lies to the right). A small K means it favors reactants (lies to the left).

The Reaction Quotient (Q)

The reaction quotient, Q, has the exact same formula as K, but you can calculate it at any point in the reaction, not just at equilibrium. It’s a snapshot. By comparing Q to K, you can predict which way the reaction will shift to reach equilibrium.

If Q < K, the reactant concentration is too high. The reaction will shift right (forward) to make more products.
If Q > K, the product concentration is too high. The reaction will shift left (backward) to make more reactants.
If Q = K, you’re already at equilibrium.

Relating Kp and Kc

You don’t need to calculate both from scratch. They are related by the ideal gas law. The formula is a direct consequence of P = (n/V)RT = [concentration]RT.

Kp = Kc (RT)^Δng

Here, R is the ideal gas constant (usually 0.0821 L atm/mol K), T is the absolute temperature in Kelvin, and Δng is the change in the number of moles of gas species only (moles of gaseous products - moles of gaseous reactants).

The Degree of Dissociation (α)

Alpha (α) is simply the fraction of a reactant that has dissociated at equilibrium. It’s a number between 0 and 1. If you start with ‘C’ moles of a reactant and α is the degree of dissociation, then at equilibrium, Cα moles have reacted, and C(1-α) moles remain. This is the foundation of the ICE (Initial, Change, Equilibrium) table, your single most powerful tool for solving equilibrium problems.

Gibbs Free Energy and Equilibrium

This connects equilibrium to thermodynamics. The standard Gibbs free energy change, ΔG°, tells you about the spontaneity of a reaction under standard conditions. Its relationship with K is fundamental.

ΔG° = -RT ln K

If K > 1, ln K is positive, and ΔG° is negative (spontaneous under standard conditions). If K < 1, ln K is negative, and ΔG° is positive (non-spontaneous under standard conditions). This formula is the bridge between two major chapters.

The van’t Hoff Equation

This equation tells you how the equilibrium constant K changes with temperature. It’s a quantitative application of Le Chatelier’s principle. The integrated form, which is what you’ll use most often for calculations, is:

ln(K2/K1) = (ΔH°/R) × (1/T1 - 1/T2)

Here, ΔH° is the standard enthalpy of reaction. For an endothermic reaction (ΔH° > 0), increasing temperature increases K. For an exothermic reaction (ΔH° < 0), increasing temperature decreases K.

Which formula, when

This is the strategic part. Knowing the formulas is useless if you don’t know when to apply them.

Situation	Formula to use	Watch out for
Given equilibrium concentrations, find K.	`Kc = [Products]^p / [Reactants]^r`	Use molarity (mol/L), not just moles.
Given equilibrium partial pressures, find K.	`Kp = (P products)^p / (P reactants)^r`	Use pressures in atm or bar consistently.
Need to convert between Kp and Kc.	`Kp = Kc (RT)^Δng`	`Δng` is for GASES ONLY. T must be in Kelvin.
Predicting the direction of a reaction.	Calculate Q and compare it to K.	Make sure you use the initial/current concentrations for Q.
Started with initial amounts, need equilibrium amounts.	Use an ICE table with the variable ‘x’ or ‘α’.	The change ‘x’ is always multiplied by the stoichiometric coefficient.
Given total pressure and α for a dissociation.	`Kp = (α^2 × Ptotal) / (1 - α^2)` (for A <=> B+C)	This specific form is only for a 1 -> 2 mole dissociation. Derive for others.
Temperature is changed from T1 to T2.	`ln(K2/K1) = (ΔH°/R) × (1/T1 - 1/T2)`	`ΔH°` must be in J/mol if R is 8.314 J/mol·K.
Connecting equilibrium to spontaneity.	`ΔG° = -RT ln K`	This is for `ΔG°` (standard), not `ΔG` at any random condition.

Mistakes that cost the most marks

Every year, students lose marks on the same handful of silly mistakes. Burn this list into your brain.

Forgetting coefficients become powers. You write 2[A] instead of [A]^2. This is a fatal, fundamental error.
- Why it happens: Rushing. Brain fart.
- The fix: After writing any K expression, double-check it against the balanced equation. Every coefficient must be an exponent.
Messing up Δng. You include solids, liquids, or aqueous species when calculating the change in moles of gas.
- Why it happens: Not paying attention to the state symbols (g), (l), (s), (aq).
- The fix: Circle every (g) in the equation. Count product moles. Count reactant moles. Subtract. That’s your Δng.
Using moles instead of concentration in Kc. The expression for Kc requires moles per liter (molarity).
- Why it happens: The ICE table gives you moles. You forget to divide by the volume of the container.
- The fix: Always ask: “What is the volume?” If it’s not 1 liter, you must divide the equilibrium moles by V.
Inert gas confusion. You can’t remember the difference between adding an inert gas at constant volume vs. constant pressure.
- Why it happens: It’s a nuanced point.
- The fix: Constant Volume: Total pressure increases, but partial pressures of reactants/products don’t change. No shift. Constant Pressure: To keep total P constant, the volume must increase. This lowers all partial pressures, shifting the equilibrium towards the side with more moles of gas.
Thinking a catalyst changes K. A catalyst speeds up both the forward and reverse reactions equally.
- Why it happens: Confusing kinetics (speed) with thermodynamics (position of equilibrium).
- The fix: A catalyst helps you reach equilibrium faster. It does NOT change the value of K or the final equilibrium position.
Including pure solids or liquids in the K expression. Their “concentration” or activity is considered constant (and equal to 1).
- Why it happens: Writing the expression mechanically without thinking.
- The fix: Before you write the K expression, cross out any species marked (s) or (l). They do not appear in the numerator or denominator.

Two worked examples (our own)

Let’s make this real.

Example 1: Calculating Kc and Kp

Problem: For the reaction N2(g) + 3H2(g) <=> 2NH3(g), a 2.0 L container at 400 K is found to contain 0.40 moles of N2, 0.50 moles of H2, and 0.20 moles of NH3 at equilibrium. Calculate Kc and Kp. (Use R = 0.08 L atm/mol K)

Solution:

Find concentrations:
- [N2] = 0.40 mol / 2.0 L = 0.20 M
- [H2] = 0.50 mol / 2.0 L = 0.25 M
- [NH3] = 0.20 mol / 2.0 L = 0.10 M
Choose formula and calculate Kc:
- Formula: Kc = [NH3]^2 / ([N2] [H2]^3)
- Kc = (0.10)^2 / ((0.20) × (0.25)^3)
- Kc = (0.01) / (0.20 × 0.015625)
- Kc = 0.01 / 0.003125 = 3.2
Choose formula and calculate Kp:
- Formula: Kp = Kc (RT)^Δng
- First, find Δng: (moles of gas products) – (moles of gas reactants) = 2 - (1 + 3) = -2.
- Kp = 3.2 × (0.08 × 400)^(-2)
- Kp = 3.2 × (32)^(-2)
- Kp = 3.2 / (32 × 32) = 3.2 / 1024 = 0.003125

Answer: Kc = 3.2 and Kp = 0.003125.

Example 2: Degree of Dissociation

Problem: 2 moles of PCl5(g) are placed in a container and heated. The reaction is PCl5(g) <=> PCl3(g) + Cl2(g). At equilibrium, the degree of dissociation (α) is found to be 0.4. The total pressure is 2 atm. Calculate Kp.

Solution:

Set up an ICE table using α:
- Reaction: PCl5(g) <=> PCl3(g) + Cl2(g)
- Initial moles: 2, 0, 0
- Change in moles: -2α, +2α, +2α
- Equilibrium moles: 2(1-α), 2α, 2α
Calculate moles of each gas at equilibrium:
- α = 0.4
- Moles PCl5 = 2(1 - 0.4) = 2 × 0.6 = 1.2 mol
- Moles PCl3 = 2 × 0.4 = 0.8 mol
- Moles Cl2 = 2 × 0.4 = 0.8 mol
- Total moles = 1.2 + 0.8 + 0.8 = 2.8 mol
Calculate partial pressures:
- Partial Pressure = Mole Fraction × Total Pressure
- P(PCl5) = (1.2 / 2.8) × 2 atm = 0.857 atm
- P(PCl3) = (0.8 / 2.8) × 2 atm = 0.571 atm
- P(Cl2) = (0.8 / 2.8) × 2 atm = 0.571 atm
Choose formula and calculate Kp:
- Formula: Kp = (P(PCl3) × P(Cl2)) / P(PCl5)
- Kp = (0.571 × 0.571) / 0.857
- Kp = 0.326 / 0.857 = 0.38

Answer: Kp = 0.38.

How to use these sheets with a problem book

Don’t just read passively. Integrate these tools into your workflow.

Prime: Before starting a problem set, spend 10 minutes reading the Formula Sheet. This loads the necessary tools into your short-term memory.
Attack: Attempt the problems from your book (like N Awasthi). Solve them as if you’re in an exam. Don’t look at solutions yet.
Debug: For every question you get wrong, open the Mistakes and Traps sheet. Identify which specific trap you fell into. Understanding why you were wrong is more important than knowing the right answer.
Consolidate: Once a week, or before a test, use the Cheat Sheet for a rapid 15-minute review of the entire chapter’s strategy.

A three-session plan for chemical equilibrium

This is a big chapter. Break it down into focused 60-90 minute sessions.

Session 1: Concepts and Tools. Your goal is to understand the “what” and “why”. Read the NCERT chapter. Watch a good lecture. Then, read our “plain words” section above and the Formula Sheet. Don’t solve many problems yet; just focus on understanding the rules of the game.
Session 2: Core Problem Solving. Now, you execute. Open your problem book and start with the Level 1 or basic questions. Use an ICE table for everything. If you get stuck, it’s okay to look at a solved example to understand the pattern. Your goal is to build confidence and competence with the main problem types.
Session 3: Advanced Strategy and Revision. Tackle the harder problems—those involving multiple equilibria, van’t Hoff equation, or tricky Le Chatelier scenarios. After the session, use our Cheat Sheet to revise the entire chapter’s logic in 15 minutes. This locks in the learning.

Quick self-test

Test yourself. No peeking.

For the exothermic reaction 2SO2(g) + O2(g) <=> 2SO3(g), what happens to K and the position of equilibrium if you increase the temperature?
In a 1L flask, 0.5 mol of A and 0.5 mol of B react via A(g) + B(g) <=> 2C(g). At equilibrium, 0.2 mol of A remains. What is the value of Kc?
What does an equilibrium constant K = 1.5 × 10^-12 tell you about a reaction?
If the Kp for N2O4(g) <=> 2NO2(g) is X, what is the Kp for NO2(g) <=> 1/2 N2O4(g)?
If you add an inert gas like Argon to an equilibrium mixture at constant pressure, which way will the equilibrium PCl5(g) <=> PCl3(g) + Cl2(g) shift?

Answers:

K will decrease, and the equilibrium will shift to the left (reactants).
[A]eq=0.2, [B]eq=0.2, [C]eq=2×(0.5-0.2)=0.6. Kc = (0.6)^2 / (0.2×0.2) = 0.36 / 0.04 = 9.
The reaction strongly favors the reactants. At equilibrium, the mixture will be almost entirely reactants.
The reaction is reversed (1/X) and halved (sqrt). So the new Kp is 1/sqrt(X) or X^(-1/2).
To the right, towards the side with more moles of gas (2 moles vs 1 mole).

FAQ

What is the relation between Kp and Kc?
They are related by the formula Kp = Kc(RT)^Δng. The key is Δng, the change in moles of gas only. If Δng is zero, Kp equals Kc.

Does a catalyst change the equilibrium constant?
No. Absolutely not. A catalyst increases the rate of both the forward and reverse reactions equally. It makes the system reach equilibrium faster, but it does not change the final ratio of products to reactants (K).

What happens when an inert gas is added at constant volume?
Nothing happens to the equilibrium. The total pressure increases, but the partial pressures (and concentrations) of the reacting gases do not change. Since Q doesn’t change, there’s no shift.

How do I use the reaction quotient Q?
Calculate Q using the current, non-equilibrium concentrations or pressures in the same expression as K. If Q < K, the reaction shifts right (towards products). If Q > K, it shifts left (towards reactants). It’s a GPS for your reaction.

Are these PDFs free to download?
Yes, all our resource sheets are completely free for personal use. The goal is to get the best tools into your hands. Just don’t sell them or re-host them.

What’s the real difference between adding an inert gas at constant volume vs. constant pressure?
At constant volume, you’re just adding more molecules to the box, increasing total pressure, but the “space” for each original molecule hasn’t changed. At constant pressure, the system must expand its volume to accommodate the new gas molecules. This expansion dilutes everything, causing a shift towards the side with more moles to counteract the dilution.

When can I use the small-alpha approximation (1-α ≈ 1)?
You can try to use it when K is very small (typically less than 10^-4). The rule is: solve for α assuming it’s small. Then, check if your result is valid. If α is less than 5% (α < 0.05) of the initial concentration, the approximation is good. If not, you must solve the full quadratic equation.

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