Reaction and equilibrium expression
A student sets up the following equation: \(0.113=\dfrac{(2x)^2}{0.100-x}\). The structure matches the equilibrium \(N_2O_4(g)\rightleftharpoons 2NO_2(g)\) in a constant-volume system with initial concentrations \([N_2O_4]_0=0.100\ \text{M}\) and \([NO_2]_0=0\ \text{M}\), and with \(K_c=0.113\) at \(25^\circ\text{C}\).
\[ K_c=\frac{[NO_2]^2}{[N_2O_4]} \]
The factor \((2x)^2\) reflects the stoichiometric coefficient 2 for \(NO_2\), and the denominator \((0.100-x)\) reflects consumption of \(N_2O_4\). The variable \(x\) has units of molarity when concentrations are used.
ICE-table relationships
The concentration changes consistent with \(N_2O_4(g)\rightleftharpoons 2NO_2(g)\) are represented by a single extent variable \(x\).
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| \(N_2O_4\) | \(0.100\) | \(-x\) | \(0.100-x\) |
| \(NO_2\) | \(0\) | \(+2x\) | \(2x\) |
\[ K_c=0.113=\frac{(2x)^2}{0.100-x}=\frac{4x^2}{0.100-x} \]
Algebraic solution and root selection
Multiplication by \((0.100-x)\) yields a quadratic equation in \(x\).
\[ 0.113(0.100-x)=4x^2 \] \[ 0.0113-0.113x=4x^2 \] \[ 4x^2+0.113x-0.0113=0 \]
The quadratic formula gives:
\[ x=\frac{-0.113\pm\sqrt{(0.113)^2-4(4)(-0.0113)}}{2\cdot 4} \] \[ x=\frac{-0.113\pm\sqrt{0.012769+0.1808}}{8} =\frac{-0.113\pm\sqrt{0.193569}}{8} \] \[ x=\frac{-0.113\pm 0.43996}{8} \]
The physically meaningful root keeps \((0.100-x)\) positive, giving:
\[ x=\frac{-0.113+0.43996}{8}\approx 0.0409\ \text{M} \]
Equilibrium concentrations
Substitution into the equilibrium expressions gives:
\[ [N_2O_4]_{\text{eq}}=0.100-x=0.100-0.0409\approx 0.0591\ \text{M} \] \[ [NO_2]_{\text{eq}}=2x=2(0.0409)\approx 0.0817\ \text{M} \]
The magnitude \(x\approx 0.0409\ \text{M}\) is about \(41\%\) of the initial \(0.100\ \text{M}\), so small-change approximations (for example, treating \(0.100-x\approx 0.100\)) are not justified for this data set.
Visualization: where the student’s equation comes from
The curve represents \(Q_c(x)=\dfrac{(2x)^2}{0.100-x}\) over \(0\le x\le 0.080\). The horizontal line is \(K_c=0.113\), and the intersection marks the equilibrium extent \(x\approx 0.0409\).
Common pitfalls
- Stoichiometric power and factor: the exponent 2 on \([NO_2]\) and the factor \(2x\) come from the coefficient 2 in \(N_2O_4\rightleftharpoons 2NO_2\).
- Initial product present: a nonzero \([NO_2]_0\) changes \([NO_2]_{\text{eq}}\) to \([NO_2]_0+2x\), producing a different algebraic equation.
- Domain restrictions: the constraint \(0.100-x>0\) enforces \(x<0.100\ \text{M}\) so concentrations remain physical.
- Approximation misuse: treating \(0.100-x\approx 0.100\) is unreliable when \(x\) is not small compared with 0.100.
Result summary
The student’s equilibrium setup leads to \(4x^2+0.113x-0.0113=0\) with the physical root \(x\approx 0.0409\ \text{M}\). The equilibrium concentrations are \([N_2O_4]_{\text{eq}}\approx 0.0591\ \text{M}\) and \([NO_2]_{\text{eq}}\approx 0.0817\ \text{M}\).