Dynamic Equilibrium Reversible Reactions Many reactions are reversible: reactants form products, and products can re-form reactants. This is Represented with a double arrow:
A + B ⇌ C + D \mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D} A + B ⇌ C + D Closed Systems Equilibrium can only be established in a closed system — one where no matter can enter or leave.
Dynamic Nature At equilibrium:
Both forward and reverse reactions continue to occur. The rates of the forward and reverse reactions are equal. The concentrations of all species remain constant (not necessarily equal). Macroscopic properties (colour, pressure, pH) are constant. Characteristics of Equilibrium Property Description Forward rate Equals reverse rate Concentrations Constant (but not necessarily equal) Can be approached From either direction Dynamic Both reactions continue Closed system Required
The Equilibrium Constant K c K_c K c For the reaction a A + b B ⇌ c C + d D a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D} a A + b B ⇌ c C + d D
K c = [ C ] c [ D ] d [ A ] a [ B ] b K_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} K c = [ A ] a [ B ] b [ C ] c [ D ] d Only gases and aqueous species are included. Pure solids and pure liquids are NOT included (their activity = 1 = 1 = 1 Square brackets denote equilibrium concentrations in mol/L. K c K_c K c K p K_p K p For gaseous reactions:
K p = ( p C ) c ( p D ) d ( p A ) a ( p B ) b K_p = \frac{(p_C)^c(p_D)^d}{(p_A)^a(p_B)^b} K p = ( p A ) a ( p B ) b ( p C ) c ( p D ) d Where p X p_X p X X X X
Relationship Between K c K_c K c K p K_p K p K p = K c ( R T ) Δ n K_p = K_c(RT)^{\Delta n} K p = K c ( R T ) Δ n Where Δ n = ( m o l e s o f g a s e o u s p r o d u c t s ) − ( m o l e s o f g a s e o u s r e a c t a n t s ) \Delta n = (\mathrm{moles of gaseous products}) - (\mathrm{moles of gaseous reactants}) Δ n = ( molesofgaseousproducts ) − ( molesofgaseousreactants )
Example
For N2 _2 2 2 _2 2 ⇌ \rightleftharpoons ⇌ 3 _3 3 Δ n = 2 − 4 = − 2 \Delta n = 2 - 4 = -2 Δ n = 2 − 4 = − 2
K p = K c ( R T ) − 2 = K c ( R T ) 2 K_p = K_c(RT)^{-2} = \frac{K_c}{(RT)^2} K p = K c ( R T ) − 2 = ( R T ) 2 K c The Reaction Quotient (Q Q Q The reaction quotient has the same form as K c K_c K c current (non-equilibrium) Concentrations:
Q c = [ C ] c [ D ] d [ A ] a [ B ] b Q_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} Q c = [ A ] a [ B ] b [ C ] c [ D ] d Comparison Meaning Q < K Q \lt K Q < K Reaction proceeds forward (more products) Q = K Q = K Q = K System is at equilibrium Q > K Q \gt K Q > K Reaction proceeds in reverse (more reactants)
Le Chatelier”s Principle Statement If a system at equilibrium is subjected to a change, the system will shift to counteract that change And restore equilibrium.
Effect of Concentration Changes Change System Response Increase [reactant] Shift to products (right) Decrease [reactant] Shift to reactants (left) Increase [product] Shift to reactants (left) Decrease [product] Shift to products (right)
Effect of Pressure Changes (Gases) Change System Response Increase pressure Shift to side with fewer moles of gas Decrease pressure Shift to side with more moles of gas
Important : Changing pressure by adding an inert gas does NOT shift the equilibrium (partial Pressures of reacting gases are unchanged).
Effect of Temperature Changes Change System Response Effect on K K K Increase temperature Shift in endothermic direction K K K Decrease temperature Shift in exothermic direction K K K
Le Chatelier’s principle does NOT apply to the equilibrium constant. The equilibrium constant only Changes with temperature. Changes in concentration, pressure, or adding a catalyst do NOT change K K K
Effect of a Catalyst A catalyst increases the rate of both forward and reverse reactions equally. It does NOT shift the equilibrium position. It does NOT change the value of K K K It helps the system reach equilibrium faster. Equilibrium Calculations ICE Tables ICE (Initial, Change, Equilibrium) tables are used to organise equilibrium calculations.
Example
For the reaction H2 _2 2 2 _2 2 ⇌ \rightleftharpoons ⇌ K c = 50.5 K_c = 50.5 K c = 50.5 448 ° C 448\degree\mathrm{C} 448° C
If 1.0 m o l 1.0\mathrm{ mol} 1.0 mol 2 _2 2 1.0 m o l 1.0\mathrm{ mol} 1.0 mol 2 _2 2 1.0 L 1.0\mathrm{ L} 1.0 L
H2 _2 2 I2 _2 2 HI Initial 1.0 1.0 0 Change − x -x − x − x -x − x + 2 x +2x + 2 x Equilibrium 1.0 − x 1.0-x 1.0 − x 1.0 − x 1.0-x 1.0 − x 2 x 2x 2 x
K c = ( 2 x ) 2 ( 1.0 − x ) ( 1.0 − x ) = 4 x 2 ( 1.0 − x ) 2 = 50.5 K_c = \frac{(2x)^2}{(1.0-x)(1.0-x)} = \frac{4x^2}{(1.0-x)^2} = 50.5 K c = ( 1.0 − x ) ( 1.0 − x ) ( 2 x ) 2 = ( 1.0 − x ) 2 4 x 2 = 50.5 2 x 1.0 − x = 50.5 = 7.11 \frac{2x}{1.0-x} = \sqrt{50.5} = 7.11 1.0 − x 2 x = 50.5 = 7.11 2 x = 7.11 − 7.11 x 2x = 7.11 - 7.11x 2 x = 7.11 − 7.11 x 9.11 x = 7.11 ⟹ x = 0.781 9.11x = 7.11 \implies x = 0.781 9.11 x = 7.11 ⟹ x = 0.781 Equilibrium concentrations: [H2 _2 2 2 _2 2 0.219 m o l / L 0.219\mathrm{ mol/L} 0.219 mol/L 1.562 m o l / L 1.562\mathrm{ mol/L} 1.562 mol/L
Calculating K K K Determine equilibrium concentrations (using ICE table if needed). Substitute into the equilibrium expression. Calculate K K K Industrial Applications The Haber Process N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 N H 3 ( g ) Δ H = − 92 k J / m o l \mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{NH}_3(\mathrm{g}) \quad \Delta H = -92\mathrm{ kJ/mol} N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) Δ H = − 92 kJ/mol Condition Le Chatelier Prediction Industrial Choice Temperature Low (exothermic) favours products 400 − − 500 ° C 400\mathrm{--}500\degree\mathrm{C} 400 − − 500° C Pressure High (fewer moles of gas on product side) 150 − − 300 a t m 150\mathrm{--}300\mathrm{ atm} 150 − − 300 atm Catalyst Does not change position but speeds up rate Iron catalyst
2 S O 2 ( g ) + O 2 ( g ) ⇌ 2 S O 3 ( g ) Δ H = − 197 k J / m o l 2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g}) \quad \Delta H = -197\mathrm{ kJ/mol} 2 SO 2 ( g ) + O 2 ( g ) ⇌ 2 SO 3 ( g ) Δ H = − 197 kJ/mol Condition Le Chatelier Prediction Industrial Choice Temperature Low (exothermic) 400 − − 450 ° C 400\mathrm{--}450\degree\mathrm{C} 400 − − 450° C Pressure High (fewer moles of gas) 1 − − 2 a t m 1\mathrm{--}2\mathrm{ atm} 1 − − 2 atm Catalyst V2 _2 2 5 _5 5 V2 _2 2 5 _5 5
Acid-Base Equilibrium The pH Scale p H = − log [ H + ] \mathrm{pH} = -\log[\mathrm{H}^+] pH = − log [ H + ] pH Description < 7 \lt 7 < 7 Acidic = 7 = 7 = 7 Neutral (at 25 ° C 25\degree\mathrm{C} 25° C > 7 \gt 7 > 7 Basic (alkaline)
Strong and Weak Acids Property Strong Acid Weak Acid Dissociation Complete Partial [ H + ] [\mathrm{H}^+] [ H + ] c c c [ H + ] = c [\mathrm{H}^+] = c [ H + ] = c [ H + ] < c [\mathrm{H}^+] \lt c [ H + ] < c Equilibrium No equilibrium Equilibrium established pH Lower (for same c c c Higher (for same c c c Examples HCl, HNO3 _3 3 2 _2 2 4 _4 4 CH3 _3 3
Acid Dissociation Constant (K a K_a K a For a weak acid HA:
H A ⇌ H + + A − \mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^- HA ⇌ H + + A − K a = [ H + ] [ A − ] [ H A ] K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} K a = [ HA ] [ H + ] [ A − ] Base Dissociation Constant (K b K_b K b For a weak base B:
B + H 2 O ⇌ B H + + O H − \mathrm{B} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{BH}^+ + \mathrm{OH}^- B + H 2 O ⇌ BH + + OH − K b = [ B H + ] [ O H − ] [ B ] K_b = \frac{[\mathrm{BH}^+][\mathrm{OH}^-]}{[\mathrm{B}]} K b = [ B ] [ BH + ] [ OH − ] Ionic Product of Water (K w K_w K w H 2 O ⇌ H + + O H − \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^+ + \mathrm{OH}^- H 2 O ⇌ H + + OH − K w = [ H + ] [ O H − ] = 1.0 × 10 − 14 ( a t 25 ° C ) K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14}\mathrm{ (at }25\degree\mathrm{C)} K w = [ H + ] [ OH − ] = 1.0 × 1 0 − 14 ( at 25° C ) pK a K_a K a K b K_b K b p K a = − log K a \mathrm{p}K_a = -\log K_a p K a = − log K a p K b = − log K b \mathrm{p}K_b = -\log K_b p K b = − log K b Relationship Between K a K_a K a K b K_b K b For a conjugate acid-base pair:
K a × K b = K w = 1.0 × 10 − 14 K_a \times K_b = K_w = 1.0 \times 10^{-14} K a × K b = K w = 1.0 × 1 0 − 14 p K a + p K b = 14 \mathrm{p}K_a + \mathrm{p}K_b = 14 p K a + p K b = 14 pH of Weak Acids For a weak acid HA of concentration c c c
[ H + ] = K a × c [\mathrm{H}^+] = \sqrt{K_a \times c} [ H + ] = K a × c p H = 1 2 ( p K a − log c ) \mathrm{pH} = \frac{1}{2}(\mathrm{p}K_a - \log c) pH = 2 1 ( p K a − log c ) (approximation valid when K a K_a K a [ H A ] e q ≈ c [\mathrm{HA}]_{\mathrm{eq}} \approx c [ HA ] eq ≈ c
Example
Calculate the pH of 0.10 M 0.10\mathrm{ M} 0.10 M K a = 1.8 × 10 − 5 K_a = 1.8 \times 10^{-5} K a = 1.8 × 1 0 − 5
[ H + ] = 1.8 × 10 − 5 × 0.10 = 1.8 × 10 − 6 = 1.34 × 10 − 3 m o l / L [\mathrm{H}^+] = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}\mathrm{ mol/L} [ H + ] = 1.8 × 1 0 − 5 × 0.10 = 1.8 × 1 0 − 6 = 1.34 × 1 0 − 3 mol/L p H = − log ( 1.34 × 10 − 3 ) = 2.87 \mathrm{pH} = -\log(1.34 \times 10^{-3}) = 2.87 pH = − log ( 1.34 × 1 0 − 3 ) = 2.87 Buffers Definition A buffer solution resists changes in pH when small amounts of acid or base are added. It Consists of:
A weak acid and its conjugate base, OR A weak base and its conjugate acid. Henderson-Hasselbalch Equation p H = p K a + log [ A − ] [ H A ] \mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} pH = p K a + log [ HA ] [ A − ] Buffer Capacity The buffer is most effective when:
[ H A ] = [ A − ] ⟹ p H = p K a [\mathrm{HA}] = [\mathrm{A}^-] \implies \mathrm{pH} = \mathrm{p}K_a [ HA ] = [ A − ] ⟹ pH = p K a A buffer works best within ± 1 \pm 1 ± 1 p K a \mathrm{p}K_a p K a
How Buffers Work Adding acid (H+ ^+ + : the conjugate base (A− ^- − + ^+ + Adding base (OH− ^- − : the weak acid (HA) reacts with the added OH− ^- − − ^- − 2 _2 2 Example
A buffer contains 0.20 M 0.20\mathrm{ M} 0.20 M 3 _3 3 p K a = 4.76 \mathrm{p}K_a = 4.76 p K a = 4.76 0.30 M 0.30\mathrm{ M} 0.30 M 3 _3 3
p H = 4.76 + log ( 0.30 0.20 ) = 4.76 + log ( 1.5 ) = 4.76 + 0.18 = 4.94 \mathrm{pH} = 4.76 + \log\!\left(\frac{0.30}{0.20}\right) = 4.76 + \log(1.5) = 4.76 + 0.18 = 4.94 pH = 4.76 + log ( 0.20 0.30 ) = 4.76 + log ( 1.5 ) = 4.76 + 0.18 = 4.94 Neutralisation and Indicators Strong Acid + Strong Base p H = 7 a t e q u i v a l e n c e p o i n t \mathrm{pH} = 7 \mathrm{ at equivalence point} pH = 7 atequivalencepoint Weak Acid + Strong Base p H > 7 a t e q u i v a l e n c e p o i n t \mathrm{pH} \gt 7 \mathrm{ at equivalence point} pH > 7 atequivalencepoint Strong Acid + Weak Base p H < 7 a t e q u i v a l e n c e p o i n t \mathrm{pH} \lt 7 \mathrm{ at equivalence point} pH < 7 atequivalencepoint Indicators An acid-base indicator is a weak acid that has different colours in its protonated and deprotonated Forms.
Indicator pH Range Colour Change Methyl orange 3.1—4.4 Red → \to → Bromothymol blue 6.0—7.6 Yellow → \to → Phenolphthalein 8.3—10.0 Colourless → \to →
Solubility Product (K s p K_{sp} K s p Definition For a sparingly soluble salt M a X b \mathrm{M}_a\mathrm{X}_b M a X b
M a X b ( s ) ⇌ a M b + ( a q ) + b X a − ( a q ) \mathrm{M}_a\mathrm{X}_b(s) \rightleftharpoons a\mathrm{M}^{b+}(aq) + b\mathrm{X}^{a-}(aq) M a X b ( s ) ⇌ a M b + ( a q ) + b X a − ( a q ) K_`\{sp}` = [\mathrm{M}^{b+}]^a[\mathrm{X}^{a-}]^b Common K s p K_{sp} K s p Salt K s p K_{sp} K s p AgCl 1.8 × 10 − 10 1.8 \times 10^{-10} 1.8 × 1 0 − 10 AgBr 5.0 × 10 − 13 5.0 \times 10^{-13} 5.0 × 1 0 − 13 AgI 8.3 × 10 − 17 8.3 \times 10^{-17} 8.3 × 1 0 − 17 BaSO4 _4 4 1.1 × 10 − 10 1.1 \times 10^{-10} 1.1 × 1 0 − 10 CaCO3 _3 3 3.4 × 10 − 9 3.4 \times 10^{-9} 3.4 × 1 0 − 9 PbCl2 _2 2 1.7 × 10 − 5 1.7 \times 10^{-5} 1.7 × 1 0 − 5
Predicting Precipitation Compare the ion product (Q Q Q K s p K_{sp} K s p
Condition Result Q < K s p Q \lt K_{sp} Q < K s p Unsaturated (no precipitate) Q = K s p Q = K_{sp} Q = K s p Saturated (at equilibrium) Q > K s p Q \gt K_{sp} Q > K s p Supersaturated (precipitate forms)
Common Ion Effect The solubility of a salt decreases when a common ion is present.
Example
The K s p K_{sp} K s p 1.8 × 10 − 10 1.8 \times 10^{-10} 1.8 × 1 0 − 10
(a) Pure water:
S 2 = 1.8 × 10 − 10 ⟹ s = 1.34 × 10 − 5 m o l / L S^2 = 1.8 \times 10^{-10} \implies s = 1.34 \times 10^{-5}\mathrm{ mol/L} S 2 = 1.8 × 1 0 − 10 ⟹ s = 1.34 × 1 0 − 5 mol/L (b) 0.10 M 0.10\mathrm{ M} 0.10 M
[ A g + ] [ C l − ] = s × 0.10 = 1.8 × 10 − 10 [\mathrm{Ag}^+][\mathrm{Cl}^-] = s \times 0.10 = 1.8 \times 10^{-10} [ Ag + ] [ Cl − ] = s × 0.10 = 1.8 × 1 0 − 10 S = 1.8 × 10 − 10 0.10 = 1.8 × 10 − 9 m o l / L S = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9}\mathrm{ mol/L} S = 0.10 1.8 × 1 0 − 10 = 1.8 × 1 0 − 9 mol/L The solubility is much lower due to the common ion effect.
IB Exam-Style Questions Question 1 (Paper 1 style) For the equilibrium 2 S O 2 ( g ) + O 2 ( g ) ⇌ 2 S O 3 ( g ) 2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g}) 2 SO 2 ( g ) + O 2 ( g ) ⇌ 2 SO 3 ( g )
The product side has 2 moles of gas, the reactant side has 3. The equilibrium shifts to the right (fewer moles of gas), increasing [SO3 _3 3 K p K_p K p
Question 2 (Paper 2 style) For the reaction P C l 5 ( g ) ⇌ P C l 3 ( g ) + C l 2 ( g ) \mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) PCl 5 ( g ) ⇌ PCl 3 ( g ) + Cl 2 ( g ) 1.0 m o l 1.0\mathrm{ mol} 1.0 mol 5 _5 5 2.0 L 2.0\mathrm{ L} 2.0 L 250 ° C 250\degree\mathrm{C} 250° C 0.40 m o l 0.40\mathrm{ mol} 0.40 mol 2 _2 2
(a) Calculate K c K_c K c
PCl5 _5 5 PCl3 _3 3 Cl2 _2 2 Initial 0.50 0 0 Change − 0.20 -0.20 − 0.20 + 0.20 +0.20 + 0.20 + 0.20 +0.20 + 0.20 Equilibrium 0.30 0.20 0.20
K c = [ P C l 3 ] [ C l 2 ] [ P C l 5 ] = ( 0.20 ) ( 0.20 ) 0.30 = 0.04 0.30 = 0.133 K_c = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]} = \frac{(0.20)(0.20)}{0.30} = \frac{0.04}{0.30} = 0.133 K c = [ PCl 5 ] [ PCl 3 ] [ Cl 2 ] = 0.30 ( 0.20 ) ( 0.20 ) = 0.30 0.04 = 0.133 (b) Calculate the percentage dissociation of PCl5 _5 5
% d i s s o c i a t i o n = 0.20 0.50 × 100 % = 40 % \%\mathrm{ dissociation} = \frac{0.20}{0.50} \times 100\% = 40\% % dissociation = 0.50 0.20 × 100% = 40% Question 3 (Paper 2 style) Calculate the pH of 0.050 M 0.050\mathrm{ M} 0.050 M 3 _3 3 K b = 1.8 × 10 − 5 K_b = 1.8 \times 10^{-5} K b = 1.8 × 1 0 − 5
[ O H − ] = K b × c = 1.8 × 10 − 5 × 0.050 = 9.0 × 10 − 7 = 9.49 × 10 − 4 [\mathrm{OH}^-] = \sqrt{K_b \times c} = \sqrt{1.8 \times 10^{-5} \times 0.050} = \sqrt{9.0 \times 10^{-7}} = 9.49 \times 10^{-4} [ OH − ] = K b × c = 1.8 × 1 0 − 5 × 0.050 = 9.0 × 1 0 − 7 = 9.49 × 1 0 − 4 p O H = − log ( 9.49 × 10 − 4 ) = 3.02 \mathrm{pOH} = -\log(9.49 \times 10^{-4}) = 3.02 pOH = − log ( 9.49 × 1 0 − 4 ) = 3.02 p H = 14 − 3.02 = 10.98 \mathrm{pH} = 14 - 3.02 = 10.98 pH = 14 − 3.02 = 10.98 Question 4 (Paper 1 style) Which salt is most soluble?
A. AgCl (K s p = 1.8 × 10 − 10 K_{sp} = 1.8 \times 10^{-10} K s p = 1.8 × 1 0 − 10 K s p = 5.0 × 10 − 13 K_{sp} = 5.0 \times 10^{-13} K s p = 5.0 × 1 0 − 13 K s p = 8.3 × 10 − 17 K_{sp} = 8.3 \times 10^{-17} K s p = 8.3 × 1 0 − 17 4 _4 4 K s p = 1.1 × 10 − 10 K_{sp} = 1.1 \times 10^{-10} K s p = 1.1 × 1 0 − 10
Answer: D — BaSO4 _4 4 K s p K_{sp} K s p K s p K_{sp} K s p 4 _4 4
Summary Concept Formula K c K_c K c K c = [ C ] c [ D ] d [ A ] a [ B ] b K_c = \dfrac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} K c = [ A ] a [ B ] b [ C ] c [ D ] d K p K_p K p K p = K c ( R T ) Δ n K_p = K_c(RT)^{\Delta n} K p = K c ( R T ) Δ n pH p H = − log [ H + ] \mathrm{pH} = -\log[\mathrm{H}^+] pH = − log [ H + ] K w K_w K w K w = [ H + ] [ O H − ] = 10 − 14 K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 10^{-14} K w = [ H + ] [ OH − ] = 1 0 − 14 Henderson-Hasselbalch p H = p K a + log [ A − ] [ H A ] \mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\mathrm{A}^-]}{[\mathrm{HA}]} pH = p K a + log [ HA ] [ A − ] K s p K_{sp} K s p K s p = [ M b + ] a [ X a − ] b K_{sp} = [\mathrm{M}^{b+}]^a[\mathrm{X}^{a-}]^b K s p = [ M b + ] a [ X a − ] b
For equilibrium calculations, always set up an ICE table. For Le Chatelier questions, be precise About what changes and what stays the same (only K K K K s p K_{sp} K s p
Equilibrium: Extended Topics The van’t Hoff Equation The effect of temperature on the equilibrium constant:
ln ( K 2 K 1 ) = − Δ H ∘ R ( 1 T 2 − 1 T 1 ) \ln\!\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) ln ( K 1 K 2 ) = − R Δ H ∘ ( T 2 1 − T 1 1 ) This is analogous to the Arrhenius equation.
Example
For the reaction N2 _2 2 4 _4 4 ⇌ \rightleftharpoons ⇌ 2 _2 2 Δ H ∘ = + 57 k J / m o l \Delta H^\circ = +57\mathrm{ kJ/mol} Δ H ∘ = + 57 kJ/mol K = 0.115 K = 0.115 K = 0.115 298 K 298\mathrm{ K} 298 K K K K 350 K 350\mathrm{ K} 350 K
ln ( K 2 0.115 ) = − 57000 8.314 ( 1 350 − 1 298 ) \ln\!\left(\frac{K_2}{0.115}\right) = -\frac{57000}{8.314}\left(\frac{1}{350} - \frac{1}{298}\right) ln ( 0.115 K 2 ) = − 8.314 57000 ( 350 1 − 298 1 ) = − 6856 × ( − 0.000500 ) = 3.428 = -6856 \times (-0.000500) = 3.428 = − 6856 × ( − 0.000500 ) = 3.428 K 2 0.115 = e 3.428 = 30.8 \frac{K_2}{0.115} = e^{3.428} = 30.8 0.115 K 2 = e 3.428 = 30.8 K 2 = 3.54 K_2 = 3.54 K 2 = 3.54 As expected for an endothermic reaction, K K K
Quantitative Le Chatelier Calculations When a change is made to a system at equilibrium, a new equilibrium is established. The new Concentrations can be found by setting up a new ICE table.
Example
For the reaction PCl5 _5 5 ⇌ \rightleftharpoons ⇌ 3 _3 3 2 _2 2 K c = 0.0211 K_c = 0.0211 K c = 0.0211 500 K 500\mathrm{ K} 500 K
At equilibrium in a 1 L 1\mathrm{ L} 1 L 5 _5 5 = 0.200 M = 0.200\mathrm{ M} = 0.200 M 3 _3 3 = [ C l 2 ] = [Cl_2] = [ C l 2 ] = 0.0650 M = 0.0650\mathrm{ M} = 0.0650 M
If 0.100 m o l 0.100\mathrm{ mol} 0.100 mol 2 _2 2
After adding Cl2 _2 2 5 _5 5 = 0.200 = 0.200 = 0.200 3 _3 3 = 0.0650 = 0.0650 = 0.0650 2 _2 2 = 0.165 = 0.165 = 0.165
Q c = ( 0.0650 ) ( 0.165 ) 0.200 = 0.0536 > K c = 0.0211 Q_c = \frac{(0.0650)(0.165)}{0.200} = 0.0536 \gt K_c = 0.0211 Q c = 0.200 ( 0.0650 ) ( 0.165 ) = 0.0536 > K c = 0.0211 The reaction shifts left. Let x x x 2 _2 2
PCl5 _5 5 PCl3 _3 3 Cl2 _2 2 Initial 0.200 0.0650 0.165 Change + x +x + x − x -x − x − x -x − x Equilibrium 0.200 + x 0.200+x 0.200 + x 0.0650 − x 0.0650-x 0.0650 − x 0.165 − x 0.165-x 0.165 − x
( 0.0650 − x ) ( 0.165 − x ) 0.200 + x = 0.0211 \frac{(0.0650-x)(0.165-x)}{0.200+x} = 0.0211 0.200 + x ( 0.0650 − x ) ( 0.165 − x ) = 0.0211 ( 0.0650 − x ) ( 0.165 − x ) = 0.0211 ( 0.200 + x ) (0.0650-x)(0.165-x) = 0.0211(0.200+x) ( 0.0650 − x ) ( 0.165 − x ) = 0.0211 ( 0.200 + x ) 0.01073 − 0.2300 x + x 2 = 0.004220 + 0.0211 x 0.01073 - 0.2300x + x^2 = 0.004220 + 0.0211x 0.01073 − 0.2300 x + x 2 = 0.004220 + 0.0211 x X 2 − 0.2511 x + 0.00651 = 0 X^2 - 0.2511x + 0.00651 = 0 X 2 − 0.2511 x + 0.00651 = 0 X = 0.2511 ± 0.06305 − 0.02604 2 = 0.2511 ± 0.1923 2 X = \frac{0.2511 \pm \sqrt{0.06305 - 0.02604}}{2} = \frac{0.2511 \pm 0.1923}{2} X = 2 0.2511 ± 0.06305 − 0.02604 = 2 0.2511 ± 0.1923 x = 0.2217 x = 0.2217 x = 0.2217 x = 0.0294 x = 0.0294 x = 0.0294
Since x ≤ 0.0650 x \le 0.0650 x ≤ 0.0650 x = 0.0294 x = 0.0294 x = 0.0294
New equilibrium: [PCl5 _5 5 = 0.229 M = 0.229\mathrm{ M} = 0.229 M 3 _3 3 = 0.0356 M = 0.0356\mathrm{ M} = 0.0356 M 2 _2 2 = 0.136 M = 0.136\mathrm{ M} = 0.136 M
Acid-Base Extended: Polyprotic Acids Polyprotic acids can donate more than one proton.
Carbonic acid (H2 _2 2 3 _3 3
H 2 C O 3 ⇌ H + + H C O 3 − K a 1 = 4.3 × 10 − 7 \mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^- \quad K_{a1} = 4.3 \times 10^{-7} H 2 CO 3 ⇌ H + + HCO 3 − K a 1 = 4.3 × 1 0 − 7 H C O 3 − ⇌ H + + C O 3 2 − K a 2 = 4.8 × 10 − 11 \mathrm{HCO}_3^- \rightleftharpoons \mathrm{H}^+ + \mathrm{CO}_3^{2-} \quad K_{a2} = 4.8 \times 10^{-11} HCO 3 − ⇌ H + + CO 3 2 − K a 2 = 4.8 × 1 0 − 11 Note: K a 1 ≫ K a 2 K_{a1} \gg K_{a2} K a 1 ≫ K a 2
pH of Salt Solutions Salt of strong acid + strong base: neutral (pH = 7). Salt of strong acid + weak base: acidic (pH < \lt < Salt of weak acid + strong base: basic (pH > \gt > Salt of weak acid + weak base: depends on relative K a K_a K a K b K_b K b Example
Calculate the pH of 0.10 M 0.10\mathrm{ M} 0.10 M 3 _3 3 K a K_a K a 3 _3 3 = 1.8 × 10 − 5 = 1.8 \times 10^{-5} = 1.8 × 1 0 − 5
K b = K w K a = 1.0 × 10 − 14 1.8 × 10 − 5 = 5.56 × 10 − 10 K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} K b = K a K w = 1.8 × 1 0 − 5 1.0 × 1 0 − 14 = 5.56 × 1 0 − 10 [ O H − ] = K b × c = 5.56 × 10 − 10 × 0.10 = 7.46 × 10 − 6 [\mathrm{OH}^-] = \sqrt{K_b \times c} = \sqrt{5.56 \times 10^{-10} \times 0.10} = 7.46 \times 10^{-6} [ OH − ] = K b × c = 5.56 × 1 0 − 10 × 0.10 = 7.46 × 1 0 − 6 p O H = 5.13 , p H = 14 − 5.13 = 8.87 \mathrm{pOH} = 5.13, \quad \mathrm{pH} = 14 - 5.13 = 8.87 pOH = 5.13 , pH = 14 − 5.13 = 8.87 The solution is basic, as expected for the salt of a weak acid and strong base.
Additional IB Exam-Style Questions Question 5 (Paper 2 style) For the equilibrium: 2 S O 2 ( g ) + O 2 ( g ) ⇌ 2 S O 3 ( g ) 2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g}) 2 SO 2 ( g ) + O 2 ( g ) ⇌ 2 SO 3 ( g ) Δ H = − 198 k J / m o l \Delta H = -198\mathrm{ kJ/mol} Δ H = − 198 kJ/mol
(a) Explain the effect of increasing temperature on the equilibrium yield of SO3 _3 3
Since the forward reaction is exothermic (Δ H < 0 \Delta H \lt 0 Δ H < 0 3 _3 3 K K K ln K ∝ − Δ H / T \ln K \propto -\Delta H / T ln K ∝ − Δ H / T T T T K K K
(b) Explain the effect of increasing pressure on the equilibrium yield of SO3 _3 3
There are 3 moles of gas on the left and 2 on the right. Increasing pressure shifts the equilibrium To the right (fewer moles), increasing the yield of SO3 _3 3
(c) Explain why a catalyst does not change the equilibrium yield.
A catalyst increases the rate of both forward and reverse reactions equally, so the equilibrium Position is unchanged. It only helps the system reach equilibrium faster.
Question 6 (Paper 1 style) What is the pH of a 0.010 M 0.010\mathrm{ M} 0.010 M 2 _2 2
Ba(OH)2 _2 2 2 _2 2 → \to → 2 + ^{2+} 2 + − ^- −
[ O H − ] = 2 × 0.010 = 0.020 M [\mathrm{OH}^-] = 2 \times 0.010 = 0.020\mathrm{ M} [ OH − ] = 2 × 0.010 = 0.020 M p O H = − log ( 0.020 ) = 1.70 \mathrm{pOH} = -\log(0.020) = 1.70 pOH = − log ( 0.020 ) = 1.70 p H = 14 − 1.70 = 12.30 \mathrm{pH} = 14 - 1.70 = 12.30 pH = 14 − 1.70 = 12.30 Question 7 (Paper 2 style) The solubility of PbI2 _2 2 25 ° C 25\degree\mathrm{C} 25° C 1.4 × 10 − 3 m o l / L 1.4 \times 10^{-3}\mathrm{ mol/L} 1.4 × 1 0 − 3 mol/L
(a) Calculate K s p K_{sp} K s p 2 _2 2
P b I 2 ( s ) ⇌ P b 2 + ( a q ) + 2 I − ( a q ) \mathrm{PbI}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{I}^-(aq) PbI 2 ( s ) ⇌ Pb 2 + ( a q ) + 2 I − ( a q ) [ P b 2 + ] = 1.4 × 10 − 3 M , [ I − ] = 2 ( 1.4 × 10 − 3 ) = 2.8 × 10 − 3 M [\mathrm{Pb}^{2+}] = 1.4 \times 10^{-3}\mathrm{ M}, \quad [\mathrm{I}^-] = 2(1.4 \times 10^{-3}) = 2.8 \times 10^{-3}\mathrm{ M} [ Pb 2 + ] = 1.4 × 1 0 − 3 M , [ I − ] = 2 ( 1.4 × 1 0 − 3 ) = 2.8 × 1 0 − 3 M K_`\{sp}` = [\mathrm{Pb}^{2+}][\mathrm{I}^-]^2 = (1.4 \times 10^{-3})(2.8 \times 10^{-3})^2 = (1.4 \times 10^{-3})(7.84 \times 10^{-6}) = 1.1 × 10 − 8 = 1.1 \times 10^{-8} = 1.1 × 1 0 − 8 (b) Will a precipitate form when 50 m L 50\mathrm{ mL} 50 mL 0.010 M 0.010\mathrm{ M} 0.010 M 3 _3 3 2 _2 2 50 m L 50\mathrm{ mL} 50 mL 0.020 M 0.020\mathrm{ M} 0.020 M
After mixing (volumes double, concentrations halve):
[ P b 2 + ] = 0.005 M , [ I − ] = 0.010 M [\mathrm{Pb}^{2+}] = 0.005\mathrm{ M}, \quad [\mathrm{I}^-] = 0.010\mathrm{ M} [ Pb 2 + ] = 0.005 M , [ I − ] = 0.010 M Q = [ P b 2 + ] [ I − ] 2 = ( 0.005 ) ( 0.010 ) 2 = 5.0 × 10 − 7 Q = [\mathrm{Pb}^{2+}][\mathrm{I}^-]^2 = (0.005)(0.010)^2 = 5.0 \times 10^{-7} Q = [ Pb 2 + ] [ I − ] 2 = ( 0.005 ) ( 0.010 ) 2 = 5.0 × 1 0 − 7 Since Q = 5.0 × 10 − 7 > K s p = 1.1 × 10 − 8 Q = 5.0 \times 10^{-7} \gt K_{sp} = 1.1 \times 10^{-8} Q = 5.0 × 1 0 − 7 > K s p = 1.1 × 1 0 − 8 2 _2 2
Question 8 (Paper 1 style) Which indicator would be most suitable for the titration of a weak acid (CH3 _3 3
A. Methyl orange (pH range 3.1—4.4) B. Bromothymol blue (pH range 6.0—7.6) C. Phenolphthalein (pH Range 8.3—10.0)
Answer: C. A weak acid-strong base titration has an equivalence point with pH > \gt >
Practice Problems Question 1: ICE Table and $K_c$ Calculation For the reaction N 2 O 4 ( g ) ⇌ 2 N O 2 ( g ) \mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g) N 2 O 4 ( g ) ⇌ 2 NO 2 ( g ) 1.00 m o l 1.00\mathrm{ mol} 1.00 mol N 2 O 4 \mathrm{N}_2\mathrm{O}_4 N 2 O 4 2.00 L 2.00\mathrm{ L} 2.00 L 350 K 350\mathrm{ K} 350 K N O 2 \mathrm{NO}_2 NO 2 0.200 m o l / L 0.200\mathrm{ mol/L} 0.200 mol/L K c K_c K c
Answer Initial [ N 2 O 4 ] = 1.00 / 2.00 = 0.500 m o l / L [\mathrm{N}_2\mathrm{O}_4] = 1.00 / 2.00 = 0.500\mathrm{ mol/L} [ N 2 O 4 ] = 1.00/2.00 = 0.500 mol/L
N 2 O 4 \mathrm{N}_2\mathrm{O}_4 N 2 O 4 N O 2 \mathrm{NO}_2 NO 2 Initial 0.500 0.500 0.500 0 0 0 Change − 0.100 -0.100 − 0.100 + 0.200 +0.200 + 0.200 Equilibrium 0.400 0.400 0.400 0.200 0.200 0.200
Since [ N O 2 ] e q = 0.200 m o l / L [\mathrm{NO}_2]_{\mathrm{eq}} = 0.200\mathrm{ mol/L} [ NO 2 ] eq = 0.200 mol/L 2 x 2x 2 x x = 0.100 x = 0.100 x = 0.100
K c = [ N O 2 ] 2 [ N 2 O 4 ] = ( 0.200 ) 2 0.400 = 0.0400 0.400 = 0.100 K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} = \frac{(0.200)^2}{0.400} = \frac{0.0400}{0.400} = 0.100 K c = [ N 2 O 4 ] [ NO 2 ] 2 = 0.400 ( 0.200 ) 2 = 0.400 0.0400 = 0.100
Question 2: Le Chatelier's Principle For the exothermic reaction N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 N H 3 ( g ) \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g) N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) N H 3 \mathrm{NH}_3 NH 3
(a) Increasing pressure
(b) Increasing temperature
(c) Adding a catalyst
Answer (a) There are 4 moles of gas on the left and 2 on the right. Increasing pressure shifts the Equilibrium to the right (fewer moles of gas), increasing the yield of N H 3 \mathrm{NH}_3 NH 3
(b) The forward reaction is exothermic (Δ H < 0 \Delta H \lt 0 Δ H < 0 decreasing the yield of N H 3 \mathrm{NH}_3 NH 3
(c) A catalyst increases the rate of both forward and reverse reactions equally. It does not Change the equilibrium position or the yield of N H 3 \mathrm{NH}_3 NH 3
Question 3: Buffer pH Calculation A buffer solution is prepared by mixing 100 m L 100\mathrm{ mL} 100 mL 0.20 M 0.20\mathrm{ M} 0.20 M C H 3 C O O H \mathrm{CH}_3\mathrm{COOH} CH 3 COOH p K a = 4.76 \mathrm{p}K_a = 4.76 p K a = 4.76 50 m L 50\mathrm{ mL} 50 mL 0.20 M 0.20\mathrm{ M} 0.20 M N a O H \mathrm{NaOH} NaOH
Answer The N a O H \mathrm{NaOH} NaOH C H 3 C O O H \mathrm{CH}_3\mathrm{COOH} CH 3 COOH
n ( C H 3 C O O H ) i n i t i a l = 0.100 × 0.20 = 0.0200 m o l n(\mathrm{CH}_3\mathrm{COOH})_{\mathrm{initial}} = 0.100 \times 0.20 = 0.0200\mathrm{ mol} n ( CH 3 COOH ) initial = 0.100 × 0.20 = 0.0200 mol
n ( N a O H ) = 0.050 × 0.20 = 0.0100 m o l n(\mathrm{NaOH}) = 0.050 \times 0.20 = 0.0100\mathrm{ mol} n ( NaOH ) = 0.050 × 0.20 = 0.0100 mol
After reaction:
n ( C H 3 C O O H ) r e m a i n i n g = 0.0200 − 0.0100 = 0.0100 m o l n(\mathrm{CH}_3\mathrm{COOH})_{\mathrm{remaining}} = 0.0200 - 0.0100 = 0.0100\mathrm{ mol} n ( CH 3 COOH ) remaining = 0.0200 − 0.0100 = 0.0100 mol
n ( C H 3 C O O − ) f o r m e d = 0.0100 m o l n(\mathrm{CH}_3\mathrm{COO}^-)_{\mathrm{formed}} = 0.0100\mathrm{ mol} n ( CH 3 COO − ) formed = 0.0100 mol
Total volume = 150 m L = 0.150 L 150\mathrm{ mL} = 0.150\mathrm{ L} 150 mL = 0.150 L
[ C H 3 C O O H ] = 0.0100 0.150 = 0.0667 M [\mathrm{CH}_3\mathrm{COOH}] = \frac{0.0100}{0.150} = 0.0667\mathrm{ M} [ CH 3 COOH ] = 0.150 0.0100 = 0.0667 M
[ C H 3 C O O − ] = 0.0100 0.150 = 0.0667 M [\mathrm{CH}_3\mathrm{COO}^-] = \frac{0.0100}{0.150} = 0.0667\mathrm{ M} [ CH 3 COO − ] = 0.150 0.0100 = 0.0667 M
p H = p K a + log [ A − ] [ H A ] = 4.76 + log 0.0667 0.0667 = 4.76 + log ( 1 ) = 4.76 \mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} = 4.76 + \log\frac{0.0667}{0.0667} = 4.76 + \log(1) = 4.76 pH = p K a + log [ HA ] [ A − ] = 4.76 + log 0.0667 0.0667 = 4.76 + log ( 1 ) = 4.76
Question 4: Solubility Product and Common Ion Effect The K s p K_{sp} K s p P b C l 2 \mathrm{PbCl}_2 PbCl 2 1.7 × 10 − 5 1.7 \times 10^{-5} 1.7 × 1 0 − 5 25 ° C 25\degree\mathrm{C} 25° C
(a) Calculate the molar solubility of P b C l 2 \mathrm{PbCl}_2 PbCl 2
(b) Calculate the molar solubility of P b C l 2 \mathrm{PbCl}_2 PbCl 2 0.10 M 0.10\mathrm{ M} 0.10 M N a C l \mathrm{NaCl} NaCl
Answer (a) Let s s s P b C l 2 \mathrm{PbCl}_2 PbCl 2
P b C l 2 ( s ) ⇌ P b 2 + ( a q ) + 2 C l − ( a q ) \mathrm{PbCl}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{Cl}^-(aq) PbCl 2 ( s ) ⇌ Pb 2 + ( a q ) + 2 Cl − ( a q )
K s p = [ P b 2 + ] [ C l − ] 2 = s × ( 2 s ) 2 = 4 s 3 K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2 = s \times (2s)^2 = 4s^3 K s p = [ Pb 2 + ] [ Cl − ] 2 = s × ( 2 s ) 2 = 4 s 3
s 3 = 1.7 × 10 − 5 4 = 4.25 × 10 − 6 s^3 = \frac{1.7 \times 10^{-5}}{4} = 4.25 \times 10^{-6} s 3 = 4 1.7 × 1 0 − 5 = 4.25 × 1 0 − 6
s = 1.62 × 10 − 2 m o l / L s = 1.62 \times 10^{-2}\mathrm{ mol/L} s = 1.62 × 1 0 − 2 mol/L
(b) In 0.10 M 0.10\mathrm{ M} 0.10 M N a C l \mathrm{NaCl} NaCl [ C l − ] i n i t i a l = 0.10 M [\mathrm{Cl}^-]_{\mathrm{initial}} = 0.10\mathrm{ M} [ Cl − ] initial = 0.10 M
K s p = [ P b 2 + ] [ C l − ] 2 = s × ( 0.10 + 2 s ) 2 ≈ s × ( 0.10 ) 2 K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2 = s \times (0.10 + 2s)^2 \approx s \times (0.10)^2 K s p = [ Pb 2 + ] [ Cl − ] 2 = s × ( 0.10 + 2 s ) 2 ≈ s × ( 0.10 ) 2
s = 1.7 × 10 − 5 0.010 = 1.7 × 10 − 3 m o l / L s = \frac{1.7 \times 10^{-5}}{0.010} = 1.7 \times 10^{-3}\mathrm{ mol/L} s = 0.010 1.7 × 1 0 − 5 = 1.7 × 1 0 − 3 mol/L
The solubility decreases significantly due to the common ion effect.
Question 5: Weak Acid pH Calculate the pH of a 0.050 M 0.050\mathrm{ M} 0.050 M H F \mathrm{HF} HF K a = 6.8 × 10 − 4 K_a = 6.8 \times 10^{-4} K a = 6.8 × 1 0 − 4
Answer [ H + ] = K a × c = 6.8 × 10 − 4 × 0.050 = 3.4 × 10 − 5 = 5.83 × 10 − 3 m o l / L [\mathrm{H}^+] = \sqrt{K_a \times c} = \sqrt{6.8 \times 10^{-4} \times 0.050} = \sqrt{3.4 \times 10^{-5}} = 5.83 \times 10^{-3}\mathrm{ mol/L} [ H + ] = K a × c = 6.8 × 1 0 − 4 × 0.050 = 3.4 × 1 0 − 5 = 5.83 × 1 0 − 3 mol/L
p H = − log ( 5.83 × 10 − 3 ) = 2.23 \mathrm{pH} = -\log(5.83 \times 10^{-3}) = 2.23 pH = − log ( 5.83 × 1 0 − 3 ) = 2.23
Related Content at Other Levels Common Pitfalls Misapplying Le Chatelier’s principle — it predicts the direction of change, not the extent.
Confusing K c K_c K c K p K_p K p K c K_c K c K p K_p K p
Confusing the terms ‘molar’ and ‘molecular’ — molar refers to per mole (mol − 1 \text{mol}^{-1} mol − 1
Assuming that a strong acid always has a lower pH than a weak acid without considering concentration.
Forgetting to balance equations before performing calculations — always check that atoms and charges balance on both sides.
Misidentifying the limiting reagent — compare mole ratios rather than comparing masses.
Cross-References Topic Site Link [Chemical Equilibrium] A-Level View [Chemical Equilibrium] IB View [Chemical Equilibrium] DSE View
Worked Examples Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.