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Equilibrium Formulas For JEE 2027
Understanding Equilibrium requires clarity on both chemical and ionic equilibrium. Focus on equilibrium constants, Le Chatelier’s principle, acid–base equilibria, pH, buffer solutions, and solubility product. Use a JEE Formula Sheet PDF to revise key equations, paying attention to the assumptions and conditions required for each formula.
Strengthen your preparation by solving previous-year questions on reaction direction, equilibrium composition, and precipitation. Take a Free JEE Mains mock test to assess how accurately you apply these concepts under time limits. Review errors in calculations, approximations, and formula selection to make your revision more focused and improve your confidence in this chapter.
Chemical Equilibrium
Equilibrium Constant $$K_c$$
For the reaction: $$aA + bB \rightleftharpoons cC + dD$$
$$$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$$
- $$K_c$$ is constant at a given temperature
- $$K_c$$ changes only with temperature
- Pure solids and pure liquids are not included (activity = 1)
$$K_p$$ and Its Relation to $$K_c$$
$$$K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$$$
$$$K_p = K_c (RT)^{\Delta n_g}$$$
where $$\Delta n_g = (c + d) - (a + b)$$ = change in moles of gas.
- If $$\Delta n_g = 0$$: $$K_p = K_c$$
- If $$\Delta n_g > 0$$: $$K_p > K_c$$
- If $$\Delta n_g < 0$$: $$K_p < K_c$$
Worked Example: $$K_p$$ from $$K_c$$
For $$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$, $$K_c = 0.5$$ at 400 K. Find $$K_p$$.
$$\Delta n_g = 2 - 4 = -2$$
$$K_p = 0.5 \times (0.0821 \times 400)^{-2} = 0.5 \times (32.84)^{-2}$$
Rules for Manipulating Equilibrium Constants
- Reaction reversed: $$K' = \dfrac{1}{K}$$
- Reaction multiplied by $$n$$: $$K' = K^n$$
- Reaction divided by $$n$$: $$K' = K^{1/n}$$
- Two reactions added: $$K' = K_1 \times K_2$$
Reaction Quotient
Reaction Quotient and Predicting Direction
$$$Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \quad \text{(using current concentrations)}$$$
- $$Q < K$$: Reaction proceeds forward
- $$Q = K$$: System is at equilibrium
- $$Q > K$$: Reaction proceeds backward
Le Chatelier's Principle
Effects of Different Changes on Equilibrium
| Change Applied | Shift Direction | Effect on $$K$$ |
|---|---|---|
| Add reactant | Forward | No change |
| Remove product | Forward | No change |
| Add product | Backward | No change |
| Increase pressure | Toward fewer moles of gas | No change |
| Increase $$T$$ (exothermic) | Backward | $$K$$ decreases |
| Increase $$T$$ (endothermic) | Forward | $$K$$ increases |
| Add catalyst | No shift | No change |
| Add inert gas (const. $$V$$) | No shift | No change |
Important Note
A catalyst does not shift equilibrium — it speeds up both forward and reverse reactions equally. Temperature is the only factor that changes $$K$$.
Degree of Dissociation
Calculating $$K$$ from Degree of Dissociation ($$\alpha$$)
For $$\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$$, starting with $$a$$ moles in volume $$V$$:
$$$K_c = \frac{a\alpha^2}{V(1-\alpha)}$$$
Total moles at equilibrium = $$a(1 + \alpha)$$
Worked Example: Degree of Dissociation
1 mol of PCl$$_5$$ in a 1 L container, $$\alpha = 0.2$$. Find $$K_c$$.
At equilibrium: PCl$$_5$$ = 0.8 mol, PCl$$_3$$ = 0.2 mol, Cl$$_2$$ = 0.2 mol
$$K_c = \dfrac{(0.2)(0.2)}{0.8} = \dfrac{0.04}{0.8} = \textbf{0.05}$$
Ionic Equilibrium: pH and pO
Theories of Acids and Bases
| Theory | Acid | Base |
|---|---|---|
| Arrhenius | Gives H$$^+$$ in water | Gives OH$$^-$$ in water |
| Brønsted–Lowry | Proton (H$$^+$$) donor | Proton (H$$^+$$) acceptor |
| Lewis | Electron pair acceptor | Electron pair donor |
Ionic Product of Water
$$$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \quad \text{at 25°C}$$$
In pure water: $$[\text{H}^+] = [\text{OH}^-] = 1.0 \times 10^{-7}$$ M
pH and pOH
$$$\text{pH} = -\log[\text{H}^+] \qquad \text{pOH} = -\log[\text{OH}^-]$$$
$$$\text{pH} + \text{pOH} = 14 \quad \text{(at 25°C)}$$$
- pH $$< 7$$: Acidic
- pH $$= 7$$: Neutral
- pH $$> 7$$: Basic
Worked Example: pH of Strong Acid
Find the pH of 0.01 M HCl.
HCl ionizes completely: $$[\text{H}^+] = 0.01 = 10^{-2}$$ M
$$\text{pH} = -\log(10^{-2}) = \textbf{2}$$
Weak Acid and Weak Base Ionization
Ionization of Weak Acid ($$K_a$$)
For HA $$\rightleftharpoons$$ H$$^+$$ + A$$^-$$:
$$$K_a = \frac{c\alpha^2}{1 - \alpha}$$$
When $$\alpha \ll 1$$:
$$$K_a \approx c\alpha^2 \qquad \alpha \approx \sqrt{K_a / c} \qquad [\text{H}^+] = \sqrt{K_a \cdot c}$$$
Ionization of Weak Base ($$K_b$$)
When $$\alpha \ll 1$$:
$$$K_b \approx c\alpha^2 \qquad [\text{OH}^-] = \sqrt{K_b \cdot c}$$$
Conjugate pair relation:
$$$K_a \times K_b = K_w = 10^{-14} \qquad \text{p}K_a + \text{p}K_b = 14$$$
Worked Example: pH of Weak Acid
Find the pH of 0.1 M acetic acid ($$K_a = 1.8 \times 10^{-5}$$).
$$[\text{H}^+] = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}$$ M
$$\text{pH} = -\log(1.34 \times 10^{-3}) = 3 - 0.127 = \textbf{2.87}$$
Tip: Ostwald's dilution law: on dilution, $$c$$ decreases so $$\alpha$$ increases (more ionization) but $$[\text{H}^+]$$ decreases (pH increases toward 7). JEE tests this frequently.
Buffer Solutions
Henderson–Hasselbalch Equation
For acidic buffer (weak acid + salt):
$$$\text{pH} = \text{p}K_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}$$$
For basic buffer (weak base + salt):
$$$\text{pOH} = \text{p}K_b + \log \frac{[\text{Salt}]}{[\text{Base}]}$$$
Worked Example: Buffer pH
pH of buffer with 0.1 M CH$$_3$$COOH and 0.1 M CH$$_3$$COONa ($$\text{p}K_a = 4.74$$).
$$\text{pH} = 4.74 + \log\dfrac{0.1}{0.1} = 4.74 + 0 = \textbf{4.74}$$
When [Salt] = [Acid], pH equals pK$$_a$$.
Solubility Product
Solubility Product Calculations
For a salt $$A_xB_y$$ with molar solubility $$s$$:
$$$K_{sp} = x^x \cdot y^y \cdot s^{(x+y)}$$$
Common cases:
- AB type (AgCl): $$K_{sp} = s^2$$, so $$s = \sqrt{K_{sp}}$$
- AB$$_2$$ type (PbCl$$_2$$): $$K_{sp} = 4s^3$$, so $$s = (K_{sp}/4)^{1/3}$$
- A$$_2$$B type (Ag$$_2$$CrO$$_4$$): $$K_{sp} = 4s^3$$
- AB$$_3$$ type: $$K_{sp} = 27s^4$$
Predicting Precipitation
- IP $$< K_{sp}$$: No precipitate (unsaturated)
- IP $$= K_{sp}$$: At equilibrium (saturated)
- IP $$> K_{sp}$$: Precipitation occurs
Worked Example: Solubility and Common Ion Effect
$$K_{sp}$$ of AgCl = $$1.8 \times 10^{-10}$$. Find solubility in pure water and in 0.1 M NaCl.
In pure water: $$s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$$ M
In 0.1 M NaCl: $$s = K_{sp}/0.1 = 1.8 \times 10^{-9}$$ M
Solubility drops by ~7500 times due to the common ion effect.
Hydrolysis of Salts
pH Formulas for Salt Hydrolysis
Weak acid + Strong base (e.g., CH$$_3$$COONa):
$$$\text{pH} = 7 + \frac{1}{2}\text{p}K_a + \frac{1}{2}\log c$$$
Strong acid + Weak base (e.g., NH$$_4$$Cl):
$$$\text{pH} = 7 - \frac{1}{2}\text{p}K_b - \frac{1}{2}\log c$$$
Weak acid + Weak base (e.g., CH$$_3$$COONH$$_4$$):
$$$\text{pH} = 7 + \frac{1}{2}\text{p}K_a - \frac{1}{2}\text{p}K_b$$$
Worked Example: Salt Hydrolysis pH
Find pH of 0.1 M sodium acetate ($$\text{p}K_a = 4.74$$).
$$\text{pH} = 7 + \frac{1}{2}(4.74) + \frac{1}{2}\log(0.1) = 7 + 2.37 - 0.5 = \textbf{8.87}$$
Summary of Key Formulas
Quick Reference — Chemical Equilibrium
- $$K_c = \dfrac{[\text{Products}]}{[\text{Reactants}]}$$ (each raised to stoichiometric power)
- $$K_p = K_c(RT)^{\Delta n_g}$$
- Reversed: $$K' = 1/K$$; multiplied by $$n$$: $$K' = K^n$$
Quick Reference — Ionic Equilibrium
- $$K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$$; pH + pOH = 14
- Weak acid: $$[\text{H}^+] = \sqrt{K_a \cdot c}$$; Weak base: $$[\text{OH}^-] = \sqrt{K_b \cdot c}$$
- $$K_a \times K_b = K_w$$ for conjugate pairs
- Henderson–Hasselbalch: $$\text{pH} = \text{p}K_a + \log\dfrac{[\text{Salt}]}{[\text{Acid}]}$$
- $$K_{sp} = [A^{y+}]^x[B^{x-}]^y$$
Equilibrium Formulas For JEE 2027: Conclusion
A strong understanding of Equilibrium comes from connecting equilibrium constants, reaction quotient, and Le Chatelier’s principle with their applications. For ionic equilibrium, practise pH calculations, buffer problems, and solubility product questions, checking the assumptions behind each calculation. Use a JEE Mains Chemistry formula sheet to revise key equations and understand when to apply them.
When revising JEE Mains Chemistry important chapters, give Equilibrium regular practice time. Solve previous-year questions, review mistakes, and revisit concepts that cause confusion, such as predicting reaction direction or deciding when precipitation occurs. Consistent revision and timed practice can help you approach equilibrium problems with greater accuracy and confidence.
