Chemical EquilibriumJEE Main

Concept

At equilibrium the forward and reverse rates are equal, so concentrations stop changing. The equilibrium constant $K_c$ fixes how far a reaction proceeds; for a dissociation it sets the degree of dissociation $\alpha$.

Key formula

Derivation

For $A \rightleftharpoons B + C$ starting at $C_0$, equilibrium gives $[A] = C_0(1-\alpha)$ and $[B]=[C]=C_0\alpha$. Substituting into $K_c$ yields $K_c = \dfrac{C_0\alpha^2}{1-\alpha}$, a quadratic in $\alpha$.

Le Chatelier's principle predicts the shift: raising $C_0$ (pressure) suppresses $\alpha$, while a larger $K_c$ (often from higher temperature for endothermic reactions) increases it.

Scenarios to explore

• Chemical Equilibrium — Degree of dissociation from Kc.

Real-world applications

• Industrial yields (Haber, contact processes).
• Weak-acid/base dissociation and solubility equilibria.
• Ostwald's dilution law (α rises on dilution).

JEE exam tips

• Ostwald's law: for small $\alpha$, $\alpha \approx \sqrt{K_c/C_0}$ — dilution increases dissociation.
• $Q < K_c$ shifts forward, $Q > K_c$ shifts backward, $Q = K_c$ is equilibrium.

Common mistakes

• Forgetting $[B]$ and $[C]$ are each $C_0\alpha$, so the numerator is $(C_0\alpha)^2$.
• Using the approximation $K_c \approx C_0\alpha^2$ when $\alpha$ is not small.
• Confusing $K_c$ (constant at fixed T) with $Q$ (varies until equilibrium).

Exam traps to avoid

• $K_c$ depends only on temperature — adding catalyst or changing $C_0$ does not change it.
• Degree of dissociation rises on dilution even though $K_c$ stays fixed.