📋 IB Content Statements (R3.2)

This topic covers the following HL syllabus points from the IB Chemistry 2025 guide:

  • R3.2.1 (HL): The equilibrium law expression: $K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ for the reaction $aA + bB \rightleftharpoons cC + dD$.
  • R3.2.2 (HL): The magnitude of $K_c$ indicates the extent of reaction: $K_c \gg 1$ (mostly products), $K_c \ll 1$ (mostly reactants).
  • R3.2.3 (HL): The reaction quotient ($Q$) uses non-equilibrium concentrations and predicts the direction in which equilibrium will be reached.
  • R3.2.4 (HL): $K_c$ is only affected by changes in temperature. Changes in concentration, pressure, or catalyst do not change $K_c$.
  • R3.2.5 (HL): The relationship between $\Delta G^\theta$ and $K$: $\Delta G^\theta = -RT \ln K$.

📐 Equilibrium Constant ($K_c$)

The Equilibrium Expression

For the reaction: $aA + bB \rightleftharpoons cC + dD$

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

where $[X]$ represents the equilibrium concentration in $mol \ dm^{-3}$.

Key Rules

  • Products go on top (numerator).
  • Reactants go on bottom (denominator).
  • Powers are the stoichiometric coefficients.
  • Pure solids and pure liquids are omitted from the expression (their concentrations are constant).

Magnitude of $K_c$

Value of $K_c$ Position of Equilibrium Interpretation
$K_c \gg 1$ Far to the right Mostly products at equilibrium. Reaction goes almost to completion.
$K_c \approx 1$ Roughly central Significant amounts of both reactants and products.
$K_c \ll 1$ Far to the left Mostly reactants. Reaction barely proceeds.

⚠️ Only temperature changes $K_c$. Adding a catalyst, changing concentration, or changing pressure shifts the position but not the value of $K_c$.

Effect of Temperature on $K_c$

Reaction Type Increase Temperature Effect on $K_c$
Exothermic ($\Delta H < 0$) Shifts left (endothermic direction) $K_c$ decreases
Endothermic ($\Delta H > 0$) Shifts right (endothermic direction) $K_c$ increases

🧊 ICE Tables

Method: Initial → Change → Equilibrium

The ICE table organizes equilibrium calculations. For each species, track the initial concentration, the change that occurs, and the equilibrium concentration.

Worked Example

$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$. Initially, 1.0 mol $H_2$ and 1.0 mol $I_2$ are placed in a 1.0 dm³ flask. At equilibrium, $[HI] = 1.56$ mol/dm³.

$H_2$ $I_2$ $2HI$
Initial 1.00 1.00 0
Change $-x$ $-x$ $+2x$
Equilibrium $1.00 - x$ $1.00 - x$ $2x = 1.56$

Solving: $2x = 1.56 → x = 0.78$. $[H_2]_{eq} = [I_2]_{eq} = 1.00 - 0.78 = 0.22$ mol/dm³.

$$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(1.56)^2}{(0.22)(0.22)} = \frac{2.434}{0.0484} = 50.3$$

ICE table example showing Initial-Change-Equilibrium calculations for 2HI decomposition reaction

📊 Reaction Quotient ($Q$)

What is $Q$?

The reaction quotient ($Q$) is calculated using the same expression as $K_c$, but with non-equilibrium concentrations. It tells you which direction the reaction must shift to reach equilibrium.

Comparison Meaning Shift Direction
$Q < K_c$ Product/reactant ratio is too small — not enough products → Right (toward products)
$Q > K_c$ Product/reactant ratio is too large — too many products ← Left (toward reactants)
$Q = K_c$ System is at equilibrium No shift

⚡ Equilibrium & Gibbs Free Energy

The Relationship

$$\Delta G^\theta = -RT \ln K$$

  • $\Delta G^\theta$ = standard Gibbs free energy change (J/mol)
  • $R$ = 8.314 J K⁻¹ mol⁻¹
  • $T$ = temperature in Kelvin
  • $K$ = equilibrium constant
If… Then… Interpretation
$K > 1$ $\Delta G^\theta < 0$ Reaction is spontaneous under standard conditions (products favored)
$K = 1$ $\Delta G^\theta = 0$ System is at equilibrium under standard conditions
$K < 1$ $\Delta G^\theta > 0$ Reaction is non-spontaneous under standard conditions (reactants favored)
Gibbs Free Energy vs reaction progress graph showing U-shaped curve with equilibrium at minimum point

🧠 Memory Aids

🔤 $K_c$ Expression — "Products Over Reactants" (POR)

Products are always on top, reactants on bottom. Powers are the stoichiometric coefficients. Omit solids and liquids.

🔤 $Q$ vs $K$ — "Q Chases K"

Imagine $Q$ trying to catch $K$:

  • If $Q < K$ → $Q$ needs to increase → make more products → shift right
  • If $Q > K$ → $Q$ needs to decrease → make more reactants → shift left

🔤 ICE Table — "I See Equilibrium"

Initial concentrations → Change (use $x$ and stoichiometric ratios: $-x$ for consumed, $+x$ for formed) → Equilibrium concentrations. Substitute into $K_c$ expression.

🔤 $\Delta G$ and $K$ — "Negative G = Go" / "Positive G = No Go"

$\Delta G^\theta < 0$ (negative) → $K> 1$ → products favored → reaction "goes". $\Delta G^\theta > 0$ (positive) → $K < 1$ → reactants favored → "doesn't go" .

🌍 Real-World Applications

🏭 Ammonia Production — $K_c$ and Compromise Conditions

Context: The Haber Process ($K_c$ at 25°C is about $6 \times 10^5$, but at 450°C drops to about 0.5).

Science: For this exothermic reaction, $K_c$ is huge at low temperature (products strongly favored). But the rate is impractically slow. At 450°C, $K_c ≈ 0.5$ (roughly equal reactants and products), but the rate is fast enough for industrial production. This shows how $K_c$ decreases with temperature for exothermic reactions ($\Delta G^\theta$ becomes less negative).

Impact: Industrial chemists must balance thermodynamic favorability ($K$) against kinetic feasibility (rate).

🔋 Battery Design — $\Delta G$ and Cell Voltage

Context: The relationship $\Delta G^\theta = -nFE^\theta$ connects thermodynamics to electrochemistry.

Science: Since $\Delta G^\theta = -RT \ln K$ and $\Delta G^\theta = -nFE^\theta$, we can link $K$ to cell voltage: a large $K$ means a large, positive $E^\theta$ and a spontaneous cell reaction. Battery chemists use this to predict whether a cell will work before building it.

Impact: This is how lithium-ion batteries were designed — by calculating $\Delta G$ and $K$ for candidate electrode reactions.

💊 Drug Solubility — $K_{sp}$ in Pharmaceuticals

Context: The solubility product ($K_{sp}$) is a special form of $K_c$ for sparingly soluble salts.

Science: For $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$, $K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 \times 10^{-10}$ (very small). This extremely low solubility makes barium sulfate safe to drink for X-ray contrast imaging, despite barium ions being toxic — $Q$ stays above $K_{sp}$ so it remains solid in the gut.

Impact: Millions of patients safely drink barium sulfate suspensions for medical imaging every year.

⚠️ Common Mistakes

  • ❌ Including solids or pure liquids in $K_c$ → ✅ Only aqueous and gaseous species appear in the $K_c$ expression. Solids and pure liquids have constant concentrations and are omitted.
  • ❌ "Increasing concentration changes $K_c$" → ✅ Only temperature changes $K_c$. Adding more reactant shifts the position of equilibrium but the value of $K_c$ stays the same.
  • ❌ Confusing $Q$ and $K$ → ✅ $K$ uses equilibrium concentrations. $Q$ uses any concentrations (at the current moment) and is compared to $K$ to predict direction of shift.
  • ❌ Using kJ in the $\Delta G = -RT \ln K$ calculation → ✅ $R = 8.314$ J K⁻¹ mol⁻¹, so $\Delta G^\theta$ from this equation is in J/mol. Convert to kJ by dividing by 1000.
  • ❌ ICE table: wrong stoichiometric ratios → ✅ The "Change" row must follow the stoichiometric ratio. If the coefficient is 2, the change is $±2x$, not $±x$.

📝 Exam-Style Questions

Question 1: At 700 K, the reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$ has $K_c = 4.3 \times 10^6$. Comment on the position of equilibrium. [1 mark]

Mark Scheme:

  • [1 mark] $K_c \gg 1$, so equilibrium lies far to the right (products strongly favored).
Question 2: Write the $K_c$ expression for: $4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g)$. [1 mark]

Mark Scheme:

  • [1 mark] $K_c = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}$
Question 3: Calculate $\Delta G^\theta$ for a reaction with $K_c = 1.5$ at 298 K. ($R = 8.314$ J K⁻¹ mol⁻¹). [2 marks]

Mark Scheme:

  • [1 mark] $\Delta G^\theta = -RT \ln K = -(8.314)(298) \ln(1.5)$.
  • [1 mark] $= -(2477.6)(0.405) = -1004$ J/mol = −1.0 kJ/mol.
Question 4: For $A + B \rightleftharpoons C + D$, given $[A]=0.1$, $[B]=0.1$, $[C]=0.5$, $[D]=0.5$ and $K_c=50$. Calculate $Q$ and predict shift. [2 marks]

Mark Scheme:

  • [1 mark] $Q = \frac{(0.5)(0.5)}{(0.1)(0.1)} = 25$.
  • [1 mark] $Q (25) < K_c (50)$ → shifts right (toward products).
Question 5: State the effect of adding a catalyst on the value of $K_c$. [1 mark]

Mark Scheme:

  • [1 mark] No effect. $K_c$ is only affected by temperature.
Question 6: 0.50 mol of $PCl_5$ is placed in a 1.0 dm³ flask at 250°C. At equilibrium, 0.16 mol of $Cl_2$ is present. $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$. Calculate $K_c$. [3 marks]

Mark Scheme:

  • [1 mark] ICE: $[Cl_2]_{eq} = 0.16$, so $x = 0.16$. $[PCl_3]_{eq} = 0.16$. $[PCl_5]_{eq} = 0.50 - 0.16 = 0.34$.
  • [1 mark] $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.16)(0.16)}{0.34}$.
  • [1 mark] $K_c = 0.075$.
Question 7: Explain why $K_c$ for an exothermic reaction decreases when temperature increases. [2 marks]

Mark Scheme:

  • [1 mark] Increasing temperature favors the endothermic (reverse) direction → equilibrium shifts left.
  • [1 mark] More reactants and fewer products at equilibrium → the ratio $\frac{[products]}{[reactants]}$ decreases → $K_c$ decreases.
Question 8: If $\Delta G^\theta$ for a reaction is $-10.5$ kJ/mol at 298 K, calculate $K$. [2 marks]

Mark Scheme:

  • [1 mark] $-10500 = -(8.314)(298) \ln K$ → $\ln K = \frac{10500}{2477.6} = 4.24$.
  • [1 mark] $K = e^{4.24} = \mathbf{69.4}$.