IB DP Chemistry - R2.3.6 Quantifying equilibrium concentrations - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R2.3.6 Quantifying equilibrium concentrations – Study Notes – New Syllabus
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Reactivity 2.3.6 – Quantifying Equilibrium Composition Using the Equilibrium Law
Reactivity 2.3.6 – Quantifying Equilibrium Composition Using the Equilibrium Law
The equilibrium law allows chemists to calculate the concentrations or partial pressures of species in a chemical system at equilibrium using the equilibrium constant \( K \). This is especially useful when we are given initial concentrations and need to determine the equilibrium composition of the mixture.
1. The General Equilibrium Expression
For a reaction:
\( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is:
\( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)
For gaseous systems in terms of partial pressures:
\( K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \)
Note:
Only species in the gas or aqueous phase are included. Solids and pure liquids are not part of the equilibrium expression.
2. Solving Equilibrium Problems Using ICE Tables
To solve problems involving \( K \), we often use the ICE Table method:
- I – Initial concentrations (before any reaction)
- C – Change in concentration (due to shift toward equilibrium)
- E – Equilibrium concentrations
We typically follow this process:
- Write the balanced equation and the expression for \( K \).
- Set up an ICE table using algebraic variables (like \( x \)) for changes.
- Use the value of \( K \) to solve for \( x \).
- Calculate the equilibrium concentrations using the values found.
Assumptions: If \( K \) is very small (e.g., \( K < 10^{-4} \)), the change in concentration might be negligible and approximations can be used (with validation).
Example
For the reaction:
\( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \quad K_c = 49.0 \) at 450°C.
Initially, [H2] = 0.200 mol/L, [I2] = 0.200 mol/L, [HI] = 0.000 mol/L.
Calculate the equilibrium concentrations.
▶️Answer/Explanation
Set up ICE table
H2 + I2 ⇌ 2HI
| [H2] | [I2] | [HI] | |
|---|---|---|---|
| Initial (I) | 0.200 | 0.200 | 0.000 |
| Change (C) | –x | –x | +2x |
| Equilibrium (E) | 0.200–x | 0.200–x | 2x |
Plug into \( K_c \) expression:
\( K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x)^2}{(0.200–x)(0.200–x)} = 49.0 \)
\( \frac{4x^2}{(0.200–x)^2} = 49.0 \)
Solve the equation:
Take square root: \( \frac{2x}{0.200–x} = 7 \)
\( 2x = 7(0.200–x) = 1.4 – 7x \)
\( 2x + 7x = 1.4 \Rightarrow 9x = 1.4 \Rightarrow x = 0.156 \)
Calculate equilibrium concentrations:
[H2] = [I2] = 0.200 – 0.156 = 0.044 mol/L
[HI] = 2 × 0.156 = 0.312 mol/L
Example
For the reaction:
\( \text{SO}_2(g) + \text{NO}_2(g) \rightleftharpoons \text{SO}_3(g) + \text{NO}(g) \quad K_c = 1.5 \)
At equilibrium, [SO2] = 0.30 mol/L, [NO2] = 0.20 mol/L, [NO] = 0.15 mol/L.
Calculate the equilibrium concentration of SO3.
▶️Answer/Explanation
Use the equilibrium law:
\( K_c = \frac{[SO_3][NO]}{[SO_2][NO_2]} = 1.5 \)
Let [SO3] = x
\( \frac{x \cdot 0.15}{0.30 \cdot 0.20} = 1.5 \Rightarrow \frac{0.15x}{0.06} = 1.5 \)
\( 0.15x = 0.09 \Rightarrow x = \frac{0.09}{0.15} = 0.60 \)
Answer: [SO3] = 0.60 mol/L
Approximation Strategy: When \( K \ll 1 \)
If the equilibrium constant is very small (typically \( K < 10^{-4} \)), the reaction hardly proceeds in the forward direction. This means the change in concentration of the reactants is minimal, and we can apply this simplification:
Approximation:
$[\text{Reactant}]_{\text{initial}} \approx [\text{Reactant}]_{\text{equilibrium}} $
Why is this helpful?
- It simplifies the math by eliminating a quadratic equation.
- It speeds up problem-solving without significantly affecting accuracy (as long as you check afterward).
When can you use this?
- When \( K \ll 1 \) and the change in concentration \( x \) is much smaller than the initial concentration.
- After solving for \( x \), you should verify that \( \frac{x}{[\text{initial}]} \times 100\% < 5\% \) to confirm the approximation was valid.
Example
The dissociation of hydrogen cyanide is represented by:
\( \text{HCN}(aq) \rightleftharpoons \text{H}^+(aq) + \text{CN}^-(aq) \)
\( K_c = 6.2 \times 10^{-10} \) at 25°C.
Calculate the concentrations of all species at equilibrium if 0.100 mol/L of HCN is initially dissolved in water.
▶️Answer/Explanation
Step 1 – Set up ICE table:
| [HCN] | [H+] | [CN–] | |
|---|---|---|---|
| Initial | 0.100 | 0 | 0 |
| Change | –x | +x | +x |
| Equilibrium | 0.100–x | x | x |
Set up the equilibrium expression:
\( K_c = \frac{[H^+][CN^-]}{[HCN]} = \frac{x^2}{0.100–x} \)
Apply the approximation:
Since \( K \ll 1 \), we assume \( 0.100 – x \approx 0.100 \), so:
\( \frac{x^2}{0.100} = 6.2 \times 10^{-10} \)
\( x^2 = 6.2 \times 10^{-11} \)
\( x = \sqrt{6.2 \times 10^{-11}} \approx 7.87 \times 10^{-6} \)
Validate the approximation:
Change in HCN = \( x = 7.87 \times 10^{-6} \)
Percent change = \( \frac{7.87 \times 10^{-6}}{0.100} \times 100\% \approx 0.0079\% \)
This is much less than 5%, so the approximation is valid.
Final concentrations:
[HCN] ≈ 0.100 mol/L
[H+] = [CN–] = \( 7.87 \times 10^{-6} \) mol/L
Example
The weak acid ethanoic acid dissociates in water:
\( \text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{H}^+(aq) + \text{CH}_3\text{COO}^-(aq) \)
At 25°C, the acid dissociation constant is \( K_c = 1.8 \times 10^{-5} \).
Calculate the concentrations of all species at equilibrium if the initial [CH3COOH] = 0.200 mol/L.
▶️Answer/Explanation
Step 1 – Set up ICE table:
| [CH3COOH] | [H+] | [CH3COO–] | |
|---|---|---|---|
| Initial | 0.200 | 0 | 0 |
| Change | –x | +x | +x |
| Equilibrium | 0.200–x | x | x |
Write the equilibrium expression:
\( K_c = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x^2}{0.200–x} \)
Use the approximation:
Since \( K \ll 1 \), assume \( 0.200–x \approx 0.200 \):
\( \frac{x^2}{0.200} = 1.8 \times 10^{-5} \Rightarrow x^2 = 3.6 \times 10^{-6} \Rightarrow x = \sqrt{3.6 \times 10^{-6}} \approx 1.9 \times 10^{-3} \)
Validate the approximation:
\( \frac{1.9 \times 10^{-3}}{0.200} \times 100\% = 0.95\% \) — valid.
Final equilibrium concentrations:
[CH3COOH] ≈ 0.200 mol/L
[H+] = [CH3COO–] = \( 1.9 \times 10^{-3} \) mol/L