2 NO2(g) <-> N2O4(g) (dark brown to colorless) The dimerization of NO2(g), an exothermic process, is — Equilibrium Chemistry Question

STEPS:

1. Calculate the initial concentrations:
The dimerization of nitrogen dioxide gas is represented by the equilibrium equation:
$$2\ \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$$
Because the volume of the rigid reaction vessel is exactly $1.00\text{ L}$, the initial molarity of the reactant is equal to its initial moles:
$$[\text{NO}_2]_0 = \frac{0.0300\text{ mol}}{1.00\text{ L}} = 0.0300\text{ M}$$
Initially, no product is added, so the starting concentration of the product is:
$$[\text{N}_2\text{O}_4]_0 = 0\text{ M}$$

2. Determine the equilibrium concentrations using an ICE (Initial, Change, Equilibrium) table:
Let $x$ represent the molarity of $\text{N}_2\text{O}_4$ produced at equilibrium. Since the stoichiometric ratio of reactant to product is $2:1$, every mole of $\text{N}_2\text{O}_4$ that is formed requires the consumption of two moles of $\text{NO}_2$.
$$\begin{array}{lccc}\text{ICE}& 2{\text{ NO}}_2(g)& \rightleftharpoons & {\text{N}}_2{\text{O}}_4(g)\\ \mathbf{\text{I}}\text{nitial}& 0.0300\text{ M}& & 0\text{ M}\\ \mathbf{\text{C}}\text{hange}& -2x& & +x\\ \mathbf{\text{E}}\text{quilibrium}& 0.0300-2x& & x\end{array}$$
\begin{array}{l|ccc}
\text{ICE} & 2\text{ NO}_2(g) & \rightleftharpoons & \text{N}_2\text{O}_4(g) \\
\hline
\text{\textbf{I}nitial} & 0.0300\text{ M} & & 0\text{ M} \\
\text{\textbf{C}hange} & -2x & & +x \\
\text{\textbf{E}quilibrium} & 0.0300 - 2x & & x \\
\end{array}

3. Solve for the change variable ($x$):
We are given that at equilibrium, the vessel contains $0.00500\text{ mol}$ of $\text{N}_2\text{O}_4(g)$. Since the volume is $1.00\text{ L}$, the equilibrium concentration of product is:
$$[\text{N}_2\text{O}_4]_{eq} = x = 0.00500\text{ M}$$

4. Calculate the equilibrium concentration of the reactant, $[\text{NO}_2]_{eq}$:
Substitute the value of $x$ into the equilibrium expression for $[\text{NO}_2]$:
$$[\text{NO}_2]_{eq} = 0.0300\text{ M} - 2(0.00500\text{ M})$$
$$[\text{NO}_2]_{eq} = 0.0300\text{ M} - 0.0100\text{ M} = \mathbf{0.0200\text{ M}}$$

5. Formulate the equilibrium constant expression ($K_c$):
Based on the law of mass action, the equilibrium constant expression for this reaction is:
$$K_c = \frac{[\text{N}_2\text{O}_4]_{eq}}{[\text{NO}_2]_{eq}^2}$$

6. Compute the numerical value of $K_c$:
Substitute the calculated equilibrium concentrations into the expression:
$$K_c = \frac{0.00500}{(0.0200)^2}$$
Because calculators are not permitted on Section I of the AP Exam, simplify the arithmetic using scientific notation:
$$K_c = \frac{5.00 \times 10^{-3}}{(2.00 \times 10^{-2})^2} = \frac{5.00 \times 10^{-3}}{4.00 \times 10^{-4}}$$
$$K_c = \frac{5.00}{4.00} \times 10^1 = 1.25 \times 10 = \mathbf{12.5}$$
This mathematically identifies Option D as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect (0.0800): This is the reciprocal of the correct equilibrium constant ($1 / 12.5 = 0.0800$). A student will arrive at this value if they write the equilibrium expression upside down (reactants over products), calculating $K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}$.
• Option B is incorrect (0.250): A student will obtain this value if they fail to square the reactant concentration in the denominator, formulating the expression incorrectly as $K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]} = \frac{0.00500}{0.0200} = 0.250$.
• Option C is incorrect (4.00): This value is obtained if a student mistakenly uses the inverted non-squared expression, calculating $\frac{[\text{NO}_2]}{[\text{N}_2\text{O}_4]} = \frac{0.0200}{0.00500} = 4.00$.