NEET Chemistry: Equilibrium – Complete Revision Guide, Formulas & Solved PYQs

Introduction to Equilibrium for NEET UG

Equilibrium is one of the most vital physical chemistry topics in the NTA NEET UG syllabus. Divided into Chemical Equilibrium and Ionic Equilibrium, this chapter bridges fundamental concepts of thermodynamics, chemical kinetics, and solution chemistry. Understanding dynamic equilibrium is essential for predicting reaction directions, evaluating yield in industrial processes (like Haber's synthesis of ammonia), and explaining biological processes such as oxygen transport by hemoglobin and blood buffer maintenance.

NEET Weightage & Expected Questions

Equilibrium carries high weightage in the NEET examination, accounting for 3 to 4 questions (12-16 marks) annually across Class 11 Physical Chemistry:

• Chemical Equilibrium: 1-2 Questions (Focus on $K_p$ vs $K_c$, Le Chatelier's Principle, and Degree of Dissociation $\alpha$).
• Ionic Equilibrium: 2 Questions (Focus on pH calculations, Buffer Solutions, Salt Hydrolysis, and Solubility Product $K_{sp}$).

Securing full marks in this chapter requires mastering formula applications, logarithmic calculations for pH, and avoiding algebraic traps.

Core Concepts & Key Mechanisms Explained

1. Dynamic Nature of Chemical Equilibrium

At equilibrium, the rate of forward reaction ($R_f$) equals the rate of backward reaction ($R_b$). The concentrations of reactants and products remain constant over time, though the reaction remains dynamic on a molecular scale.

For a reversible gaseous reaction: $aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$

The equilibrium constants are defined as:

$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ and $K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$

The relation between $K_p$ and $K_c$ is given by:

$K_p = K_c(RT)^{\Delta n_g}$

where $\Delta n_g = (c + d) - (a + b)$ represents the change in moles of gaseous species, $R = 0.0821\text{ L atm K}^{-1}\text{mol}^{-1}$, and $T$ is temperature in Kelvin.

2. Reaction Quotient ($Q$) and Le Chatelier's Principle

The Reaction Quotient ($Q$) has the same mathematical expression as $K_{eq}$, but is evaluated at non-equilibrium states:

• If $Q < K_{eq}$: Reaction proceeds in the forward direction.
• If $Q = K_{eq}$: Reaction is at equilibrium.
• If $Q > K_{eq}$: Reaction proceeds in the reverse direction.

Le Chatelier's Principle: If a system at equilibrium is subjected to a change in concentration, pressure, temperature, or volume, the system shifts in a direction that counteracts the applied change.

• Concentration: Adding reactants shifts equilibrium forward; adding products shifts equilibrium backward.
• Pressure/Volume: Increasing total pressure shifts equilibrium toward the side with fewer gaseous moles ($\Delta n_g$).
• Temperature: Increasing temperature shifts equilibrium toward the endothermic direction ($\Delta H > 0$).
• Inert Gas Addition: At constant volume, inert gas has no effect. At constant pressure, inert gas addition shifts equilibrium toward greater gaseous moles.

3. Ionic Equilibrium & pH Calculations

According to Ostwald's Dilution Law for a weak electrolyte ($HA$):

$HA \rightleftharpoons H^+ + A^-$

$K_a = \frac{C\alpha^2}{1 - \alpha} \approx C\alpha^2 \implies \alpha = \sqrt{\frac{K_a}{C}}$

Hydronium ion concentration: $[H^+] = C\alpha = \sqrt{K_a C}$

pH is defined as: $\text{pH} = -\log_{10}[H^+]$

For water at $25^\circ\text{C}$: $K_w = [H^+][OH^-] = 10^{-14}$, giving $\text{pH} + \text{pOH} = 14$.

4. Salt Hydrolysis & Buffer Solutions

Depending on the constituent acid and base, salts behave differently in water:

• Weak Acid + Strong Base (e.g., $CH_3COONa$): Basic solution ($ \text{pH} > 7$). $\text{pH} = 7 + \frac{1}{2}\text{pK}_a + \frac{1}{2}\log C$
• Strong Acid + Weak Base (e.g., $NH_4Cl$): Acidic solution ($ \text{pH} < 7$). $\text{pH} = 7 - \frac{1}{2}\text{pK}_b - \frac{1}{2}\log C$
• Weak Acid + Weak Base (e.g., $CH_3COONH_4$): Independent of concentration. $\text{pH} = 7 + \frac{1}{2}\text{pK}_a - \frac{1}{2}\text{pK}_b$

Buffer Solutions: Solutions that resist changes in pH upon addition of small amounts of acid or base.

• Acidic Buffer: Weak acid + its conjugate salt (e.g., $CH_3COOH + CH_3COONa$). Handerson-Hasselbalch Equation: $\text{pH} = \text{pK}_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}$
• Basic Buffer: Weak base + its conjugate salt (e.g., $NH_4OH + NH_4Cl$). $\text{pOH} = \text{pK}_b + \log\frac{[\text{Salt}]}{[\text{Base}]}$

5. Solubility Product ($K_{sp}$) & Precipitation

For a sparingly soluble salt $A_x B_y \rightleftharpoons x A^{y+} + y B^{x-}$:

$K_{sp} = [A^{y+}]^x [B^{x-}]^y = (xs)^x (ys)^y = x^x y^y s^{x+y}$

• If Ionic Product ($I_{IP}$) $> K_{sp}$: Precipitation occurs.
• If $I_{IP} = K_{sp}$: Saturated solution at equilibrium.
• If $I_{IP} < K_{sp}$: Unsaturated solution (no precipitate).

Important Formulas & Key Terms Table

Solved Step-by-Step Previous Years Questions (PYQs)

PYQ 1 (NEET): For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $K_c = 64$ at a specific temperature. What is the value of $K_c'$ for $NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$?

Solution:

1. Original reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ with constant $K_c$.

2. Reverse reaction: $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$ has constant $K_c'' = \frac{1}{K_c}$.

3. Multiplying the reversed equation by $\frac{1}{2}$ gives the target equation: $NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$.

4. The new equilibrium constant $K_c' = (K_c'')^{1/2} = \left(\frac{1}{K_c}\right)^{1/2} = \frac{1}{\sqrt{64}} = \frac{1}{8} = 0.125$.

Correct Answer: $0.125$ (or $1/8$)

PYQ 2 (NEET): Find the pH of a solution formed by mixing $50\text{ mL}$ of $0.1\text{ M } CH_3COOH$ and $50\text{ mL}$ of $0.05\text{ M } NaOH$. Given $\text{pK}_a(CH_3COOH) = 4.74$.

Solution:

1. Calculate millimoles of reactants:

Millimoles of $CH_3COOH = 50 \times 0.1 = 5\text{ mmol}$

Millimoles of $NaOH = 50 \times 0.05 = 2.5\text{ mmol}$

2. Reaction: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

• Initial mmol: $5\text{ mmol } CH_3COOH$, $2.5\text{ mmol } NaOH$, $0\text{ mmol } CH_3COONa$
• Final mmol: $2.5\text{ mmol } CH_3COOH$, $0\text{ mmol } NaOH$, $2.5\text{ mmol } CH_3COONa$

3. The resulting system contains a weak acid and its salt, forming an acidic buffer.

4. Using Henderson-Hasselbalch equation:

$\text{pH} = \text{pK}_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} = 4.74 + \log\left(\frac{2.5}{2.5}\right) = 4.74 + \log(1) = 4.74 + 0 = 4.74$.

Correct Answer: $4.74$

PYQ 3 (NEET): The solubility of $AgCl$ in water is $1.0 \times 10^{-5}\text{ mol/L}$. What is its solubility in $0.1\text{ M } AgNO_3$ solution due to the common ion effect?

Solution:

1. Find $K_{sp}$ of $AgCl$ from pure water solubility ($s = 1.0 \times 10^{-5}$):

$K_{sp} = s^2 = (1.0 \times 10^{-5})^2 = 1.0 \times 10^{-10}$.

2. In $0.1\text{ M } AgNO_3$, $[Ag^+] = s' + 0.1 \approx 0.1\text{ M}$ (since new solubility $s' \ll 0.1$).

3. Express $K_{sp} = [Ag^+][Cl^-] = (0.1)(s')$.

4. Solve for $s'$: $s' = \frac{1.0 \times 10^{-10}}{0.1} = 1.0 \times 10^{-9}\text{ mol/L}$.

Correct Answer: $1.0 \times 10^{-9}\text{ mol/L}$

Common NEET Traps & Mistakes to Avoid

• Ignoring Concentration Changes in Very Dilute Acids: For $10^{-8}\text{ M } HCl$, students often wrongly answer $\text{pH} = 8$. Remember that acid solutions cannot have a basic pH! Auto-ionization of water must be included ($[H^+]_{water} = 10^{-7}\text{ M}$), leading to total $[H^+] = 1.08 \times 10^{-7}\text{ M}$ and $\text{pH} \approx 6.96$.
• Confusing Temperature Dependencies: Equilibrium constants ($K_p, K_c, K_{sp}, K_w$) change only with temperature. Catalyst addition, pressure changes, or initial concentration modifications do NOT alter equilibrium constants.
• Forgetting Gas Mole Changes ($\Delta n_g$): Always count only gaseous moles when calculating $\Delta n_g$. Solids and pure liquids have zero contribution to $\Delta n_g$.
• Incorrect $K_{sp}$ Expressions for Multi-Ion Salts: For $Al_2(SO_4)_3$, $x=2$ and $y=3$. The correct formula is $K_{sp} = 2^2 \cdot 3^3 \cdot s^{2+3} = 108 s^5$, not simply $s^5$ or $6s^5$.

High-Yield NEET Practice MCQs with Answer Keys

Q1. Which of the following conditions favors maximum yield of $NH_3$ in Haber's Process: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92.4\text{ kJ}$?

A) High temperature and high pressure
B) Low temperature and low pressure
C) Low temperature and high pressure
D) High temperature and low pressure

Answer: C
Explanation: Exothermic reactions ($\\Delta H < 0$) are favored by lower temperatures. Since the forward reaction reduces gaseous moles from 4 to 2, high pressure shifts equilibrium forward according to Le Chatelier's Principle.

Q2. Calculate the pH of a $0.01\text{ M } Ba(OH)_2$ solution assuming complete dissociation.

A) 12.0
B) 12.3
C) 1.7
D) 2.0

Answer: B
Explanation: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$. Thus $[OH^-] = 2 \times 0.01 = 0.02 = 2 \times 10^{-2}\text{ M}$.
$\text{pOH} = -\log(2 \times 10^{-2}) = 2 - \log 2 = 2 - 0.301 = 1.699 \approx 1.7$.
$\text{pH} = 14 - \text{pOH} = 14 - 1.7 = 12.3$.

Q3. Conjugate base of $HPO_4^{2-}$ is:

A) $H_2PO_4^-$
B) $PO_4^{3-}$
C) $H_3PO_4$
D) $H_2PO_4^{2-}$

Answer: B
Explanation: A conjugate base is formed by removing a proton ($H^+$) from an acid. Removing $H^+$ from $HPO_4^{2-}$ yields $PO_4^{3-}$.

Q4. The $K_{sp}$ of $Zr_3(PO_4)_4$ in terms of its solubility ($s$) is:

A) $108 s^7$
B) $6912 s^7$
C) $144 s^5$
D) $256 s^7$

Answer: B
Explanation: $Zr_3(PO_4)_4 \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}$.
$K_{sp} = (3s)^3 (4s)^4 = 27s^3 \times 256s^4 = 6912 s^7$.

Q5. Addition of $NH_4Cl$ to an aqueous solution of $NH_4OH$ results in:

A) Increase in $[OH^-]$
B) Decrease in pH
C) Increase in degree of ionization of $NH_4OH$
D) No change in pH

Answer: B
Explanation: Adding $NH_4Cl$ supplies $NH_4^+$ ions, which suppress the ionization of weak base $NH_4OH$ due to the Common Ion Effect. This decreases $[OH^-]$, making the solution less basic and lowering the pH.

Summary & Final NEET Revision Tips

• Master Logarithmic Shortcuts: Memorize values like $\log 2 = 0.301$, $\log 3 = 0.477$, $\log 5 = 0.699$, and $\log 7 = 0.845$ to solve pH questions within seconds during the exam.
• Identify Buffer Systems Instantly: Look for partial neutralization of weak species (e.g., Weak Acid + half-equivalent Strong Base creates a buffer with maximum capacity where $\text{pH} = \text{pK}_a$).
• Remember Unit Dependencies: $K_c$ units are $(\text{mol L}^{-1})^{\Delta n_g}$ while $K_p$ units are $(\text{atm})^{\Delta n_g}$. If $\Delta n_g = 0$, both equilibrium constants are dimensionless and $K_p = K_c$.
• Focus on Temperature Trends: Endothermic reactions ($\\Delta H > 0$) show increased $K_{eq}$ at higher temperatures, whereas exothermic reactions ($\\Delta H < 0$) show decreased $K_{eq}$ at higher temperatures.