Formula sheet: Equilibrium

Equilibrium12 min readLesson 1 of 8

1.1 One idea, several forms Equilibrium is the state in which the rates of the forward and reverse processes are equal, so the measurable properties stop changing even though both processes continue. The same condition is written with concentrations (Kc), partial pressures (Kp) or, for acids and bases, with hydrogen ion activity (pH). Every formula below is one of those forms. 1.2 Chemical equilibrium

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

Law of mass action for a general reaction at equilibrium

K_{p} = K_{c} (R T)^{Δ n_{g}} Δ n_{g} = (moles of gaseous products) - (moles of gaseous reactants)

Relation between the pressure and concentration constants

Q = \frac{[ products ]}{[ reactants ]} at any moment Q < K right, Q > K left, Q = K equilibrium

Reaction quotient and the direction test

Δ G^{0} = - R T ln K Δ G = Δ G^{0} + R T ln Q

Equilibrium constant and Gibbs energy

ln \frac{K _{2}}{K _{1}} = \frac{Δ H ^{0}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

Temperature dependence of the equilibrium constant (van't Hoff equation)

1.3 Ionic equilibrium

2 H_{2} O ⇌ H_{3} O^{+} + O H^{-} K_{w} = [H^{+}] [O H^{-}] = 1 0^{- 14} at 298 K

Autoionisation of water and the ionic product

p H = - lo g [H^{+}] pO H = - lo g [O H^{-}] p H + pO H = 14

pH, pOH and their sum at 298 K

K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]} K_{b} = \frac{[ B H ^{+} ] [ O H ^{-} ]}{[ B ]} K_{a} K_{b} = K_{w}

Acid and base dissociation constants and the relation with water

[H^{+}] = K_{a} c α = \frac{K _{a}}{c}

Weak acid approximation and Ostwald dilution law

1.4 Buffers and salts

p H = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}

Henderson-Hasselbalch equation for an acidic buffer

pO H = p K_{b} + lo g \frac{[ salt ]}{[ base ]}

Buffer of a weak base and its salt

K_{h} = \frac{K _{w}}{K _{a}} (salt of weak acid and strong base) K_{h} = \frac{K _{w}}{K _{b}} (salt of strong acid and weak base)

Hydrolysis constant and pH for different salt types

1.5 Solubility product

A_{x} B_{y} (s) ⇌ x A^{y +} + y B^{x -} K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}

Solubility product of a sparingly soluble salt

A B : K_{s p} = s^{2} A B_{2} : K_{s p} = 4 s^{3} A_{2} B : K_{s p} = 4 s^{3} A_{3} B_{2} : K_{s p} = 108 s^{5}

Solubility product from solubility s

K_{s p} = (s + C) (s) \approx C s when C ≫ s

Common ion effect for a salt AB with added A+ at concentration C

1.6 Reading the sheet in an exam The first three relations belong together: write the balanced equation, write Kc from it, then convert to Kp only if gases are involved and the question gives pressures. The fourth relation answers the direction question: compute Q from the current concentrations and compare it with K. The ionic relations form a separate tool kit: begin from Kw, find the hydrogen ion concentration, then move to pH, and use Ka or Kb only for weak species. The buffer and solubility relations are specialised forms of the same equilibrium idea, so they are best memorised together with the situation that calls for each. 1.7 Units and approximations Every concentration constant is written in mol L^-1 and every pressure constant in the pressure unit used for the partial pressures, so the units of Kc and Kp are usually omitted in chemical work. The weak electrolyte formulas assume that the degree of dissociation is small; when it is not, the exact quadratic must be solved. The pH values in this sheet are for 298 K unless the question states another temperature. 1.8 Constants worth remembering Quantity Value at 298 K Ionic product of water Kw 1.0 x 10^-14 pH of pure water 7.0 pKa of acetic acid 4.74 pKb of ammonia 4.74 Relation Ka Kb = Kw Kp and Kc relation Kp = Kc (RT)^delta n_g

Exam

Always balance the equation before writing an equilibrium constant expression, and never include pure solids or pure liquids in the expression. Their activity is taken as one. Gases, dissolved species and ions do appear, with gases entered as partial pressures in Kp.

Balance the equation before writing Kc or Kp.
Convert between Kc and Kp with delta n_g of gases only.
Use Q against K to decide the direction of a reaction.
State clearly whether a change is at constant volume or constant pressure before applying Le Chatelier.
Write pH from hydrogen ion concentration, and remember pH + pOH = 14 at 298 K.
For buffers, use the Henderson equation with the ratio of salt to acid.
For Ksp, write the expression from the dissociation equation before substituting solubility.

in hinglish

Equilibrium ka core: forward aur reverse process ki rates equal ho jaati hain, properties constant dikhti hain. Law of mass action se Kc likho, gases ke liye Kp = Kc (RT)^Δn_g, aur kisi bhi moment par Q nikaal kar Q vs K se direction decide karo. Le Chatelier kehta hai ki change ka effect kam karne ki direction mein equilibrium shift hota hai; temperature ka effect K par padta hai, pressure/volume ka effect sirf gas moles par. ΔG° = -RT ln K se thermodynamics aur equilibrium jud jaate hain. Ionic equilibrium mein Kw = [H+][OH-] = 10^-14, pH + pOH = 14, weak acid ke liye [H+] = \sqrt(Ka c) aur α = \sqrt(Ka/c). Buffer ke liye Henderson equation pH = pKa + log([salt]/[acid]). Salt hydrolysis mein Kh = Kw/Ka ya Kw/Kb. Ksp sparingly soluble salts ke liye likho aur common ion effect se solubility ghatne ka reason samjho. Numerical mein equation balance, units aur approximations check karo.