Equilibrium Class 11 Formulas

This page collects the equilibrium class 11 formulas from the rationalised NCERT Chemistry Part I textbook (chapter 6) into one place: the equilibrium constant \( K_c \), its pressure form \( K_p \), the reaction quotient \( Q \), the Gibbs-energy link \( \Delta G^\ominus = -RT\ln K \), and the ionic-equilibrium set — \( K_w \), pH, \( pOH \), \( K_a \), \( K_b \) and degree of ionisation \( \alpha \).

Each formula is grouped under the textbook topic that introduces it, followed by a table of symbol meanings with units, a when-to-use guide, three original worked examples and chapter-specific mistakes to avoid. For the derivations and full explanations behind these equations, see the class 11 chemistry formulas hub; the wider chemistry formulas index links every chapter’s sheet.

Formulas at a Glance

All Formulas, Grouped by Topic

Equilibria in Physical Processes

Vapour pressure is constant at a given temperature, and a saturated solution has a constant solute concentration. For gas dissolved in liquid, Henry’s law states that the mass of a gas dissolved in a given mass of solvent at any temperature is proportional to the pressure of the gas above the solvent (NCERT, p. 172):

\[ CO_2(g) \rightleftharpoons CO_2(aq) \qquad \text{and} \qquad \frac{[CO_2(aq)]}{[CO_2(g)]} = \text{constant at fixed } T \]

Law of Chemical Equilibrium

For the general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium law states that the product of the equilibrium concentrations of products, each raised to its stoichiometric coefficient, divided by the product of the reactant concentrations raised to their coefficients, is a constant at a given temperature (NCERT, p. 175):

\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]

Concentrations in square brackets are always equilibrium values; phase symbols \( (g), (l), (s), (aq) \) are dropped when writing the expression.

Reverse reaction: if the equation is written the other way, the constant is the reciprocal.

\[ K’_c = \frac{1}{K_c} \]

Equation multiplied by \( n \): the constant is raised to that power (NCERT, p. 177).

\[ K”_c = (K_c)^n \]

Equilibrium Constant in Gaseous Systems

For reactions involving gases it is often more convenient to express the constant in partial pressures. Because \( p = [\text{gas}]RT \) with \( R = 0.0831 \) bar L mol⁻¹ K⁻¹, the two constants are related by (NCERT, p. 178):

\[ K_p = K_c(RT)^{\Delta n} \]

\[ \Delta n = (\text{moles of gaseous products}) – (\text{moles of gaseous reactants}) \]

For \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \), \( \Delta n = 0 \) so \( K_p = K_c \). Pressure must be quoted in bar, the standard state for pressure.

Heterogeneous Equilibria

Pure solids and pure liquids have a constant molar concentration, independent of the amount present, so they are left out of the equilibrium expression (NCERT, p. 180). For the thermal decomposition of calcium carbonate:

\[ CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) \qquad K_c = [CO_2], \; K_p = p_{CO_2} \]

Similarly, for \( Ag_2O(s) + 2HNO_3(aq) \rightleftharpoons 2AgNO_3(aq) + H_2O(l) \), \[ K_c = \frac{[AgNO_3]^2}{[HNO_3]^2} \]

Reaction Quotient: Extent and Direction

The reaction quotient has the same form as \( K_c \), but the concentrations are those at some arbitrary time \( t \), not necessarily at equilibrium (NCERT, p. 183):

\[ Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]

• \( Q_c \lt K_c \) — net reaction proceeds forward (toward products).
• \( Q_c \gt K_c \) — net reaction proceeds in reverse (toward reactants).
• \( Q_c = K_c \) — the mixture is already at equilibrium.

The magnitude of \( K \) also predicts the extent of reaction (NCERT, p. 182): if \( K_c \gt 10^3 \) products predominate and the reaction runs nearly to completion; if \( K_c \lt 10^{-3} \) reactants predominate; between \( 10^{-3} \) and \( 10^3 \), appreciable amounts of both are present. A large \( K \) says nothing about the rate at which equilibrium is reached.

Equilibrium Constant and Gibbs Energy

The constant is tied to thermodynamics through (NCERT, p. 184):

\[ \Delta G = \Delta G^\ominus + RT\ln Q \]

At equilibrium \( \Delta G = 0 \) and \( Q = K \), which gives:

\[ \Delta G^\ominus = -RT\ln K \qquad \text{or} \qquad K = e^{-\Delta G^\ominus / RT} \]

So \( \Delta G^\ominus \lt 0 \) implies \( K \gt 1 \) (products predominate), and \( \Delta G^\ominus \gt 0 \) implies \( K \lt 1 \) (reactants predominate).

Ionic Product of Water and the pH Scale

Water auto-ionises: \( 2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq) \). Because water is a pure liquid, \( [H_2O] \) is constant and is folded into the constant called the ionic product of water (NCERT, p. 194):

\[ K_w = [H^+][OH^-] = 1\times 10^{-14} \text{ M}^2 \quad \text{at 298 K} \]

The pH scale converts hydrogen-ion concentration to a logarithmic number; for dilute solutions (below 0.01 M) activity equals molarity:

\[ pH = -\log[H^+] \qquad pOH = -\log[OH^-] \qquad pK_w = pH + pOH = 14 \text{ at 298 K} \]

Because the scale is logarithmic, a change of one pH unit means \( [H^+] \) changes by a factor of 10.

Ionization of Weak Acids

For a weak acid \( HX \rightleftharpoons H^+ + X^- \), the acid ionisation constant is (NCERT, p. 196):

\[ K_a = \frac{[H^+][X^-]}{[HX]} \qquad pK_a = -\log K_a \]

With initial concentration \( c \) and degree of ionisation \( \alpha \):

\[ K_a = \frac{c\alpha^2}{1-\alpha} \]

The larger \( K_a \), the stronger the acid at that temperature. Percent dissociation measures strength directly (NCERT, p. 197):

\[ \text{Percent dissociation} = \frac{[HA]_{\text{dissociated}}}{[HA]_{\text{initial}}} \times 100\% \]

Ionization of Weak Bases

For a weak base \( MOH \rightleftharpoons M^+ + OH^- \), the base ionisation constant is (NCERT, p. 198):

\[ K_b = \frac{[M^+][OH^-]}{[MOH]} \qquad pK_b = -\log K_b \]

Again with degree of ionisation \( \alpha \):

\[ K_b = \frac{c\alpha^2}{1-\alpha} \]

What Each Symbol Means

When to Use Each Formula

Worked Examples

Example 1: Computing \( K_c \) directly

For \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) at 450 K, the equilibrium concentrations are \( [N_2] = 0.20 \) M, \( [H_2] = 0.60 \) M and \( [NH_3] = 0.050 \) M. Find \( K_c \).

Example 2: Working backwards — degree of ionisation and pH of a weak acid

A 0.10 M solution of a weak acid HA has \( K_a = 4.0\times 10^{-6} \). Find the degree of ionisation \( \alpha \) and the pH.

Example 3: Converting between \( K_c \) and \( K_p \)

For \( 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \), \( K_c = 3.0\times 10^{-3} \) at 800 K. Calculate \( K_p \).

Common Mistakes to Avoid

Frequently Asked Questions

When do I use \( Q \) instead of \( K \)?

Use \( Q \) whenever the system is not necessarily at equilibrium — you have concentrations measured at some arbitrary time and want to predict which way the reaction will shift. \( K \) is used only when the concentrations are genuine equilibrium values.

Comparing them decides the direction: \( Q \lt K \) moves forward, \( Q \gt K \) moves in reverse, \( Q = K \) means equilibrium has been reached.

Why are solids and pure liquids left out of the equilibrium expression?

Because their molar concentration is constant regardless of how much is present. Including a constant in both nature and the expression would not change the ratio, so it is absorbed into the equilibrium constant. That is why the \( CaCO_3 \) decomposition reduces to \( K_p = p_{CO_2} \) — the only species whose amount can actually vary is the gas.

What is the difference between \( K_a \) and \( pK_a \), and when is the approximation valid?

\( pK_a = -\log K_a \) is just the logarithmic way of quoting the same quantity — a smaller \( pK_a \) means a larger \( K_a \), hence a stronger acid.

The approximation \( K_a \approx c\alpha^2 \), which avoids solving the quadratic \( K_a = c\alpha^2/(1-\alpha) \), is valid only when \( \alpha \) is small (roughly \( \alpha \lt 0.05 \), which follows from \( K_a \ll c \)). If the constant is not much smaller than the concentration, solve the full quadratic.

How are pH and pOH related?

From \( K_w = [H^+][OH^-] = 10^{-14} \) at 298 K, taking negative logs on both sides gives \( pH + pOH = 14 \). So in any aqueous solution you can go from one to the other directly: if \( pH = 3.20 \), then \( pOH = 10.80 \) and \( [OH^-] = 10^{-10.8} \) M.

For the worked solutions to the exercises of this chapter, the accompanying class 11 chemistry formulas material pairs each sheet with practice problems. The complete rationalised chapters are published on the official NCERT site, ncert.nic.in, where you can verify any value quoted here.

Reference: NCERT Class 11 Chemistry textbook, chapter Equilibrium.