Problem
Apply ice table chemistry (the Initial–Change–Equilibrium method) to the equilibrium below and determine the equilibrium concentrations.
Reaction: N2O4(g) ⇄ 2 NO2(g)
Given: \(K_c = 0.36\) at a fixed temperature; initially \([\text{N}_2\text{O}_4]_0 = 0.200\,\text{M}\), \([\text{NO}_2]_0 = 0.000\,\text{M}\).
Find: \([\text{N}_2\text{O}_4]_{\text{eq}}\) and \([\text{NO}_2]_{\text{eq}}\).
Step 1: Write the equilibrium-constant expression
For N2O4(g) ⇄ 2 NO2(g), the equilibrium constant in concentration form is:
\[ K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \]
Step 2: Build the ICE table
Because \([\text{NO}_2]_0 = 0.000\,\text{M}\), the reaction must proceed forward to create NO2. Let \(x\) be the amount (in M) of N2O4 that dissociates.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| N2O4 | 0.200 | \(-x\) | \(0.200 - x\) |
| NO2 | 0.000 | \( +2x\) | \(2x\) |
Step 3: Substitute ICE expressions into \(K_c\)
Insert \([\text{NO}_2]_{\text{eq}} = 2x\) and \([\text{N}_2\text{O}_4]_{\text{eq}} = 0.200 - x\) into the equilibrium expression:
\[ 0.36 = \frac{(2x)^2}{0.200 - x} = \frac{4x^2}{0.200 - x} \]
Step 4: Solve for \(x\) (quadratic)
Multiply both sides by \((0.200 - x)\) and rearrange to standard form:
\[ 0.36\,(0.200 - x) = 4x^2 \]
\[ 0.072 - 0.36x = 4x^2 \]
\[ 4x^2 + 0.36x - 0.072 = 0 \]
Apply the quadratic formula with \(a = 4\), \(b = 0.36\), \(c = -0.072\):
\[ x = \frac{-b \pm \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a} = \frac{-0.36 \pm \sqrt{(0.36)^2 - 4 \cdot 4 \cdot (-0.072)}}{2 \cdot 4} \]
\[ x = \frac{-0.36 \pm \sqrt{0.1296 + 1.152}}{8} = \frac{-0.36 \pm \sqrt{1.2816}}{8} = \frac{-0.36 \pm 1.1320777}{8} \]
Two mathematical roots appear, but only the physically meaningful one keeps concentrations nonnegative:
\[ x_1 = \frac{-0.36 + 1.1320777}{8} \approx 0.09651 \quad (\text{valid}) \]
\[ x_2 = \frac{-0.36 - 1.1320777}{8} \approx -0.18651 \quad (\text{reject: negative change}) \]
Step 5: Compute equilibrium concentrations
Use the ICE table expressions:
\[ [\text{NO}_2]_{\text{eq}} = 2x = 2 \cdot 0.09651 \approx 0.1930\,\text{M} \]
\[ [\text{N}_2\text{O}_4]_{\text{eq}} = 0.200 - x = 0.200 - 0.09651 \approx 0.1035\,\text{M} \]
Equilibrium concentrations: \([\text{NO}_2]_{\text{eq}} \approx 0.193\,\text{M}\), \([\text{N}_2\text{O}_4]_{\text{eq}} \approx 0.103\,\text{M}\).
Step 6: Quick verification
Substitute the equilibrium values back into the equilibrium expression:
\[ K_c = \frac{(0.1930)^2}{0.1035} \approx 0.36 \]
The result matches the given \(K_c\), confirming that the ICE table chemistry setup and root choice are consistent.
Summary of the ICE-table method
In ice table chemistry, the “Initial” row lists starting concentrations, the “Change” row follows stoichiometric coefficients using a single variable, and the “Equilibrium” row is substituted into the equilibrium-constant expression \(K_c\). Solving the resulting equation yields \(x\), and substituting back gives the equilibrium concentrations.