Equilibrium — Practice Questions with Answers

At equilibrium, the forward and reverse reactions do not stop — they continue occurring continuously, but at exactly equal rates, so the concentrations of reactants and products remain constant over time, even though molecular-level reactions are still actively taking place in both directions.

This ongoing, continuous molecular activity (despite unchanging macroscopic concentrations) is precisely why chemical equilibrium is described as "dynamic," rather than "static" (which would imply the reaction had genuinely stopped altogether).

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, temperature, pressure, or volume, the equilibrium will shift in the direction that tends to counteract (partially oppose) the applied change, establishing a new equilibrium position.

This principle is widely applied to predict how an equilibrium system will respond to various external changes/stresses, and is particularly important in industrial chemistry for optimising reaction conditions (temperature, pressure, concentration) to maximise desired product yield.

Arrhenius concept: An acid produces $H^+$ ions, and a base produces $OH^-$ ions, in aqueous solution — limited specifically to aqueous systems.

Bronsted-Lowry concept: An acid is a proton ($H^+$) donor, and a base is a proton acceptor — broader, applicable beyond just aqueous solutions.

Lewis concept: An acid is an electron-pair acceptor, and a base is an electron-pair donor — the broadest definition, encompassing species that don't even necessarily involve protons at all.

$pH = -\log_{10}[H^+]$, a measure of the hydrogen ion concentration in a solution.

$pOH = -\log_{10}[OH^-]$, a measure of the hydroxide ion concentration.

At 25°C: $pH + pOH = 14$ (derived from the water ionisation constant $K_w = [H^+][OH^-] = 10^{-14}$ at this temperature).

An acetic acid ($CH_3COOH$)/sodium acetate ($CH_3COONa$) buffer contains both the weak acid and its conjugate base (acetate ion) in significant, comparable quantities.

If a small amount of strong acid ($H^+$) is added, the excess acetate ions neutralise it (forming more acetic acid), preventing a large pH drop. If a small amount of strong base ($OH^-$) is added, the acetic acid neutralises it (forming more acetate and water), preventing a large pH rise — this dual capacity to neutralise either added acid or base, using the buffer's two components, is precisely why buffers effectively resist significant pH change upon small additions of acid or base.

For an exothermic reaction (releasing heat, $\Delta H<0$), increasing temperature can be thought of as adding "heat" as if it were a product of the reaction.

By Le Chatelier's principle, the equilibrium shifts to counteract this added heat — since heat is effectively a product of an exothermic reaction, the equilibrium shifts backward (towards reactants), reducing the amount of product formed and decreasing the equilibrium constant $K_c$ at the higher temperature.

Degree of dissociation ($\alpha$) is the fraction of the total amount of a weak electrolyte that actually dissociates into ions at equilibrium, compared to the initial total amount present.

For a simple weak electrolyte $AB \rightleftharpoons A^++B^-$ with initial concentration $C$, using Ostwald's dilution law: $K_a = \dfrac{C\alpha^2}{1-\alpha}$ — for weak electrolytes with small $\alpha$ (where $1-\alpha\approx1$), this simplifies to $K_a \approx C\alpha^2$, or $\alpha \approx \sqrt{K_a/C}$.

Acetic acid partially dissociates: $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$. Adding sodium acetate introduces additional acetate ions ($CH_3COO^-$) — an ion already present in (and common to) the acetic acid dissociation equilibrium.

By Le Chatelier's principle, this increased acetate ion concentration shifts the acetic acid dissociation equilibrium backward (towards the undissociated acid), suppressing further dissociation of acetic acid, and correspondingly reducing the $H^+$ ion concentration compared to what it would be in a solution of acetic acid alone — this suppression of a weak electrolyte's ionisation due to the addition of a common ion is precisely the common ion effect.

$K_c$ is the equilibrium constant expressed in terms of molar concentrations of reactants and products; $K_p$ is the equilibrium constant expressed in terms of partial pressures of gaseous reactants and products.

They are related by: $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the difference between the moles of gaseous products and gaseous reactants in the balanced equation, $R$ is the gas constant, and $T$ is absolute temperature.

Sodium acetate dissociates completely into $Na^+$ and $CH_3COO^-$ ions in water. The acetate ion, being the conjugate base of a weak acid (acetic acid), undergoes hydrolysis — reacting with water to a small extent: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.

This hydrolysis reaction produces excess $OH^-$ ions in solution (since $Na^+$, being the conjugate acid of a strong base, does not undergo significant hydrolysis), making the overall solution basic — this is precisely why salts of a weak acid and strong base produce basic solutions upon dissolution in water.

$pH = -\log_{10}[H^+] = -\log_{10}(1\times10^{-4})$

$pH = -(-4) = \mathbf{4}$

$K_c = \dfrac{[NO_2]^2}{[N_2O_4]}$

$K_c = \dfrac{(0.02)^2}{0.05} = \dfrac{0.0004}{0.05}$

$K_c = \mathbf{0.008}$

Since HCl is a strong acid, it dissociates completely: $[H^+] = 0.01\text{ M} = 1\times10^{-2}\text{ M}$

$pH = -\log_{10}(1\times10^{-2}) = \mathbf{2}$

Using $\alpha \approx \sqrt{K_a/C}$ (valid for weak acids with small $\alpha$):

$\alpha = \sqrt{\dfrac{1.6\times10^{-5}}{0.1}} = \sqrt{1.6\times10^{-4}}$

$\alpha = 0.0126$ (or 1.26%)

$[H^+] = C\alpha = 0.1\times0.0126 = \mathbf{1.26\times10^{-3}\text{ M}}$

Henderson-Hasselbalch equation: $pH = pK_a + \log\left(\dfrac{[\text{salt}]}{[\text{acid}]}\right)$

$pK_a = -\log(1.8\times10^{-5}) \approx 4.745$

Since $[\text{salt}]=[\text{acid}]=0.1$ M, the ratio is 1, and $\log(1)=0$:

$pH = 4.745 + 0 = \mathbf{4.745}$

$K_c = \dfrac{[SO_3]^2}{[SO_2]^2[O_2]}$

$4 = \dfrac{[SO_3]^2}{(0.5)^2(0.5)} = \dfrac{[SO_3]^2}{0.125}$

$[SO_3]^2 = 4\times0.125 = 0.5$

$[SO_3] = \sqrt{0.5} \approx \mathbf{0.707\text{ M}}$

$AgCl \rightleftharpoons Ag^+ + Cl^-$. If solubility is $s$ mol/L, then $[Ag^+]=[Cl^-]=s$

$K_{sp} = [Ag^+][Cl^-] = s^2$

$s = \sqrt{K_{sp}} = \sqrt{1.8\times10^{-10}}$

$s \approx \mathbf{1.34\times10^{-5}\text{ mol/L}}$

$[OH^-] = 0.001\text{ M} = 1\times10^{-3}\text{ M}$

$pOH = -\log(1\times10^{-3}) = 3$

$pH = 14 - pOH = 14 - 3 = \mathbf{11}$; $\mathbf{pOH = 3}$

Let $x$ = amount of A converted to B at equilibrium: $[A]=1-x$, $[B]=x$

$K_c = \dfrac{[B]}{[A]} = \dfrac{x}{1-x} = 2$

$x = 2(1-x) = 2-2x \Rightarrow 3x=2 \Rightarrow x=0.667$

$[A] = 1-0.667 = \mathbf{0.333\text{ M}}$; $[B] = \mathbf{0.667\text{ M}}$

$BOH \rightleftharpoons B^+ + OH^-$. Degree of dissociation: $\alpha = \dfrac{[OH^-]}{C} = \dfrac{2\times10^{-3}}{0.2} = 0.01$

Using $K_b \approx C\alpha^2$ (valid for small $\alpha$): $K_b = 0.2\times(0.01)^2 = 0.2\times0.0001$

$K_b = \mathbf{2\times10^{-5}}$